Mit's Undergraduate String Theory Project

String Thermodynamics and Black Holes 8.251 Course Project Igor Sylvester May 13, 2005

Quick Calculations QC16.1 Solution of Quick Calculation 16.1 Using F = E − T S, (16.39), and (16.40), E = F + TS π2 1 π2 1 − T k = − ~ω0 6 β 2 3 !2 π 2 kT ~ω0 . E = 6 ~ω0

kT ~ω0

!

QC16.2 Solution of Quick Calculation 16.2 Using the asymptotic expansion of p24 (N ), we get √ p24 (N + 1) 2−1/2 (N + 1)−27/4 exp(4π N + 1) √ ≈ p24 (N ) 2−1/2 N −27/4 exp(4π N )   −27/4   √  √ N +1 exp 4π N +1− N . ≈ N For large N , (N + 1)/N ≈ 1, so    √  √ p24 (N + 1) N +1− N . ≈ exp 4π p24 (N ) √ Pulling out a factor of 1/ N in the exponent gives us, " #  p24 (N + 1) 4π p ≈ exp √ N (N + 1) − N . p24 (N ) N For large N , p

1 N (N + 1) ≈ N + . 2 1

Thus, p24 (N + 1) ≈ exp p24 (N )

2π √ N

! .

QC16.3 Solution of Quick Calculation 16.3 For N = 1, there is only one partition, namely {1b }, where b runs from 1 to 24. We have to choose one element out of 24, thus p24 (1) = 24. The partitions of N = 2 are {2b1 },

{1b1 , 1b2 }.

For the first partition, we can choose 24 elements. There are two cases for the second partition. If b1 6= b2 , we can choose out of 24 elements for the first element, and out of 23 for the second element. Since the two elements can be in any order, we must divide by 2. If b1 = b2 , then there are 24 such partitions. Thus, p24 (2) = 24 + (24 × 23/2 + 24) = 324. The partitions of N = 3 are {3b1 },

{2b1 , 1b2 },

{1b1 , 1b2 , 1b3 }.

The first case has 24 partitions. The second case also has 24 × 24 elements because none of the 2b1 ’s is equal to any of the 1b2 ’s. So, we can pick any b1 and b2 for both elements. For the three-element partition, there are four cases. For all bi ’s different, there are   24 24! = 2024 = 3!(24 − 3)! 3 such partitions. For only two equal indices, there are   24 2 = 828 2 partitions. The factor of 2 takes care of all the ways of choosing two elements with equal indices. The last case, b1 = b2 = b3 , has 24 partitions. Thus, "  #   24 24 +2 + 24 = 3200. p24 (3) = 24 + 24 × 24 + 3 2 2

The partitions of N = 4 are {4b1 },

{3b1 , 1b2 },

{2b1 , 2b2 },

{2b1 , 1b2 , 1b3 },

{1b1 , 1b2 , 1b3 , 1b4 }.

Using the results obtained previously for the first three partitions and generalizing the counting for the partition with four elements, we get "  # "   # "  #     24 24 24 24 24 p24 (4) = 24 + 242 + + 24 + 24 + 242 + +3 +3 + 24 2 2 4 3 2 = 25 650. QC16.4 Solution of Quick Calculation 16.4 A spherical uniform mass is a black hole if its radius R satisfies R < RS , where RS = 2GM/c2 is the Schwarzchild radius. Since the density of the spherical object is ρ=

3M M = , V 3πR3

we have RS > R 2GM M > ρR. 2 c V A little rearranging gives us R> p

c 8πGρ/3

.

QC16.5 Solution of Quick Calculation 16.5 Since E = ~ω and ω = 2πc/λ, λ = 2π~c/E. Using E = k T¯H , the ratio of the wavelength to the radius of the black hole is λ 2π~c/k T¯H = = 8π 2 ≈ 80. 2 R 2GM/c QC16.6 Solution of Quick Calculation 16.6 Recall (16.145) R4 = gY2 M N. α02 The ’t Hooft coupling is λ = gY2 M N , and the IIB theory expansion parameter is λ0 = α02 /R4 . Hence, λ0 λ = 1. 3

QC16.7 Solution of Quick Calculation 16.7 Since λ is fixed, λ0 is fixed (see QC 16.5). Then, (16.145), R4 = gY2 M N 02 α tells us that R is fixed (α0 is a constant). Then, the free parameter of the Yang-Mills theory is N and the free parameter of IIB theory is g. Since g = gY2 M , R4 = gN. α02 Then, g = 1/N . Thus, if we make an expansion in terms of 1/N in Yang-Mills theory, it corresponds to an expansion in terms of g in IIB theory.

16.1 Review of statistical mechanics (a) If we make an infinitesimal change dV to the volume of a system quasistatically, then the energy E(V ) of the system increases by dE =

∂E dV. ∂V

Then, the work 6dW done on the system is the mean value of dE in the thermal equilibrium distribution: P −βEi ∂E P −βEi e dV dE e ∂V i . (1) 6dW = P −βEi = iP −βE i ie ie We can write the numerator in terms of the partition function as follows: X i

e−βEi

∂E 1 ∂ X −βEi 1 ∂Z =− . e =− ∂V β ∂V i β ∂V

Thus, (??) becomes 6dW =

1 ∂Z 1 ∂ ln Z dV = dV. βZ ∂V β ∂V

The work element 6dW can be expressed in terms of the pressure p by 6dW = p dV Thus, the pressure in terms of the partition function is p=

1 ∂ ln Z . β ∂V 4

(b) The partition function of the system is Z = Z(β, V ). Let us consider the differential d ln Z =

∂ ln Z ∂ ln Z dV + dβ. ∂V ∂β

Since E=−

∂ ln Z , ∂β

6dW =

1 ∂ ln Z dV, β ∂V

we have d ln Z = β6dW − Edβ = β6dW − d(Eβ) + βdE. Rearranging terms gives d(ln Z + Eβ) = β6dW + βdE. The heat of the system is defined by 6dQ ≡ d 6 W + dE. Thus, d(ln Z + Eβ) = β6dQ.

(2)

The second law of thermodynamics states that the heat 6dQ of the system is related to the entropy by 6dQ = T dS. Thus, (??) becomes d(ln Z + Eβ) =

1 dS k

Integrating the differentials and making use of the definition of the free energy, F ≡ E − T S, gives us F = −kT ln S.

16.2 Fermionic violin string and counting unequal partitions 5

(a) We start with the definition of the partition function Z from (16.27), !   Y ∞ X ∞ X Eα ~ω0 `n` Z = exp − = exp − . kT kT α `=1 n =0 `

Since n` = 0 or 1, ! 1 + exp − ~ω0 `n`  . Z = kT `=1 ∞ Y



(1)

Then, the free energy is F = −kT ln Z = −

∞ X

 ln 1 + exp −

`=1

!

~ω0 `n`  . kT

In the high temperature limit, kT  ~ω0 , so each term in the sum differs very little from the previous one. Then, we approximate the sum by an integral  ! Z ∞ ~ω0 `  d` ln 1 + exp − . F ≈ −kT kT 1 Using x =

~ω0 `, kT

we get (kT )2 F ≈− ~ω0

∞

Z

dx ln(1 + e−x ).

0

Using the expansion 1 1 1 ln(1 + y) = y − y 2 + y 3 − y 4 + . . . , 2 3 4 which is valid for 0 ≤ t ≤ 1, we have   Z (kT )2 ∞ 1 −2x 1 −3x 1 −4x −x F ≈ − dx e − e + e − e + ... ~ω0 0 2 3 4   (kT )2 1 1 1 1 − 2 + 2 − 2 + ... . ≈ − ~ω0 2 3 4 In order to evaluate the infinite sum, consider (16.38): ζ(2) = 1 +

1 1 1 π2 + + + . . . = . 22 32 42 6 6

Pulling out a factor of 4 from the right-hand side gives   1 1 1 1 ζ(2) = 4 2 + 2 + 2 + 2 + . . . . 2 4 6 8 We now realize that subtracting twice the sum in the right-hand side from the sum representation for ζ(2) gives     1 ζ(2) 1 1 1 1 = 1 + 2 + 2 + 2 + ... − 2 2 + 2 + ... ζ(2) − 2 4 2 3 4 2 4 1 1 1 = 1 − 2 + 2 − 2 + .... 2 3 4 This is exactly the sum that we need and its value is ζ(2)/2 = π 2 /12. Thus, the free energy in the high temperature limit is F =−

π 2 (kT )2 . 12~ω0

(b) We have computed the free energy F . Then, we can calculate q(n) by noting that it is equivalent to the number of states Ω that defines the entropy through the relations: S=−

∂F = k ln Ω. ∂T

The entropy, in the high temperature limit, is S = k

π 2 kT . 6 ~ω0

(2)

The energy is given by ∂ ln Z ∂ π2 1 ∂ = (βF ) = − E= ∂β ∂β 12 ~ω0 ∂β

1 β

! ,

(3)

which gives 2

E =

π 12

kT ~ω0

!2 ~ω0 .

Since E = ~ω0 N , combining (??) and (??) yields r r 1 E N S(E) = kπ = k2π . 3 ~ω0 12 7

(4)

(5)

Comparing this result with "  # E S(E) = k ln q(N ) = k ln q ~ω0 gives the partition of fermionic numbers in the high temperature limit r N . ln q(n) = 2π 12 (c) Now, instead of using the non-relativistic formula E = ~ω0 N , we use p E = N ⊥ /α0 .

(6)

(7)

The number of states Ω(E) equals q24 (N ⊥ ). For large energy, N ⊥ is also large, and using (??) r √ N ⊥ × 24 = k2 2N ⊥ . S(E) = k2π 12 Making use of the number-energy relation (??) gives us √ S(E) = k2π 2α0 E. We can compute the Hagedorn temperature by using the relation 1 ∂S = . T ∂E Thus, kTH =

1 √

2π 2α0

.

√ The Hagedorn temperature for a fermionic string is 1/ 2 times the Hagedorn temperature for a bosonic string.

16.3 Generating functions for partitions We have ∞ Y

∞

X 1 p(n) xn . = n 1 − x n=1 n=0 8

Using the expansion ∞ X 1 = ym, 1 − y m=0

about y = 0 gives ∞ X ∞ ∞ Y X nm x = p(n) xn n=1 m=0

n=0     = 1 + x + x2 + . . . 1 + x2 + x4 + . . . 1 + x3 + x6 + . . . . . .

(1)

= 1 + x + 2x2 + 3x3 + 5x4 + . . .

(2)

We now evaluate p(n) for four different values of n by matching the coefficients of the x’s. For the case of n = 1, there is only one term in (??) that contributes to x. It comes from multiplying the x in the first factor and all 1’s in the rest the of the factors. Thus, p(1) = 1. For the case of n = 2, there are two terms that contribute to x2 : first, by multiplying the x2 term in the first factor with the 1’s; second, by multiplying the x2 term in the second factor with the 1’s of the rest of the factors. Thus, p(2) = 2. For the case of n = 3 and n = 4, we use the power series in (??) to read the coefficients of xn and obtain p(3) = 3 and p(4) = 5. The generating function works in general for n because collecting the factors of xn amounts to counting the different ways we can build a partition for n. Since exponents add when multiplying x’s, each possible way of getting a xn term is a partition of n. The generating function for unequal partitions q(n) is given by ∞ Y n=1

n

(1 + x ) =

∞ X

p(n) xn

n=0

= (1 + x)(1 + x2 )(1 + x3 )(1 + x4 ) . . . (1 + xk ) . . . = 1 + x + x2 + 2x3 + 2x4 + . . . Thus, q(1) = 1, q(2) = 1, q(3) = 2, and q(4) = 2.

16.4 Counting of generalized partitions The partition function is defined by Z=

X

e−βEα .

α

In order to compute the partition function, we must sum over all bosonic and fermionic numbers:   ∞ X b X X X ~ω0 (q) `n`  , exp − ZT = kT (1) (b) (1) (f ) `=0 q=1 nr ,...,nr

ns ,...,ns

9

(q)

(q)

where r, s = 1, 2, . . . , ∞, the nr run over all non-zero integers and ns = 0 or 1. The sums separate, thus       ∞ ∞ ∞ ∞ ~ω0 X (1)  Y X ~ω0 X (1)  Y X ZT =  exp − `n`   `n exp − . kT `=0 kT `=0 ` r=1 (1) s=1 (1) nr

ns

We recognize the left term as Z b and the right term as (Z 0 )f . Thus, ZT = (Z)b (Z 0 )f . In general, whenever we take two independent systems and consider them as one, the total partition function is the product of the partition functions of the constituent systems. We can compute ln P (N ; b, f ) by evaluating the free energy F = −kT ln ZT h i = −kT ln (Z)b (Z 0 )f   0 = −kT b ln Z + f ln Z = bFb + f Ff , where Fb and Ff correspond to the free energy of bosonic and fermionic strings. We can then compute the entropy ∂Fb ∂Ff ∂F = −b −f ∂T ∂T ∂T = bSb + f Sf ,

S = −

where Sb and Sf are the entropies of the bosonic and fermionic strings. Using S = k ln P (N ; b, f ),

Sb = k ln p,

Sf = k ln q,

and the approximate expressions for p(n) and q(n) gives us v ! u uN f b+ . ln P (N ; b, f ) ≈ 2π t 6 2

16.5 Open superstring Hagedorn temperature (a) Let us consider the NS sector. The most general state in the NS sector is given by (13.95) |Ψi =

9 Y ∞  Y I=2 n=1

I α−n

9 λn ,I Y

Y 

J=2 r= 1 , 3 ,... 2 2

10

bJ−r

ρr ,J

|N Si ⊗ |p+ , p~T i.

For each coordinate, there are 8 bosonic operators α−n and 8 fermionic operators b−r . Since we sum over 8 coordinates, the number of states is 8P (N ⊥ ; 8, 8). Since superstrings have supersymmetry, the total number of states is twice that of the NS sector. Thus, the number of states is 16P (N ⊥ ; 8, 8). (b) The number of states of an open superstring with energy given by α0 E 2 = N ⊥ is √ Ω(E) = 16P (N ⊥ ; 8, 8) = 16P ( α0 E 2 ; 8, 8). Using the expression for ln P (N ; b, f ) derived in Problem 16.4, we get √ ln Ω(E) ≈ 2π 2α0 E. Since S(E) = k ln Ω(E) and 1 ∂S = , T ∂E we get kTH =

1 √

2π 2α0

.

Thus, the Hagedorn temperature of the open superstring is a factor of the Hagerdorn temperature of the bosonic string.

√

2 larger than

16.6 Partition function of the relativistic particle The partition function of the relativistic particle (16.83) reads: Z dd~u −βm√1+~u2 2 d Z(m ) = V m e . (2π)d We wish to express the integral in terms of modified Bessel functions √  1 ν Z ∞ π 2z   e−z cosh t sinh2ν (t) dt. Kν (z) = 1 0 Γ ν+2 Let ~u2 = u2 . Using spherical coordinates with radial coordinate u, we have dd~u = ud−1 dΩdu, where dΩ is the infinitesimal element of solid angle in d dimensions (think of the case for d = 2, 3). The integral then becomes  d Z √ m 2 2 Z(m ) = V ud−1 e−βm 1+u dΩdu. 2π 11

√ Let u = sinh v, then 1 + u2 = cosh v, and du = cosh v dv. Performing the change of variables in the integral gives us  d Z m 2 Z(m ) = V cosh(v) sinhd−1 (v) e−βm cosh v dΩdv. 2π We can factor this integral into a radial integral and an angular integral:  d Z ∞ Z m d−1 2 −βm cosh v Z(m ) = V cosh(v) sinh (v) e dv dΩ. 2π 0 We recognize the second integral as the volume of the unit (d − 1)-sphere, S d−1 . Making use of (3.51), vol(S d−1 ) =

2π d/2  , Γ d2

we have 2

Z(m ) = V



m 2π

d

2π d/2   Γ d2

Z

∞

cosh(v) sinhd−1 (v) e−βm cosh v dv.

0

Using ∂ Kν (z) ∂z (z/2)ν

 ν   √ Z ∞ 2 ν π 0  = cosh(t) sinh2ν (t) e−z cosh t dt, Kν (z) − Kν (z) = −  1 z z 0 Γ ν+ 2

with 2ν = d − 1 and z = βm yields the partition function in terms of modified Bessel functions: ! d2 " # r 2βm d − 1 m Z(m2 ) = V K d−1 (βm) − K 0d−1 (βm) . (1) 2 2 π 2πβ 2βm We need the asymptotic expansion of the modified Bessel functions (and derivatives thereof), # # r " r " 2 2 π 4ν − 1 π 4ν + 3 Kν (z) ∼ e−z 1+ + ... , Kν0 (z) ∼ e−z 1+ + ... . 2z 8z 2z 8z To first order in the low temperature limit, βm  1, we can neglect the first term in (??). Then, using only the first term in the asymptotic expansion for Kν0 (z) gives us Z(m2 ) ∼ V e−βm 12

m 2πβ

! d2 ,

which corresponds to the asymptotic expansion given in (16.89). We obtain the first nontrivial correction to the partition function by considering the asymptotic expansion for Kz (z), to first order, and the expansion for Kn u0 (z), to second order. We then have Z(m2 ) ∼ V e−βm

m 2πβ

! d2 "

# d(d − 2) . 1+ 8z

16.7 Corrections to the temperature/energy relation in the microcanonical ensemble The more accurate version for the number of partitions is  √  1 −27/4 p24 (N ) ≈ √ N exp 4π N . 2

(1)

The entropy is given by √ 1 27 S = ln p24 (N ) = − ln N + 4π N − ln 2. k 4 2

(2)

Using the number-energy relation α0 E 2 = N ⊥ gives √ 27 1 S(E) 27 = − ln E + 4π α0 E − ln α0 − ln 2. k 2 4 2 Thus, the temperature-energy relation is given by √ 27 1 ∂S = 4π α0 − k ∂E 2E 2E √ . T (E) = k(8π α0 E − 27) 1 kT

=

A plot of T (E) is shown below. Note that the temperature approaches the Hagedorn temperature from above. The specific heat, at constant volume, of the string is given by ! ∂E 27k √ CV = =− . ∂T 2(4π αkT − 1)2 V

Thus, in the high energy regime, strings have a negative heat capacity. We now compute the entropy S(E), in a different way, by considering continuous (rather than discrete) energies. We now have Ω(E) dE = p24 (N ) dN. 13

Then, the entropy is given by dN S(E) = k ln Ω(E) = k ln p24 (N ) dE

! .

Using the relation number-energy relation α0 E 2 = N ⊥ , and the approximate expression for p24 (N ) we get S(E) = ln p24 (N ) + 2α0 E. k Compare this expression for the entropy with (??). Thus, we obtain the extra term 2α0 E when we consider continuous energies. The number of states of an open string with energy E on a Dq-brane is   √ Ω(E) ≈ E −γ exp 4π α0 E , where γ = (25 − q)/2. The entropy is given by √ S(E) = ln Ω(E) = −γ ln E + 4π α0 E. k . Then, the temperature-energy relation is √ 1 1 ∂S γ = = − + 4π α0 . kT k ∂E E The energy as a function of temperature is given by  −1 1 1 E(T ) = γk − , kTH kT √ where TH = (4π α0 k)−1 . Using the continuous energy formulation, the heat capacity is ! ∂E γk √ =− CV = ∂T (4π α0 kT − 1)2 V  −2 γ 1 1 = − 2 − kT kT kTH γ3k = − 2 2. E T

16.8 Long string are entropically favored 14

(a) The number of states available to a string with energy E is given by pb (α0E + 1) ≈ pb (α0E). Then, the ratio of the number of states of a string with energy E0 to the number of states of two distinguishable strings with energy E0 /2 is √  −γ pb (N0 ) 1 16 βN0−γ eδ N √ = . = pb (N0 /4)2 β N0 β 2 (N0 /4)−2γ e2δ N0 /4 The change in entropy in this process is   N0 ∆S = γ ln − ln β. k 16 For N0  1, the first term in the right-hand side dominates. Since γ > 0, the entropy ∆S > 0. (b) Consider now a string withpenergy E0 and two string with energy E1 and E2 satisfying E0 = E1 + E2 . Since E ≈ N/α0, the numbers of the string are related by p p p N0 = N 1 + N 2 . (1) The ratio of the number of states available to the string with energy E0 to the number of states available to the distinguishable strings of energy E1 and E2 is given by √  −γ βN0−γ eδ N0 1 N0 pb (N0 ) =  . √  √  = pb (N1 )pb (N2 ) β N1 N2 βN1−γ eγ N1 βN2−γ eδ N2 The change in entropy is given by N 1 N2 ∆S = γ ln − ln β. k N0

(2)

The expression for the entropy (??) is positive for large N1 and N2 if the argument of the logarithm is larger than 1 (remember that γ > 0). Then, we require that N 1 N2 N0 < N1 N 2 .

1 < N0 Using (??) gives us

p N1 + N2 + 2 N1 N2 < N 1 N2 , which is true for large N1 and N2 because the right-hand side is quadratic and the left hand side is linear. Thus, the combination of two very energetic open strings into one string is entropically favored. 15

(c) The number of available states increases if the change in entropy is positive. The resultant string has N0 = 9. The two original string each have E1 = E0 /2. Since α0E02 = 8, α0E12 = 2. So, we have N0 = 9 and N1 = 3. Then, using the results from QC 16.3, the number of times that the number of available states is increases in this process is p24 (9) ≈ 14. p24 (3)2 The change in the entropy is given by p24 (9) ∆S = ln ≈ 2.6. k p24 (3)2

16.9 Estimating the size of a string state (a) A random walk of N steps in a d-dimensional space has an average value for the traveled distance squared of N , for √ large N . If we consider a string of length L made out of pieces each of length `s = α0 , then the number of pieces of the√string is L/`s . Since the root-mean-square of the traveled distance of a random walk is N , the string “size” is p √ Rstr (M ) ∼ N ∼ L/`s . √ In order for Rstr to have units of length, we multiply by `s , Then, Rstr ∼ L`s . Using (16.127) M ∼ T0 L ∼ `s =

√

L , α0

α0 , and α0 M = N , we obtain Rstr ∼ M 1/2 `3/2 ∼ N 1/4 `s . s

(1)

(b) We require that RSch > Rstr , where RSch is the Schwarzchild radius. Using our result for Rstr from the previous part of this problem, and the definition of RSch , we get 2GM > M 1/2 `s3/2 . c2 ¯ ∼ G−2 `3 . Using G ∼ g 2 `2 , and mP = (g`s )−1 , we obtain Thus, M s s ¯ ∼ 1 ∼ mP . M g 4 `s g3 16

(2)

¯ is The Schwarzchild radius for an object with mass M 2 2 ¯ ∼ g `s ∼ `s . RSch ∼ GM g 4 `s g2

Making use of (16.64), ¯ 2 α0 ∼ N ≈M

1 `2s = . g 8 `2s g8

The planck mass is mP ≈ 2.2 × 10−8 Kg. Then, using (??) gives us −8 ¯ ∼ 2.2 × 10 Kg = 2.2 × 10−2 Kg. M (0.01)3

(c) Let us assume that the entropy of the black hole is constant while we dial down the string coupling constant g. Making use of (16.133), we have S0 ∼ g02 `2s M02 = g∗2 `2s M∗2 , k

R∗ ∼ g∗2 `2s M∗ .

(3)

We do not expect these results to hold when the black hole becomes smaller than the string length, so let us fix R∗ = `s as the minimum radius for which (??) can be trusted. The condition R∗ = `s tells us that M∗ ∼

1 g∗2 `s

.

Using (16.133), g02 `2s M02 , we get g 2 `2s M 2 ∼ `s M∗2 . Making use of (??), we can write N `4s ∼ M∗2 `6s . Hence, N∼

M∗2 `2s

∼

g 4 `4s M 4

 =

M mP

4 .

Using (??) again and G ∼ g 2 `2s , we obtain the following expression for Rstr : Rstr ∼

GM M `s R ∼ M g`2s ∼ ∼ . mP g g 17

Thus, the length of the string is L ∼ α0 M ∼

α 0 mP Rstr . `s

Recalling that mP = 2.2 × 10−8 Kg and the mass of the sun Msun = 1.99 × 1030 Kg, the black hole at the center of our galaxy has !4  4 M 2.6 × 106 Msun N∼ = = 3 × 10177 . mP mP

18