STRING THEORY AND M-THEORY
1 1. RENORMALIZATION IN STRING THEORY
Problem 1. Consider a closed string theory in 26-dimensional spacetime, where the critical
dimension is Dcrit = 26. The worldsheet theory after quantization leads to a 2D quantum field
theory with a beta function given by β(g) = −Rg2, where Ris a constant related to the string
tension.
a) If the coupling constant gflows from a weak coupling regime g1to a strong coupling regime
g2, where g1> g2, determine the change in the renormalized string coupling G.
b) Find the renormalized string coupling Gat the weak coupling regime g1.
c) Given that the string tension is 2πα′= 1, compute the value of the constant R.
Solution 1.
a) The change in the renormalized string coupling Gis given by:
∆G=G(g2)−G(g1) = −ln g2
g1.
b) At the weak coupling regime g1, the renormalized string coupling G(g1)is related to the bare
string coupling g1by G(g1) = ln(g1).
c) Using the value of the string tension 2πα′= 1 and the relation β(g) = −Rg2, we have:
1 = −R1
4πα′2
=−R1
4π2
.
Solving for R, we find R=1
16π2.
I’m glad you’re interested in practicing numerical problems related to String Theory and M-
Theory. Here’s a problem for you:
2 2. CONFORMAL SYMMETRY IN M-THEORY
Problem: Consider a closed string moving in 5-dimensional spacetime with a radius of com-
pactification R. If the mode number of the first excited vibrational mode is n= 2, calculate the
energy of this mode in terms of the string tension parameter α′.
Additional information: The energy of a closed string in Dspacetime dimensions is given by
the formula:
E=n
Rrα′
2
a) Calculate the energy of the first excited vibrational mode for R= 2 and α′= 1.
Solution:
a) The energy of the first excited vibrational mode is given by:
E=2
2r1
2= 1
Therefore, for R= 2 and α′= 1, the energy of the first excited vibrational mode of the closed
string is E= 1.
3 3. QUANTUM GRAVITY IN STRING THEORY
Problem 3. Consider an open string moving in 4-dimensional spacetime. The string has tension
T= 1 and mass per unit length µ= 2.
a) Find the speed of propagation of waves on this string.
b) Calculate the energy stored in a segment of string of length L= 3.
c) If the string is stretched with a force of F= 5, determine the amplitude of the standing wave
that could form on the string.
Solution 3.
a) The speed of propagation of waves on a string is given by
v=sT
µ.
Substitute T= 1 and µ= 2:
v=r1
2=√2
2.
Therefore, the speed of propagation of waves on this string is √2
2.
b) The energy stored in a segment of string of length Lis given by
E=1
2µv2L.
Substitute µ= 2,v=√2
2, and L= 3:
E=1
2×2× √2
2!2
×3 = 3
2.
Therefore, the energy stored in a segment of string of length 3 is 3
2.
c) The amplitude of the standing wave that could form on the string under the applied force F
is given by
A=F
2πv2.
Substitute F= 5 and v=√2
2:
A=5
2π×√2
22=5
2π×1
2
=5√2
π.
Therefore, the amplitude of the standing wave that could form on the string is 5√2
π.
4 4. BLACK HOLES AND STRING THEORY
Problem 4. Consider a black hole in four-dimensional spacetime described by the Schwarzschild
metric:
ds2=−1−2GM
c2rc2dt2+1−2GM
c2r−1
dr2+r2dΩ2
where Mis the mass of the black hole, cis the speed of light, Gis the gravitational constant,
and dΩ2=dθ2+sin2θdϕ2in spherical coordinates (t, r, θ, ϕ).
a) Find the event horizon radius Rhorizon of this black hole.
b) Determine the Schwarzschild radius RSin terms of M,G, and c.
c) Given that the mass of the black hole is M= 2 ×1030 kg, calculate the mass of the black
hole in units of solar masses (M⊙= 1.989 ×1030 kg).
Solution 4.
a) To find the event horizon radius, we set the metric coefficient of dt2to zero at the event
horizon. So, 1−2GM
c2Rhorizon = 0. Solving for Rhorizon gives:
Rhorizon =2GM
c2
b) The Schwarzschild radius is defined as the radius at which the metric becomes singular. It
is given by RS=2GM
c2.
c) Substituting M= 2 ×1030 kg into the Schwarzschild radius formula, we have:
RS=2G(2 ×1030)
c2=4×1030G
c2
Now, we can express the mass of the black hole in terms of solar masses by dividing by the
mass of the Sun:
4×1030G
c2÷1.989 ×1030 =2G
c2
Therefore, the mass of the black hole is 2solar masses.
5 5. TACHYON CONDENSATION IN STRING THEORY
Problem 5. Consider a closed bosonic string theory where the endpoint of the string coor-
dinates are subject to Neumann boundary conditions. The first excited level of the closed string
contains a tachyon with mass m2=−2
α′.
a) Calculate the momentum of the tachyon state in the string theory.
b) Show that the mass of the tachyon state in the open bosonic string theory is m= 0.
c) Interpret the result in relation to tachyon condensation.
Solution 5.
a) The mass-squared of a state in a closed string theory is given by the level matching condition:
m2=4
α′(N−1)
where Nis the occupation number operator for the state. For the first excited level, N= 1 and
m2=−2
α′. Substituting these values into the equation above, we have:
−2
α′=4
α′(1 −1)
−2=0
This equation is a contradiction, indicating that there is no physical state with a mass-squared of
−2
α′for the first excited level. So, the tachyon state does not exist.
b) In the open bosonic string theory, the mass-squared of a state is given by:
m2=2
α′(N−1)
For the tachyon state at the first excited level in the open string, N= 1 and m2=−1
α′. Taking the
square root of this, we get m= 0. Therefore, the mass of the tachyon state in the open bosonic
string theory is zero.
c) The result m= 0 for the open string tachyon state implies that it is a massless state. In string
theory, a tachyon has negative mass squared and indicates an instability in the theory. Tachyon
condensation is a process where this unstable state "condenses" to a minimum energy state. In
this case, the tachyon state in the open bosonic string theory having a mass of zero suggests that
it represents the minimum energy state after tachyon condensation has occurred. This process
helps stabilize the theory by eliminating the instability caused by the presence of the tachyon.
6 6. EXTRA DIMENSIONS IN M-THEORY
Problem 6. Consider a scenario in M-Theory where there are 7 spatial dimensions and 3
temporal dimensions. The size of the compactified extra dimensions is given by R= 10−17 meters.
Calculate the compactification scale in GeV.
Solution 6.
a) The compactification scale Mccan be calculated using the formula for the compactified extra
dimensions:
Mc=1
R
Substitute R= 10−17 meters into the formula:
Mc=1
10−17 = 1017 m−1
b) To convert the compactification scale from meters to GeV, we need to use the relation 1GeV =
1.97 ×10−16 m−1.
Let’s convert the compactification scale:
Mc= 1017 ×1.97 ×10−16 = 1.97 ×10 GeV = 19.7GeV
Therefore, the compactification scale in GeV is 19.7 GeV.
7 7. BRANE DYNAMICS IN STRING THEORY
Problem 7. Consider a type IIB superstring theory in 10 dimensions with D3-branes. The
tension of the D3-brane is given by T=1
(2π)3(α′)2, where α′is the string length parameter. Calculate
the energy density stored in a D3-brane of length L= 10√α′.
Solution 7. a) The energy density stored in a D3-brane can be calculated by dividing its energy
by its volume. The energy Eof a D3-brane is given by E=T×V, where Tis the tension of the
D3-brane and Vis its volume. Since the D3-brane extends in 3 spatial dimensions, the volume V
is L3= (10√α′)3= 1000α′3/2.
Substitute T=1
(2π)3(α′)2and V= 1000α′3/2into the equation:
E=1
(2π)3(α′)2×1000α′3/2=1000
(2π)3α′1/2
Thus, the energy stored in the D3-brane is 1000
(2π)3α′1/2.
b) The energy density is given by the energy per unit volume. We divide the energy Eby the
volume Vto find the energy density u=E
V.
Substitute E=1000
(2π)3α′1/2and V= 1000α′3/2into the equation:
u=1000/(2π)3α′1/2
1000α′3/2=1
(2π)3α′2
Therefore, the energy density stored in the D3-brane is 1
(2π)3α′2.
8 8. SUPERSYMMETRY BREAKING IN M-THEORY
Problem 8. Consider a supersymmetric M-Theory compactified on a 2-torus with radii R1and
R2. The volume of the torus is given by V=R1R2. Suppose that supersymmetry is broken by the
flux through the torus such that the gravitino mass term is generated.
[Given: The gravitino mass term is given by m3/2=e⟨G⟩, where Gis the flux. Also, we have
⟨G⟩ =−2 ln(V), where ⟨G⟩ denotes the expectation value of the flux.]
a) Show that the gravitino mass term m3/2in terms of the radii R1and R2.
b) If R1= 2 and R2= 3, calculate the gravitino mass m3/2.
Solution 8.
a) To find the gravitino mass term m3/2in terms of the radii R1and R2, we first need to express
the volume Vin terms of R1and R2:
V=R1·R2
Next, we can find the expectation value of the flux ⟨G⟩:
⟨G⟩ =−2 ln(V) = −2 ln(R1·R2) = −2 ln(R1)−2 ln(R2)
Substitute this into the expression for the gravitino mass term:
m3/2=e⟨G⟩ =e−2 ln(R1)−2 ln(R2)=e−2 ln(R1)·e−2 ln(R2)=1
R2
1·1
R2
2
=1
R2
1R2
2
=1
V2
b) Given R1= 2 and R2= 3, the volume V=R1·R2= 2 ·3=6. Therefore, the gravitino mass
term m3/2is:
m3/2=1
V2=1
62=1
36 = 0.0278
I. Problem 9. Consider a Type IIA superstring theory with a D6-brane wrapped on a compact
3-torus with sides of length L. The tension of the D6-brane is given by T6=1
(2π)6(α′)4. Suppose
the compactified space is a cube, compute the energy density UD6 of the D6-brane in terms of L.
Hint: The energy density is defined as the energy per unit volume.
II. Problem 10. In Type IIB superstring theory, a D3-brane has tension T3=1
(2π)3(α′)2. If this
theory is in a 10-dimensional spacetime with a toroidal compactification down to 6 dimensions, and
the 3-brane extends along all 3 of those compact dimensions, calculate the energy density UD3 of
the D3-brane in terms of the compactification radii Ri.
Hint: The energy density is defined as the energy per unit volume.
III. Problem 11. In M-Theory, consider a M5-brane wrapped on a torus with radii R1and R2. If
the tension of the brane is T5=1
(2π)5(ℓp)3, find the energy density UM5 of the M5-brane in terms of
the torus radii.
Hint: The energy density is defined as the energy per unit volume.
9 10. COSMOLOGY AND STRING/M-THEORY
Problem 10. Consider a string theory model in a universe with extra dimensions compactified
on a torus. The radius of the torus is R, and the string tension is T. The compactified dimensions
are described by an effective field theory with a massless scalar field ϕ, whose potential is given
by V(ϕ) = 1
2m2ϕ2.
a) Show that the effective tension in the compactified dimensions, Teff , is given by Tef f =
T e−2ϕ.
b) Determine the equation of motion for the scalar field ϕ.
c) Find the minimum of the potential V(ϕ).
Solution 10.
a) The effective tension in the compactified dimensions is given by Teff =T e−2ϕ. Let’s derive
this expression:
In string theory, the effective tension depends on the string coupling as Teff =T gs, where
gs=e−2ϕ. Therefore, Teff =T e−2ϕ.
b) The equation of motion for the scalar field ϕis given by:
d
dt ∂L
∂˙
ϕ−∂L
∂ϕ = 0
where the Lagrangian L=1
2˙
ϕ2−V(ϕ).
d
dt ∂L
∂˙
ϕ−∂L
∂ϕ =¨
ϕ+m2ϕ= 0
Therefore, the equation of motion for the scalar field ϕis ¨
ϕ+m2ϕ= 0.
c) To find the minimum of the potential V(ϕ), we need to solve dV
dϕ = 0:
dV
dϕ =m2ϕ= 0
This implies that the minimum of the potential occurs at ϕ= 0.
10 11. DUALITIES IN M-THEORY
Problem 11. Consider two particular string theories, Type IIA and Type IIB, related by T-duality.
Let the radius of a circle in Type IIA theory be R. If the number of fundamental strings winding
around the circle is n, find the dual circle radius in Type IIB theory.
Solution 11. Given: Radius of circle in Type IIA theory, R, and number of winding fundamental
strings, n.
In Type IIA theory, the momentum along the circle is given by P=n
R.
By T-duality, the winding number in Type IIB string theory is the same as the momentum in Type
IIA theory, and vice versa. Therefore, in Type IIB theory, the radius of the dual circle is related to
the momentum by Rdual =α′
R.
Substitute P=n
Rinto the formula for the dual radius in Type IIB theory:
Rdual =α′
R=α′
n/P =α′R
n
Hence, the dual circle radius in Type IIB theory is α′R
n.
11 12. INTEGRABILITY IN STRING THEORY
Problem 12. Consider a closed bosonic string moving in a background with a constant mag-
netic field Bin the z-direction. The equation of motion for the string is given by the Nambu-Goto
action
S=−T
2Zdσdτ√−h hab∂aXµ∂bXνGµν
where Tis the tension of the string, his the determinant of the world-sheet metric hab,Xµ=
Xµ(τ, σ)are the embedding coordinates of the string world-sheet, and Gµν is the background metric
tensor.
Given that the background metric is flat Minkowski space with Gµν =ηµν and the magnetic field
is B=Bz, where Bzis constant, compute the equation of motion for the string in this background.
Solution 12.
The equation of motion for the string can be derived by varying the action with respect to the
embedding coordinates Xµ.
Let’s denote ∂aXµ≡∂aXµ(τ, σ).
The variation of the action with respect to Xµgives the equation of motion:
δS
δXµ=−T
2Zdσdτ√−hhab∂a∂bXνηνµ = 0
Expanding the terms and using the fact that the world-sheet metric is hab =diag(−1,1), we get:
TZdσdτ ∂2
τXν−∂2
σXνηνµ = 0
The equations of motion are then given by:
a) In the µ= 0 direction:
∂2
τX0−∂2
σX0= 0
b) In the µ= 1 direction:
∂2
τX1−∂2
σX1= 0
c) In the µ= 2 direction:
∂2
τX2−∂2
σX2= 0
These equations of motion describe the dynamics of the string in the background of a constant
magnetic field B=Bz.
12 13. CAUSALITY AND STRING/M-THEORY
Problem 13. Consider a closed string propagating in a spacetime described by D= 10
dimensions. The string moves in a geometry where the background metric is given by ds2=
−dt2+dx2
1+ (dx2)2+···+ (dx8)2+ (dx9)2+ (dx10)2, where x10 is the compactified spatial direction
with a radius R.
a) Calculate the maximum energy Emax of a closed string state that can propagate in this ge-
ometry without creating a closed timelike curve.
b) Find the minimum allowed radius Rmin of the compactified spatial direction that ensures
causality is not violated in this spacetime.
Solution 13.
a) The maximum energy Emax of a closed string state can be found using the formula:
Emax =1
α′rN
2
where Nis the level of the state and α′is the Regge slope parameter. Since we are dealing with
a closed string in D= 10 dimensions, the critical dimension is D= 10, and the Regge slope
parameter is α′= 1/2πT .
For a closed string moving in a compactified dimension with radius R, the contribution to the
mass in the compact direction is n/R where nis the winding number. To avoid closed timelike
curves, we must have Emax =n/R. Equating these two expressions for Emax, we get:
n
R=1
α′rN
2
n
R=2πT
2rN
2
nR = 2πT √2N
R=2πT √2N
n
Therefore, the maximum energy is achieved at n= 1 and N= 2, leading to:
Emax =1
α′rN
2=1
α′r2
2=1
α′=√2πT
2
b) To ensure that causality is not violated in this spacetime, we must have the inequality Rmin ≥
2πR to avoid closed timelike curves. Substituting in the expression for Rmin from part (a), we get:
Rmin =2πT p2(1)
1= 2π√2πT ≥2πR
√2πT ≥R
2πT ≥R2
R2−2πT ≤0
Hence, the minimum allowed radius Rmin of the compactified spatial direction is R=√2πT .
13 14. HOLOGRAPHY IN M-THEORY
Problem 14. Consider a spacetime described by M-theory with 11 dimensions. The holo-
graphic principle states that the information of a region of space can be encoded on its boundary.
Suppose we have a 4-dimensional hypercube with side length L.
a) Calculate the volume of this hypercube in terms of L.
b) According to the holographic principle, what is the size of the boundary that encodes all the
information of this hypercube?
c) If the hypercube is in an 11-dimensional spacetime in M-theory, how many spatial dimensions
are "compactified"?
Solution 14.
a) The volume of a 4-dimensional hypercube with side length Lis given by V=L4.
b) The boundary of the hypercube is a 3-dimensional cube with sides of length L. The total
surface area of a cube is given by A= 6L2. Hence, for a 4-dimensional hypercube, the size of the
boundary that encodes all the information is A= 6L2.
c) In M-theory with 11 dimensions, the 4 spatial dimensions of the hypercube and the 1 time
dimension are known. This leaves 11 −4−1=6spatial dimensions "compactified".
I’m glad to help generate numerical problem questions for you. Could you please specify the
topic within String Theory and M-Theory that you would like the problem to be based on?
14 16. DARK MATTER AND STRING/M-THEORY
Problem 16. Consider a string theory scenario where a closed string moving in a compact
spatial dimension of radius Rinteracts with dark matter particles gravitationally. The string coupling
constant is given by gs= 0.1.
a) If the mass of the dark matter particle is m= 10−22 eV/c2, calculate the gravitational force
between the closed string and a dark matter particle located at a distance r= 1 mm.
b) If the string coupling constant gsis changed to 0.05, how does this affect the gravitational
force between the closed string and the dark matter particle?
Solution 16.
a) The gravitational force between the closed string and the dark matter particle can be calcu-
lated using Newton’s law of gravitation:
F=G·mdark matter ·mstring
r2
where Gis the gravitational constant (6.67430 ×10−11 m3kg−1s−2), mdark matter is the mass of
the dark matter particle (in kg), mstring is the mass of the closed string (in kg), and ris the distance
between the closed string and the dark matter particle (in meters).
Given m= 10−22 eV/c2, we need to convert this to kilograms by using the conversion factor
1eV/c2= 1.783 ×10−36 kg. Thus, mdark matter = 10−22 ×1.783 ×10−36 = 1.783 ×10−58 kg.
Plugging in the values G= 6.67430 ×10−11 m3kg−1s−2,mdark matter = 1.783 ×10−58 kg,
mstring =? (mass of the string is not provided but let’s assume it to be of the same order of magnitude
as a dark matter particle), and r= 1 mm = 0.001 m, we can find the gravitational force.
b) If the string coupling constant gsis changed to 0.05, we can re-calculate the gravitational
force using the new value of gsin the formula above. Let’s assume that the mass of the dark matter
particle remains the same.
Solution 16.
a) Given: G= 6.67430 ×10−11 m3kg−1s−2,mdark matter = 1.783 ×10−58 kg, mstring = 1.783 ×
10−58 kg (assumed to be of the same order), r= 0.001 m
Plugging in the values, we get:
F=6.67430 ×10−11 ×1.783 ×10−58 ×1.783 ×10−58
(0.001)2
⇒F=2.0004 ×10−124
10−6
⇒F= 2.0004 ×10−118 N
Thus, the gravitational force between the closed string and a dark matter particle located at a
distance of 1 mm is approximately 2.0004 ×10−118 N.
b) If gs= 0.05, we can re-calculate the gravitational force using the updated value of gsin the
formula. Plugging in the new value, we can find the new gravitational force.
I’m sorry, but it seems that I cannot generate numerical problems for String Theory and M-
Theory as they are primarily theoretical and mathematical physics concepts that do not involve
numerical computations. Would you like a conceptual problem or a theoretical problem instead?
15 18. VACUUM ENERGY IN M-THEORY
Problem 18. Consider a particular compactification of M-theory on a 5-dimensional torus with
each side having length R. The vacuum energy in this scenario is given by ρ=−1
(2π)5R9.
a) Calculate the vacuum energy ρin GeV4when R= 1 cm. b) Find the value of Rin m for
which the vacuum energy is ρ=−1GeV4.
Solution 18. a) To calculate the vacuum energy in GeV4, we need to convert the length R=
1cm to meters and then substitute it into the given formula.
Given: 1 cm = 10−2m
Plugging this value into the formula for vacuum energy: ρ=−1
(2π)5(10−2)9
ρ=−1
(2π)5(10−18)
ρ=−1
(2π)5(10−18)
ρ=−1
(2π)5×10−18
ρ=−1
(2π)5×10−18
ρ≈ −8.53254 ×1065 GeV4
Therefore, the vacuum energy when R= 1 cm is approximately −8.53254 ×1065 GeV4.
b) To find the value of Rin meters for which the vacuum energy is ρ=−1GeV4, we set the
vacuum energy formula equal to −1and solve for R:
−1 = −1
(2π)5R9
R9= (2π)5
R=9
p(2π)5
R≈3.28626 ×10−16 m
Therefore, the value of Rin meters for which the vacuum energy is ρ=−1GeV4is approxi-
mately 3.28626 ×10−16 m.
16 19. SOLITONS IN STRING THEORY
Problem 19. Consider a D-brane in type IIA superstring theory with tension T. Suppose the
D-brane extends in pspatial dimensions. The energy density Uper unit p-dimensional volume of
the D-brane is given by U=cT , where cis a constant.
a) If the D-brane has p= 3 spatial dimensions and the tension is T= 5 ×10−2, find the energy
density U.
b) Now, let’s consider another D-brane with tension T= 0.1and energy density U= 0.4.
Determine the number of spatial dimensions pin which this D-brane extends.
Solution 19.
a) Given p= 3 and T= 5 ×10−2, we use the formula U=cT to find the energy density U:
U=c×5×10−2= 0.05c
So, the energy density Ufor the D-brane with p= 3 spatial dimensions and tension T= 5×10−2
is 0.05c.
b) For this case, we have T= 0.1and U= 0.4. Using the same formula for energy density,
U=cT , we can solve for cfirst:
c=U
T=0.4
0.1= 4
Now, we substitute c= 4 back into the equation U=cT :
0.4=4×p
Solving for p, we find:
p=0.4
4= 0.1
Thus, the D-brane extends in p= 0.1spatial dimensions.
I. Let’s formulate a numerical problem in the context of String Theory and M-Theory:
17 20. ADS/CFT CORRESPONDENCE IN M-THEORY
Problem 20. Consider a type IIA string theory on an Anti-de Sitter space (AdS) with a five-
dimensional radius RAdS = 10. Calculate the corresponding conformal field theory (CFT) central
charge using the AdS/CFT correspondence formula:
c=3L
2G
where Lis the AdS radius and Gis Newton’s constant. Take the value of Newton’s constant
G= 6.71 ×10−11 m3kg−1s−2.
Solution 20. Given: - AdS radius RAdS = 10 - Newton’s constant G= 6.71 ×10−11 m3kg−1
s−2
The CFT central charge can be calculated using the AdS/CFT correspondence formula:
c=3L
2G
Plugging in the values:
c=3×10
2×6.71 ×10−11
c=30
13.42 ×10−11
c=30
1.342 ×10−10
c=30
1.342
c≈22.36
Therefore, the central charge of the corresponding conformal field theory is approximately
22.36.
I. Entanglement Entropy in String Theory:
18 21. Entanglement Entropy in String Theory
Problem 21. Consider a 1+1 dimensional conformal field theory described by a CFT with
central charge c= 1. Let’s calculate the entanglement entropy of a subsystem in this theory.
[Additional context: The entanglement entropy Sof a subsystem in a CFT is given by the formula
S=c
3log L
ϵ, where cis the central charge, Lis the length of the subsystem, and ϵis a short-
distance cutoff.]
a) Suppose we have a subsystem of length L= 2 in this CFT. Calculate the entanglement
entropy when the short-distance cutoff ϵ= 0.1.
b) Now, consider another CFT described by a different conformal field theory with central charge
c= 2. If we have a subsystem of length L= 3 in this CFT, what is the entanglement entropy when
the short-distance cutoff ϵ= 0.01?
Solution 21.
a) We are given c= 1,L= 2, and ϵ= 0.1. Plugging these values into the formula for entangle-
ment entropy, we get:
S=1
3log 2
0.1=1
3log 20 ≈1
3×2.9957 ≈0.9986
Therefore, the entanglement entropy when L= 2spaceunits and ϵ= 0.1is approximately
0.9986.
b) For the new CFT with c= 2,L= 3, and ϵ= 0.01, we apply the formula:
S=2
3log 3
0.01=2
3log 300 ≈2
3×5.7038 ≈3.8026
Therefore, the entanglement entropy when L= 3spaceunits and ϵ= 0.01 is approximately
3.8026.
I can certainly help with that! Let’s proceed with a numerical problem for String Theory and
M-Theory.
19 22. SPACE-TIME SINGULARITIES IN M-THEORY
Problem 22. Consider a compactified M-theory scenario where the compact spatial dimension
has a radius of R= 10−17 meters. If an object is traveling at a speed of 0.9c(where cis the speed
of light) in this compactified dimension, determine the time duration (in seconds) it takes for the
object to travel halfway around the compactified dimension.
Solution 22. a) The circumference of the compact spatial dimension is given by 2πR. There-
fore, the distance required to travel halfway around the dimension is πR.
b) The speed of the object is 0.9c, where c= 3 ×108m/s. Hence, the object travels a distance
of 0.9c×t=πR in time t, where tis the time duration we want to find.
Using the equation 0.9c×t=πR, we can solve for t:
0.9×3×108m/s ×t=π×10−17 m
0.9×3×108t=π×10−17
2.7×108t= 3.1416 ×10−17
t=3.1416 ×10−17
2.7×108
t≈1.163 ×10−25 s
Therefore, it takes approximately 1.163 ×10−25 seconds for the object to travel halfway around
the compactified dimension.
20 23. DEFORMATIONS OF STRING THEORY
Problem 23. Consider a closed bosonic string of length Lmoving in Dspacetime dimensions.
The string is perturbed by a small deformation given by the following equation of motion:
∂2
∂τ 2−∂2
∂σ2Xµ(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ)
where Xµ(τ, σ)denotes the string embedding coordinates, µ, ν = 0,1, . . . , D −1,ϵis a small
parameter, and α′is the Regge slope parameter.
a) Compute the equation of motion for X0(τ, σ)and X1(τ, σ).
b) Determine the general solution for X0(τ, σ)and X1(τ, σ).
c) Given that the string is parameterized by (τ, σ)such that 0≤σ≤π, find the normal mode
frequencies for X0(τ, σ)and X1(τ, σ).
Solution 23.
a) The equation of motion can be split into separate equations for X0(τ, σ)and X1(τ, σ)by
setting µ= 0 and µ= 1 respectively:
∂2
∂τ 2X0(τ, σ)−∂2
∂σ2X0(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ) = ϵα′sin(2σ)∂0X1(τ, σ)
∂2
∂τ 2X1(τ, σ)−∂2
∂σ2X1(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ) = ϵα′sin(2σ)∂1X0(τ, σ)
b) To find the general solutions, we solve the wave equation for X0(τ, σ)and X1(τ, σ). The
solutions will have the form:
X0(τ, σ) = A(σ)eiω0τ+B(σ)e−iω0τ
X1(τ, σ) = C(σ)eiω1τ+D(σ)e−iω1τ
where A(σ),B(σ),C(σ), and D(σ)are functions of σ, and ω0and ω1are the normal mode
frequencies.
c) To find the normal mode frequencies, we substitute the solutions back into the equations of
motion and solve for ω0and ω1. After normalization, the normal mode frequencies are given by:
ω0=1
Lrn2−ϵα′n2
2
ω1=1
Lrn2+ϵα′n2
2
where n= 1,2,3, . . . represents the mode number.
21 24. QUANTUM FIELD THEORY LIMIT OF M-THEORY
Problem 24. Consider a closed string in 10 dimensions with a winding mode wrapping around
a circle of radius R. The momentum mode of the string has energy E=n
R, where nis an integer.
The winding mode has energy E=mR
α′, where mis an integer and α′is the Regge slope.
a) If the winding mode has energy E= 2π/α′and the momentum mode has energy E= 1/√α′,
find the values of mand n.
b) Determine the total energy ETof the closed string in terms of α′.
Solution 24.
a) Since the winding mode has energy E=mR
α′=2π
α′, we have:
mR
α′=2π
α′
mR = 2π
Similarly, for the momentum mode with energy E=n
R=1
√α′, we get:
n
R=1
√α′
nR =√α′
From these two equations, we can see that m= 1 and n= 1.
b) The total energy ETof the closed string is given by:
ET=n
R+mR
α′=1
R+R
α′=1 + R2/α′
R
Therefore, the total energy of the closed string in terms of α′is ET=1 + R2/α′
R.
22 25. TOPOLOGICAL ASPECTS OF STRING THEORY
Problem 25. Consider a closed oriented Riemann surface Σgof genus g. The Euler charac-
teristic of Σgis given by χ(Σg)=2−2g.
a) Calculate the Euler characteristic for a torus (g= 1).
b) Calculate the Euler characteristic for a sphere with two handles (g= 2).
c) Calculate the Euler characteristic for a surface with genus g= 3.
Solution 25.
a) For a torus (g= 1), the Euler characteristic is given by χ(Σ1) = 2 −2·1 = 2 −2 = 0.
Therefore, the Euler characteristic for a torus is 0.
b) For a sphere with two handles (g= 2), the Euler characteristic is χ(Σ2) = 2−2·2 = 2−4 = −2.
Hence, the Euler characteristic for a sphere with two handles is -2.
c) For a surface with genus g= 3, the Euler characteristic is χ(Σ3)=2−2·3=2−6 = −4.
Therefore, the Euler characteristic for a surface of genus 3 is -4.
3 3. QUANTUM GRAVITY IN STRING THEORY
Problem 3. Consider an open string moving in 4-dimensional spacetime. The string has tension
T= 1 and mass per unit length µ= 2.
a) Find the speed of propagation of waves on this string.
b) Calculate the energy stored in a segment of string of length L= 3.
c) If the string is stretched with a force of F= 5, determine the amplitude of the standing wave
that could form on the string.
Solution 3.
a) The speed of propagation of waves on a string is given by
v=sT
µ.
Substitute T= 1 and µ= 2:
v=r1
2=√2
2.
Therefore, the speed of propagation of waves on this string is √2
2.
b) The energy stored in a segment of string of length Lis given by
E=1
2µv2L.
Substitute µ= 2,v=√2
2, and L= 3:
E=1
2×2× √2
2!2
×3 = 3
2.
Therefore, the energy stored in a segment of string of length 3 is 3
2.
c) The amplitude of the standing wave that could form on the string under the applied force F
is given by
A=F
2πv2.
Substitute F= 5 and v=√2
2:
A=5
2π×√2
22=5
2π×1
2
=5√2
π.
Therefore, the amplitude of the standing wave that could form on the string is 5√2
π.
4 4. BLACK HOLES AND STRING THEORY
Problem 4. Consider a black hole in four-dimensional spacetime described by the Schwarzschild
metric:
ds2=−1−2GM
c2rc2dt2+1−2GM
c2r−1
dr2+r2dΩ2
where Mis the mass of the black hole, cis the speed of light, Gis the gravitational constant,
and dΩ2=dθ2+sin2θdϕ2in spherical coordinates (t, r, θ, ϕ).
a) Find the event horizon radius Rhorizon of this black hole.
b) Determine the Schwarzschild radius RSin terms of M,G, and c.
c) Given that the mass of the black hole is M= 2 ×1030 kg, calculate the mass of the black
hole in units of solar masses (M⊙= 1.989 ×1030 kg).
Solution 4.
a) To find the event horizon radius, we set the metric coefficient of dt2to zero at the event
horizon. So, 1−2GM
c2Rhorizon = 0. Solving for Rhorizon gives:
Rhorizon =2GM
c2
b) The Schwarzschild radius is defined as the radius at which the metric becomes singular. It
is given by RS=2GM
c2.
c) Substituting M= 2 ×1030 kg into the Schwarzschild radius formula, we have:
RS=2G(2 ×1030)
c2=4×1030G
c2
Now, we can express the mass of the black hole in terms of solar masses by dividing by the
mass of the Sun:
4×1030G
c2÷1.989 ×1030 =2G
c2
Therefore, the mass of the black hole is 2solar masses.
5 5. TACHYON CONDENSATION IN STRING THEORY
Problem 5. Consider a closed bosonic string theory where the endpoint of the string coor-
dinates are subject to Neumann boundary conditions. The first excited level of the closed string
contains a tachyon with mass m2=−2
α′.
a) Calculate the momentum of the tachyon state in the string theory.
b) Show that the mass of the tachyon state in the open bosonic string theory is m= 0.
c) Interpret the result in relation to tachyon condensation.
Solution 5.
a) The mass-squared of a state in a closed string theory is given by the level matching condition:
m2=4
α′(N−1)
where Nis the occupation number operator for the state. For the first excited level, N= 1 and
m2=−2
α′. Substituting these values into the equation above, we have:
−2
α′=4
α′(1 −1)
−2=0
This equation is a contradiction, indicating that there is no physical state with a mass-squared of
−2
α′for the first excited level. So, the tachyon state does not exist.
b) In the open bosonic string theory, the mass-squared of a state is given by:
m2=2
α′(N−1)
For the tachyon state at the first excited level in the open string, N= 1 and m2=−1
α′. Taking the
square root of this, we get m= 0. Therefore, the mass of the tachyon state in the open bosonic
string theory is zero.
c) The result m= 0 for the open string tachyon state implies that it is a massless state. In string
theory, a tachyon has negative mass squared and indicates an instability in the theory. Tachyon
condensation is a process where this unstable state "condenses" to a minimum energy state. In
this case, the tachyon state in the open bosonic string theory having a mass of zero suggests that
it represents the minimum energy state after tachyon condensation has occurred. This process
helps stabilize the theory by eliminating the instability caused by the presence of the tachyon.
6 6. EXTRA DIMENSIONS IN M-THEORY
Problem 6. Consider a scenario in M-Theory where there are 7 spatial dimensions and 3
temporal dimensions. The size of the compactified extra dimensions is given by R= 10−17 meters.
Calculate the compactification scale in GeV.
Solution 6.
a) The compactification scale Mccan be calculated using the formula for the compactified extra
dimensions:
Mc=1
R
Substitute R= 10−17 meters into the formula:
Mc=1
10−17 = 1017 m−1
b) To convert the compactification scale from meters to GeV, we need to use the relation 1GeV =
1.97 ×10−16 m−1.
Let’s convert the compactification scale:
Mc= 1017 ×1.97 ×10−16 = 1.97 ×10 GeV = 19.7GeV
Therefore, the compactification scale in GeV is 19.7 GeV.
7 7. BRANE DYNAMICS IN STRING THEORY
Problem 7. Consider a type IIB superstring theory in 10 dimensions with D3-branes. The
tension of the D3-brane is given by T=1
(2π)3(α′)2, where α′is the string length parameter. Calculate
the energy density stored in a D3-brane of length L= 10√α′.
Solution 7. a) The energy density stored in a D3-brane can be calculated by dividing its energy
by its volume. The energy Eof a D3-brane is given by E=T×V, where Tis the tension of the
D3-brane and Vis its volume. Since the D3-brane extends in 3 spatial dimensions, the volume V
is L3= (10√α′)3= 1000α′3/2.
Substitute T=1
(2π)3(α′)2and V= 1000α′3/2into the equation:
E=1
(2π)3(α′)2×1000α′3/2=1000
(2π)3α′1/2
Thus, the energy stored in the D3-brane is 1000
(2π)3α′1/2.
b) The energy density is given by the energy per unit volume. We divide the energy Eby the
volume Vto find the energy density u=E
V.
Substitute E=1000
(2π)3α′1/2and V= 1000α′3/2into the equation:
u=1000/(2π)3α′1/2
1000α′3/2=1
(2π)3α′2
Therefore, the energy density stored in the D3-brane is 1
(2π)3α′2.
8 8. SUPERSYMMETRY BREAKING IN M-THEORY
Problem 8. Consider a supersymmetric M-Theory compactified on a 2-torus with radii R1and
R2. The volume of the torus is given by V=R1R2. Suppose that supersymmetry is broken by the
flux through the torus such that the gravitino mass term is generated.
[Given: The gravitino mass term is given by m3/2=e⟨G⟩, where Gis the flux. Also, we have
⟨G⟩ =−2 ln(V), where ⟨G⟩ denotes the expectation value of the flux.]
a) Show that the gravitino mass term m3/2in terms of the radii R1and R2.
b) If R1= 2 and R2= 3, calculate the gravitino mass m3/2.
Solution 8.
a) To find the gravitino mass term m3/2in terms of the radii R1and R2, we first need to express
the volume Vin terms of R1and R2:
V=R1·R2
Next, we can find the expectation value of the flux ⟨G⟩:
⟨G⟩ =−2 ln(V) = −2 ln(R1·R2) = −2 ln(R1)−2 ln(R2)
Substitute this into the expression for the gravitino mass term:
m3/2=e⟨G⟩ =e−2 ln(R1)−2 ln(R2)=e−2 ln(R1)·e−2 ln(R2)=1
R2
1·1
R2
2
=1
R2
1R2
2
=1
V2
b) Given R1= 2 and R2= 3, the volume V=R1·R2= 2 ·3=6. Therefore, the gravitino mass
term m3/2is:
m3/2=1
V2=1
62=1
36 = 0.0278
I. Problem 9. Consider a Type IIA superstring theory with a D6-brane wrapped on a compact
3-torus with sides of length L. The tension of the D6-brane is given by T6=1
(2π)6(α′)4. Suppose
the compactified space is a cube, compute the energy density UD6 of the D6-brane in terms of L.
Hint: The energy density is defined as the energy per unit volume.
II. Problem 10. In Type IIB superstring theory, a D3-brane has tension T3=1
(2π)3(α′)2. If this
theory is in a 10-dimensional spacetime with a toroidal compactification down to 6 dimensions, and
the 3-brane extends along all 3 of those compact dimensions, calculate the energy density UD3 of
the D3-brane in terms of the compactification radii Ri.
Hint: The energy density is defined as the energy per unit volume.
III. Problem 11. In M-Theory, consider a M5-brane wrapped on a torus with radii R1and R2. If
the tension of the brane is T5=1
(2π)5(ℓp)3, find the energy density UM5 of the M5-brane in terms of
the torus radii.
Hint: The energy density is defined as the energy per unit volume.
9 10. COSMOLOGY AND STRING/M-THEORY
Problem 10. Consider a string theory model in a universe with extra dimensions compactified
on a torus. The radius of the torus is R, and the string tension is T. The compactified dimensions
are described by an effective field theory with a massless scalar field ϕ, whose potential is given
by V(ϕ) = 1
2m2ϕ2.
a) Show that the effective tension in the compactified dimensions, Teff , is given by Tef f =
T e−2ϕ.
b) Determine the equation of motion for the scalar field ϕ.
c) Find the minimum of the potential V(ϕ).
Solution 10.
a) The effective tension in the compactified dimensions is given by Teff =T e−2ϕ. Let’s derive
this expression:
In string theory, the effective tension depends on the string coupling as Teff =T gs, where
gs=e−2ϕ. Therefore, Teff =T e−2ϕ.
b) The equation of motion for the scalar field ϕis given by:
d
dt ∂L
∂˙
ϕ−∂L
∂ϕ = 0
where the Lagrangian L=1
2˙
ϕ2−V(ϕ).
d
dt ∂L
∂˙
ϕ−∂L
∂ϕ =¨
ϕ+m2ϕ= 0
Therefore, the equation of motion for the scalar field ϕis ¨
ϕ+m2ϕ= 0.
c) To find the minimum of the potential V(ϕ), we need to solve dV
dϕ = 0:
dV
dϕ =m2ϕ= 0
This implies that the minimum of the potential occurs at ϕ= 0.
10 11. DUALITIES IN M-THEORY
Problem 11. Consider two particular string theories, Type IIA and Type IIB, related by T-duality.
Let the radius of a circle in Type IIA theory be R. If the number of fundamental strings winding
around the circle is n, find the dual circle radius in Type IIB theory.
Solution 11. Given: Radius of circle in Type IIA theory, R, and number of winding fundamental
strings, n.
In Type IIA theory, the momentum along the circle is given by P=n
R.
By T-duality, the winding number in Type IIB string theory is the same as the momentum in Type
IIA theory, and vice versa. Therefore, in Type IIB theory, the radius of the dual circle is related to
the momentum by Rdual =α′
R.
Substitute P=n
Rinto the formula for the dual radius in Type IIB theory:
Rdual =α′
R=α′
n/P =α′R
n
Hence, the dual circle radius in Type IIB theory is α′R
n.
11 12. INTEGRABILITY IN STRING THEORY
Problem 12. Consider a closed bosonic string moving in a background with a constant mag-
netic field Bin the z-direction. The equation of motion for the string is given by the Nambu-Goto
action
S=−T
2Zdσdτ√−h hab∂aXµ∂bXνGµν
where Tis the tension of the string, his the determinant of the world-sheet metric hab,Xµ=
Xµ(τ, σ)are the embedding coordinates of the string world-sheet, and Gµν is the background metric
tensor.
Given that the background metric is flat Minkowski space with Gµν =ηµν and the magnetic field
is B=Bz, where Bzis constant, compute the equation of motion for the string in this background.
Solution 12.
The equation of motion for the string can be derived by varying the action with respect to the
embedding coordinates Xµ.
Let’s denote ∂aXµ≡∂aXµ(τ, σ).
The variation of the action with respect to Xµgives the equation of motion:
δS
δXµ=−T
2Zdσdτ√−hhab∂a∂bXνηνµ = 0
Expanding the terms and using the fact that the world-sheet metric is hab =diag(−1,1), we get:
TZdσdτ ∂2
τXν−∂2
σXνηνµ = 0
The equations of motion are then given by:
a) In the µ= 0 direction:
∂2
τX0−∂2
σX0= 0
b) In the µ= 1 direction:
∂2
τX1−∂2
σX1= 0
c) In the µ= 2 direction:
∂2
τX2−∂2
σX2= 0
These equations of motion describe the dynamics of the string in the background of a constant
magnetic field B=Bz.
12 13. CAUSALITY AND STRING/M-THEORY
Problem 13. Consider a closed string propagating in a spacetime described by D= 10
dimensions. The string moves in a geometry where the background metric is given by ds2=
−dt2+dx2
1+ (dx2)2+···+ (dx8)2+ (dx9)2+ (dx10)2, where x10 is the compactified spatial direction
with a radius R.
a) Calculate the maximum energy Emax of a closed string state that can propagate in this ge-
ometry without creating a closed timelike curve.
b) Find the minimum allowed radius Rmin of the compactified spatial direction that ensures
causality is not violated in this spacetime.
Solution 13.
a) The maximum energy Emax of a closed string state can be found using the formula:
Emax =1
α′rN
2
where Nis the level of the state and α′is the Regge slope parameter. Since we are dealing with
a closed string in D= 10 dimensions, the critical dimension is D= 10, and the Regge slope
parameter is α′= 1/2πT .
For a closed string moving in a compactified dimension with radius R, the contribution to the
mass in the compact direction is n/R where nis the winding number. To avoid closed timelike
curves, we must have Emax =n/R. Equating these two expressions for Emax, we get:
n
R=1
α′rN
2
n
R=2πT
2rN
2
nR = 2πT √2N
R=2πT √2N
n
Therefore, the maximum energy is achieved at n= 1 and N= 2, leading to:
Emax =1
α′rN
2=1
α′r2
2=1
α′=√2πT
2
b) To ensure that causality is not violated in this spacetime, we must have the inequality Rmin ≥
2πR to avoid closed timelike curves. Substituting in the expression for Rmin from part (a), we get:
Rmin =2πT p2(1)
1= 2π√2πT ≥2πR
√2πT ≥R
2πT ≥R2
R2−2πT ≤0
Hence, the minimum allowed radius Rmin of the compactified spatial direction is R=√2πT .
13 14. HOLOGRAPHY IN M-THEORY
Problem 14. Consider a spacetime described by M-theory with 11 dimensions. The holo-
graphic principle states that the information of a region of space can be encoded on its boundary.
Suppose we have a 4-dimensional hypercube with side length L.
a) Calculate the volume of this hypercube in terms of L.
b) According to the holographic principle, what is the size of the boundary that encodes all the
information of this hypercube?
c) If the hypercube is in an 11-dimensional spacetime in M-theory, how many spatial dimensions
are "compactified"?
Solution 14.
a) The volume of a 4-dimensional hypercube with side length Lis given by V=L4.
b) The boundary of the hypercube is a 3-dimensional cube with sides of length L. The total
surface area of a cube is given by A= 6L2. Hence, for a 4-dimensional hypercube, the size of the
boundary that encodes all the information is A= 6L2.
c) In M-theory with 11 dimensions, the 4 spatial dimensions of the hypercube and the 1 time
dimension are known. This leaves 11 −4−1=6spatial dimensions "compactified".
I’m glad to help generate numerical problem questions for you. Could you please specify the
topic within String Theory and M-Theory that you would like the problem to be based on?
14 16. DARK MATTER AND STRING/M-THEORY
Problem 16. Consider a string theory scenario where a closed string moving in a compact
spatial dimension of radius Rinteracts with dark matter particles gravitationally. The string coupling
constant is given by gs= 0.1.
a) If the mass of the dark matter particle is m= 10−22 eV/c2, calculate the gravitational force
between the closed string and a dark matter particle located at a distance r= 1 mm.
b) If the string coupling constant gsis changed to 0.05, how does this affect the gravitational
force between the closed string and the dark matter particle?
Solution 16.
a) The gravitational force between the closed string and the dark matter particle can be calcu-
lated using Newton’s law of gravitation:
F=G·mdark matter ·mstring
r2
where Gis the gravitational constant (6.67430 ×10−11 m3kg−1s−2), mdark matter is the mass of
the dark matter particle (in kg), mstring is the mass of the closed string (in kg), and ris the distance
between the closed string and the dark matter particle (in meters).
Given m= 10−22 eV/c2, we need to convert this to kilograms by using the conversion factor
1eV/c2= 1.783 ×10−36 kg. Thus, mdark matter = 10−22 ×1.783 ×10−36 = 1.783 ×10−58 kg.
Plugging in the values G= 6.67430 ×10−11 m3kg−1s−2,mdark matter = 1.783 ×10−58 kg,
mstring =? (mass of the string is not provided but let’s assume it to be of the same order of magnitude
as a dark matter particle), and r= 1 mm = 0.001 m, we can find the gravitational force.
b) If the string coupling constant gsis changed to 0.05, we can re-calculate the gravitational
force using the new value of gsin the formula above. Let’s assume that the mass of the dark matter
particle remains the same.
Solution 16.
a) Given: G= 6.67430 ×10−11 m3kg−1s−2,mdark matter = 1.783 ×10−58 kg, mstring = 1.783 ×
10−58 kg (assumed to be of the same order), r= 0.001 m
Plugging in the values, we get:
F=6.67430 ×10−11 ×1.783 ×10−58 ×1.783 ×10−58
(0.001)2
⇒F=2.0004 ×10−124
10−6
⇒F= 2.0004 ×10−118 N
Thus, the gravitational force between the closed string and a dark matter particle located at a
distance of 1 mm is approximately 2.0004 ×10−118 N.
b) If gs= 0.05, we can re-calculate the gravitational force using the updated value of gsin the
formula. Plugging in the new value, we can find the new gravitational force.
I’m sorry, but it seems that I cannot generate numerical problems for String Theory and M-
Theory as they are primarily theoretical and mathematical physics concepts that do not involve
numerical computations. Would you like a conceptual problem or a theoretical problem instead?
15 18. VACUUM ENERGY IN M-THEORY
Problem 18. Consider a particular compactification of M-theory on a 5-dimensional torus with
each side having length R. The vacuum energy in this scenario is given by ρ=−1
(2π)5R9.
a) Calculate the vacuum energy ρin GeV4when R= 1 cm. b) Find the value of Rin m for
which the vacuum energy is ρ=−1GeV4.
Solution 18. a) To calculate the vacuum energy in GeV4, we need to convert the length R=
1cm to meters and then substitute it into the given formula.
Given: 1 cm = 10−2m
Plugging this value into the formula for vacuum energy: ρ=−1
(2π)5(10−2)9
ρ=−1
(2π)5(10−18)
ρ=−1
(2π)5(10−18)
ρ=−1
(2π)5×10−18
ρ=−1
(2π)5×10−18
ρ≈ −8.53254 ×1065 GeV4
Therefore, the vacuum energy when R= 1 cm is approximately −8.53254 ×1065 GeV4.
b) To find the value of Rin meters for which the vacuum energy is ρ=−1GeV4, we set the
vacuum energy formula equal to −1and solve for R:
−1 = −1
(2π)5R9
R9= (2π)5
R=9
p(2π)5
R≈3.28626 ×10−16 m
Therefore, the value of Rin meters for which the vacuum energy is ρ=−1GeV4is approxi-
mately 3.28626 ×10−16 m.
16 19. SOLITONS IN STRING THEORY
Problem 19. Consider a D-brane in type IIA superstring theory with tension T. Suppose the
D-brane extends in pspatial dimensions. The energy density Uper unit p-dimensional volume of
the D-brane is given by U=cT , where cis a constant.
a) If the D-brane has p= 3 spatial dimensions and the tension is T= 5 ×10−2, find the energy
density U.
b) Now, let’s consider another D-brane with tension T= 0.1and energy density U= 0.4.
Determine the number of spatial dimensions pin which this D-brane extends.
Solution 19.
a) Given p= 3 and T= 5 ×10−2, we use the formula U=cT to find the energy density U:
U=c×5×10−2= 0.05c
So, the energy density Ufor the D-brane with p= 3 spatial dimensions and tension T= 5×10−2
is 0.05c.
b) For this case, we have T= 0.1and U= 0.4. Using the same formula for energy density,
U=cT , we can solve for cfirst:
c=U
T=0.4
0.1= 4
Now, we substitute c= 4 back into the equation U=cT :
0.4=4×p
Solving for p, we find:
p=0.4
4= 0.1
Thus, the D-brane extends in p= 0.1spatial dimensions.
I. Let’s formulate a numerical problem in the context of String Theory and M-Theory:
17 20. ADS/CFT CORRESPONDENCE IN M-THEORY
Problem 20. Consider a type IIA string theory on an Anti-de Sitter space (AdS) with a five-
dimensional radius RAdS = 10. Calculate the corresponding conformal field theory (CFT) central
charge using the AdS/CFT correspondence formula:
c=3L
2G
where Lis the AdS radius and Gis Newton’s constant. Take the value of Newton’s constant
G= 6.71 ×10−11 m3kg−1s−2.
Solution 20. Given: - AdS radius RAdS = 10 - Newton’s constant G= 6.71 ×10−11 m3kg−1
s−2
The CFT central charge can be calculated using the AdS/CFT correspondence formula:
c=3L
2G
Plugging in the values:
c=3×10
2×6.71 ×10−11
c=30
13.42 ×10−11
c=30
1.342 ×10−10
c=30
1.342
c≈22.36
Therefore, the central charge of the corresponding conformal field theory is approximately
22.36.
I. Entanglement Entropy in String Theory:
18 21. Entanglement Entropy in String Theory
Problem 21. Consider a 1+1 dimensional conformal field theory described by a CFT with
central charge c= 1. Let’s calculate the entanglement entropy of a subsystem in this theory.
[Additional context: The entanglement entropy Sof a subsystem in a CFT is given by the formula
S=c
3log L
ϵ, where cis the central charge, Lis the length of the subsystem, and ϵis a short-
distance cutoff.]
a) Suppose we have a subsystem of length L= 2 in this CFT. Calculate the entanglement
entropy when the short-distance cutoff ϵ= 0.1.
b) Now, consider another CFT described by a different conformal field theory with central charge
c= 2. If we have a subsystem of length L= 3 in this CFT, what is the entanglement entropy when
the short-distance cutoff ϵ= 0.01?
Solution 21.
a) We are given c= 1,L= 2, and ϵ= 0.1. Plugging these values into the formula for entangle-
ment entropy, we get:
S=1
3log 2
0.1=1
3log 20 ≈1
3×2.9957 ≈0.9986
Therefore, the entanglement entropy when L= 2spaceunits and ϵ= 0.1is approximately
0.9986.
b) For the new CFT with c= 2,L= 3, and ϵ= 0.01, we apply the formula:
S=2
3log 3
0.01=2
3log 300 ≈2
3×5.7038 ≈3.8026
Therefore, the entanglement entropy when L= 3spaceunits and ϵ= 0.01 is approximately
3.8026.
I can certainly help with that! Let’s proceed with a numerical problem for String Theory and
M-Theory.
19 22. SPACE-TIME SINGULARITIES IN M-THEORY
Problem 22. Consider a compactified M-theory scenario where the compact spatial dimension
has a radius of R= 10−17 meters. If an object is traveling at a speed of 0.9c(where cis the speed
of light) in this compactified dimension, determine the time duration (in seconds) it takes for the
object to travel halfway around the compactified dimension.
Solution 22. a) The circumference of the compact spatial dimension is given by 2πR. There-
fore, the distance required to travel halfway around the dimension is πR.
b) The speed of the object is 0.9c, where c= 3 ×108m/s. Hence, the object travels a distance
of 0.9c×t=πR in time t, where tis the time duration we want to find.
Using the equation 0.9c×t=πR, we can solve for t:
0.9×3×108m/s ×t=π×10−17 m
0.9×3×108t=π×10−17
2.7×108t= 3.1416 ×10−17
t=3.1416 ×10−17
2.7×108
t≈1.163 ×10−25 s
Therefore, it takes approximately 1.163 ×10−25 seconds for the object to travel halfway around
the compactified dimension.
20 23. DEFORMATIONS OF STRING THEORY
Problem 23. Consider a closed bosonic string of length Lmoving in Dspacetime dimensions.
The string is perturbed by a small deformation given by the following equation of motion:
∂2
∂τ 2−∂2
∂σ2Xµ(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ)
where Xµ(τ, σ)denotes the string embedding coordinates, µ, ν = 0,1, . . . , D −1,ϵis a small
parameter, and α′is the Regge slope parameter.
a) Compute the equation of motion for X0(τ, σ)and X1(τ, σ).
b) Determine the general solution for X0(τ, σ)and X1(τ, σ).
c) Given that the string is parameterized by (τ, σ)such that 0≤σ≤π, find the normal mode
frequencies for X0(τ, σ)and X1(τ, σ).
Solution 23.
a) The equation of motion can be split into separate equations for X0(τ, σ)and X1(τ, σ)by
setting µ= 0 and µ= 1 respectively:
∂2
∂τ 2X0(τ, σ)−∂2
∂σ2X0(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ) = ϵα′sin(2σ)∂0X1(τ, σ)
∂2
∂τ 2X1(τ, σ)−∂2
∂σ2X1(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ) = ϵα′sin(2σ)∂1X0(τ, σ)
b) To find the general solutions, we solve the wave equation for X0(τ, σ)and X1(τ, σ). The
solutions will have the form:
X0(τ, σ) = A(σ)eiω0τ+B(σ)e−iω0τ
X1(τ, σ) = C(σ)eiω1τ+D(σ)e−iω1τ
where A(σ),B(σ),C(σ), and D(σ)are functions of σ, and ω0and ω1are the normal mode
frequencies.
c) To find the normal mode frequencies, we substitute the solutions back into the equations of
motion and solve for ω0and ω1. After normalization, the normal mode frequencies are given by:
ω0=1
Lrn2−ϵα′n2
2
ω1=1
Lrn2+ϵα′n2
2
where n= 1,2,3, . . . represents the mode number.
21 24. QUANTUM FIELD THEORY LIMIT OF M-THEORY
Problem 24. Consider a closed string in 10 dimensions with a winding mode wrapping around
a circle of radius R. The momentum mode of the string has energy E=n
R, where nis an integer.
The winding mode has energy E=mR
α′, where mis an integer and α′is the Regge slope.
a) If the winding mode has energy E= 2π/α′and the momentum mode has energy E= 1/√α′,
find the values of mand n.
b) Determine the total energy ETof the closed string in terms of α′.
Solution 24.
a) Since the winding mode has energy E=mR
α′=2π
α′, we have:
mR
α′=2π
α′
mR = 2π
Similarly, for the momentum mode with energy E=n
R=1
√α′, we get:
n
R=1
√α′
nR =√α′
From these two equations, we can see that m= 1 and n= 1.
b) The total energy ETof the closed string is given by:
ET=n
R+mR
α′=1
R+R
α′=1 + R2/α′
R
Therefore, the total energy of the closed string in terms of α′is ET=1 + R2/α′
R.
22 25. TOPOLOGICAL ASPECTS OF STRING THEORY
Problem 25. Consider a closed oriented Riemann surface Σgof genus g. The Euler charac-
teristic of Σgis given by χ(Σg)=2−2g.
a) Calculate the Euler characteristic for a torus (g= 1).
b) Calculate the Euler characteristic for a sphere with two handles (g= 2).
c) Calculate the Euler characteristic for a surface with genus g= 3.
Solution 25.
a) For a torus (g= 1), the Euler characteristic is given by χ(Σ1) = 2 −2·1 = 2 −2 = 0.
Therefore, the Euler characteristic for a torus is 0.
b) For a sphere with two handles (g= 2), the Euler characteristic is χ(Σ2) = 2−2·2 = 2−4 = −2.
Hence, the Euler characteristic for a sphere with two handles is -2.
c) For a surface with genus g= 3, the Euler characteristic is χ(Σ3)=2−2·3=2−6 = −4.
Therefore, the Euler characteristic for a surface of genus 3 is -4.
3 3. QUANTUM GRAVITY IN STRING THEORY
Problem 3. Consider an open string moving in 4-dimensional spacetime. The string has tension
T= 1 and mass per unit length µ= 2.
a) Find the speed of propagation of waves on this string.
b) Calculate the energy stored in a segment of string of length L= 3.
c) If the string is stretched with a force of F= 5, determine the amplitude of the standing wave
that could form on the string.
Solution 3.
a) The speed of propagation of waves on a string is given by
v=sT
µ.
Substitute T= 1 and µ= 2:
v=r1
2=√2
2.
Therefore, the speed of propagation of waves on this string is √2
2.
b) The energy stored in a segment of string of length Lis given by
E=1
2µv2L.
Substitute µ= 2,v=√2
2, and L= 3:
E=1
2×2× √2
2!2
×3 = 3
2.
Therefore, the energy stored in a segment of string of length 3 is 3
2.
c) The amplitude of the standing wave that could form on the string under the applied force F
is given by
A=F
2πv2.
Substitute F= 5 and v=√2
2:
A=5
2π×√2
22=5
2π×1
2
=5√2
π.
Therefore, the amplitude of the standing wave that could form on the string is 5√2
π.
4 4. BLACK HOLES AND STRING THEORY
Problem 4. Consider a black hole in four-dimensional spacetime described by the Schwarzschild
metric:
ds2=−1−2GM
c2rc2dt2+1−2GM
c2r−1
dr2+r2dΩ2
where Mis the mass of the black hole, cis the speed of light, Gis the gravitational constant,
and dΩ2=dθ2+sin2θdϕ2in spherical coordinates (t, r, θ, ϕ).
a) Find the event horizon radius Rhorizon of this black hole.
b) Determine the Schwarzschild radius RSin terms of M,G, and c.
c) Given that the mass of the black hole is M= 2 ×1030 kg, calculate the mass of the black
hole in units of solar masses (M⊙= 1.989 ×1030 kg).
Solution 4.
a) To find the event horizon radius, we set the metric coefficient of dt2to zero at the event
horizon. So, 1−2GM
c2Rhorizon = 0. Solving for Rhorizon gives:
Rhorizon =2GM
c2
b) The Schwarzschild radius is defined as the radius at which the metric becomes singular. It
is given by RS=2GM
c2.
c) Substituting M= 2 ×1030 kg into the Schwarzschild radius formula, we have:
RS=2G(2 ×1030)
c2=4×1030G
c2
Now, we can express the mass of the black hole in terms of solar masses by dividing by the
mass of the Sun:
4×1030G
c2÷1.989 ×1030 =2G
c2
Therefore, the mass of the black hole is 2solar masses.
5 5. TACHYON CONDENSATION IN STRING THEORY
Problem 5. Consider a closed bosonic string theory where the endpoint of the string coor-
dinates are subject to Neumann boundary conditions. The first excited level of the closed string
contains a tachyon with mass m2=−2
α′.
a) Calculate the momentum of the tachyon state in the string theory.
b) Show that the mass of the tachyon state in the open bosonic string theory is m= 0.
c) Interpret the result in relation to tachyon condensation.
Solution 5.
a) The mass-squared of a state in a closed string theory is given by the level matching condition:
m2=4
α′(N−1)
where Nis the occupation number operator for the state. For the first excited level, N= 1 and
m2=−2
α′. Substituting these values into the equation above, we have:
−2
α′=4
α′(1 −1)
−2=0
This equation is a contradiction, indicating that there is no physical state with a mass-squared of
−2
α′for the first excited level. So, the tachyon state does not exist.
b) In the open bosonic string theory, the mass-squared of a state is given by:
m2=2
α′(N−1)
For the tachyon state at the first excited level in the open string, N= 1 and m2=−1
α′. Taking the
square root of this, we get m= 0. Therefore, the mass of the tachyon state in the open bosonic
string theory is zero.
c) The result m= 0 for the open string tachyon state implies that it is a massless state. In string
theory, a tachyon has negative mass squared and indicates an instability in the theory. Tachyon
condensation is a process where this unstable state "condenses" to a minimum energy state. In
this case, the tachyon state in the open bosonic string theory having a mass of zero suggests that
it represents the minimum energy state after tachyon condensation has occurred. This process
helps stabilize the theory by eliminating the instability caused by the presence of the tachyon.
6 6. EXTRA DIMENSIONS IN M-THEORY
Problem 6. Consider a scenario in M-Theory where there are 7 spatial dimensions and 3
temporal dimensions. The size of the compactified extra dimensions is given by R= 10−17 meters.
Calculate the compactification scale in GeV.
Solution 6.
a) The compactification scale Mccan be calculated using the formula for the compactified extra
dimensions:
Mc=1
R
Substitute R= 10−17 meters into the formula:
Mc=1
10−17 = 1017 m−1
b) To convert the compactification scale from meters to GeV, we need to use the relation 1GeV =
1.97 ×10−16 m−1.
Let’s convert the compactification scale:
Mc= 1017 ×1.97 ×10−16 = 1.97 ×10 GeV = 19.7GeV
Therefore, the compactification scale in GeV is 19.7 GeV.
7 7. BRANE DYNAMICS IN STRING THEORY
Problem 7. Consider a type IIB superstring theory in 10 dimensions with D3-branes. The
tension of the D3-brane is given by T=1
(2π)3(α′)2, where α′is the string length parameter. Calculate
the energy density stored in a D3-brane of length L= 10√α′.
Solution 7. a) The energy density stored in a D3-brane can be calculated by dividing its energy
by its volume. The energy Eof a D3-brane is given by E=T×V, where Tis the tension of the
D3-brane and Vis its volume. Since the D3-brane extends in 3 spatial dimensions, the volume V
is L3= (10√α′)3= 1000α′3/2.
Substitute T=1
(2π)3(α′)2and V= 1000α′3/2into the equation:
E=1
(2π)3(α′)2×1000α′3/2=1000
(2π)3α′1/2
Thus, the energy stored in the D3-brane is 1000
(2π)3α′1/2.
b) The energy density is given by the energy per unit volume. We divide the energy Eby the
volume Vto find the energy density u=E
V.
Substitute E=1000
(2π)3α′1/2and V= 1000α′3/2into the equation:
u=1000/(2π)3α′1/2
1000α′3/2=1
(2π)3α′2
Therefore, the energy density stored in the D3-brane is 1
(2π)3α′2.
8 8. SUPERSYMMETRY BREAKING IN M-THEORY
Problem 8. Consider a supersymmetric M-Theory compactified on a 2-torus with radii R1and
R2. The volume of the torus is given by V=R1R2. Suppose that supersymmetry is broken by the
flux through the torus such that the gravitino mass term is generated.
[Given: The gravitino mass term is given by m3/2=e⟨G⟩, where Gis the flux. Also, we have
⟨G⟩ =−2 ln(V), where ⟨G⟩ denotes the expectation value of the flux.]
a) Show that the gravitino mass term m3/2in terms of the radii R1and R2.
b) If R1= 2 and R2= 3, calculate the gravitino mass m3/2.
Solution 8.
a) To find the gravitino mass term m3/2in terms of the radii R1and R2, we first need to express
the volume Vin terms of R1and R2:
V=R1·R2
Next, we can find the expectation value of the flux ⟨G⟩:
⟨G⟩ =−2 ln(V) = −2 ln(R1·R2) = −2 ln(R1)−2 ln(R2)
Substitute this into the expression for the gravitino mass term:
m3/2=e⟨G⟩ =e−2 ln(R1)−2 ln(R2)=e−2 ln(R1)·e−2 ln(R2)=1
R2
1·1
R2
2
=1
R2
1R2
2
=1
V2
b) Given R1= 2 and R2= 3, the volume V=R1·R2= 2 ·3=6. Therefore, the gravitino mass
term m3/2is:
m3/2=1
V2=1
62=1
36 = 0.0278
I. Problem 9. Consider a Type IIA superstring theory with a D6-brane wrapped on a compact
3-torus with sides of length L. The tension of the D6-brane is given by T6=1
(2π)6(α′)4. Suppose
the compactified space is a cube, compute the energy density UD6 of the D6-brane in terms of L.
Hint: The energy density is defined as the energy per unit volume.
II. Problem 10. In Type IIB superstring theory, a D3-brane has tension T3=1
(2π)3(α′)2. If this
theory is in a 10-dimensional spacetime with a toroidal compactification down to 6 dimensions, and
the 3-brane extends along all 3 of those compact dimensions, calculate the energy density UD3 of
the D3-brane in terms of the compactification radii Ri.
Hint: The energy density is defined as the energy per unit volume.
III. Problem 11. In M-Theory, consider a M5-brane wrapped on a torus with radii R1and R2. If
the tension of the brane is T5=1
(2π)5(ℓp)3, find the energy density UM5 of the M5-brane in terms of
the torus radii.
Hint: The energy density is defined as the energy per unit volume.
9 10. COSMOLOGY AND STRING/M-THEORY
Problem 10. Consider a string theory model in a universe with extra dimensions compactified
on a torus. The radius of the torus is R, and the string tension is T. The compactified dimensions
are described by an effective field theory with a massless scalar field ϕ, whose potential is given
by V(ϕ) = 1
2m2ϕ2.
a) Show that the effective tension in the compactified dimensions, Teff , is given by Tef f =
T e−2ϕ.
b) Determine the equation of motion for the scalar field ϕ.
c) Find the minimum of the potential V(ϕ).
Solution 10.
a) The effective tension in the compactified dimensions is given by Teff =T e−2ϕ. Let’s derive
this expression:
In string theory, the effective tension depends on the string coupling as Teff =T gs, where
gs=e−2ϕ. Therefore, Teff =T e−2ϕ.
b) The equation of motion for the scalar field ϕis given by:
d
dt ∂L
∂˙
ϕ−∂L
∂ϕ = 0
where the Lagrangian L=1
2˙
ϕ2−V(ϕ).
d
dt ∂L
∂˙
ϕ−∂L
∂ϕ =¨
ϕ+m2ϕ= 0
Therefore, the equation of motion for the scalar field ϕis ¨
ϕ+m2ϕ= 0.
c) To find the minimum of the potential V(ϕ), we need to solve dV
dϕ = 0:
dV
dϕ =m2ϕ= 0
This implies that the minimum of the potential occurs at ϕ= 0.
10 11. DUALITIES IN M-THEORY
Problem 11. Consider two particular string theories, Type IIA and Type IIB, related by T-duality.
Let the radius of a circle in Type IIA theory be R. If the number of fundamental strings winding
around the circle is n, find the dual circle radius in Type IIB theory.
Solution 11. Given: Radius of circle in Type IIA theory, R, and number of winding fundamental
strings, n.
In Type IIA theory, the momentum along the circle is given by P=n
R.
By T-duality, the winding number in Type IIB string theory is the same as the momentum in Type
IIA theory, and vice versa. Therefore, in Type IIB theory, the radius of the dual circle is related to
the momentum by Rdual =α′
R.
Substitute P=n
Rinto the formula for the dual radius in Type IIB theory:
Rdual =α′
R=α′
n/P =α′R
n
Hence, the dual circle radius in Type IIB theory is α′R
n.
11 12. INTEGRABILITY IN STRING THEORY
Problem 12. Consider a closed bosonic string moving in a background with a constant mag-
netic field Bin the z-direction. The equation of motion for the string is given by the Nambu-Goto
action
S=−T
2Zdσdτ√−h hab∂aXµ∂bXνGµν
where Tis the tension of the string, his the determinant of the world-sheet metric hab,Xµ=
Xµ(τ, σ)are the embedding coordinates of the string world-sheet, and Gµν is the background metric
tensor.
Given that the background metric is flat Minkowski space with Gµν =ηµν and the magnetic field
is B=Bz, where Bzis constant, compute the equation of motion for the string in this background.
Solution 12.
The equation of motion for the string can be derived by varying the action with respect to the
embedding coordinates Xµ.
Let’s denote ∂aXµ≡∂aXµ(τ, σ).
The variation of the action with respect to Xµgives the equation of motion:
δS
δXµ=−T
2Zdσdτ√−hhab∂a∂bXνηνµ = 0
Expanding the terms and using the fact that the world-sheet metric is hab =diag(−1,1), we get:
TZdσdτ ∂2
τXν−∂2
σXνηνµ = 0
The equations of motion are then given by:
a) In the µ= 0 direction:
∂2
τX0−∂2
σX0= 0
b) In the µ= 1 direction:
∂2
τX1−∂2
σX1= 0
c) In the µ= 2 direction:
∂2
τX2−∂2
σX2= 0
These equations of motion describe the dynamics of the string in the background of a constant
magnetic field B=Bz.
12 13. CAUSALITY AND STRING/M-THEORY
Problem 13. Consider a closed string propagating in a spacetime described by D= 10
dimensions. The string moves in a geometry where the background metric is given by ds2=
−dt2+dx2
1+ (dx2)2+···+ (dx8)2+ (dx9)2+ (dx10)2, where x10 is the compactified spatial direction
with a radius R.
a) Calculate the maximum energy Emax of a closed string state that can propagate in this ge-
ometry without creating a closed timelike curve.
b) Find the minimum allowed radius Rmin of the compactified spatial direction that ensures
causality is not violated in this spacetime.
Solution 13.
a) The maximum energy Emax of a closed string state can be found using the formula:
Emax =1
α′rN
2
where Nis the level of the state and α′is the Regge slope parameter. Since we are dealing with
a closed string in D= 10 dimensions, the critical dimension is D= 10, and the Regge slope
parameter is α′= 1/2πT .
For a closed string moving in a compactified dimension with radius R, the contribution to the
mass in the compact direction is n/R where nis the winding number. To avoid closed timelike
curves, we must have Emax =n/R. Equating these two expressions for Emax, we get:
n
R=1
α′rN
2
n
R=2πT
2rN
2
nR = 2πT √2N
R=2πT √2N
n
Therefore, the maximum energy is achieved at n= 1 and N= 2, leading to:
Emax =1
α′rN
2=1
α′r2
2=1
α′=√2πT
2
b) To ensure that causality is not violated in this spacetime, we must have the inequality Rmin ≥
2πR to avoid closed timelike curves. Substituting in the expression for Rmin from part (a), we get:
Rmin =2πT p2(1)
1= 2π√2πT ≥2πR
√2πT ≥R
2πT ≥R2
R2−2πT ≤0
Hence, the minimum allowed radius Rmin of the compactified spatial direction is R=√2πT .
13 14. HOLOGRAPHY IN M-THEORY
Problem 14. Consider a spacetime described by M-theory with 11 dimensions. The holo-
graphic principle states that the information of a region of space can be encoded on its boundary.
Suppose we have a 4-dimensional hypercube with side length L.
a) Calculate the volume of this hypercube in terms of L.
b) According to the holographic principle, what is the size of the boundary that encodes all the
information of this hypercube?
c) If the hypercube is in an 11-dimensional spacetime in M-theory, how many spatial dimensions
are "compactified"?
Solution 14.
a) The volume of a 4-dimensional hypercube with side length Lis given by V=L4.
b) The boundary of the hypercube is a 3-dimensional cube with sides of length L. The total
surface area of a cube is given by A= 6L2. Hence, for a 4-dimensional hypercube, the size of the
boundary that encodes all the information is A= 6L2.
c) In M-theory with 11 dimensions, the 4 spatial dimensions of the hypercube and the 1 time
dimension are known. This leaves 11 −4−1=6spatial dimensions "compactified".
I’m glad to help generate numerical problem questions for you. Could you please specify the
topic within String Theory and M-Theory that you would like the problem to be based on?
14 16. DARK MATTER AND STRING/M-THEORY
Problem 16. Consider a string theory scenario where a closed string moving in a compact
spatial dimension of radius Rinteracts with dark matter particles gravitationally. The string coupling
constant is given by gs= 0.1.
a) If the mass of the dark matter particle is m= 10−22 eV/c2, calculate the gravitational force
between the closed string and a dark matter particle located at a distance r= 1 mm.
b) If the string coupling constant gsis changed to 0.05, how does this affect the gravitational
force between the closed string and the dark matter particle?
Solution 16.
a) The gravitational force between the closed string and the dark matter particle can be calcu-
lated using Newton’s law of gravitation:
F=G·mdark matter ·mstring
r2
where Gis the gravitational constant (6.67430 ×10−11 m3kg−1s−2), mdark matter is the mass of
the dark matter particle (in kg), mstring is the mass of the closed string (in kg), and ris the distance
between the closed string and the dark matter particle (in meters).
Given m= 10−22 eV/c2, we need to convert this to kilograms by using the conversion factor
1eV/c2= 1.783 ×10−36 kg. Thus, mdark matter = 10−22 ×1.783 ×10−36 = 1.783 ×10−58 kg.
Plugging in the values G= 6.67430 ×10−11 m3kg−1s−2,mdark matter = 1.783 ×10−58 kg,
mstring =? (mass of the string is not provided but let’s assume it to be of the same order of magnitude
as a dark matter particle), and r= 1 mm = 0.001 m, we can find the gravitational force.
b) If the string coupling constant gsis changed to 0.05, we can re-calculate the gravitational
force using the new value of gsin the formula above. Let’s assume that the mass of the dark matter
particle remains the same.
Solution 16.
a) Given: G= 6.67430 ×10−11 m3kg−1s−2,mdark matter = 1.783 ×10−58 kg, mstring = 1.783 ×
10−58 kg (assumed to be of the same order), r= 0.001 m
Plugging in the values, we get:
F=6.67430 ×10−11 ×1.783 ×10−58 ×1.783 ×10−58
(0.001)2
⇒F=2.0004 ×10−124
10−6
⇒F= 2.0004 ×10−118 N
Thus, the gravitational force between the closed string and a dark matter particle located at a
distance of 1 mm is approximately 2.0004 ×10−118 N.
b) If gs= 0.05, we can re-calculate the gravitational force using the updated value of gsin the
formula. Plugging in the new value, we can find the new gravitational force.
I’m sorry, but it seems that I cannot generate numerical problems for String Theory and M-
Theory as they are primarily theoretical and mathematical physics concepts that do not involve
numerical computations. Would you like a conceptual problem or a theoretical problem instead?
15 18. VACUUM ENERGY IN M-THEORY
Problem 18. Consider a particular compactification of M-theory on a 5-dimensional torus with
each side having length R. The vacuum energy in this scenario is given by ρ=−1
(2π)5R9.
a) Calculate the vacuum energy ρin GeV4when R= 1 cm. b) Find the value of Rin m for
which the vacuum energy is ρ=−1GeV4.
Solution 18. a) To calculate the vacuum energy in GeV4, we need to convert the length R=
1cm to meters and then substitute it into the given formula.
Given: 1 cm = 10−2m
Plugging this value into the formula for vacuum energy: ρ=−1
(2π)5(10−2)9
ρ=−1
(2π)5(10−18)
ρ=−1
(2π)5(10−18)
ρ=−1
(2π)5×10−18
ρ=−1
(2π)5×10−18
ρ≈ −8.53254 ×1065 GeV4
Therefore, the vacuum energy when R= 1 cm is approximately −8.53254 ×1065 GeV4.
b) To find the value of Rin meters for which the vacuum energy is ρ=−1GeV4, we set the
vacuum energy formula equal to −1and solve for R:
−1 = −1
(2π)5R9
R9= (2π)5
R=9
p(2π)5
R≈3.28626 ×10−16 m
Therefore, the value of Rin meters for which the vacuum energy is ρ=−1GeV4is approxi-
mately 3.28626 ×10−16 m.
16 19. SOLITONS IN STRING THEORY
Problem 19. Consider a D-brane in type IIA superstring theory with tension T. Suppose the
D-brane extends in pspatial dimensions. The energy density Uper unit p-dimensional volume of
the D-brane is given by U=cT , where cis a constant.
a) If the D-brane has p= 3 spatial dimensions and the tension is T= 5 ×10−2, find the energy
density U.
b) Now, let’s consider another D-brane with tension T= 0.1and energy density U= 0.4.
Determine the number of spatial dimensions pin which this D-brane extends.
Solution 19.
a) Given p= 3 and T= 5 ×10−2, we use the formula U=cT to find the energy density U:
U=c×5×10−2= 0.05c
So, the energy density Ufor the D-brane with p= 3 spatial dimensions and tension T= 5×10−2
is 0.05c.
b) For this case, we have T= 0.1and U= 0.4. Using the same formula for energy density,
U=cT , we can solve for cfirst:
c=U
T=0.4
0.1= 4
Now, we substitute c= 4 back into the equation U=cT :
0.4=4×p
Solving for p, we find:
p=0.4
4= 0.1
Thus, the D-brane extends in p= 0.1spatial dimensions.
I. Let’s formulate a numerical problem in the context of String Theory and M-Theory:
17 20. ADS/CFT CORRESPONDENCE IN M-THEORY
Problem 20. Consider a type IIA string theory on an Anti-de Sitter space (AdS) with a five-
dimensional radius RAdS = 10. Calculate the corresponding conformal field theory (CFT) central
charge using the AdS/CFT correspondence formula:
c=3L
2G
where Lis the AdS radius and Gis Newton’s constant. Take the value of Newton’s constant
G= 6.71 ×10−11 m3kg−1s−2.
Solution 20. Given: - AdS radius RAdS = 10 - Newton’s constant G= 6.71 ×10−11 m3kg−1
s−2
The CFT central charge can be calculated using the AdS/CFT correspondence formula:
c=3L
2G
Plugging in the values:
c=3×10
2×6.71 ×10−11
c=30
13.42 ×10−11
c=30
1.342 ×10−10
c=30
1.342
c≈22.36
Therefore, the central charge of the corresponding conformal field theory is approximately
22.36.
I. Entanglement Entropy in String Theory:
18 21. Entanglement Entropy in String Theory
Problem 21. Consider a 1+1 dimensional conformal field theory described by a CFT with
central charge c= 1. Let’s calculate the entanglement entropy of a subsystem in this theory.
[Additional context: The entanglement entropy Sof a subsystem in a CFT is given by the formula
S=c
3log L
ϵ, where cis the central charge, Lis the length of the subsystem, and ϵis a short-
distance cutoff.]
a) Suppose we have a subsystem of length L= 2 in this CFT. Calculate the entanglement
entropy when the short-distance cutoff ϵ= 0.1.
b) Now, consider another CFT described by a different conformal field theory with central charge
c= 2. If we have a subsystem of length L= 3 in this CFT, what is the entanglement entropy when
the short-distance cutoff ϵ= 0.01?
Solution 21.
a) We are given c= 1,L= 2, and ϵ= 0.1. Plugging these values into the formula for entangle-
ment entropy, we get:
S=1
3log 2
0.1=1
3log 20 ≈1
3×2.9957 ≈0.9986
Therefore, the entanglement entropy when L= 2spaceunits and ϵ= 0.1is approximately
0.9986.
b) For the new CFT with c= 2,L= 3, and ϵ= 0.01, we apply the formula:
S=2
3log 3
0.01=2
3log 300 ≈2
3×5.7038 ≈3.8026
Therefore, the entanglement entropy when L= 3spaceunits and ϵ= 0.01 is approximately
3.8026.
I can certainly help with that! Let’s proceed with a numerical problem for String Theory and
M-Theory.
19 22. SPACE-TIME SINGULARITIES IN M-THEORY
Problem 22. Consider a compactified M-theory scenario where the compact spatial dimension
has a radius of R= 10−17 meters. If an object is traveling at a speed of 0.9c(where cis the speed
of light) in this compactified dimension, determine the time duration (in seconds) it takes for the
object to travel halfway around the compactified dimension.
Solution 22. a) The circumference of the compact spatial dimension is given by 2πR. There-
fore, the distance required to travel halfway around the dimension is πR.
b) The speed of the object is 0.9c, where c= 3 ×108m/s. Hence, the object travels a distance
of 0.9c×t=πR in time t, where tis the time duration we want to find.
Using the equation 0.9c×t=πR, we can solve for t:
0.9×3×108m/s ×t=π×10−17 m
0.9×3×108t=π×10−17
2.7×108t= 3.1416 ×10−17
t=3.1416 ×10−17
2.7×108
t≈1.163 ×10−25 s
Therefore, it takes approximately 1.163 ×10−25 seconds for the object to travel halfway around
the compactified dimension.
20 23. DEFORMATIONS OF STRING THEORY
Problem 23. Consider a closed bosonic string of length Lmoving in Dspacetime dimensions.
The string is perturbed by a small deformation given by the following equation of motion:
∂2
∂τ 2−∂2
∂σ2Xµ(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ)
where Xµ(τ, σ)denotes the string embedding coordinates, µ, ν = 0,1, . . . , D −1,ϵis a small
parameter, and α′is the Regge slope parameter.
a) Compute the equation of motion for X0(τ, σ)and X1(τ, σ).
b) Determine the general solution for X0(τ, σ)and X1(τ, σ).
c) Given that the string is parameterized by (τ, σ)such that 0≤σ≤π, find the normal mode
frequencies for X0(τ, σ)and X1(τ, σ).
Solution 23.
a) The equation of motion can be split into separate equations for X0(τ, σ)and X1(τ, σ)by
setting µ= 0 and µ= 1 respectively:
∂2
∂τ 2X0(τ, σ)−∂2
∂σ2X0(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ) = ϵα′sin(2σ)∂0X1(τ, σ)
∂2
∂τ 2X1(τ, σ)−∂2
∂σ2X1(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ) = ϵα′sin(2σ)∂1X0(τ, σ)
b) To find the general solutions, we solve the wave equation for X0(τ, σ)and X1(τ, σ). The
solutions will have the form:
X0(τ, σ) = A(σ)eiω0τ+B(σ)e−iω0τ
X1(τ, σ) = C(σ)eiω1τ+D(σ)e−iω1τ
where A(σ),B(σ),C(σ), and D(σ)are functions of σ, and ω0and ω1are the normal mode
frequencies.
c) To find the normal mode frequencies, we substitute the solutions back into the equations of
motion and solve for ω0and ω1. After normalization, the normal mode frequencies are given by:
ω0=1
Lrn2−ϵα′n2
2
ω1=1
Lrn2+ϵα′n2
2
where n= 1,2,3, . . . represents the mode number.
21 24. QUANTUM FIELD THEORY LIMIT OF M-THEORY
Problem 24. Consider a closed string in 10 dimensions with a winding mode wrapping around
a circle of radius R. The momentum mode of the string has energy E=n
R, where nis an integer.
The winding mode has energy E=mR
α′, where mis an integer and α′is the Regge slope.
a) If the winding mode has energy E= 2π/α′and the momentum mode has energy E= 1/√α′,
find the values of mand n.
b) Determine the total energy ETof the closed string in terms of α′.
Solution 24.
a) Since the winding mode has energy E=mR
α′=2π
α′, we have:
mR
α′=2π
α′
mR = 2π
Similarly, for the momentum mode with energy E=n
R=1
√α′, we get:
n
R=1
√α′
nR =√α′
From these two equations, we can see that m= 1 and n= 1.
b) The total energy ETof the closed string is given by:
ET=n
R+mR
α′=1
R+R
α′=1 + R2/α′
R
Therefore, the total energy of the closed string in terms of α′is ET=1 + R2/α′
R.
22 25. TOPOLOGICAL ASPECTS OF STRING THEORY
Problem 25. Consider a closed oriented Riemann surface Σgof genus g. The Euler charac-
teristic of Σgis given by χ(Σg)=2−2g.
a) Calculate the Euler characteristic for a torus (g= 1).
b) Calculate the Euler characteristic for a sphere with two handles (g= 2).
c) Calculate the Euler characteristic for a surface with genus g= 3.
Solution 25.
a) For a torus (g= 1), the Euler characteristic is given by χ(Σ1) = 2 −2·1 = 2 −2 = 0.
Therefore, the Euler characteristic for a torus is 0.
b) For a sphere with two handles (g= 2), the Euler characteristic is χ(Σ2) = 2−2·2 = 2−4 = −2.
Hence, the Euler characteristic for a sphere with two handles is -2.
c) For a surface with genus g= 3, the Euler characteristic is χ(Σ3)=2−2·3=2−6 = −4.
Therefore, the Euler characteristic for a surface of genus 3 is -4.
3 3. QUANTUM GRAVITY IN STRING THEORY
Problem 3. Consider an open string moving in 4-dimensional spacetime. The string has tension
T= 1 and mass per unit length µ= 2.
a) Find the speed of propagation of waves on this string.
b) Calculate the energy stored in a segment of string of length L= 3.
c) If the string is stretched with a force of F= 5, determine the amplitude of the standing wave
that could form on the string.
Solution 3.
a) The speed of propagation of waves on a string is given by
v=sT
µ.
Substitute T= 1 and µ= 2:
v=r1
2=√2
2.
Therefore, the speed of propagation of waves on this string is √2
2.
b) The energy stored in a segment of string of length Lis given by
E=1
2µv2L.
Substitute µ= 2,v=√2
2, and L= 3:
E=1
2×2× √2
2!2
×3 = 3
2.
Therefore, the energy stored in a segment of string of length 3 is 3
2.
c) The amplitude of the standing wave that could form on the string under the applied force F
is given by
A=F
2πv2.
Substitute F= 5 and v=√2
2:
A=5
2π×√2
22=5
2π×1
2
=5√2
π.
Therefore, the amplitude of the standing wave that could form on the string is 5√2
π.
4 4. BLACK HOLES AND STRING THEORY
Problem 4. Consider a black hole in four-dimensional spacetime described by the Schwarzschild
metric:
ds2=−1−2GM
c2rc2dt2+1−2GM
c2r−1
dr2+r2dΩ2
where Mis the mass of the black hole, cis the speed of light, Gis the gravitational constant,
and dΩ2=dθ2+sin2θdϕ2in spherical coordinates (t, r, θ, ϕ).
a) Find the event horizon radius Rhorizon of this black hole.
b) Determine the Schwarzschild radius RSin terms of M,G, and c.
c) Given that the mass of the black hole is M= 2 ×1030 kg, calculate the mass of the black
hole in units of solar masses (M⊙= 1.989 ×1030 kg).
Solution 4.
a) To find the event horizon radius, we set the metric coefficient of dt2to zero at the event
horizon. So, 1−2GM
c2Rhorizon = 0. Solving for Rhorizon gives:
Rhorizon =2GM
c2
b) The Schwarzschild radius is defined as the radius at which the metric becomes singular. It
is given by RS=2GM
c2.
c) Substituting M= 2 ×1030 kg into the Schwarzschild radius formula, we have:
RS=2G(2 ×1030)
c2=4×1030G
c2
Now, we can express the mass of the black hole in terms of solar masses by dividing by the
mass of the Sun:
4×1030G
c2÷1.989 ×1030 =2G
c2
Therefore, the mass of the black hole is 2solar masses.
5 5. TACHYON CONDENSATION IN STRING THEORY
Problem 5. Consider a closed bosonic string theory where the endpoint of the string coor-
dinates are subject to Neumann boundary conditions. The first excited level of the closed string
contains a tachyon with mass m2=−2
α′.
a) Calculate the momentum of the tachyon state in the string theory.
b) Show that the mass of the tachyon state in the open bosonic string theory is m= 0.
c) Interpret the result in relation to tachyon condensation.
Solution 5.
a) The mass-squared of a state in a closed string theory is given by the level matching condition:
m2=4
α′(N−1)
where Nis the occupation number operator for the state. For the first excited level, N= 1 and
m2=−2
α′. Substituting these values into the equation above, we have:
−2
α′=4
α′(1 −1)
−2=0
This equation is a contradiction, indicating that there is no physical state with a mass-squared of
−2
α′for the first excited level. So, the tachyon state does not exist.
b) In the open bosonic string theory, the mass-squared of a state is given by:
m2=2
α′(N−1)
For the tachyon state at the first excited level in the open string, N= 1 and m2=−1
α′. Taking the
square root of this, we get m= 0. Therefore, the mass of the tachyon state in the open bosonic
string theory is zero.
c) The result m= 0 for the open string tachyon state implies that it is a massless state. In string
theory, a tachyon has negative mass squared and indicates an instability in the theory. Tachyon
condensation is a process where this unstable state "condenses" to a minimum energy state. In
this case, the tachyon state in the open bosonic string theory having a mass of zero suggests that
it represents the minimum energy state after tachyon condensation has occurred. This process
helps stabilize the theory by eliminating the instability caused by the presence of the tachyon.
6 6. EXTRA DIMENSIONS IN M-THEORY
Problem 6. Consider a scenario in M-Theory where there are 7 spatial dimensions and 3
temporal dimensions. The size of the compactified extra dimensions is given by R= 10−17 meters.
Calculate the compactification scale in GeV.
Solution 6.
a) The compactification scale Mccan be calculated using the formula for the compactified extra
dimensions:
Mc=1
R
Substitute R= 10−17 meters into the formula:
Mc=1
10−17 = 1017 m−1
b) To convert the compactification scale from meters to GeV, we need to use the relation 1GeV =
1.97 ×10−16 m−1.
Let’s convert the compactification scale:
Mc= 1017 ×1.97 ×10−16 = 1.97 ×10 GeV = 19.7GeV
Therefore, the compactification scale in GeV is 19.7 GeV.
7 7. BRANE DYNAMICS IN STRING THEORY
Problem 7. Consider a type IIB superstring theory in 10 dimensions with D3-branes. The
tension of the D3-brane is given by T=1
(2π)3(α′)2, where α′is the string length parameter. Calculate
the energy density stored in a D3-brane of length L= 10√α′.
Solution 7. a) The energy density stored in a D3-brane can be calculated by dividing its energy
by its volume. The energy Eof a D3-brane is given by E=T×V, where Tis the tension of the
D3-brane and Vis its volume. Since the D3-brane extends in 3 spatial dimensions, the volume V
is L3= (10√α′)3= 1000α′3/2.
Substitute T=1
(2π)3(α′)2and V= 1000α′3/2into the equation:
E=1
(2π)3(α′)2×1000α′3/2=1000
(2π)3α′1/2
Thus, the energy stored in the D3-brane is 1000
(2π)3α′1/2.
b) The energy density is given by the energy per unit volume. We divide the energy Eby the
volume Vto find the energy density u=E
V.
Substitute E=1000
(2π)3α′1/2and V= 1000α′3/2into the equation:
u=1000/(2π)3α′1/2
1000α′3/2=1
(2π)3α′2
Therefore, the energy density stored in the D3-brane is 1
(2π)3α′2.
8 8. SUPERSYMMETRY BREAKING IN M-THEORY
Problem 8. Consider a supersymmetric M-Theory compactified on a 2-torus with radii R1and
R2. The volume of the torus is given by V=R1R2. Suppose that supersymmetry is broken by the
flux through the torus such that the gravitino mass term is generated.
[Given: The gravitino mass term is given by m3/2=e⟨G⟩, where Gis the flux. Also, we have
⟨G⟩ =−2 ln(V), where ⟨G⟩ denotes the expectation value of the flux.]
a) Show that the gravitino mass term m3/2in terms of the radii R1and R2.
b) If R1= 2 and R2= 3, calculate the gravitino mass m3/2.
Solution 8.
a) To find the gravitino mass term m3/2in terms of the radii R1and R2, we first need to express
the volume Vin terms of R1and R2:
V=R1·R2
Next, we can find the expectation value of the flux ⟨G⟩:
⟨G⟩ =−2 ln(V) = −2 ln(R1·R2) = −2 ln(R1)−2 ln(R2)
Substitute this into the expression for the gravitino mass term:
m3/2=e⟨G⟩ =e−2 ln(R1)−2 ln(R2)=e−2 ln(R1)·e−2 ln(R2)=1
R2
1·1
R2
2
=1
R2
1R2
2
=1
V2
b) Given R1= 2 and R2= 3, the volume V=R1·R2= 2 ·3=6. Therefore, the gravitino mass
term m3/2is:
m3/2=1
V2=1
62=1
36 = 0.0278
I. Problem 9. Consider a Type IIA superstring theory with a D6-brane wrapped on a compact
3-torus with sides of length L. The tension of the D6-brane is given by T6=1
(2π)6(α′)4. Suppose
the compactified space is a cube, compute the energy density UD6 of the D6-brane in terms of L.
Hint: The energy density is defined as the energy per unit volume.
II. Problem 10. In Type IIB superstring theory, a D3-brane has tension T3=1
(2π)3(α′)2. If this
theory is in a 10-dimensional spacetime with a toroidal compactification down to 6 dimensions, and
the 3-brane extends along all 3 of those compact dimensions, calculate the energy density UD3 of
the D3-brane in terms of the compactification radii Ri.
Hint: The energy density is defined as the energy per unit volume.
III. Problem 11. In M-Theory, consider a M5-brane wrapped on a torus with radii R1and R2. If
the tension of the brane is T5=1
(2π)5(ℓp)3, find the energy density UM5 of the M5-brane in terms of
the torus radii.
Hint: The energy density is defined as the energy per unit volume.
9 10. COSMOLOGY AND STRING/M-THEORY
Problem 10. Consider a string theory model in a universe with extra dimensions compactified
on a torus. The radius of the torus is R, and the string tension is T. The compactified dimensions
are described by an effective field theory with a massless scalar field ϕ, whose potential is given
by V(ϕ) = 1
2m2ϕ2.
a) Show that the effective tension in the compactified dimensions, Teff , is given by Tef f =
T e−2ϕ.
b) Determine the equation of motion for the scalar field ϕ.
c) Find the minimum of the potential V(ϕ).
Solution 10.
a) The effective tension in the compactified dimensions is given by Teff =T e−2ϕ. Let’s derive
this expression:
In string theory, the effective tension depends on the string coupling as Teff =T gs, where
gs=e−2ϕ. Therefore, Teff =T e−2ϕ.
b) The equation of motion for the scalar field ϕis given by:
d
dt ∂L
∂˙
ϕ−∂L
∂ϕ = 0
where the Lagrangian L=1
2˙
ϕ2−V(ϕ).
d
dt ∂L
∂˙
ϕ−∂L
∂ϕ =¨
ϕ+m2ϕ= 0
Therefore, the equation of motion for the scalar field ϕis ¨
ϕ+m2ϕ= 0.
c) To find the minimum of the potential V(ϕ), we need to solve dV
dϕ = 0:
dV
dϕ =m2ϕ= 0
This implies that the minimum of the potential occurs at ϕ= 0.
10 11. DUALITIES IN M-THEORY
Problem 11. Consider two particular string theories, Type IIA and Type IIB, related by T-duality.
Let the radius of a circle in Type IIA theory be R. If the number of fundamental strings winding
around the circle is n, find the dual circle radius in Type IIB theory.
Solution 11. Given: Radius of circle in Type IIA theory, R, and number of winding fundamental
strings, n.
In Type IIA theory, the momentum along the circle is given by P=n
R.
By T-duality, the winding number in Type IIB string theory is the same as the momentum in Type
IIA theory, and vice versa. Therefore, in Type IIB theory, the radius of the dual circle is related to
the momentum by Rdual =α′
R.
Substitute P=n
Rinto the formula for the dual radius in Type IIB theory:
Rdual =α′
R=α′
n/P =α′R
n
Hence, the dual circle radius in Type IIB theory is α′R
n.
11 12. INTEGRABILITY IN STRING THEORY
Problem 12. Consider a closed bosonic string moving in a background with a constant mag-
netic field Bin the z-direction. The equation of motion for the string is given by the Nambu-Goto
action
S=−T
2Zdσdτ√−h hab∂aXµ∂bXνGµν
where Tis the tension of the string, his the determinant of the world-sheet metric hab,Xµ=
Xµ(τ, σ)are the embedding coordinates of the string world-sheet, and Gµν is the background metric
tensor.
Given that the background metric is flat Minkowski space with Gµν =ηµν and the magnetic field
is B=Bz, where Bzis constant, compute the equation of motion for the string in this background.
Solution 12.
The equation of motion for the string can be derived by varying the action with respect to the
embedding coordinates Xµ.
Let’s denote ∂aXµ≡∂aXµ(τ, σ).
The variation of the action with respect to Xµgives the equation of motion:
δS
δXµ=−T
2Zdσdτ√−hhab∂a∂bXνηνµ = 0
Expanding the terms and using the fact that the world-sheet metric is hab =diag(−1,1), we get:
TZdσdτ ∂2
τXν−∂2
σXνηνµ = 0
The equations of motion are then given by:
a) In the µ= 0 direction:
∂2
τX0−∂2
σX0= 0
b) In the µ= 1 direction:
∂2
τX1−∂2
σX1= 0
c) In the µ= 2 direction:
∂2
τX2−∂2
σX2= 0
These equations of motion describe the dynamics of the string in the background of a constant
magnetic field B=Bz.
12 13. CAUSALITY AND STRING/M-THEORY
Problem 13. Consider a closed string propagating in a spacetime described by D= 10
dimensions. The string moves in a geometry where the background metric is given by ds2=
−dt2+dx2
1+ (dx2)2+···+ (dx8)2+ (dx9)2+ (dx10)2, where x10 is the compactified spatial direction
with a radius R.
a) Calculate the maximum energy Emax of a closed string state that can propagate in this ge-
ometry without creating a closed timelike curve.
b) Find the minimum allowed radius Rmin of the compactified spatial direction that ensures
causality is not violated in this spacetime.
Solution 13.
a) The maximum energy Emax of a closed string state can be found using the formula:
Emax =1
α′rN
2
where Nis the level of the state and α′is the Regge slope parameter. Since we are dealing with
a closed string in D= 10 dimensions, the critical dimension is D= 10, and the Regge slope
parameter is α′= 1/2πT .
For a closed string moving in a compactified dimension with radius R, the contribution to the
mass in the compact direction is n/R where nis the winding number. To avoid closed timelike
curves, we must have Emax =n/R. Equating these two expressions for Emax, we get:
n
R=1
α′rN
2
n
R=2πT
2rN
2
nR = 2πT √2N
R=2πT √2N
n
Therefore, the maximum energy is achieved at n= 1 and N= 2, leading to:
Emax =1
α′rN
2=1
α′r2
2=1
α′=√2πT
2
b) To ensure that causality is not violated in this spacetime, we must have the inequality Rmin ≥
2πR to avoid closed timelike curves. Substituting in the expression for Rmin from part (a), we get:
Rmin =2πT p2(1)
1= 2π√2πT ≥2πR
√2πT ≥R
2πT ≥R2
R2−2πT ≤0
Hence, the minimum allowed radius Rmin of the compactified spatial direction is R=√2πT .
13 14. HOLOGRAPHY IN M-THEORY
Problem 14. Consider a spacetime described by M-theory with 11 dimensions. The holo-
graphic principle states that the information of a region of space can be encoded on its boundary.
Suppose we have a 4-dimensional hypercube with side length L.
a) Calculate the volume of this hypercube in terms of L.
b) According to the holographic principle, what is the size of the boundary that encodes all the
information of this hypercube?
c) If the hypercube is in an 11-dimensional spacetime in M-theory, how many spatial dimensions
are "compactified"?
Solution 14.
a) The volume of a 4-dimensional hypercube with side length Lis given by V=L4.
b) The boundary of the hypercube is a 3-dimensional cube with sides of length L. The total
surface area of a cube is given by A= 6L2. Hence, for a 4-dimensional hypercube, the size of the
boundary that encodes all the information is A= 6L2.
c) In M-theory with 11 dimensions, the 4 spatial dimensions of the hypercube and the 1 time
dimension are known. This leaves 11 −4−1=6spatial dimensions "compactified".
I’m glad to help generate numerical problem questions for you. Could you please specify the
topic within String Theory and M-Theory that you would like the problem to be based on?
14 16. DARK MATTER AND STRING/M-THEORY
Problem 16. Consider a string theory scenario where a closed string moving in a compact
spatial dimension of radius Rinteracts with dark matter particles gravitationally. The string coupling
constant is given by gs= 0.1.
a) If the mass of the dark matter particle is m= 10−22 eV/c2, calculate the gravitational force
between the closed string and a dark matter particle located at a distance r= 1 mm.
b) If the string coupling constant gsis changed to 0.05, how does this affect the gravitational
force between the closed string and the dark matter particle?
Solution 16.
a) The gravitational force between the closed string and the dark matter particle can be calcu-
lated using Newton’s law of gravitation:
F=G·mdark matter ·mstring
r2
where Gis the gravitational constant (6.67430 ×10−11 m3kg−1s−2), mdark matter is the mass of
the dark matter particle (in kg), mstring is the mass of the closed string (in kg), and ris the distance
between the closed string and the dark matter particle (in meters).
Given m= 10−22 eV/c2, we need to convert this to kilograms by using the conversion factor
1eV/c2= 1.783 ×10−36 kg. Thus, mdark matter = 10−22 ×1.783 ×10−36 = 1.783 ×10−58 kg.
Plugging in the values G= 6.67430 ×10−11 m3kg−1s−2,mdark matter = 1.783 ×10−58 kg,
mstring =? (mass of the string is not provided but let’s assume it to be of the same order of magnitude
as a dark matter particle), and r= 1 mm = 0.001 m, we can find the gravitational force.
b) If the string coupling constant gsis changed to 0.05, we can re-calculate the gravitational
force using the new value of gsin the formula above. Let’s assume that the mass of the dark matter
particle remains the same.
Solution 16.
a) Given: G= 6.67430 ×10−11 m3kg−1s−2,mdark matter = 1.783 ×10−58 kg, mstring = 1.783 ×
10−58 kg (assumed to be of the same order), r= 0.001 m
Plugging in the values, we get:
F=6.67430 ×10−11 ×1.783 ×10−58 ×1.783 ×10−58
(0.001)2
⇒F=2.0004 ×10−124
10−6
⇒F= 2.0004 ×10−118 N
Thus, the gravitational force between the closed string and a dark matter particle located at a
distance of 1 mm is approximately 2.0004 ×10−118 N.
b) If gs= 0.05, we can re-calculate the gravitational force using the updated value of gsin the
formula. Plugging in the new value, we can find the new gravitational force.
I’m sorry, but it seems that I cannot generate numerical problems for String Theory and M-
Theory as they are primarily theoretical and mathematical physics concepts that do not involve
numerical computations. Would you like a conceptual problem or a theoretical problem instead?
15 18. VACUUM ENERGY IN M-THEORY
Problem 18. Consider a particular compactification of M-theory on a 5-dimensional torus with
each side having length R. The vacuum energy in this scenario is given by ρ=−1
(2π)5R9.
a) Calculate the vacuum energy ρin GeV4when R= 1 cm. b) Find the value of Rin m for
which the vacuum energy is ρ=−1GeV4.
Solution 18. a) To calculate the vacuum energy in GeV4, we need to convert the length R=
1cm to meters and then substitute it into the given formula.
Given: 1 cm = 10−2m
Plugging this value into the formula for vacuum energy: ρ=−1
(2π)5(10−2)9
ρ=−1
(2π)5(10−18)
ρ=−1
(2π)5(10−18)
ρ=−1
(2π)5×10−18
ρ=−1
(2π)5×10−18
ρ≈ −8.53254 ×1065 GeV4
Therefore, the vacuum energy when R= 1 cm is approximately −8.53254 ×1065 GeV4.
b) To find the value of Rin meters for which the vacuum energy is ρ=−1GeV4, we set the
vacuum energy formula equal to −1and solve for R:
−1 = −1
(2π)5R9
R9= (2π)5
R=9
p(2π)5
R≈3.28626 ×10−16 m
Therefore, the value of Rin meters for which the vacuum energy is ρ=−1GeV4is approxi-
mately 3.28626 ×10−16 m.
16 19. SOLITONS IN STRING THEORY
Problem 19. Consider a D-brane in type IIA superstring theory with tension T. Suppose the
D-brane extends in pspatial dimensions. The energy density Uper unit p-dimensional volume of
the D-brane is given by U=cT , where cis a constant.
a) If the D-brane has p= 3 spatial dimensions and the tension is T= 5 ×10−2, find the energy
density U.
b) Now, let’s consider another D-brane with tension T= 0.1and energy density U= 0.4.
Determine the number of spatial dimensions pin which this D-brane extends.
Solution 19.
a) Given p= 3 and T= 5 ×10−2, we use the formula U=cT to find the energy density U:
U=c×5×10−2= 0.05c
So, the energy density Ufor the D-brane with p= 3 spatial dimensions and tension T= 5×10−2
is 0.05c.
b) For this case, we have T= 0.1and U= 0.4. Using the same formula for energy density,
U=cT , we can solve for cfirst:
c=U
T=0.4
0.1= 4
Now, we substitute c= 4 back into the equation U=cT :
0.4=4×p
Solving for p, we find:
p=0.4
4= 0.1
Thus, the D-brane extends in p= 0.1spatial dimensions.
I. Let’s formulate a numerical problem in the context of String Theory and M-Theory:
17 20. ADS/CFT CORRESPONDENCE IN M-THEORY
Problem 20. Consider a type IIA string theory on an Anti-de Sitter space (AdS) with a five-
dimensional radius RAdS = 10. Calculate the corresponding conformal field theory (CFT) central
charge using the AdS/CFT correspondence formula:
c=3L
2G
where Lis the AdS radius and Gis Newton’s constant. Take the value of Newton’s constant
G= 6.71 ×10−11 m3kg−1s−2.
Solution 20. Given: - AdS radius RAdS = 10 - Newton’s constant G= 6.71 ×10−11 m3kg−1
s−2
The CFT central charge can be calculated using the AdS/CFT correspondence formula:
c=3L
2G
Plugging in the values:
c=3×10
2×6.71 ×10−11
c=30
13.42 ×10−11
c=30
1.342 ×10−10
c=30
1.342
c≈22.36
Therefore, the central charge of the corresponding conformal field theory is approximately
22.36.
I. Entanglement Entropy in String Theory:
18 21. Entanglement Entropy in String Theory
Problem 21. Consider a 1+1 dimensional conformal field theory described by a CFT with
central charge c= 1. Let’s calculate the entanglement entropy of a subsystem in this theory.
[Additional context: The entanglement entropy Sof a subsystem in a CFT is given by the formula
S=c
3log L
ϵ, where cis the central charge, Lis the length of the subsystem, and ϵis a short-
distance cutoff.]
a) Suppose we have a subsystem of length L= 2 in this CFT. Calculate the entanglement
entropy when the short-distance cutoff ϵ= 0.1.
b) Now, consider another CFT described by a different conformal field theory with central charge
c= 2. If we have a subsystem of length L= 3 in this CFT, what is the entanglement entropy when
the short-distance cutoff ϵ= 0.01?
Solution 21.
a) We are given c= 1,L= 2, and ϵ= 0.1. Plugging these values into the formula for entangle-
ment entropy, we get:
S=1
3log 2
0.1=1
3log 20 ≈1
3×2.9957 ≈0.9986
Therefore, the entanglement entropy when L= 2spaceunits and ϵ= 0.1is approximately
0.9986.
b) For the new CFT with c= 2,L= 3, and ϵ= 0.01, we apply the formula:
S=2
3log 3
0.01=2
3log 300 ≈2
3×5.7038 ≈3.8026
Therefore, the entanglement entropy when L= 3spaceunits and ϵ= 0.01 is approximately
3.8026.
I can certainly help with that! Let’s proceed with a numerical problem for String Theory and
M-Theory.
19 22. SPACE-TIME SINGULARITIES IN M-THEORY
Problem 22. Consider a compactified M-theory scenario where the compact spatial dimension
has a radius of R= 10−17 meters. If an object is traveling at a speed of 0.9c(where cis the speed
of light) in this compactified dimension, determine the time duration (in seconds) it takes for the
object to travel halfway around the compactified dimension.
Solution 22. a) The circumference of the compact spatial dimension is given by 2πR. There-
fore, the distance required to travel halfway around the dimension is πR.
b) The speed of the object is 0.9c, where c= 3 ×108m/s. Hence, the object travels a distance
of 0.9c×t=πR in time t, where tis the time duration we want to find.
Using the equation 0.9c×t=πR, we can solve for t:
0.9×3×108m/s ×t=π×10−17 m
0.9×3×108t=π×10−17
2.7×108t= 3.1416 ×10−17
t=3.1416 ×10−17
2.7×108
t≈1.163 ×10−25 s
Therefore, it takes approximately 1.163 ×10−25 seconds for the object to travel halfway around
the compactified dimension.
20 23. DEFORMATIONS OF STRING THEORY
Problem 23. Consider a closed bosonic string of length Lmoving in Dspacetime dimensions.
The string is perturbed by a small deformation given by the following equation of motion:
∂2
∂τ 2−∂2
∂σ2Xµ(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ)
where Xµ(τ, σ)denotes the string embedding coordinates, µ, ν = 0,1, . . . , D −1,ϵis a small
parameter, and α′is the Regge slope parameter.
a) Compute the equation of motion for X0(τ, σ)and X1(τ, σ).
b) Determine the general solution for X0(τ, σ)and X1(τ, σ).
c) Given that the string is parameterized by (τ, σ)such that 0≤σ≤π, find the normal mode
frequencies for X0(τ, σ)and X1(τ, σ).
Solution 23.
a) The equation of motion can be split into separate equations for X0(τ, σ)and X1(τ, σ)by
setting µ= 0 and µ= 1 respectively:
∂2
∂τ 2X0(τ, σ)−∂2
∂σ2X0(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ) = ϵα′sin(2σ)∂0X1(τ, σ)
∂2
∂τ 2X1(τ, σ)−∂2
∂σ2X1(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ) = ϵα′sin(2σ)∂1X0(τ, σ)
b) To find the general solutions, we solve the wave equation for X0(τ, σ)and X1(τ, σ). The
solutions will have the form:
X0(τ, σ) = A(σ)eiω0τ+B(σ)e−iω0τ
X1(τ, σ) = C(σ)eiω1τ+D(σ)e−iω1τ
where A(σ),B(σ),C(σ), and D(σ)are functions of σ, and ω0and ω1are the normal mode
frequencies.
c) To find the normal mode frequencies, we substitute the solutions back into the equations of
motion and solve for ω0and ω1. After normalization, the normal mode frequencies are given by:
ω0=1
Lrn2−ϵα′n2
2
ω1=1
Lrn2+ϵα′n2
2
where n= 1,2,3, . . . represents the mode number.
21 24. QUANTUM FIELD THEORY LIMIT OF M-THEORY
Problem 24. Consider a closed string in 10 dimensions with a winding mode wrapping around
a circle of radius R. The momentum mode of the string has energy E=n
R, where nis an integer.
The winding mode has energy E=mR
α′, where mis an integer and α′is the Regge slope.
a) If the winding mode has energy E= 2π/α′and the momentum mode has energy E= 1/√α′,
find the values of mand n.
b) Determine the total energy ETof the closed string in terms of α′.
Solution 24.
a) Since the winding mode has energy E=mR
α′=2π
α′, we have:
mR
α′=2π
α′
mR = 2π
Similarly, for the momentum mode with energy E=n
R=1
√α′, we get:
n
R=1
√α′
nR =√α′
From these two equations, we can see that m= 1 and n= 1.
b) The total energy ETof the closed string is given by:
ET=n
R+mR
α′=1
R+R
α′=1 + R2/α′
R
Therefore, the total energy of the closed string in terms of α′is ET=1 + R2/α′
R.
22 25. TOPOLOGICAL ASPECTS OF STRING THEORY
Problem 25. Consider a closed oriented Riemann surface Σgof genus g. The Euler charac-
teristic of Σgis given by χ(Σg)=2−2g.
a) Calculate the Euler characteristic for a torus (g= 1).
b) Calculate the Euler characteristic for a sphere with two handles (g= 2).
c) Calculate the Euler characteristic for a surface with genus g= 3.
Solution 25.
a) For a torus (g= 1), the Euler characteristic is given by χ(Σ1) = 2 −2·1 = 2 −2 = 0.
Therefore, the Euler characteristic for a torus is 0.
b) For a sphere with two handles (g= 2), the Euler characteristic is χ(Σ2) = 2−2·2 = 2−4 = −2.
Hence, the Euler characteristic for a sphere with two handles is -2.
c) For a surface with genus g= 3, the Euler characteristic is χ(Σ3)=2−2·3=2−6 = −4.
Therefore, the Euler characteristic for a surface of genus 3 is -4.
3 3. QUANTUM GRAVITY IN STRING THEORY
Problem 3. Consider an open string moving in 4-dimensional spacetime. The string has tension
T= 1 and mass per unit length µ= 2.
a) Find the speed of propagation of waves on this string.
b) Calculate the energy stored in a segment of string of length L= 3.
c) If the string is stretched with a force of F= 5, determine the amplitude of the standing wave
that could form on the string.
Solution 3.
a) The speed of propagation of waves on a string is given by
v=sT
µ.
Substitute T= 1 and µ= 2:
v=r1
2=√2
2.
Therefore, the speed of propagation of waves on this string is √2
2.
b) The energy stored in a segment of string of length Lis given by
E=1
2µv2L.
Substitute µ= 2,v=√2
2, and L= 3:
E=1
2×2× √2
2!2
×3 = 3
2.
Therefore, the energy stored in a segment of string of length 3 is 3
2.
c) The amplitude of the standing wave that could form on the string under the applied force F
is given by
A=F
2πv2.
Substitute F= 5 and v=√2
2:
A=5
2π×√2
22=5
2π×1
2
=5√2
π.
Therefore, the amplitude of the standing wave that could form on the string is 5√2
π.
4 4. BLACK HOLES AND STRING THEORY
Problem 4. Consider a black hole in four-dimensional spacetime described by the Schwarzschild
metric:
ds2=−1−2GM
c2rc2dt2+1−2GM
c2r−1
dr2+r2dΩ2
where Mis the mass of the black hole, cis the speed of light, Gis the gravitational constant,
and dΩ2=dθ2+sin2θdϕ2in spherical coordinates (t, r, θ, ϕ).
a) Find the event horizon radius Rhorizon of this black hole.
b) Determine the Schwarzschild radius RSin terms of M,G, and c.
c) Given that the mass of the black hole is M= 2 ×1030 kg, calculate the mass of the black
hole in units of solar masses (M⊙= 1.989 ×1030 kg).
Solution 4.
a) To find the event horizon radius, we set the metric coefficient of dt2to zero at the event
horizon. So, 1−2GM
c2Rhorizon = 0. Solving for Rhorizon gives:
Rhorizon =2GM
c2
b) The Schwarzschild radius is defined as the radius at which the metric becomes singular. It
is given by RS=2GM
c2.
c) Substituting M= 2 ×1030 kg into the Schwarzschild radius formula, we have:
RS=2G(2 ×1030)
c2=4×1030G
c2
Now, we can express the mass of the black hole in terms of solar masses by dividing by the
mass of the Sun:
4×1030G
c2÷1.989 ×1030 =2G
c2
Therefore, the mass of the black hole is 2solar masses.
5 5. TACHYON CONDENSATION IN STRING THEORY
Problem 5. Consider a closed bosonic string theory where the endpoint of the string coor-
dinates are subject to Neumann boundary conditions. The first excited level of the closed string
contains a tachyon with mass m2=−2
α′.
a) Calculate the momentum of the tachyon state in the string theory.
b) Show that the mass of the tachyon state in the open bosonic string theory is m= 0.
c) Interpret the result in relation to tachyon condensation.
Solution 5.
a) The mass-squared of a state in a closed string theory is given by the level matching condition:
m2=4
α′(N−1)
where Nis the occupation number operator for the state. For the first excited level, N= 1 and
m2=−2
α′. Substituting these values into the equation above, we have:
−2
α′=4
α′(1 −1)
−2=0
This equation is a contradiction, indicating that there is no physical state with a mass-squared of
−2
α′for the first excited level. So, the tachyon state does not exist.
b) In the open bosonic string theory, the mass-squared of a state is given by:
m2=2
α′(N−1)
For the tachyon state at the first excited level in the open string, N= 1 and m2=−1
α′. Taking the
square root of this, we get m= 0. Therefore, the mass of the tachyon state in the open bosonic
string theory is zero.
c) The result m= 0 for the open string tachyon state implies that it is a massless state. In string
theory, a tachyon has negative mass squared and indicates an instability in the theory. Tachyon
condensation is a process where this unstable state "condenses" to a minimum energy state. In
this case, the tachyon state in the open bosonic string theory having a mass of zero suggests that
it represents the minimum energy state after tachyon condensation has occurred. This process
helps stabilize the theory by eliminating the instability caused by the presence of the tachyon.
6 6. EXTRA DIMENSIONS IN M-THEORY
Problem 6. Consider a scenario in M-Theory where there are 7 spatial dimensions and 3
temporal dimensions. The size of the compactified extra dimensions is given by R= 10−17 meters.
Calculate the compactification scale in GeV.
Solution 6.
a) The compactification scale Mccan be calculated using the formula for the compactified extra
dimensions:
Mc=1
R
Substitute R= 10−17 meters into the formula:
Mc=1
10−17 = 1017 m−1
b) To convert the compactification scale from meters to GeV, we need to use the relation 1GeV =
1.97 ×10−16 m−1.
Let’s convert the compactification scale:
Mc= 1017 ×1.97 ×10−16 = 1.97 ×10 GeV = 19.7GeV
Therefore, the compactification scale in GeV is 19.7 GeV.
7 7. BRANE DYNAMICS IN STRING THEORY
Problem 7. Consider a type IIB superstring theory in 10 dimensions with D3-branes. The
tension of the D3-brane is given by T=1
(2π)3(α′)2, where α′is the string length parameter. Calculate
the energy density stored in a D3-brane of length L= 10√α′.
Solution 7. a) The energy density stored in a D3-brane can be calculated by dividing its energy
by its volume. The energy Eof a D3-brane is given by E=T×V, where Tis the tension of the
D3-brane and Vis its volume. Since the D3-brane extends in 3 spatial dimensions, the volume V
is L3= (10√α′)3= 1000α′3/2.
Substitute T=1
(2π)3(α′)2and V= 1000α′3/2into the equation:
E=1
(2π)3(α′)2×1000α′3/2=1000
(2π)3α′1/2
Thus, the energy stored in the D3-brane is 1000
(2π)3α′1/2.
b) The energy density is given by the energy per unit volume. We divide the energy Eby the
volume Vto find the energy density u=E
V.
Substitute E=1000
(2π)3α′1/2and V= 1000α′3/2into the equation:
u=1000/(2π)3α′1/2
1000α′3/2=1
(2π)3α′2
Therefore, the energy density stored in the D3-brane is 1
(2π)3α′2.
8 8. SUPERSYMMETRY BREAKING IN M-THEORY
Problem 8. Consider a supersymmetric M-Theory compactified on a 2-torus with radii R1and
R2. The volume of the torus is given by V=R1R2. Suppose that supersymmetry is broken by the
flux through the torus such that the gravitino mass term is generated.
[Given: The gravitino mass term is given by m3/2=e⟨G⟩, where Gis the flux. Also, we have
⟨G⟩ =−2 ln(V), where ⟨G⟩ denotes the expectation value of the flux.]
a) Show that the gravitino mass term m3/2in terms of the radii R1and R2.
b) If R1= 2 and R2= 3, calculate the gravitino mass m3/2.
Solution 8.
a) To find the gravitino mass term m3/2in terms of the radii R1and R2, we first need to express
the volume Vin terms of R1and R2:
V=R1·R2
Next, we can find the expectation value of the flux ⟨G⟩:
⟨G⟩ =−2 ln(V) = −2 ln(R1·R2) = −2 ln(R1)−2 ln(R2)
Substitute this into the expression for the gravitino mass term:
m3/2=e⟨G⟩ =e−2 ln(R1)−2 ln(R2)=e−2 ln(R1)·e−2 ln(R2)=1
R2
1·1
R2
2
=1
R2
1R2
2
=1
V2
b) Given R1= 2 and R2= 3, the volume V=R1·R2= 2 ·3=6. Therefore, the gravitino mass
term m3/2is:
m3/2=1
V2=1
62=1
36 = 0.0278
I. Problem 9. Consider a Type IIA superstring theory with a D6-brane wrapped on a compact
3-torus with sides of length L. The tension of the D6-brane is given by T6=1
(2π)6(α′)4. Suppose
the compactified space is a cube, compute the energy density UD6 of the D6-brane in terms of L.
Hint: The energy density is defined as the energy per unit volume.
II. Problem 10. In Type IIB superstring theory, a D3-brane has tension T3=1
(2π)3(α′)2. If this
theory is in a 10-dimensional spacetime with a toroidal compactification down to 6 dimensions, and
the 3-brane extends along all 3 of those compact dimensions, calculate the energy density UD3 of
the D3-brane in terms of the compactification radii Ri.
Hint: The energy density is defined as the energy per unit volume.
III. Problem 11. In M-Theory, consider a M5-brane wrapped on a torus with radii R1and R2. If
the tension of the brane is T5=1
(2π)5(ℓp)3, find the energy density UM5 of the M5-brane in terms of
the torus radii.
Hint: The energy density is defined as the energy per unit volume.
9 10. COSMOLOGY AND STRING/M-THEORY
Problem 10. Consider a string theory model in a universe with extra dimensions compactified
on a torus. The radius of the torus is R, and the string tension is T. The compactified dimensions
are described by an effective field theory with a massless scalar field ϕ, whose potential is given
by V(ϕ) = 1
2m2ϕ2.
a) Show that the effective tension in the compactified dimensions, Teff , is given by Tef f =
T e−2ϕ.
b) Determine the equation of motion for the scalar field ϕ.
c) Find the minimum of the potential V(ϕ).
Solution 10.
a) The effective tension in the compactified dimensions is given by Teff =T e−2ϕ. Let’s derive
this expression:
In string theory, the effective tension depends on the string coupling as Teff =T gs, where
gs=e−2ϕ. Therefore, Teff =T e−2ϕ.
b) The equation of motion for the scalar field ϕis given by:
d
dt ∂L
∂˙
ϕ−∂L
∂ϕ = 0
where the Lagrangian L=1
2˙
ϕ2−V(ϕ).
d
dt ∂L
∂˙
ϕ−∂L
∂ϕ =¨
ϕ+m2ϕ= 0
Therefore, the equation of motion for the scalar field ϕis ¨
ϕ+m2ϕ= 0.
c) To find the minimum of the potential V(ϕ), we need to solve dV
dϕ = 0:
dV
dϕ =m2ϕ= 0
This implies that the minimum of the potential occurs at ϕ= 0.
10 11. DUALITIES IN M-THEORY
Problem 11. Consider two particular string theories, Type IIA and Type IIB, related by T-duality.
Let the radius of a circle in Type IIA theory be R. If the number of fundamental strings winding
around the circle is n, find the dual circle radius in Type IIB theory.
Solution 11. Given: Radius of circle in Type IIA theory, R, and number of winding fundamental
strings, n.
In Type IIA theory, the momentum along the circle is given by P=n
R.
By T-duality, the winding number in Type IIB string theory is the same as the momentum in Type
IIA theory, and vice versa. Therefore, in Type IIB theory, the radius of the dual circle is related to
the momentum by Rdual =α′
R.
Substitute P=n
Rinto the formula for the dual radius in Type IIB theory:
Rdual =α′
R=α′
n/P =α′R
n
Hence, the dual circle radius in Type IIB theory is α′R
n.
11 12. INTEGRABILITY IN STRING THEORY
Problem 12. Consider a closed bosonic string moving in a background with a constant mag-
netic field Bin the z-direction. The equation of motion for the string is given by the Nambu-Goto
action
S=−T
2Zdσdτ√−h hab∂aXµ∂bXνGµν
where Tis the tension of the string, his the determinant of the world-sheet metric hab,Xµ=
Xµ(τ, σ)are the embedding coordinates of the string world-sheet, and Gµν is the background metric
tensor.
Given that the background metric is flat Minkowski space with Gµν =ηµν and the magnetic field
is B=Bz, where Bzis constant, compute the equation of motion for the string in this background.
Solution 12.
The equation of motion for the string can be derived by varying the action with respect to the
embedding coordinates Xµ.
Let’s denote ∂aXµ≡∂aXµ(τ, σ).
The variation of the action with respect to Xµgives the equation of motion:
δS
δXµ=−T
2Zdσdτ√−hhab∂a∂bXνηνµ = 0
Expanding the terms and using the fact that the world-sheet metric is hab =diag(−1,1), we get:
TZdσdτ ∂2
τXν−∂2
σXνηνµ = 0
The equations of motion are then given by:
a) In the µ= 0 direction:
∂2
τX0−∂2
σX0= 0
b) In the µ= 1 direction:
∂2
τX1−∂2
σX1= 0
c) In the µ= 2 direction:
∂2
τX2−∂2
σX2= 0
These equations of motion describe the dynamics of the string in the background of a constant
magnetic field B=Bz.
12 13. CAUSALITY AND STRING/M-THEORY
Problem 13. Consider a closed string propagating in a spacetime described by D= 10
dimensions. The string moves in a geometry where the background metric is given by ds2=
−dt2+dx2
1+ (dx2)2+···+ (dx8)2+ (dx9)2+ (dx10)2, where x10 is the compactified spatial direction
with a radius R.
a) Calculate the maximum energy Emax of a closed string state that can propagate in this ge-
ometry without creating a closed timelike curve.
b) Find the minimum allowed radius Rmin of the compactified spatial direction that ensures
causality is not violated in this spacetime.
Solution 13.
a) The maximum energy Emax of a closed string state can be found using the formula:
Emax =1
α′rN
2
where Nis the level of the state and α′is the Regge slope parameter. Since we are dealing with
a closed string in D= 10 dimensions, the critical dimension is D= 10, and the Regge slope
parameter is α′= 1/2πT .
For a closed string moving in a compactified dimension with radius R, the contribution to the
mass in the compact direction is n/R where nis the winding number. To avoid closed timelike
curves, we must have Emax =n/R. Equating these two expressions for Emax, we get:
n
R=1
α′rN
2
n
R=2πT
2rN
2
nR = 2πT √2N
R=2πT √2N
n
Therefore, the maximum energy is achieved at n= 1 and N= 2, leading to:
Emax =1
α′rN
2=1
α′r2
2=1
α′=√2πT
2
b) To ensure that causality is not violated in this spacetime, we must have the inequality Rmin ≥
2πR to avoid closed timelike curves. Substituting in the expression for Rmin from part (a), we get:
Rmin =2πT p2(1)
1= 2π√2πT ≥2πR
√2πT ≥R
2πT ≥R2
R2−2πT ≤0
Hence, the minimum allowed radius Rmin of the compactified spatial direction is R=√2πT .
13 14. HOLOGRAPHY IN M-THEORY
Problem 14. Consider a spacetime described by M-theory with 11 dimensions. The holo-
graphic principle states that the information of a region of space can be encoded on its boundary.
Suppose we have a 4-dimensional hypercube with side length L.
a) Calculate the volume of this hypercube in terms of L.
b) According to the holographic principle, what is the size of the boundary that encodes all the
information of this hypercube?
c) If the hypercube is in an 11-dimensional spacetime in M-theory, how many spatial dimensions
are "compactified"?
Solution 14.
a) The volume of a 4-dimensional hypercube with side length Lis given by V=L4.
b) The boundary of the hypercube is a 3-dimensional cube with sides of length L. The total
surface area of a cube is given by A= 6L2. Hence, for a 4-dimensional hypercube, the size of the
boundary that encodes all the information is A= 6L2.
c) In M-theory with 11 dimensions, the 4 spatial dimensions of the hypercube and the 1 time
dimension are known. This leaves 11 −4−1=6spatial dimensions "compactified".
I’m glad to help generate numerical problem questions for you. Could you please specify the
topic within String Theory and M-Theory that you would like the problem to be based on?
14 16. DARK MATTER AND STRING/M-THEORY
Problem 16. Consider a string theory scenario where a closed string moving in a compact
spatial dimension of radius Rinteracts with dark matter particles gravitationally. The string coupling
constant is given by gs= 0.1.
a) If the mass of the dark matter particle is m= 10−22 eV/c2, calculate the gravitational force
between the closed string and a dark matter particle located at a distance r= 1 mm.
b) If the string coupling constant gsis changed to 0.05, how does this affect the gravitational
force between the closed string and the dark matter particle?
Solution 16.
a) The gravitational force between the closed string and the dark matter particle can be calcu-
lated using Newton’s law of gravitation:
F=G·mdark matter ·mstring
r2
where Gis the gravitational constant (6.67430 ×10−11 m3kg−1s−2), mdark matter is the mass of
the dark matter particle (in kg), mstring is the mass of the closed string (in kg), and ris the distance
between the closed string and the dark matter particle (in meters).
Given m= 10−22 eV/c2, we need to convert this to kilograms by using the conversion factor
1eV/c2= 1.783 ×10−36 kg. Thus, mdark matter = 10−22 ×1.783 ×10−36 = 1.783 ×10−58 kg.
Plugging in the values G= 6.67430 ×10−11 m3kg−1s−2,mdark matter = 1.783 ×10−58 kg,
mstring =? (mass of the string is not provided but let’s assume it to be of the same order of magnitude
as a dark matter particle), and r= 1 mm = 0.001 m, we can find the gravitational force.
b) If the string coupling constant gsis changed to 0.05, we can re-calculate the gravitational
force using the new value of gsin the formula above. Let’s assume that the mass of the dark matter
particle remains the same.
Solution 16.
a) Given: G= 6.67430 ×10−11 m3kg−1s−2,mdark matter = 1.783 ×10−58 kg, mstring = 1.783 ×
10−58 kg (assumed to be of the same order), r= 0.001 m
Plugging in the values, we get:
F=6.67430 ×10−11 ×1.783 ×10−58 ×1.783 ×10−58
(0.001)2
⇒F=2.0004 ×10−124
10−6
⇒F= 2.0004 ×10−118 N
Thus, the gravitational force between the closed string and a dark matter particle located at a
distance of 1 mm is approximately 2.0004 ×10−118 N.
b) If gs= 0.05, we can re-calculate the gravitational force using the updated value of gsin the
formula. Plugging in the new value, we can find the new gravitational force.
I’m sorry, but it seems that I cannot generate numerical problems for String Theory and M-
Theory as they are primarily theoretical and mathematical physics concepts that do not involve
numerical computations. Would you like a conceptual problem or a theoretical problem instead?
15 18. VACUUM ENERGY IN M-THEORY
Problem 18. Consider a particular compactification of M-theory on a 5-dimensional torus with
each side having length R. The vacuum energy in this scenario is given by ρ=−1
(2π)5R9.
a) Calculate the vacuum energy ρin GeV4when R= 1 cm. b) Find the value of Rin m for
which the vacuum energy is ρ=−1GeV4.
Solution 18. a) To calculate the vacuum energy in GeV4, we need to convert the length R=
1cm to meters and then substitute it into the given formula.
Given: 1 cm = 10−2m
Plugging this value into the formula for vacuum energy: ρ=−1
(2π)5(10−2)9
ρ=−1
(2π)5(10−18)
ρ=−1
(2π)5(10−18)
ρ=−1
(2π)5×10−18
ρ=−1
(2π)5×10−18
ρ≈ −8.53254 ×1065 GeV4
Therefore, the vacuum energy when R= 1 cm is approximately −8.53254 ×1065 GeV4.
b) To find the value of Rin meters for which the vacuum energy is ρ=−1GeV4, we set the
vacuum energy formula equal to −1and solve for R:
−1 = −1
(2π)5R9
R9= (2π)5
R=9
p(2π)5
R≈3.28626 ×10−16 m
Therefore, the value of Rin meters for which the vacuum energy is ρ=−1GeV4is approxi-
mately 3.28626 ×10−16 m.
16 19. SOLITONS IN STRING THEORY
Problem 19. Consider a D-brane in type IIA superstring theory with tension T. Suppose the
D-brane extends in pspatial dimensions. The energy density Uper unit p-dimensional volume of
the D-brane is given by U=cT , where cis a constant.
a) If the D-brane has p= 3 spatial dimensions and the tension is T= 5 ×10−2, find the energy
density U.
b) Now, let’s consider another D-brane with tension T= 0.1and energy density U= 0.4.
Determine the number of spatial dimensions pin which this D-brane extends.
Solution 19.
a) Given p= 3 and T= 5 ×10−2, we use the formula U=cT to find the energy density U:
U=c×5×10−2= 0.05c
So, the energy density Ufor the D-brane with p= 3 spatial dimensions and tension T= 5×10−2
is 0.05c.
b) For this case, we have T= 0.1and U= 0.4. Using the same formula for energy density,
U=cT , we can solve for cfirst:
c=U
T=0.4
0.1= 4
Now, we substitute c= 4 back into the equation U=cT :
0.4=4×p
Solving for p, we find:
p=0.4
4= 0.1
Thus, the D-brane extends in p= 0.1spatial dimensions.
I. Let’s formulate a numerical problem in the context of String Theory and M-Theory:
17 20. ADS/CFT CORRESPONDENCE IN M-THEORY
Problem 20. Consider a type IIA string theory on an Anti-de Sitter space (AdS) with a five-
dimensional radius RAdS = 10. Calculate the corresponding conformal field theory (CFT) central
charge using the AdS/CFT correspondence formula:
c=3L
2G
where Lis the AdS radius and Gis Newton’s constant. Take the value of Newton’s constant
G= 6.71 ×10−11 m3kg−1s−2.
Solution 20. Given: - AdS radius RAdS = 10 - Newton’s constant G= 6.71 ×10−11 m3kg−1
s−2
The CFT central charge can be calculated using the AdS/CFT correspondence formula:
c=3L
2G
Plugging in the values:
c=3×10
2×6.71 ×10−11
c=30
13.42 ×10−11
c=30
1.342 ×10−10
c=30
1.342
c≈22.36
Therefore, the central charge of the corresponding conformal field theory is approximately
22.36.
I. Entanglement Entropy in String Theory:
18 21. Entanglement Entropy in String Theory
Problem 21. Consider a 1+1 dimensional conformal field theory described by a CFT with
central charge c= 1. Let’s calculate the entanglement entropy of a subsystem in this theory.
[Additional context: The entanglement entropy Sof a subsystem in a CFT is given by the formula
S=c
3log L
ϵ, where cis the central charge, Lis the length of the subsystem, and ϵis a short-
distance cutoff.]
a) Suppose we have a subsystem of length L= 2 in this CFT. Calculate the entanglement
entropy when the short-distance cutoff ϵ= 0.1.
b) Now, consider another CFT described by a different conformal field theory with central charge
c= 2. If we have a subsystem of length L= 3 in this CFT, what is the entanglement entropy when
the short-distance cutoff ϵ= 0.01?
Solution 21.
a) We are given c= 1,L= 2, and ϵ= 0.1. Plugging these values into the formula for entangle-
ment entropy, we get:
S=1
3log 2
0.1=1
3log 20 ≈1
3×2.9957 ≈0.9986
Therefore, the entanglement entropy when L= 2spaceunits and ϵ= 0.1is approximately
0.9986.
b) For the new CFT with c= 2,L= 3, and ϵ= 0.01, we apply the formula:
S=2
3log 3
0.01=2
3log 300 ≈2
3×5.7038 ≈3.8026
Therefore, the entanglement entropy when L= 3spaceunits and ϵ= 0.01 is approximately
3.8026.
I can certainly help with that! Let’s proceed with a numerical problem for String Theory and
M-Theory.
19 22. SPACE-TIME SINGULARITIES IN M-THEORY
Problem 22. Consider a compactified M-theory scenario where the compact spatial dimension
has a radius of R= 10−17 meters. If an object is traveling at a speed of 0.9c(where cis the speed
of light) in this compactified dimension, determine the time duration (in seconds) it takes for the
object to travel halfway around the compactified dimension.
Solution 22. a) The circumference of the compact spatial dimension is given by 2πR. There-
fore, the distance required to travel halfway around the dimension is πR.
b) The speed of the object is 0.9c, where c= 3 ×108m/s. Hence, the object travels a distance
of 0.9c×t=πR in time t, where tis the time duration we want to find.
Using the equation 0.9c×t=πR, we can solve for t:
0.9×3×108m/s ×t=π×10−17 m
0.9×3×108t=π×10−17
2.7×108t= 3.1416 ×10−17
t=3.1416 ×10−17
2.7×108
t≈1.163 ×10−25 s
Therefore, it takes approximately 1.163 ×10−25 seconds for the object to travel halfway around
the compactified dimension.
20 23. DEFORMATIONS OF STRING THEORY
Problem 23. Consider a closed bosonic string of length Lmoving in Dspacetime dimensions.
The string is perturbed by a small deformation given by the following equation of motion:
∂2
∂τ 2−∂2
∂σ2Xµ(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ)
where Xµ(τ, σ)denotes the string embedding coordinates, µ, ν = 0,1, . . . , D −1,ϵis a small
parameter, and α′is the Regge slope parameter.
a) Compute the equation of motion for X0(τ, σ)and X1(τ, σ).
b) Determine the general solution for X0(τ, σ)and X1(τ, σ).
c) Given that the string is parameterized by (τ, σ)such that 0≤σ≤π, find the normal mode
frequencies for X0(τ, σ)and X1(τ, σ).
Solution 23.
a) The equation of motion can be split into separate equations for X0(τ, σ)and X1(τ, σ)by
setting µ= 0 and µ= 1 respectively:
∂2
∂τ 2X0(τ, σ)−∂2
∂σ2X0(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ) = ϵα′sin(2σ)∂0X1(τ, σ)
∂2
∂τ 2X1(τ, σ)−∂2
∂σ2X1(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ) = ϵα′sin(2σ)∂1X0(τ, σ)
b) To find the general solutions, we solve the wave equation for X0(τ, σ)and X1(τ, σ). The
solutions will have the form:
X0(τ, σ) = A(σ)eiω0τ+B(σ)e−iω0τ
X1(τ, σ) = C(σ)eiω1τ+D(σ)e−iω1τ
where A(σ),B(σ),C(σ), and D(σ)are functions of σ, and ω0and ω1are the normal mode
frequencies.
c) To find the normal mode frequencies, we substitute the solutions back into the equations of
motion and solve for ω0and ω1. After normalization, the normal mode frequencies are given by:
ω0=1
Lrn2−ϵα′n2
2
ω1=1
Lrn2+ϵα′n2
2
where n= 1,2,3, . . . represents the mode number.
21 24. QUANTUM FIELD THEORY LIMIT OF M-THEORY
Problem 24. Consider a closed string in 10 dimensions with a winding mode wrapping around
a circle of radius R. The momentum mode of the string has energy E=n
R, where nis an integer.
The winding mode has energy E=mR
α′, where mis an integer and α′is the Regge slope.
a) If the winding mode has energy E= 2π/α′and the momentum mode has energy E= 1/√α′,
find the values of mand n.
b) Determine the total energy ETof the closed string in terms of α′.
Solution 24.
a) Since the winding mode has energy E=mR
α′=2π
α′, we have:
mR
α′=2π
α′
mR = 2π
Similarly, for the momentum mode with energy E=n
R=1
√α′, we get:
n
R=1
√α′
nR =√α′
From these two equations, we can see that m= 1 and n= 1.
b) The total energy ETof the closed string is given by:
ET=n
R+mR
α′=1
R+R
α′=1 + R2/α′
R
Therefore, the total energy of the closed string in terms of α′is ET=1 + R2/α′
R.
22 25. TOPOLOGICAL ASPECTS OF STRING THEORY
Problem 25. Consider a closed oriented Riemann surface Σgof genus g. The Euler charac-
teristic of Σgis given by χ(Σg)=2−2g.
a) Calculate the Euler characteristic for a torus (g= 1).
b) Calculate the Euler characteristic for a sphere with two handles (g= 2).
c) Calculate the Euler characteristic for a surface with genus g= 3.
Solution 25.
a) For a torus (g= 1), the Euler characteristic is given by χ(Σ1) = 2 −2·1 = 2 −2 = 0.
Therefore, the Euler characteristic for a torus is 0.
b) For a sphere with two handles (g= 2), the Euler characteristic is χ(Σ2) = 2−2·2 = 2−4 = −2.
Hence, the Euler characteristic for a sphere with two handles is -2.
c) For a surface with genus g= 3, the Euler characteristic is χ(Σ3)=2−2·3=2−6 = −4.
Therefore, the Euler characteristic for a surface of genus 3 is -4.
3 3. QUANTUM GRAVITY IN STRING THEORY
Problem 3. Consider an open string moving in 4-dimensional spacetime. The string has tension
T= 1 and mass per unit length µ= 2.
a) Find the speed of propagation of waves on this string.
b) Calculate the energy stored in a segment of string of length L= 3.
c) If the string is stretched with a force of F= 5, determine the amplitude of the standing wave
that could form on the string.
Solution 3.
a) The speed of propagation of waves on a string is given by
v=sT
µ.
Substitute T= 1 and µ= 2:
v=r1
2=√2
2.
Therefore, the speed of propagation of waves on this string is √2
2.
b) The energy stored in a segment of string of length Lis given by
E=1
2µv2L.
Substitute µ= 2,v=√2
2, and L= 3:
E=1
2×2× √2
2!2
×3 = 3
2.
Therefore, the energy stored in a segment of string of length 3 is 3
2.
c) The amplitude of the standing wave that could form on the string under the applied force F
is given by
A=F
2πv2.
Substitute F= 5 and v=√2
2:
A=5
2π×√2
22=5
2π×1
2
=5√2
π.
Therefore, the amplitude of the standing wave that could form on the string is 5√2
π.
4 4. BLACK HOLES AND STRING THEORY
Problem 4. Consider a black hole in four-dimensional spacetime described by the Schwarzschild
metric:
ds2=−1−2GM
c2rc2dt2+1−2GM
c2r−1
dr2+r2dΩ2
where Mis the mass of the black hole, cis the speed of light, Gis the gravitational constant,
and dΩ2=dθ2+sin2θdϕ2in spherical coordinates (t, r, θ, ϕ).
a) Find the event horizon radius Rhorizon of this black hole.
b) Determine the Schwarzschild radius RSin terms of M,G, and c.
c) Given that the mass of the black hole is M= 2 ×1030 kg, calculate the mass of the black
hole in units of solar masses (M⊙= 1.989 ×1030 kg).
Solution 4.
a) To find the event horizon radius, we set the metric coefficient of dt2to zero at the event
horizon. So, 1−2GM
c2Rhorizon = 0. Solving for Rhorizon gives:
Rhorizon =2GM
c2
b) The Schwarzschild radius is defined as the radius at which the metric becomes singular. It
is given by RS=2GM
c2.
c) Substituting M= 2 ×1030 kg into the Schwarzschild radius formula, we have:
RS=2G(2 ×1030)
c2=4×1030G
c2
Now, we can express the mass of the black hole in terms of solar masses by dividing by the
mass of the Sun:
4×1030G
c2÷1.989 ×1030 =2G
c2
Therefore, the mass of the black hole is 2solar masses.
5 5. TACHYON CONDENSATION IN STRING THEORY
Problem 5. Consider a closed bosonic string theory where the endpoint of the string coor-
dinates are subject to Neumann boundary conditions. The first excited level of the closed string
contains a tachyon with mass m2=−2
α′.
a) Calculate the momentum of the tachyon state in the string theory.
b) Show that the mass of the tachyon state in the open bosonic string theory is m= 0.
c) Interpret the result in relation to tachyon condensation.
Solution 5.
a) The mass-squared of a state in a closed string theory is given by the level matching condition:
m2=4
α′(N−1)
where Nis the occupation number operator for the state. For the first excited level, N= 1 and
m2=−2
α′. Substituting these values into the equation above, we have:
−2
α′=4
α′(1 −1)
−2=0
This equation is a contradiction, indicating that there is no physical state with a mass-squared of
−2
α′for the first excited level. So, the tachyon state does not exist.
b) In the open bosonic string theory, the mass-squared of a state is given by:
m2=2
α′(N−1)
For the tachyon state at the first excited level in the open string, N= 1 and m2=−1
α′. Taking the
square root of this, we get m= 0. Therefore, the mass of the tachyon state in the open bosonic
string theory is zero.
c) The result m= 0 for the open string tachyon state implies that it is a massless state. In string
theory, a tachyon has negative mass squared and indicates an instability in the theory. Tachyon
condensation is a process where this unstable state "condenses" to a minimum energy state. In
this case, the tachyon state in the open bosonic string theory having a mass of zero suggests that
it represents the minimum energy state after tachyon condensation has occurred. This process
helps stabilize the theory by eliminating the instability caused by the presence of the tachyon.
6 6. EXTRA DIMENSIONS IN M-THEORY
Problem 6. Consider a scenario in M-Theory where there are 7 spatial dimensions and 3
temporal dimensions. The size of the compactified extra dimensions is given by R= 10−17 meters.
Calculate the compactification scale in GeV.
Solution 6.
a) The compactification scale Mccan be calculated using the formula for the compactified extra
dimensions:
Mc=1
R
Substitute R= 10−17 meters into the formula:
Mc=1
10−17 = 1017 m−1
b) To convert the compactification scale from meters to GeV, we need to use the relation 1GeV =
1.97 ×10−16 m−1.
Let’s convert the compactification scale:
Mc= 1017 ×1.97 ×10−16 = 1.97 ×10 GeV = 19.7GeV
Therefore, the compactification scale in GeV is 19.7 GeV.
7 7. BRANE DYNAMICS IN STRING THEORY
Problem 7. Consider a type IIB superstring theory in 10 dimensions with D3-branes. The
tension of the D3-brane is given by T=1
(2π)3(α′)2, where α′is the string length parameter. Calculate
the energy density stored in a D3-brane of length L= 10√α′.
Solution 7. a) The energy density stored in a D3-brane can be calculated by dividing its energy
by its volume. The energy Eof a D3-brane is given by E=T×V, where Tis the tension of the
D3-brane and Vis its volume. Since the D3-brane extends in 3 spatial dimensions, the volume V
is L3= (10√α′)3= 1000α′3/2.
Substitute T=1
(2π)3(α′)2and V= 1000α′3/2into the equation:
E=1
(2π)3(α′)2×1000α′3/2=1000
(2π)3α′1/2
Thus, the energy stored in the D3-brane is 1000
(2π)3α′1/2.
b) The energy density is given by the energy per unit volume. We divide the energy Eby the
volume Vto find the energy density u=E
V.
Substitute E=1000
(2π)3α′1/2and V= 1000α′3/2into the equation:
u=1000/(2π)3α′1/2
1000α′3/2=1
(2π)3α′2
Therefore, the energy density stored in the D3-brane is 1
(2π)3α′2.
8 8. SUPERSYMMETRY BREAKING IN M-THEORY
Problem 8. Consider a supersymmetric M-Theory compactified on a 2-torus with radii R1and
R2. The volume of the torus is given by V=R1R2. Suppose that supersymmetry is broken by the
flux through the torus such that the gravitino mass term is generated.
[Given: The gravitino mass term is given by m3/2=e⟨G⟩, where Gis the flux. Also, we have
⟨G⟩ =−2 ln(V), where ⟨G⟩ denotes the expectation value of the flux.]
a) Show that the gravitino mass term m3/2in terms of the radii R1and R2.
b) If R1= 2 and R2= 3, calculate the gravitino mass m3/2.
Solution 8.
a) To find the gravitino mass term m3/2in terms of the radii R1and R2, we first need to express
the volume Vin terms of R1and R2:
V=R1·R2
Next, we can find the expectation value of the flux ⟨G⟩:
⟨G⟩ =−2 ln(V) = −2 ln(R1·R2) = −2 ln(R1)−2 ln(R2)
Substitute this into the expression for the gravitino mass term:
m3/2=e⟨G⟩ =e−2 ln(R1)−2 ln(R2)=e−2 ln(R1)·e−2 ln(R2)=1
R2
1·1
R2
2
=1
R2
1R2
2
=1
V2
b) Given R1= 2 and R2= 3, the volume V=R1·R2= 2 ·3=6. Therefore, the gravitino mass
term m3/2is:
m3/2=1
V2=1
62=1
36 = 0.0278
I. Problem 9. Consider a Type IIA superstring theory with a D6-brane wrapped on a compact
3-torus with sides of length L. The tension of the D6-brane is given by T6=1
(2π)6(α′)4. Suppose
the compactified space is a cube, compute the energy density UD6 of the D6-brane in terms of L.
Hint: The energy density is defined as the energy per unit volume.
II. Problem 10. In Type IIB superstring theory, a D3-brane has tension T3=1
(2π)3(α′)2. If this
theory is in a 10-dimensional spacetime with a toroidal compactification down to 6 dimensions, and
the 3-brane extends along all 3 of those compact dimensions, calculate the energy density UD3 of
the D3-brane in terms of the compactification radii Ri.
Hint: The energy density is defined as the energy per unit volume.
III. Problem 11. In M-Theory, consider a M5-brane wrapped on a torus with radii R1and R2. If
the tension of the brane is T5=1
(2π)5(ℓp)3, find the energy density UM5 of the M5-brane in terms of
the torus radii.
Hint: The energy density is defined as the energy per unit volume.
9 10. COSMOLOGY AND STRING/M-THEORY
Problem 10. Consider a string theory model in a universe with extra dimensions compactified
on a torus. The radius of the torus is R, and the string tension is T. The compactified dimensions
are described by an effective field theory with a massless scalar field ϕ, whose potential is given
by V(ϕ) = 1
2m2ϕ2.
a) Show that the effective tension in the compactified dimensions, Teff , is given by Tef f =
T e−2ϕ.
b) Determine the equation of motion for the scalar field ϕ.
c) Find the minimum of the potential V(ϕ).
Solution 10.
a) The effective tension in the compactified dimensions is given by Teff =T e−2ϕ. Let’s derive
this expression:
In string theory, the effective tension depends on the string coupling as Teff =T gs, where
gs=e−2ϕ. Therefore, Teff =T e−2ϕ.
b) The equation of motion for the scalar field ϕis given by:
d
dt ∂L
∂˙
ϕ−∂L
∂ϕ = 0
where the Lagrangian L=1
2˙
ϕ2−V(ϕ).
d
dt ∂L
∂˙
ϕ−∂L
∂ϕ =¨
ϕ+m2ϕ= 0
Therefore, the equation of motion for the scalar field ϕis ¨
ϕ+m2ϕ= 0.
c) To find the minimum of the potential V(ϕ), we need to solve dV
dϕ = 0:
dV
dϕ =m2ϕ= 0
This implies that the minimum of the potential occurs at ϕ= 0.
10 11. DUALITIES IN M-THEORY
Problem 11. Consider two particular string theories, Type IIA and Type IIB, related by T-duality.
Let the radius of a circle in Type IIA theory be R. If the number of fundamental strings winding
around the circle is n, find the dual circle radius in Type IIB theory.
Solution 11. Given: Radius of circle in Type IIA theory, R, and number of winding fundamental
strings, n.
In Type IIA theory, the momentum along the circle is given by P=n
R.
By T-duality, the winding number in Type IIB string theory is the same as the momentum in Type
IIA theory, and vice versa. Therefore, in Type IIB theory, the radius of the dual circle is related to
the momentum by Rdual =α′
R.
Substitute P=n
Rinto the formula for the dual radius in Type IIB theory:
Rdual =α′
R=α′
n/P =α′R
n
Hence, the dual circle radius in Type IIB theory is α′R
n.
11 12. INTEGRABILITY IN STRING THEORY
Problem 12. Consider a closed bosonic string moving in a background with a constant mag-
netic field Bin the z-direction. The equation of motion for the string is given by the Nambu-Goto
action
S=−T
2Zdσdτ√−h hab∂aXµ∂bXνGµν
where Tis the tension of the string, his the determinant of the world-sheet metric hab,Xµ=
Xµ(τ, σ)are the embedding coordinates of the string world-sheet, and Gµν is the background metric
tensor.
Given that the background metric is flat Minkowski space with Gµν =ηµν and the magnetic field
is B=Bz, where Bzis constant, compute the equation of motion for the string in this background.
Solution 12.
The equation of motion for the string can be derived by varying the action with respect to the
embedding coordinates Xµ.
Let’s denote ∂aXµ≡∂aXµ(τ, σ).
The variation of the action with respect to Xµgives the equation of motion:
δS
δXµ=−T
2Zdσdτ√−hhab∂a∂bXνηνµ = 0
Expanding the terms and using the fact that the world-sheet metric is hab =diag(−1,1), we get:
TZdσdτ ∂2
τXν−∂2
σXνηνµ = 0
The equations of motion are then given by:
a) In the µ= 0 direction:
∂2
τX0−∂2
σX0= 0
b) In the µ= 1 direction:
∂2
τX1−∂2
σX1= 0
c) In the µ= 2 direction:
∂2
τX2−∂2
σX2= 0
These equations of motion describe the dynamics of the string in the background of a constant
magnetic field B=Bz.
12 13. CAUSALITY AND STRING/M-THEORY
Problem 13. Consider a closed string propagating in a spacetime described by D= 10
dimensions. The string moves in a geometry where the background metric is given by ds2=
−dt2+dx2
1+ (dx2)2+···+ (dx8)2+ (dx9)2+ (dx10)2, where x10 is the compactified spatial direction
with a radius R.
a) Calculate the maximum energy Emax of a closed string state that can propagate in this ge-
ometry without creating a closed timelike curve.
b) Find the minimum allowed radius Rmin of the compactified spatial direction that ensures
causality is not violated in this spacetime.
Solution 13.
a) The maximum energy Emax of a closed string state can be found using the formula:
Emax =1
α′rN
2
where Nis the level of the state and α′is the Regge slope parameter. Since we are dealing with
a closed string in D= 10 dimensions, the critical dimension is D= 10, and the Regge slope
parameter is α′= 1/2πT .
For a closed string moving in a compactified dimension with radius R, the contribution to the
mass in the compact direction is n/R where nis the winding number. To avoid closed timelike
curves, we must have Emax =n/R. Equating these two expressions for Emax, we get:
n
R=1
α′rN
2
n
R=2πT
2rN
2
nR = 2πT √2N
R=2πT √2N
n
Therefore, the maximum energy is achieved at n= 1 and N= 2, leading to:
Emax =1
α′rN
2=1
α′r2
2=1
α′=√2πT
2
b) To ensure that causality is not violated in this spacetime, we must have the inequality Rmin ≥
2πR to avoid closed timelike curves. Substituting in the expression for Rmin from part (a), we get:
Rmin =2πT p2(1)
1= 2π√2πT ≥2πR
√2πT ≥R
2πT ≥R2
R2−2πT ≤0
Hence, the minimum allowed radius Rmin of the compactified spatial direction is R=√2πT .
13 14. HOLOGRAPHY IN M-THEORY
Problem 14. Consider a spacetime described by M-theory with 11 dimensions. The holo-
graphic principle states that the information of a region of space can be encoded on its boundary.
Suppose we have a 4-dimensional hypercube with side length L.
a) Calculate the volume of this hypercube in terms of L.
b) According to the holographic principle, what is the size of the boundary that encodes all the
information of this hypercube?
c) If the hypercube is in an 11-dimensional spacetime in M-theory, how many spatial dimensions
are "compactified"?
Solution 14.
a) The volume of a 4-dimensional hypercube with side length Lis given by V=L4.
b) The boundary of the hypercube is a 3-dimensional cube with sides of length L. The total
surface area of a cube is given by A= 6L2. Hence, for a 4-dimensional hypercube, the size of the
boundary that encodes all the information is A= 6L2.
c) In M-theory with 11 dimensions, the 4 spatial dimensions of the hypercube and the 1 time
dimension are known. This leaves 11 −4−1=6spatial dimensions "compactified".
I’m glad to help generate numerical problem questions for you. Could you please specify the
topic within String Theory and M-Theory that you would like the problem to be based on?
14 16. DARK MATTER AND STRING/M-THEORY
Problem 16. Consider a string theory scenario where a closed string moving in a compact
spatial dimension of radius Rinteracts with dark matter particles gravitationally. The string coupling
constant is given by gs= 0.1.
a) If the mass of the dark matter particle is m= 10−22 eV/c2, calculate the gravitational force
between the closed string and a dark matter particle located at a distance r= 1 mm.
b) If the string coupling constant gsis changed to 0.05, how does this affect the gravitational
force between the closed string and the dark matter particle?
Solution 16.
a) The gravitational force between the closed string and the dark matter particle can be calcu-
lated using Newton’s law of gravitation:
F=G·mdark matter ·mstring
r2
where Gis the gravitational constant (6.67430 ×10−11 m3kg−1s−2), mdark matter is the mass of
the dark matter particle (in kg), mstring is the mass of the closed string (in kg), and ris the distance
between the closed string and the dark matter particle (in meters).
Given m= 10−22 eV/c2, we need to convert this to kilograms by using the conversion factor
1eV/c2= 1.783 ×10−36 kg. Thus, mdark matter = 10−22 ×1.783 ×10−36 = 1.783 ×10−58 kg.
Plugging in the values G= 6.67430 ×10−11 m3kg−1s−2,mdark matter = 1.783 ×10−58 kg,
mstring =? (mass of the string is not provided but let’s assume it to be of the same order of magnitude
as a dark matter particle), and r= 1 mm = 0.001 m, we can find the gravitational force.
b) If the string coupling constant gsis changed to 0.05, we can re-calculate the gravitational
force using the new value of gsin the formula above. Let’s assume that the mass of the dark matter
particle remains the same.
Solution 16.
a) Given: G= 6.67430 ×10−11 m3kg−1s−2,mdark matter = 1.783 ×10−58 kg, mstring = 1.783 ×
10−58 kg (assumed to be of the same order), r= 0.001 m
Plugging in the values, we get:
F=6.67430 ×10−11 ×1.783 ×10−58 ×1.783 ×10−58
(0.001)2
⇒F=2.0004 ×10−124
10−6
⇒F= 2.0004 ×10−118 N
Thus, the gravitational force between the closed string and a dark matter particle located at a
distance of 1 mm is approximately 2.0004 ×10−118 N.
b) If gs= 0.05, we can re-calculate the gravitational force using the updated value of gsin the
formula. Plugging in the new value, we can find the new gravitational force.
I’m sorry, but it seems that I cannot generate numerical problems for String Theory and M-
Theory as they are primarily theoretical and mathematical physics concepts that do not involve
numerical computations. Would you like a conceptual problem or a theoretical problem instead?
15 18. VACUUM ENERGY IN M-THEORY
Problem 18. Consider a particular compactification of M-theory on a 5-dimensional torus with
each side having length R. The vacuum energy in this scenario is given by ρ=−1
(2π)5R9.
a) Calculate the vacuum energy ρin GeV4when R= 1 cm. b) Find the value of Rin m for
which the vacuum energy is ρ=−1GeV4.
Solution 18. a) To calculate the vacuum energy in GeV4, we need to convert the length R=
1cm to meters and then substitute it into the given formula.
Given: 1 cm = 10−2m
Plugging this value into the formula for vacuum energy: ρ=−1
(2π)5(10−2)9
ρ=−1
(2π)5(10−18)
ρ=−1
(2π)5(10−18)
ρ=−1
(2π)5×10−18
ρ=−1
(2π)5×10−18
ρ≈ −8.53254 ×1065 GeV4
Therefore, the vacuum energy when R= 1 cm is approximately −8.53254 ×1065 GeV4.
b) To find the value of Rin meters for which the vacuum energy is ρ=−1GeV4, we set the
vacuum energy formula equal to −1and solve for R:
−1 = −1
(2π)5R9
R9= (2π)5
R=9
p(2π)5
R≈3.28626 ×10−16 m
Therefore, the value of Rin meters for which the vacuum energy is ρ=−1GeV4is approxi-
mately 3.28626 ×10−16 m.
16 19. SOLITONS IN STRING THEORY
Problem 19. Consider a D-brane in type IIA superstring theory with tension T. Suppose the
D-brane extends in pspatial dimensions. The energy density Uper unit p-dimensional volume of
the D-brane is given by U=cT , where cis a constant.
a) If the D-brane has p= 3 spatial dimensions and the tension is T= 5 ×10−2, find the energy
density U.
b) Now, let’s consider another D-brane with tension T= 0.1and energy density U= 0.4.
Determine the number of spatial dimensions pin which this D-brane extends.
Solution 19.
a) Given p= 3 and T= 5 ×10−2, we use the formula U=cT to find the energy density U:
U=c×5×10−2= 0.05c
So, the energy density Ufor the D-brane with p= 3 spatial dimensions and tension T= 5×10−2
is 0.05c.
b) For this case, we have T= 0.1and U= 0.4. Using the same formula for energy density,
U=cT , we can solve for cfirst:
c=U
T=0.4
0.1= 4
Now, we substitute c= 4 back into the equation U=cT :
0.4=4×p
Solving for p, we find:
p=0.4
4= 0.1
Thus, the D-brane extends in p= 0.1spatial dimensions.
I. Let’s formulate a numerical problem in the context of String Theory and M-Theory:
17 20. ADS/CFT CORRESPONDENCE IN M-THEORY
Problem 20. Consider a type IIA string theory on an Anti-de Sitter space (AdS) with a five-
dimensional radius RAdS = 10. Calculate the corresponding conformal field theory (CFT) central
charge using the AdS/CFT correspondence formula:
c=3L
2G
where Lis the AdS radius and Gis Newton’s constant. Take the value of Newton’s constant
G= 6.71 ×10−11 m3kg−1s−2.
Solution 20. Given: - AdS radius RAdS = 10 - Newton’s constant G= 6.71 ×10−11 m3kg−1
s−2
The CFT central charge can be calculated using the AdS/CFT correspondence formula:
c=3L
2G
Plugging in the values:
c=3×10
2×6.71 ×10−11
c=30
13.42 ×10−11
c=30
1.342 ×10−10
c=30
1.342
c≈22.36
Therefore, the central charge of the corresponding conformal field theory is approximately
22.36.
I. Entanglement Entropy in String Theory:
18 21. Entanglement Entropy in String Theory
Problem 21. Consider a 1+1 dimensional conformal field theory described by a CFT with
central charge c= 1. Let’s calculate the entanglement entropy of a subsystem in this theory.
[Additional context: The entanglement entropy Sof a subsystem in a CFT is given by the formula
S=c
3log L
ϵ, where cis the central charge, Lis the length of the subsystem, and ϵis a short-
distance cutoff.]
a) Suppose we have a subsystem of length L= 2 in this CFT. Calculate the entanglement
entropy when the short-distance cutoff ϵ= 0.1.
b) Now, consider another CFT described by a different conformal field theory with central charge
c= 2. If we have a subsystem of length L= 3 in this CFT, what is the entanglement entropy when
the short-distance cutoff ϵ= 0.01?
Solution 21.
a) We are given c= 1,L= 2, and ϵ= 0.1. Plugging these values into the formula for entangle-
ment entropy, we get:
S=1
3log 2
0.1=1
3log 20 ≈1
3×2.9957 ≈0.9986
Therefore, the entanglement entropy when L= 2spaceunits and ϵ= 0.1is approximately
0.9986.
b) For the new CFT with c= 2,L= 3, and ϵ= 0.01, we apply the formula:
S=2
3log 3
0.01=2
3log 300 ≈2
3×5.7038 ≈3.8026
Therefore, the entanglement entropy when L= 3spaceunits and ϵ= 0.01 is approximately
3.8026.
I can certainly help with that! Let’s proceed with a numerical problem for String Theory and
M-Theory.
19 22. SPACE-TIME SINGULARITIES IN M-THEORY
Problem 22. Consider a compactified M-theory scenario where the compact spatial dimension
has a radius of R= 10−17 meters. If an object is traveling at a speed of 0.9c(where cis the speed
of light) in this compactified dimension, determine the time duration (in seconds) it takes for the
object to travel halfway around the compactified dimension.
Solution 22. a) The circumference of the compact spatial dimension is given by 2πR. There-
fore, the distance required to travel halfway around the dimension is πR.
b) The speed of the object is 0.9c, where c= 3 ×108m/s. Hence, the object travels a distance
of 0.9c×t=πR in time t, where tis the time duration we want to find.
Using the equation 0.9c×t=πR, we can solve for t:
0.9×3×108m/s ×t=π×10−17 m
0.9×3×108t=π×10−17
2.7×108t= 3.1416 ×10−17
t=3.1416 ×10−17
2.7×108
t≈1.163 ×10−25 s
Therefore, it takes approximately 1.163 ×10−25 seconds for the object to travel halfway around
the compactified dimension.
20 23. DEFORMATIONS OF STRING THEORY
Problem 23. Consider a closed bosonic string of length Lmoving in Dspacetime dimensions.
The string is perturbed by a small deformation given by the following equation of motion:
∂2
∂τ 2−∂2
∂σ2Xµ(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ)
where Xµ(τ, σ)denotes the string embedding coordinates, µ, ν = 0,1, . . . , D −1,ϵis a small
parameter, and α′is the Regge slope parameter.
a) Compute the equation of motion for X0(τ, σ)and X1(τ, σ).
b) Determine the general solution for X0(τ, σ)and X1(τ, σ).
c) Given that the string is parameterized by (τ, σ)such that 0≤σ≤π, find the normal mode
frequencies for X0(τ, σ)and X1(τ, σ).
Solution 23.
a) The equation of motion can be split into separate equations for X0(τ, σ)and X1(τ, σ)by
setting µ= 0 and µ= 1 respectively:
∂2
∂τ 2X0(τ, σ)−∂2
∂σ2X0(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ) = ϵα′sin(2σ)∂0X1(τ, σ)
∂2
∂τ 2X1(τ, σ)−∂2
∂σ2X1(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ) = ϵα′sin(2σ)∂1X0(τ, σ)
b) To find the general solutions, we solve the wave equation for X0(τ, σ)and X1(τ, σ). The
solutions will have the form:
X0(τ, σ) = A(σ)eiω0τ+B(σ)e−iω0τ
X1(τ, σ) = C(σ)eiω1τ+D(σ)e−iω1τ
where A(σ),B(σ),C(σ), and D(σ)are functions of σ, and ω0and ω1are the normal mode
frequencies.
c) To find the normal mode frequencies, we substitute the solutions back into the equations of
motion and solve for ω0and ω1. After normalization, the normal mode frequencies are given by:
ω0=1
Lrn2−ϵα′n2
2
ω1=1
Lrn2+ϵα′n2
2
where n= 1,2,3, . . . represents the mode number.
21 24. QUANTUM FIELD THEORY LIMIT OF M-THEORY
Problem 24. Consider a closed string in 10 dimensions with a winding mode wrapping around
a circle of radius R. The momentum mode of the string has energy E=n
R, where nis an integer.
The winding mode has energy E=mR
α′, where mis an integer and α′is the Regge slope.
a) If the winding mode has energy E= 2π/α′and the momentum mode has energy E= 1/√α′,
find the values of mand n.
b) Determine the total energy ETof the closed string in terms of α′.
Solution 24.
a) Since the winding mode has energy E=mR
α′=2π
α′, we have:
mR
α′=2π
α′
mR = 2π
Similarly, for the momentum mode with energy E=n
R=1
√α′, we get:
n
R=1
√α′
nR =√α′
From these two equations, we can see that m= 1 and n= 1.
b) The total energy ETof the closed string is given by:
ET=n
R+mR
α′=1
R+R
α′=1 + R2/α′
R
Therefore, the total energy of the closed string in terms of α′is ET=1 + R2/α′
R.
22 25. TOPOLOGICAL ASPECTS OF STRING THEORY
Problem 25. Consider a closed oriented Riemann surface Σgof genus g. The Euler charac-
teristic of Σgis given by χ(Σg)=2−2g.
a) Calculate the Euler characteristic for a torus (g= 1).
b) Calculate the Euler characteristic for a sphere with two handles (g= 2).
c) Calculate the Euler characteristic for a surface with genus g= 3.
Solution 25.
a) For a torus (g= 1), the Euler characteristic is given by χ(Σ1) = 2 −2·1 = 2 −2 = 0.
Therefore, the Euler characteristic for a torus is 0.
b) For a sphere with two handles (g= 2), the Euler characteristic is χ(Σ2) = 2−2·2 = 2−4 = −2.
Hence, the Euler characteristic for a sphere with two handles is -2.
c) For a surface with genus g= 3, the Euler characteristic is χ(Σ3)=2−2·3=2−6 = −4.
Therefore, the Euler characteristic for a surface of genus 3 is -4.
3 3. QUANTUM GRAVITY IN STRING THEORY
Problem 3. Consider an open string moving in 4-dimensional spacetime. The string has tension
T= 1 and mass per unit length µ= 2.
a) Find the speed of propagation of waves on this string.
b) Calculate the energy stored in a segment of string of length L= 3.
c) If the string is stretched with a force of F= 5, determine the amplitude of the standing wave
that could form on the string.
Solution 3.
a) The speed of propagation of waves on a string is given by
v=sT
µ.
Substitute T= 1 and µ= 2:
v=r1
2=√2
2.
Therefore, the speed of propagation of waves on this string is √2
2.
b) The energy stored in a segment of string of length Lis given by
E=1
2µv2L.
Substitute µ= 2,v=√2
2, and L= 3:
E=1
2×2× √2
2!2
×3 = 3
2.
Therefore, the energy stored in a segment of string of length 3 is 3
2.
c) The amplitude of the standing wave that could form on the string under the applied force F
is given by
A=F
2πv2.
Substitute F= 5 and v=√2
2:
A=5
2π×√2
22=5
2π×1
2
=5√2
π.
Therefore, the amplitude of the standing wave that could form on the string is 5√2
π.
4 4. BLACK HOLES AND STRING THEORY
Problem 4. Consider a black hole in four-dimensional spacetime described by the Schwarzschild
metric:
ds2=−1−2GM
c2rc2dt2+1−2GM
c2r−1
dr2+r2dΩ2
where Mis the mass of the black hole, cis the speed of light, Gis the gravitational constant,
and dΩ2=dθ2+sin2θdϕ2in spherical coordinates (t, r, θ, ϕ).
a) Find the event horizon radius Rhorizon of this black hole.
b) Determine the Schwarzschild radius RSin terms of M,G, and c.
c) Given that the mass of the black hole is M= 2 ×1030 kg, calculate the mass of the black
hole in units of solar masses (M⊙= 1.989 ×1030 kg).
Solution 4.
a) To find the event horizon radius, we set the metric coefficient of dt2to zero at the event
horizon. So, 1−2GM
c2Rhorizon = 0. Solving for Rhorizon gives:
Rhorizon =2GM
c2
b) The Schwarzschild radius is defined as the radius at which the metric becomes singular. It
is given by RS=2GM
c2.
c) Substituting M= 2 ×1030 kg into the Schwarzschild radius formula, we have:
RS=2G(2 ×1030)
c2=4×1030G
c2
Now, we can express the mass of the black hole in terms of solar masses by dividing by the
mass of the Sun:
4×1030G
c2÷1.989 ×1030 =2G
c2
Therefore, the mass of the black hole is 2solar masses.
5 5. TACHYON CONDENSATION IN STRING THEORY
Problem 5. Consider a closed bosonic string theory where the endpoint of the string coor-
dinates are subject to Neumann boundary conditions. The first excited level of the closed string
contains a tachyon with mass m2=−2
α′.
a) Calculate the momentum of the tachyon state in the string theory.
b) Show that the mass of the tachyon state in the open bosonic string theory is m= 0.
c) Interpret the result in relation to tachyon condensation.
Solution 5.
a) The mass-squared of a state in a closed string theory is given by the level matching condition:
m2=4
α′(N−1)
where Nis the occupation number operator for the state. For the first excited level, N= 1 and
m2=−2
α′. Substituting these values into the equation above, we have:
−2
α′=4
α′(1 −1)
−2=0
This equation is a contradiction, indicating that there is no physical state with a mass-squared of
−2
α′for the first excited level. So, the tachyon state does not exist.
b) In the open bosonic string theory, the mass-squared of a state is given by:
m2=2
α′(N−1)
For the tachyon state at the first excited level in the open string, N= 1 and m2=−1
α′. Taking the
square root of this, we get m= 0. Therefore, the mass of the tachyon state in the open bosonic
string theory is zero.
c) The result m= 0 for the open string tachyon state implies that it is a massless state. In string
theory, a tachyon has negative mass squared and indicates an instability in the theory. Tachyon
condensation is a process where this unstable state "condenses" to a minimum energy state. In
this case, the tachyon state in the open bosonic string theory having a mass of zero suggests that
it represents the minimum energy state after tachyon condensation has occurred. This process
helps stabilize the theory by eliminating the instability caused by the presence of the tachyon.
6 6. EXTRA DIMENSIONS IN M-THEORY
Problem 6. Consider a scenario in M-Theory where there are 7 spatial dimensions and 3
temporal dimensions. The size of the compactified extra dimensions is given by R= 10−17 meters.
Calculate the compactification scale in GeV.
Solution 6.
a) The compactification scale Mccan be calculated using the formula for the compactified extra
dimensions:
Mc=1
R
Substitute R= 10−17 meters into the formula:
Mc=1
10−17 = 1017 m−1
b) To convert the compactification scale from meters to GeV, we need to use the relation 1GeV =
1.97 ×10−16 m−1.
Let’s convert the compactification scale:
Mc= 1017 ×1.97 ×10−16 = 1.97 ×10 GeV = 19.7GeV
Therefore, the compactification scale in GeV is 19.7 GeV.
7 7. BRANE DYNAMICS IN STRING THEORY
Problem 7. Consider a type IIB superstring theory in 10 dimensions with D3-branes. The
tension of the D3-brane is given by T=1
(2π)3(α′)2, where α′is the string length parameter. Calculate
the energy density stored in a D3-brane of length L= 10√α′.
Solution 7. a) The energy density stored in a D3-brane can be calculated by dividing its energy
by its volume. The energy Eof a D3-brane is given by E=T×V, where Tis the tension of the
D3-brane and Vis its volume. Since the D3-brane extends in 3 spatial dimensions, the volume V
is L3= (10√α′)3= 1000α′3/2.
Substitute T=1
(2π)3(α′)2and V= 1000α′3/2into the equation:
E=1
(2π)3(α′)2×1000α′3/2=1000
(2π)3α′1/2
Thus, the energy stored in the D3-brane is 1000
(2π)3α′1/2.
b) The energy density is given by the energy per unit volume. We divide the energy Eby the
volume Vto find the energy density u=E
V.
Substitute E=1000
(2π)3α′1/2and V= 1000α′3/2into the equation:
u=1000/(2π)3α′1/2
1000α′3/2=1
(2π)3α′2
Therefore, the energy density stored in the D3-brane is 1
(2π)3α′2.
8 8. SUPERSYMMETRY BREAKING IN M-THEORY
Problem 8. Consider a supersymmetric M-Theory compactified on a 2-torus with radii R1and
R2. The volume of the torus is given by V=R1R2. Suppose that supersymmetry is broken by the
flux through the torus such that the gravitino mass term is generated.
[Given: The gravitino mass term is given by m3/2=e⟨G⟩, where Gis the flux. Also, we have
⟨G⟩ =−2 ln(V), where ⟨G⟩ denotes the expectation value of the flux.]
a) Show that the gravitino mass term m3/2in terms of the radii R1and R2.
b) If R1= 2 and R2= 3, calculate the gravitino mass m3/2.
Solution 8.
a) To find the gravitino mass term m3/2in terms of the radii R1and R2, we first need to express
the volume Vin terms of R1and R2:
V=R1·R2
Next, we can find the expectation value of the flux ⟨G⟩:
⟨G⟩ =−2 ln(V) = −2 ln(R1·R2) = −2 ln(R1)−2 ln(R2)
Substitute this into the expression for the gravitino mass term:
m3/2=e⟨G⟩ =e−2 ln(R1)−2 ln(R2)=e−2 ln(R1)·e−2 ln(R2)=1
R2
1·1
R2
2
=1
R2
1R2
2
=1
V2
b) Given R1= 2 and R2= 3, the volume V=R1·R2= 2 ·3=6. Therefore, the gravitino mass
term m3/2is:
m3/2=1
V2=1
62=1
36 = 0.0278
I. Problem 9. Consider a Type IIA superstring theory with a D6-brane wrapped on a compact
3-torus with sides of length L. The tension of the D6-brane is given by T6=1
(2π)6(α′)4. Suppose
the compactified space is a cube, compute the energy density UD6 of the D6-brane in terms of L.
Hint: The energy density is defined as the energy per unit volume.
II. Problem 10. In Type IIB superstring theory, a D3-brane has tension T3=1
(2π)3(α′)2. If this
theory is in a 10-dimensional spacetime with a toroidal compactification down to 6 dimensions, and
the 3-brane extends along all 3 of those compact dimensions, calculate the energy density UD3 of
the D3-brane in terms of the compactification radii Ri.
Hint: The energy density is defined as the energy per unit volume.
III. Problem 11. In M-Theory, consider a M5-brane wrapped on a torus with radii R1and R2. If
the tension of the brane is T5=1
(2π)5(ℓp)3, find the energy density UM5 of the M5-brane in terms of
the torus radii.
Hint: The energy density is defined as the energy per unit volume.
9 10. COSMOLOGY AND STRING/M-THEORY
Problem 10. Consider a string theory model in a universe with extra dimensions compactified
on a torus. The radius of the torus is R, and the string tension is T. The compactified dimensions
are described by an effective field theory with a massless scalar field ϕ, whose potential is given
by V(ϕ) = 1
2m2ϕ2.
a) Show that the effective tension in the compactified dimensions, Teff , is given by Tef f =
T e−2ϕ.
b) Determine the equation of motion for the scalar field ϕ.
c) Find the minimum of the potential V(ϕ).
Solution 10.
a) The effective tension in the compactified dimensions is given by Teff =T e−2ϕ. Let’s derive
this expression:
In string theory, the effective tension depends on the string coupling as Teff =T gs, where
gs=e−2ϕ. Therefore, Teff =T e−2ϕ.
b) The equation of motion for the scalar field ϕis given by:
d
dt ∂L
∂˙
ϕ−∂L
∂ϕ = 0
where the Lagrangian L=1
2˙
ϕ2−V(ϕ).
d
dt ∂L
∂˙
ϕ−∂L
∂ϕ =¨
ϕ+m2ϕ= 0
Therefore, the equation of motion for the scalar field ϕis ¨
ϕ+m2ϕ= 0.
c) To find the minimum of the potential V(ϕ), we need to solve dV
dϕ = 0:
dV
dϕ =m2ϕ= 0
This implies that the minimum of the potential occurs at ϕ= 0.
10 11. DUALITIES IN M-THEORY
Problem 11. Consider two particular string theories, Type IIA and Type IIB, related by T-duality.
Let the radius of a circle in Type IIA theory be R. If the number of fundamental strings winding
around the circle is n, find the dual circle radius in Type IIB theory.
Solution 11. Given: Radius of circle in Type IIA theory, R, and number of winding fundamental
strings, n.
In Type IIA theory, the momentum along the circle is given by P=n
R.
By T-duality, the winding number in Type IIB string theory is the same as the momentum in Type
IIA theory, and vice versa. Therefore, in Type IIB theory, the radius of the dual circle is related to
the momentum by Rdual =α′
R.
Substitute P=n
Rinto the formula for the dual radius in Type IIB theory:
Rdual =α′
R=α′
n/P =α′R
n
Hence, the dual circle radius in Type IIB theory is α′R
n.
11 12. INTEGRABILITY IN STRING THEORY
Problem 12. Consider a closed bosonic string moving in a background with a constant mag-
netic field Bin the z-direction. The equation of motion for the string is given by the Nambu-Goto
action
S=−T
2Zdσdτ√−h hab∂aXµ∂bXνGµν
where Tis the tension of the string, his the determinant of the world-sheet metric hab,Xµ=
Xµ(τ, σ)are the embedding coordinates of the string world-sheet, and Gµν is the background metric
tensor.
Given that the background metric is flat Minkowski space with Gµν =ηµν and the magnetic field
is B=Bz, where Bzis constant, compute the equation of motion for the string in this background.
Solution 12.
The equation of motion for the string can be derived by varying the action with respect to the
embedding coordinates Xµ.
Let’s denote ∂aXµ≡∂aXµ(τ, σ).
The variation of the action with respect to Xµgives the equation of motion:
δS
δXµ=−T
2Zdσdτ√−hhab∂a∂bXνηνµ = 0
Expanding the terms and using the fact that the world-sheet metric is hab =diag(−1,1), we get:
TZdσdτ ∂2
τXν−∂2
σXνηνµ = 0
The equations of motion are then given by:
a) In the µ= 0 direction:
∂2
τX0−∂2
σX0= 0
b) In the µ= 1 direction:
∂2
τX1−∂2
σX1= 0
c) In the µ= 2 direction:
∂2
τX2−∂2
σX2= 0
These equations of motion describe the dynamics of the string in the background of a constant
magnetic field B=Bz.
12 13. CAUSALITY AND STRING/M-THEORY
Problem 13. Consider a closed string propagating in a spacetime described by D= 10
dimensions. The string moves in a geometry where the background metric is given by ds2=
−dt2+dx2
1+ (dx2)2+···+ (dx8)2+ (dx9)2+ (dx10)2, where x10 is the compactified spatial direction
with a radius R.
a) Calculate the maximum energy Emax of a closed string state that can propagate in this ge-
ometry without creating a closed timelike curve.
b) Find the minimum allowed radius Rmin of the compactified spatial direction that ensures
causality is not violated in this spacetime.
Solution 13.
a) The maximum energy Emax of a closed string state can be found using the formula:
Emax =1
α′rN
2
where Nis the level of the state and α′is the Regge slope parameter. Since we are dealing with
a closed string in D= 10 dimensions, the critical dimension is D= 10, and the Regge slope
parameter is α′= 1/2πT .
For a closed string moving in a compactified dimension with radius R, the contribution to the
mass in the compact direction is n/R where nis the winding number. To avoid closed timelike
curves, we must have Emax =n/R. Equating these two expressions for Emax, we get:
n
R=1
α′rN
2
n
R=2πT
2rN
2
nR = 2πT √2N
R=2πT √2N
n
Therefore, the maximum energy is achieved at n= 1 and N= 2, leading to:
Emax =1
α′rN
2=1
α′r2
2=1
α′=√2πT
2
b) To ensure that causality is not violated in this spacetime, we must have the inequality Rmin ≥
2πR to avoid closed timelike curves. Substituting in the expression for Rmin from part (a), we get:
Rmin =2πT p2(1)
1= 2π√2πT ≥2πR
√2πT ≥R
2πT ≥R2
R2−2πT ≤0
Hence, the minimum allowed radius Rmin of the compactified spatial direction is R=√2πT .
13 14. HOLOGRAPHY IN M-THEORY
Problem 14. Consider a spacetime described by M-theory with 11 dimensions. The holo-
graphic principle states that the information of a region of space can be encoded on its boundary.
Suppose we have a 4-dimensional hypercube with side length L.
a) Calculate the volume of this hypercube in terms of L.
b) According to the holographic principle, what is the size of the boundary that encodes all the
information of this hypercube?
c) If the hypercube is in an 11-dimensional spacetime in M-theory, how many spatial dimensions
are "compactified"?
Solution 14.
a) The volume of a 4-dimensional hypercube with side length Lis given by V=L4.
b) The boundary of the hypercube is a 3-dimensional cube with sides of length L. The total
surface area of a cube is given by A= 6L2. Hence, for a 4-dimensional hypercube, the size of the
boundary that encodes all the information is A= 6L2.
c) In M-theory with 11 dimensions, the 4 spatial dimensions of the hypercube and the 1 time
dimension are known. This leaves 11 −4−1=6spatial dimensions "compactified".
I’m glad to help generate numerical problem questions for you. Could you please specify the
topic within String Theory and M-Theory that you would like the problem to be based on?
14 16. DARK MATTER AND STRING/M-THEORY
Problem 16. Consider a string theory scenario where a closed string moving in a compact
spatial dimension of radius Rinteracts with dark matter particles gravitationally. The string coupling
constant is given by gs= 0.1.
a) If the mass of the dark matter particle is m= 10−22 eV/c2, calculate the gravitational force
between the closed string and a dark matter particle located at a distance r= 1 mm.
b) If the string coupling constant gsis changed to 0.05, how does this affect the gravitational
force between the closed string and the dark matter particle?
Solution 16.
a) The gravitational force between the closed string and the dark matter particle can be calcu-
lated using Newton’s law of gravitation:
F=G·mdark matter ·mstring
r2
where Gis the gravitational constant (6.67430 ×10−11 m3kg−1s−2), mdark matter is the mass of
the dark matter particle (in kg), mstring is the mass of the closed string (in kg), and ris the distance
between the closed string and the dark matter particle (in meters).
Given m= 10−22 eV/c2, we need to convert this to kilograms by using the conversion factor
1eV/c2= 1.783 ×10−36 kg. Thus, mdark matter = 10−22 ×1.783 ×10−36 = 1.783 ×10−58 kg.
Plugging in the values G= 6.67430 ×10−11 m3kg−1s−2,mdark matter = 1.783 ×10−58 kg,
mstring =? (mass of the string is not provided but let’s assume it to be of the same order of magnitude
as a dark matter particle), and r= 1 mm = 0.001 m, we can find the gravitational force.
b) If the string coupling constant gsis changed to 0.05, we can re-calculate the gravitational
force using the new value of gsin the formula above. Let’s assume that the mass of the dark matter
particle remains the same.
Solution 16.
a) Given: G= 6.67430 ×10−11 m3kg−1s−2,mdark matter = 1.783 ×10−58 kg, mstring = 1.783 ×
10−58 kg (assumed to be of the same order), r= 0.001 m
Plugging in the values, we get:
F=6.67430 ×10−11 ×1.783 ×10−58 ×1.783 ×10−58
(0.001)2
⇒F=2.0004 ×10−124
10−6
⇒F= 2.0004 ×10−118 N
Thus, the gravitational force between the closed string and a dark matter particle located at a
distance of 1 mm is approximately 2.0004 ×10−118 N.
b) If gs= 0.05, we can re-calculate the gravitational force using the updated value of gsin the
formula. Plugging in the new value, we can find the new gravitational force.
I’m sorry, but it seems that I cannot generate numerical problems for String Theory and M-
Theory as they are primarily theoretical and mathematical physics concepts that do not involve
numerical computations. Would you like a conceptual problem or a theoretical problem instead?
15 18. VACUUM ENERGY IN M-THEORY
Problem 18. Consider a particular compactification of M-theory on a 5-dimensional torus with
each side having length R. The vacuum energy in this scenario is given by ρ=−1
(2π)5R9.
a) Calculate the vacuum energy ρin GeV4when R= 1 cm. b) Find the value of Rin m for
which the vacuum energy is ρ=−1GeV4.
Solution 18. a) To calculate the vacuum energy in GeV4, we need to convert the length R=
1cm to meters and then substitute it into the given formula.
Given: 1 cm = 10−2m
Plugging this value into the formula for vacuum energy: ρ=−1
(2π)5(10−2)9
ρ=−1
(2π)5(10−18)
ρ=−1
(2π)5(10−18)
ρ=−1
(2π)5×10−18
ρ=−1
(2π)5×10−18
ρ≈ −8.53254 ×1065 GeV4
Therefore, the vacuum energy when R= 1 cm is approximately −8.53254 ×1065 GeV4.
b) To find the value of Rin meters for which the vacuum energy is ρ=−1GeV4, we set the
vacuum energy formula equal to −1and solve for R:
−1 = −1
(2π)5R9
R9= (2π)5
R=9
p(2π)5
R≈3.28626 ×10−16 m
Therefore, the value of Rin meters for which the vacuum energy is ρ=−1GeV4is approxi-
mately 3.28626 ×10−16 m.
16 19. SOLITONS IN STRING THEORY
Problem 19. Consider a D-brane in type IIA superstring theory with tension T. Suppose the
D-brane extends in pspatial dimensions. The energy density Uper unit p-dimensional volume of
the D-brane is given by U=cT , where cis a constant.
a) If the D-brane has p= 3 spatial dimensions and the tension is T= 5 ×10−2, find the energy
density U.
b) Now, let’s consider another D-brane with tension T= 0.1and energy density U= 0.4.
Determine the number of spatial dimensions pin which this D-brane extends.
Solution 19.
a) Given p= 3 and T= 5 ×10−2, we use the formula U=cT to find the energy density U:
U=c×5×10−2= 0.05c
So, the energy density Ufor the D-brane with p= 3 spatial dimensions and tension T= 5×10−2
is 0.05c.
b) For this case, we have T= 0.1and U= 0.4. Using the same formula for energy density,
U=cT , we can solve for cfirst:
c=U
T=0.4
0.1= 4
Now, we substitute c= 4 back into the equation U=cT :
0.4=4×p
Solving for p, we find:
p=0.4
4= 0.1
Thus, the D-brane extends in p= 0.1spatial dimensions.
I. Let’s formulate a numerical problem in the context of String Theory and M-Theory:
17 20. ADS/CFT CORRESPONDENCE IN M-THEORY
Problem 20. Consider a type IIA string theory on an Anti-de Sitter space (AdS) with a five-
dimensional radius RAdS = 10. Calculate the corresponding conformal field theory (CFT) central
charge using the AdS/CFT correspondence formula:
c=3L
2G
where Lis the AdS radius and Gis Newton’s constant. Take the value of Newton’s constant
G= 6.71 ×10−11 m3kg−1s−2.
Solution 20. Given: - AdS radius RAdS = 10 - Newton’s constant G= 6.71 ×10−11 m3kg−1
s−2
The CFT central charge can be calculated using the AdS/CFT correspondence formula:
c=3L
2G
Plugging in the values:
c=3×10
2×6.71 ×10−11
c=30
13.42 ×10−11
c=30
1.342 ×10−10
c=30
1.342
c≈22.36
Therefore, the central charge of the corresponding conformal field theory is approximately
22.36.
I. Entanglement Entropy in String Theory:
18 21. Entanglement Entropy in String Theory
Problem 21. Consider a 1+1 dimensional conformal field theory described by a CFT with
central charge c= 1. Let’s calculate the entanglement entropy of a subsystem in this theory.
[Additional context: The entanglement entropy Sof a subsystem in a CFT is given by the formula
S=c
3log L
ϵ, where cis the central charge, Lis the length of the subsystem, and ϵis a short-
distance cutoff.]
a) Suppose we have a subsystem of length L= 2 in this CFT. Calculate the entanglement
entropy when the short-distance cutoff ϵ= 0.1.
b) Now, consider another CFT described by a different conformal field theory with central charge
c= 2. If we have a subsystem of length L= 3 in this CFT, what is the entanglement entropy when
the short-distance cutoff ϵ= 0.01?
Solution 21.
a) We are given c= 1,L= 2, and ϵ= 0.1. Plugging these values into the formula for entangle-
ment entropy, we get:
S=1
3log 2
0.1=1
3log 20 ≈1
3×2.9957 ≈0.9986
Therefore, the entanglement entropy when L= 2spaceunits and ϵ= 0.1is approximately
0.9986.
b) For the new CFT with c= 2,L= 3, and ϵ= 0.01, we apply the formula:
S=2
3log 3
0.01=2
3log 300 ≈2
3×5.7038 ≈3.8026
Therefore, the entanglement entropy when L= 3spaceunits and ϵ= 0.01 is approximately
3.8026.
I can certainly help with that! Let’s proceed with a numerical problem for String Theory and
M-Theory.
19 22. SPACE-TIME SINGULARITIES IN M-THEORY
Problem 22. Consider a compactified M-theory scenario where the compact spatial dimension
has a radius of R= 10−17 meters. If an object is traveling at a speed of 0.9c(where cis the speed
of light) in this compactified dimension, determine the time duration (in seconds) it takes for the
object to travel halfway around the compactified dimension.
Solution 22. a) The circumference of the compact spatial dimension is given by 2πR. There-
fore, the distance required to travel halfway around the dimension is πR.
b) The speed of the object is 0.9c, where c= 3 ×108m/s. Hence, the object travels a distance
of 0.9c×t=πR in time t, where tis the time duration we want to find.
Using the equation 0.9c×t=πR, we can solve for t:
0.9×3×108m/s ×t=π×10−17 m
0.9×3×108t=π×10−17
2.7×108t= 3.1416 ×10−17
t=3.1416 ×10−17
2.7×108
t≈1.163 ×10−25 s
Therefore, it takes approximately 1.163 ×10−25 seconds for the object to travel halfway around
the compactified dimension.
20 23. DEFORMATIONS OF STRING THEORY
Problem 23. Consider a closed bosonic string of length Lmoving in Dspacetime dimensions.
The string is perturbed by a small deformation given by the following equation of motion:
∂2
∂τ 2−∂2
∂σ2Xµ(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ)
where Xµ(τ, σ)denotes the string embedding coordinates, µ, ν = 0,1, . . . , D −1,ϵis a small
parameter, and α′is the Regge slope parameter.
a) Compute the equation of motion for X0(τ, σ)and X1(τ, σ).
b) Determine the general solution for X0(τ, σ)and X1(τ, σ).
c) Given that the string is parameterized by (τ, σ)such that 0≤σ≤π, find the normal mode
frequencies for X0(τ, σ)and X1(τ, σ).
Solution 23.
a) The equation of motion can be split into separate equations for X0(τ, σ)and X1(τ, σ)by
setting µ= 0 and µ= 1 respectively:
∂2
∂τ 2X0(τ, σ)−∂2
∂σ2X0(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ) = ϵα′sin(2σ)∂0X1(τ, σ)
∂2
∂τ 2X1(τ, σ)−∂2
∂σ2X1(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ) = ϵα′sin(2σ)∂1X0(τ, σ)
b) To find the general solutions, we solve the wave equation for X0(τ, σ)and X1(τ, σ). The
solutions will have the form:
X0(τ, σ) = A(σ)eiω0τ+B(σ)e−iω0τ
X1(τ, σ) = C(σ)eiω1τ+D(σ)e−iω1τ
where A(σ),B(σ),C(σ), and D(σ)are functions of σ, and ω0and ω1are the normal mode
frequencies.
c) To find the normal mode frequencies, we substitute the solutions back into the equations of
motion and solve for ω0and ω1. After normalization, the normal mode frequencies are given by:
ω0=1
Lrn2−ϵα′n2
2
ω1=1
Lrn2+ϵα′n2
2
where n= 1,2,3, . . . represents the mode number.
21 24. QUANTUM FIELD THEORY LIMIT OF M-THEORY
Problem 24. Consider a closed string in 10 dimensions with a winding mode wrapping around
a circle of radius R. The momentum mode of the string has energy E=n
R, where nis an integer.
The winding mode has energy E=mR
α′, where mis an integer and α′is the Regge slope.
a) If the winding mode has energy E= 2π/α′and the momentum mode has energy E= 1/√α′,
find the values of mand n.
b) Determine the total energy ETof the closed string in terms of α′.
Solution 24.
a) Since the winding mode has energy E=mR
α′=2π
α′, we have:
mR
α′=2π
α′
mR = 2π
Similarly, for the momentum mode with energy E=n
R=1
√α′, we get:
n
R=1
√α′
nR =√α′
From these two equations, we can see that m= 1 and n= 1.
b) The total energy ETof the closed string is given by:
ET=n
R+mR
α′=1
R+R
α′=1 + R2/α′
R
Therefore, the total energy of the closed string in terms of α′is ET=1 + R2/α′
R.
22 25. TOPOLOGICAL ASPECTS OF STRING THEORY
Problem 25. Consider a closed oriented Riemann surface Σgof genus g. The Euler charac-
teristic of Σgis given by χ(Σg)=2−2g.
a) Calculate the Euler characteristic for a torus (g= 1).
b) Calculate the Euler characteristic for a sphere with two handles (g= 2).
c) Calculate the Euler characteristic for a surface with genus g= 3.
Solution 25.
a) For a torus (g= 1), the Euler characteristic is given by χ(Σ1) = 2 −2·1 = 2 −2 = 0.
Therefore, the Euler characteristic for a torus is 0.
b) For a sphere with two handles (g= 2), the Euler characteristic is χ(Σ2) = 2−2·2 = 2−4 = −2.
Hence, the Euler characteristic for a sphere with two handles is -2.
c) For a surface with genus g= 3, the Euler characteristic is χ(Σ3)=2−2·3=2−6 = −4.
Therefore, the Euler characteristic for a surface of genus 3 is -4.
3 3. QUANTUM GRAVITY IN STRING THEORY
Problem 3. Consider an open string moving in 4-dimensional spacetime. The string has tension
T= 1 and mass per unit length µ= 2.
a) Find the speed of propagation of waves on this string.
b) Calculate the energy stored in a segment of string of length L= 3.
c) If the string is stretched with a force of F= 5, determine the amplitude of the standing wave
that could form on the string.
Solution 3.
a) The speed of propagation of waves on a string is given by
v=sT
µ.
Substitute T= 1 and µ= 2:
v=r1
2=√2
2.
Therefore, the speed of propagation of waves on this string is √2
2.
b) The energy stored in a segment of string of length Lis given by
E=1
2µv2L.
Substitute µ= 2,v=√2
2, and L= 3:
E=1
2×2× √2
2!2
×3 = 3
2.
Therefore, the energy stored in a segment of string of length 3 is 3
2.
c) The amplitude of the standing wave that could form on the string under the applied force F
is given by
A=F
2πv2.
Substitute F= 5 and v=√2
2:
A=5
2π×√2
22=5
2π×1
2
=5√2
π.
Therefore, the amplitude of the standing wave that could form on the string is 5√2
π.
4 4. BLACK HOLES AND STRING THEORY
Problem 4. Consider a black hole in four-dimensional spacetime described by the Schwarzschild
metric:
ds2=−1−2GM
c2rc2dt2+1−2GM
c2r−1
dr2+r2dΩ2
where Mis the mass of the black hole, cis the speed of light, Gis the gravitational constant,
and dΩ2=dθ2+sin2θdϕ2in spherical coordinates (t, r, θ, ϕ).
a) Find the event horizon radius Rhorizon of this black hole.
b) Determine the Schwarzschild radius RSin terms of M,G, and c.
c) Given that the mass of the black hole is M= 2 ×1030 kg, calculate the mass of the black
hole in units of solar masses (M⊙= 1.989 ×1030 kg).
Solution 4.
a) To find the event horizon radius, we set the metric coefficient of dt2to zero at the event
horizon. So, 1−2GM
c2Rhorizon = 0. Solving for Rhorizon gives:
Rhorizon =2GM
c2
b) The Schwarzschild radius is defined as the radius at which the metric becomes singular. It
is given by RS=2GM
c2.
c) Substituting M= 2 ×1030 kg into the Schwarzschild radius formula, we have:
RS=2G(2 ×1030)
c2=4×1030G
c2
Now, we can express the mass of the black hole in terms of solar masses by dividing by the
mass of the Sun:
4×1030G
c2÷1.989 ×1030 =2G
c2
Therefore, the mass of the black hole is 2solar masses.
5 5. TACHYON CONDENSATION IN STRING THEORY
Problem 5. Consider a closed bosonic string theory where the endpoint of the string coor-
dinates are subject to Neumann boundary conditions. The first excited level of the closed string
contains a tachyon with mass m2=−2
α′.
a) Calculate the momentum of the tachyon state in the string theory.
b) Show that the mass of the tachyon state in the open bosonic string theory is m= 0.
c) Interpret the result in relation to tachyon condensation.
Solution 5.
a) The mass-squared of a state in a closed string theory is given by the level matching condition:
m2=4
α′(N−1)
where Nis the occupation number operator for the state. For the first excited level, N= 1 and
m2=−2
α′. Substituting these values into the equation above, we have:
−2
α′=4
α′(1 −1)
−2=0
This equation is a contradiction, indicating that there is no physical state with a mass-squared of
−2
α′for the first excited level. So, the tachyon state does not exist.
b) In the open bosonic string theory, the mass-squared of a state is given by:
m2=2
α′(N−1)
For the tachyon state at the first excited level in the open string, N= 1 and m2=−1
α′. Taking the
square root of this, we get m= 0. Therefore, the mass of the tachyon state in the open bosonic
string theory is zero.
c) The result m= 0 for the open string tachyon state implies that it is a massless state. In string
theory, a tachyon has negative mass squared and indicates an instability in the theory. Tachyon
condensation is a process where this unstable state "condenses" to a minimum energy state. In
this case, the tachyon state in the open bosonic string theory having a mass of zero suggests that
it represents the minimum energy state after tachyon condensation has occurred. This process
helps stabilize the theory by eliminating the instability caused by the presence of the tachyon.
6 6. EXTRA DIMENSIONS IN M-THEORY
Problem 6. Consider a scenario in M-Theory where there are 7 spatial dimensions and 3
temporal dimensions. The size of the compactified extra dimensions is given by R= 10−17 meters.
Calculate the compactification scale in GeV.
Solution 6.
a) The compactification scale Mccan be calculated using the formula for the compactified extra
dimensions:
Mc=1
R
Substitute R= 10−17 meters into the formula:
Mc=1
10−17 = 1017 m−1
b) To convert the compactification scale from meters to GeV, we need to use the relation 1GeV =
1.97 ×10−16 m−1.
Let’s convert the compactification scale:
Mc= 1017 ×1.97 ×10−16 = 1.97 ×10 GeV = 19.7GeV
Therefore, the compactification scale in GeV is 19.7 GeV.
7 7. BRANE DYNAMICS IN STRING THEORY
Problem 7. Consider a type IIB superstring theory in 10 dimensions with D3-branes. The
tension of the D3-brane is given by T=1
(2π)3(α′)2, where α′is the string length parameter. Calculate
the energy density stored in a D3-brane of length L= 10√α′.
Solution 7. a) The energy density stored in a D3-brane can be calculated by dividing its energy
by its volume. The energy Eof a D3-brane is given by E=T×V, where Tis the tension of the
D3-brane and Vis its volume. Since the D3-brane extends in 3 spatial dimensions, the volume V
is L3= (10√α′)3= 1000α′3/2.
Substitute T=1
(2π)3(α′)2and V= 1000α′3/2into the equation:
E=1
(2π)3(α′)2×1000α′3/2=1000
(2π)3α′1/2
Thus, the energy stored in the D3-brane is 1000
(2π)3α′1/2.
b) The energy density is given by the energy per unit volume. We divide the energy Eby the
volume Vto find the energy density u=E
V.
Substitute E=1000
(2π)3α′1/2and V= 1000α′3/2into the equation:
u=1000/(2π)3α′1/2
1000α′3/2=1
(2π)3α′2
Therefore, the energy density stored in the D3-brane is 1
(2π)3α′2.
8 8. SUPERSYMMETRY BREAKING IN M-THEORY
Problem 8. Consider a supersymmetric M-Theory compactified on a 2-torus with radii R1and
R2. The volume of the torus is given by V=R1R2. Suppose that supersymmetry is broken by the
flux through the torus such that the gravitino mass term is generated.
[Given: The gravitino mass term is given by m3/2=e⟨G⟩, where Gis the flux. Also, we have
⟨G⟩ =−2 ln(V), where ⟨G⟩ denotes the expectation value of the flux.]
a) Show that the gravitino mass term m3/2in terms of the radii R1and R2.
b) If R1= 2 and R2= 3, calculate the gravitino mass m3/2.
Solution 8.
a) To find the gravitino mass term m3/2in terms of the radii R1and R2, we first need to express
the volume Vin terms of R1and R2:
V=R1·R2
Next, we can find the expectation value of the flux ⟨G⟩:
⟨G⟩ =−2 ln(V) = −2 ln(R1·R2) = −2 ln(R1)−2 ln(R2)
Substitute this into the expression for the gravitino mass term:
m3/2=e⟨G⟩ =e−2 ln(R1)−2 ln(R2)=e−2 ln(R1)·e−2 ln(R2)=1
R2
1·1
R2
2
=1
R2
1R2
2
=1
V2
b) Given R1= 2 and R2= 3, the volume V=R1·R2= 2 ·3=6. Therefore, the gravitino mass
term m3/2is:
m3/2=1
V2=1
62=1
36 = 0.0278
I. Problem 9. Consider a Type IIA superstring theory with a D6-brane wrapped on a compact
3-torus with sides of length L. The tension of the D6-brane is given by T6=1
(2π)6(α′)4. Suppose
the compactified space is a cube, compute the energy density UD6 of the D6-brane in terms of L.
Hint: The energy density is defined as the energy per unit volume.
II. Problem 10. In Type IIB superstring theory, a D3-brane has tension T3=1
(2π)3(α′)2. If this
theory is in a 10-dimensional spacetime with a toroidal compactification down to 6 dimensions, and
the 3-brane extends along all 3 of those compact dimensions, calculate the energy density UD3 of
the D3-brane in terms of the compactification radii Ri.
Hint: The energy density is defined as the energy per unit volume.
III. Problem 11. In M-Theory, consider a M5-brane wrapped on a torus with radii R1and R2. If
the tension of the brane is T5=1
(2π)5(ℓp)3, find the energy density UM5 of the M5-brane in terms of
the torus radii.
Hint: The energy density is defined as the energy per unit volume.
9 10. COSMOLOGY AND STRING/M-THEORY
Problem 10. Consider a string theory model in a universe with extra dimensions compactified
on a torus. The radius of the torus is R, and the string tension is T. The compactified dimensions
are described by an effective field theory with a massless scalar field ϕ, whose potential is given
by V(ϕ) = 1
2m2ϕ2.
a) Show that the effective tension in the compactified dimensions, Teff , is given by Tef f =
T e−2ϕ.
b) Determine the equation of motion for the scalar field ϕ.
c) Find the minimum of the potential V(ϕ).
Solution 10.
a) The effective tension in the compactified dimensions is given by Teff =T e−2ϕ. Let’s derive
this expression:
In string theory, the effective tension depends on the string coupling as Teff =T gs, where
gs=e−2ϕ. Therefore, Teff =T e−2ϕ.
b) The equation of motion for the scalar field ϕis given by:
d
dt ∂L
∂˙
ϕ−∂L
∂ϕ = 0
where the Lagrangian L=1
2˙
ϕ2−V(ϕ).
d
dt ∂L
∂˙
ϕ−∂L
∂ϕ =¨
ϕ+m2ϕ= 0
Therefore, the equation of motion for the scalar field ϕis ¨
ϕ+m2ϕ= 0.
c) To find the minimum of the potential V(ϕ), we need to solve dV
dϕ = 0:
dV
dϕ =m2ϕ= 0
This implies that the minimum of the potential occurs at ϕ= 0.
10 11. DUALITIES IN M-THEORY
Problem 11. Consider two particular string theories, Type IIA and Type IIB, related by T-duality.
Let the radius of a circle in Type IIA theory be R. If the number of fundamental strings winding
around the circle is n, find the dual circle radius in Type IIB theory.
Solution 11. Given: Radius of circle in Type IIA theory, R, and number of winding fundamental
strings, n.
In Type IIA theory, the momentum along the circle is given by P=n
R.
By T-duality, the winding number in Type IIB string theory is the same as the momentum in Type
IIA theory, and vice versa. Therefore, in Type IIB theory, the radius of the dual circle is related to
the momentum by Rdual =α′
R.
Substitute P=n
Rinto the formula for the dual radius in Type IIB theory:
Rdual =α′
R=α′
n/P =α′R
n
Hence, the dual circle radius in Type IIB theory is α′R
n.
11 12. INTEGRABILITY IN STRING THEORY
Problem 12. Consider a closed bosonic string moving in a background with a constant mag-
netic field Bin the z-direction. The equation of motion for the string is given by the Nambu-Goto
action
S=−T
2Zdσdτ√−h hab∂aXµ∂bXνGµν
where Tis the tension of the string, his the determinant of the world-sheet metric hab,Xµ=
Xµ(τ, σ)are the embedding coordinates of the string world-sheet, and Gµν is the background metric
tensor.
Given that the background metric is flat Minkowski space with Gµν =ηµν and the magnetic field
is B=Bz, where Bzis constant, compute the equation of motion for the string in this background.
Solution 12.
The equation of motion for the string can be derived by varying the action with respect to the
embedding coordinates Xµ.
Let’s denote ∂aXµ≡∂aXµ(τ, σ).
The variation of the action with respect to Xµgives the equation of motion:
δS
δXµ=−T
2Zdσdτ√−hhab∂a∂bXνηνµ = 0
Expanding the terms and using the fact that the world-sheet metric is hab =diag(−1,1), we get:
TZdσdτ ∂2
τXν−∂2
σXνηνµ = 0
The equations of motion are then given by:
a) In the µ= 0 direction:
∂2
τX0−∂2
σX0= 0
b) In the µ= 1 direction:
∂2
τX1−∂2
σX1= 0
c) In the µ= 2 direction:
∂2
τX2−∂2
σX2= 0
These equations of motion describe the dynamics of the string in the background of a constant
magnetic field B=Bz.
12 13. CAUSALITY AND STRING/M-THEORY
Problem 13. Consider a closed string propagating in a spacetime described by D= 10
dimensions. The string moves in a geometry where the background metric is given by ds2=
−dt2+dx2
1+ (dx2)2+···+ (dx8)2+ (dx9)2+ (dx10)2, where x10 is the compactified spatial direction
with a radius R.
a) Calculate the maximum energy Emax of a closed string state that can propagate in this ge-
ometry without creating a closed timelike curve.
b) Find the minimum allowed radius Rmin of the compactified spatial direction that ensures
causality is not violated in this spacetime.
Solution 13.
a) The maximum energy Emax of a closed string state can be found using the formula:
Emax =1
α′rN
2
where Nis the level of the state and α′is the Regge slope parameter. Since we are dealing with
a closed string in D= 10 dimensions, the critical dimension is D= 10, and the Regge slope
parameter is α′= 1/2πT .
For a closed string moving in a compactified dimension with radius R, the contribution to the
mass in the compact direction is n/R where nis the winding number. To avoid closed timelike
curves, we must have Emax =n/R. Equating these two expressions for Emax, we get:
n
R=1
α′rN
2
n
R=2πT
2rN
2
nR = 2πT √2N
R=2πT √2N
n
Therefore, the maximum energy is achieved at n= 1 and N= 2, leading to:
Emax =1
α′rN
2=1
α′r2
2=1
α′=√2πT
2
b) To ensure that causality is not violated in this spacetime, we must have the inequality Rmin ≥
2πR to avoid closed timelike curves. Substituting in the expression for Rmin from part (a), we get:
Rmin =2πT p2(1)
1= 2π√2πT ≥2πR
√2πT ≥R
2πT ≥R2
R2−2πT ≤0
Hence, the minimum allowed radius Rmin of the compactified spatial direction is R=√2πT .
13 14. HOLOGRAPHY IN M-THEORY
Problem 14. Consider a spacetime described by M-theory with 11 dimensions. The holo-
graphic principle states that the information of a region of space can be encoded on its boundary.
Suppose we have a 4-dimensional hypercube with side length L.
a) Calculate the volume of this hypercube in terms of L.
b) According to the holographic principle, what is the size of the boundary that encodes all the
information of this hypercube?
c) If the hypercube is in an 11-dimensional spacetime in M-theory, how many spatial dimensions
are "compactified"?
Solution 14.
a) The volume of a 4-dimensional hypercube with side length Lis given by V=L4.
b) The boundary of the hypercube is a 3-dimensional cube with sides of length L. The total
surface area of a cube is given by A= 6L2. Hence, for a 4-dimensional hypercube, the size of the
boundary that encodes all the information is A= 6L2.
c) In M-theory with 11 dimensions, the 4 spatial dimensions of the hypercube and the 1 time
dimension are known. This leaves 11 −4−1=6spatial dimensions "compactified".
I’m glad to help generate numerical problem questions for you. Could you please specify the
topic within String Theory and M-Theory that you would like the problem to be based on?
14 16. DARK MATTER AND STRING/M-THEORY
Problem 16. Consider a string theory scenario where a closed string moving in a compact
spatial dimension of radius Rinteracts with dark matter particles gravitationally. The string coupling
constant is given by gs= 0.1.
a) If the mass of the dark matter particle is m= 10−22 eV/c2, calculate the gravitational force
between the closed string and a dark matter particle located at a distance r= 1 mm.
b) If the string coupling constant gsis changed to 0.05, how does this affect the gravitational
force between the closed string and the dark matter particle?
Solution 16.
a) The gravitational force between the closed string and the dark matter particle can be calcu-
lated using Newton’s law of gravitation:
F=G·mdark matter ·mstring
r2
where Gis the gravitational constant (6.67430 ×10−11 m3kg−1s−2), mdark matter is the mass of
the dark matter particle (in kg), mstring is the mass of the closed string (in kg), and ris the distance
between the closed string and the dark matter particle (in meters).
Given m= 10−22 eV/c2, we need to convert this to kilograms by using the conversion factor
1eV/c2= 1.783 ×10−36 kg. Thus, mdark matter = 10−22 ×1.783 ×10−36 = 1.783 ×10−58 kg.
Plugging in the values G= 6.67430 ×10−11 m3kg−1s−2,mdark matter = 1.783 ×10−58 kg,
mstring =? (mass of the string is not provided but let’s assume it to be of the same order of magnitude
as a dark matter particle), and r= 1 mm = 0.001 m, we can find the gravitational force.
b) If the string coupling constant gsis changed to 0.05, we can re-calculate the gravitational
force using the new value of gsin the formula above. Let’s assume that the mass of the dark matter
particle remains the same.
Solution 16.
a) Given: G= 6.67430 ×10−11 m3kg−1s−2,mdark matter = 1.783 ×10−58 kg, mstring = 1.783 ×
10−58 kg (assumed to be of the same order), r= 0.001 m
Plugging in the values, we get:
F=6.67430 ×10−11 ×1.783 ×10−58 ×1.783 ×10−58
(0.001)2
⇒F=2.0004 ×10−124
10−6
⇒F= 2.0004 ×10−118 N
Thus, the gravitational force between the closed string and a dark matter particle located at a
distance of 1 mm is approximately 2.0004 ×10−118 N.
b) If gs= 0.05, we can re-calculate the gravitational force using the updated value of gsin the
formula. Plugging in the new value, we can find the new gravitational force.
I’m sorry, but it seems that I cannot generate numerical problems for String Theory and M-
Theory as they are primarily theoretical and mathematical physics concepts that do not involve
numerical computations. Would you like a conceptual problem or a theoretical problem instead?
15 18. VACUUM ENERGY IN M-THEORY
Problem 18. Consider a particular compactification of M-theory on a 5-dimensional torus with
each side having length R. The vacuum energy in this scenario is given by ρ=−1
(2π)5R9.
a) Calculate the vacuum energy ρin GeV4when R= 1 cm. b) Find the value of Rin m for
which the vacuum energy is ρ=−1GeV4.
Solution 18. a) To calculate the vacuum energy in GeV4, we need to convert the length R=
1cm to meters and then substitute it into the given formula.
Given: 1 cm = 10−2m
Plugging this value into the formula for vacuum energy: ρ=−1
(2π)5(10−2)9
ρ=−1
(2π)5(10−18)
ρ=−1
(2π)5(10−18)
ρ=−1
(2π)5×10−18
ρ=−1
(2π)5×10−18
ρ≈ −8.53254 ×1065 GeV4
Therefore, the vacuum energy when R= 1 cm is approximately −8.53254 ×1065 GeV4.
b) To find the value of Rin meters for which the vacuum energy is ρ=−1GeV4, we set the
vacuum energy formula equal to −1and solve for R:
−1 = −1
(2π)5R9
R9= (2π)5
R=9
p(2π)5
R≈3.28626 ×10−16 m
Therefore, the value of Rin meters for which the vacuum energy is ρ=−1GeV4is approxi-
mately 3.28626 ×10−16 m.
16 19. SOLITONS IN STRING THEORY
Problem 19. Consider a D-brane in type IIA superstring theory with tension T. Suppose the
D-brane extends in pspatial dimensions. The energy density Uper unit p-dimensional volume of
the D-brane is given by U=cT , where cis a constant.
a) If the D-brane has p= 3 spatial dimensions and the tension is T= 5 ×10−2, find the energy
density U.
b) Now, let’s consider another D-brane with tension T= 0.1and energy density U= 0.4.
Determine the number of spatial dimensions pin which this D-brane extends.
Solution 19.
a) Given p= 3 and T= 5 ×10−2, we use the formula U=cT to find the energy density U:
U=c×5×10−2= 0.05c
So, the energy density Ufor the D-brane with p= 3 spatial dimensions and tension T= 5×10−2
is 0.05c.
b) For this case, we have T= 0.1and U= 0.4. Using the same formula for energy density,
U=cT , we can solve for cfirst:
c=U
T=0.4
0.1= 4
Now, we substitute c= 4 back into the equation U=cT :
0.4=4×p
Solving for p, we find:
p=0.4
4= 0.1
Thus, the D-brane extends in p= 0.1spatial dimensions.
I. Let’s formulate a numerical problem in the context of String Theory and M-Theory:
17 20. ADS/CFT CORRESPONDENCE IN M-THEORY
Problem 20. Consider a type IIA string theory on an Anti-de Sitter space (AdS) with a five-
dimensional radius RAdS = 10. Calculate the corresponding conformal field theory (CFT) central
charge using the AdS/CFT correspondence formula:
c=3L
2G
where Lis the AdS radius and Gis Newton’s constant. Take the value of Newton’s constant
G= 6.71 ×10−11 m3kg−1s−2.
Solution 20. Given: - AdS radius RAdS = 10 - Newton’s constant G= 6.71 ×10−11 m3kg−1
s−2
The CFT central charge can be calculated using the AdS/CFT correspondence formula:
c=3L
2G
Plugging in the values:
c=3×10
2×6.71 ×10−11
c=30
13.42 ×10−11
c=30
1.342 ×10−10
c=30
1.342
c≈22.36
Therefore, the central charge of the corresponding conformal field theory is approximately
22.36.
I. Entanglement Entropy in String Theory:
18 21. Entanglement Entropy in String Theory
Problem 21. Consider a 1+1 dimensional conformal field theory described by a CFT with
central charge c= 1. Let’s calculate the entanglement entropy of a subsystem in this theory.
[Additional context: The entanglement entropy Sof a subsystem in a CFT is given by the formula
S=c
3log L
ϵ, where cis the central charge, Lis the length of the subsystem, and ϵis a short-
distance cutoff.]
a) Suppose we have a subsystem of length L= 2 in this CFT. Calculate the entanglement
entropy when the short-distance cutoff ϵ= 0.1.
b) Now, consider another CFT described by a different conformal field theory with central charge
c= 2. If we have a subsystem of length L= 3 in this CFT, what is the entanglement entropy when
the short-distance cutoff ϵ= 0.01?
Solution 21.
a) We are given c= 1,L= 2, and ϵ= 0.1. Plugging these values into the formula for entangle-
ment entropy, we get:
S=1
3log 2
0.1=1
3log 20 ≈1
3×2.9957 ≈0.9986
Therefore, the entanglement entropy when L= 2spaceunits and ϵ= 0.1is approximately
0.9986.
b) For the new CFT with c= 2,L= 3, and ϵ= 0.01, we apply the formula:
S=2
3log 3
0.01=2
3log 300 ≈2
3×5.7038 ≈3.8026
Therefore, the entanglement entropy when L= 3spaceunits and ϵ= 0.01 is approximately
3.8026.
I can certainly help with that! Let’s proceed with a numerical problem for String Theory and
M-Theory.
19 22. SPACE-TIME SINGULARITIES IN M-THEORY
Problem 22. Consider a compactified M-theory scenario where the compact spatial dimension
has a radius of R= 10−17 meters. If an object is traveling at a speed of 0.9c(where cis the speed
of light) in this compactified dimension, determine the time duration (in seconds) it takes for the
object to travel halfway around the compactified dimension.
Solution 22. a) The circumference of the compact spatial dimension is given by 2πR. There-
fore, the distance required to travel halfway around the dimension is πR.
b) The speed of the object is 0.9c, where c= 3 ×108m/s. Hence, the object travels a distance
of 0.9c×t=πR in time t, where tis the time duration we want to find.
Using the equation 0.9c×t=πR, we can solve for t:
0.9×3×108m/s ×t=π×10−17 m
0.9×3×108t=π×10−17
2.7×108t= 3.1416 ×10−17
t=3.1416 ×10−17
2.7×108
t≈1.163 ×10−25 s
Therefore, it takes approximately 1.163 ×10−25 seconds for the object to travel halfway around
the compactified dimension.
20 23. DEFORMATIONS OF STRING THEORY
Problem 23. Consider a closed bosonic string of length Lmoving in Dspacetime dimensions.
The string is perturbed by a small deformation given by the following equation of motion:
∂2
∂τ 2−∂2
∂σ2Xµ(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ)
where Xµ(τ, σ)denotes the string embedding coordinates, µ, ν = 0,1, . . . , D −1,ϵis a small
parameter, and α′is the Regge slope parameter.
a) Compute the equation of motion for X0(τ, σ)and X1(τ, σ).
b) Determine the general solution for X0(τ, σ)and X1(τ, σ).
c) Given that the string is parameterized by (τ, σ)such that 0≤σ≤π, find the normal mode
frequencies for X0(τ, σ)and X1(τ, σ).
Solution 23.
a) The equation of motion can be split into separate equations for X0(τ, σ)and X1(τ, σ)by
setting µ= 0 and µ= 1 respectively:
∂2
∂τ 2X0(τ, σ)−∂2
∂σ2X0(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ) = ϵα′sin(2σ)∂0X1(τ, σ)
∂2
∂τ 2X1(τ, σ)−∂2
∂σ2X1(τ, σ) = ϵα′sin(2σ)∂µXν(τ, σ) = ϵα′sin(2σ)∂1X0(τ, σ)
b) To find the general solutions, we solve the wave equation for X0(τ, σ)and X1(τ, σ). The
solutions will have the form:
X0(τ, σ) = A(σ)eiω0τ+B(σ)e−iω0τ
X1(τ, σ) = C(σ)eiω1τ+D(σ)e−iω1τ
where A(σ),B(σ),C(σ), and D(σ)are functions of σ, and ω0and ω1are the normal mode
frequencies.
c) To find the normal mode frequencies, we substitute the solutions back into the equations of
motion and solve for ω0and ω1. After normalization, the normal mode frequencies are given by:
ω0=1
Lrn2−ϵα′n2
2
ω1=1
Lrn2+ϵα′n2
2
where n= 1,2,3, . . . represents the mode number.
21 24. QUANTUM FIELD THEORY LIMIT OF M-THEORY
Problem 24. Consider a closed string in 10 dimensions with a winding mode wrapping around
a circle of radius R. The momentum mode of the string has energy E=n
R, where nis an integer.
The winding mode has energy E=mR
α′, where mis an integer and α′is the Regge slope.
a) If the winding mode has energy E= 2π/α′and the momentum mode has energy E= 1/√α′,
find the values of mand n.
b) Determine the total energy ETof the closed string in terms of α′.
Solution 24.
a) Since the winding mode has energy E=mR
α′=2π
α′, we have:
mR
α′=2π
α′
mR = 2π
Similarly, for the momentum mode with energy E=n
R=1
√α′, we get:
n
R=1
√α′
nR =√α′
From these two equations, we can see that m= 1 and n= 1.
b) The total energy ETof the closed string is given by:
ET=n
R+mR
α′=1
R+R
α′=1 + R2/α′
R
Therefore, the total energy of the closed string in terms of α′is ET=1 + R2/α′
R.
22 25. TOPOLOGICAL ASPECTS OF STRING THEORY
Problem 25. Consider a closed oriented Riemann surface Σgof genus g. The Euler charac-
teristic of Σgis given by χ(Σg)=2−2g.
a) Calculate the Euler characteristic for a torus (g= 1).
b) Calculate the Euler characteristic for a sphere with two handles (g= 2).
c) Calculate the Euler characteristic for a surface with genus g= 3.
Solution 25.
a) For a torus (g= 1), the Euler characteristic is given by χ(Σ1) = 2 −2·1 = 2 −2 = 0.
Therefore, the Euler characteristic for a torus is 0.
b) For a sphere with two handles (g= 2), the Euler characteristic is χ(Σ2) = 2−2·2 = 2−4 = −2.
Hence, the Euler characteristic for a sphere with two handles is -2.
c) For a surface with genus g= 3, the Euler characteristic is χ(Σ3)=2−2·3=2−6 = −4.
Therefore, the Euler characteristic for a surface of genus 3 is -4.
