Black Hole thermodynamics

Transcription

1 Black Hole thermodynamics I Black holes evaporates I Black holes have a partition function For a Schwarzschild black hole, the famous Bekenstein-Hawking results are: T = 1 8 M S = A 4G = 4 r + 4G Note that the entropy grows like M making the partition function unstable.

2 Anicederivationatd =..(notdonehere) At d=, the energy momentum tensor has 3 components, and is fully fixed by its conservation equation T µ ; = 0 plus the trace anomaly g µ T µ = c 4 R ( equations) (1 equation) Evaluating on a black hole background (including some regularity condition at horizon), T µ takes the form of a flux at infinity with a temperature equal to Hawking s result.

3 For the +1 black hole (no rotation, reinserting G) ds = 8GM + r ` dt + 8GM + r 1 dr + r d' ` T = r + ` S = A 4G = r + 4G The entropy grows like p M,stable.

4 Quick derivation of black hole thermodynamics parameters For any black hole of the form, ds = f (r)dt + dr where f (r) has a point r = r + with, f (r + )=0, f (r) +... df (r) dr r + 6=0 we can associate a temperature: T = 1 4 f 0 (r + ) Indeed, in the vicinity of the horizon (a f 0 (r + )) f (r) ' a (r r + ), ds ' a(r r + )dt + and doing d = dr p a(r ds = a r+) we find dr p a(r r+ ) 4 dt + d ) hti = = 4 f 0 (r + )!

5 I For the +1 black hole and f (r) = 8GM + r ` T = 1 4 f 0 (r + )= r + ` I For the Schwarzschild black hole and f =1 MG r T = 1 MG 4 r+ = 1 8 MG

6 The entropy If black holes have mass M, andtemperaturet, there should be an entropy given by first law, h For the +1 black hole dm = TdS i 8MG + r + =0, T = r+, ` ` S = = Z dm T Z r d + 8G` = r + 4G r + `

7 Not convinced? Proving the first law If black holes are thermodynamical objects, these results should arise directly from the Euclidean General Relativity partition function, interpreted as a thermodynamical free energy: F ( ) e = 1 Z Dg e 1 R p 16 G g R+ ` Z 0 ' 1 e 1 R p 16 G g R+ ` Z 0 on shell This indeed works. Let us evaluate the exponent on the classical solution, properly including Z 0...

8 I The black hole has R = 6` and p g = r I The on-shell action is I = = = 1 16 G Z 1 dt 16 G 0 Z 1 dt G` 0 Z pg R + ` Z 1 r + Z 1 r + dr dr r Z 0 d' r 4 ` This clearly diverges, we need to regularize and subtract the AdS background

9 Integrating to a large radius L, and subtracting the AdS action, I I 0 = 1 Z G` 0 dt Z L r + dr r 1 G` Z 0 I For AdS the radial coordinate runs from the origin 0 dt Z L 0 dr r I Ar large r = L, both metrics must coincide with their proper periods at infinity calibrated (the same temperature), 8MG + L ` = 1+ L This condition does not imply 0 = (even for large L), ` 0

10 1 I I 0 = (L r+) 0L 4G` s 3 = 4L r+ L r+ /` + L /` 5 4G` 1+L /` 1 = 4G` r + + O(1/L) The free energy is then, F ( )= ` G Now, from the known thermodynamical expressions I The energy M = U, I The entropy M = F + ` = G! 8GM = r + ` = ` G = r + 4G L!1

11 Could we not do this calculation in a easier way? Yes, we can. The magic is provided, again, by the Chern-Simons action, I [A µ ]= k 4 Z Tr AdA + 3 A3, A = A µ dx µ G. We shall do a much more general calculation: Evaluate the Chern-Simons action for any Lie algebra on a solid torus: horizon ρ=0 0 apple t <, contractible loop t ρ φ 0 apple <, non-contractible loop The three dimensional spacetime topology can be seen as a torus < + =disc S 1.

12 The crucial point is that the radial dependence is a gauge transformation. If A is the field representing the black hole, then a new field A 0 can be built, A 0 = U 1 (r) AU(r)+U 1 (r)du(r) such that A 0 does not depend on r. I Performing this transformation is equivalent to work in the radial gauge A r =0 I Inserting the solution in the action will produce no IR divergencies I Some care is necessary with possible singularities. A constant field is not necessarily regular!

13 Interesting (not zero), regular, solutions on the solid torus Interesting solutions A µ = {A t, A r, A }SL(N, <) mustsatisfy: 1. The Chern-Simons equations of motion F µ =0. Must have a non-trivial holonomy along : PeH A d 6=1 If this holonomy was trivial, the solution can be set to zero by a gauge transformation. 3. Must have a trivial holonomy along t. PeH Atdt =1. If this holonomy is not trivial, the field will be singular.

14 Building the general solution in radial gauge A r =0 I F µ = 0 in the gauge A r =0imply A t (t, ), A (t, t A t +[A t, A ]=0. I Furthermore, for black holes, we consider static and spherically symmetric fields. That is, we take A t, A constant matrices. The equations reduce to: to be [A t, A ]=0 In summary, our game will be to find constant SL(N, <) matrices A, A t that commute, and satisfy the holonomy conditions. PeH A d = e A 6=1, PeH Atdt = e At =1

15 Chemical potentials For a given A,weseekA t such that [A t, A ]=0 The Cayley-Hamilton theorem imply A t = A + 3 A + + N A N 1 Trace Example N = 3. The condition Pe H A tdt = 1 becomes: 0 = Q (1 + ) Q (1 + ) Q 3 Q Q (1 + ) 3 Q 3, 8 = 4Q (1 + ) +6Q 3 (1 + ) Q 3. where Q =Tr(A ), Q 3 =Tr(A 3 ). These equations relate Q, Q 3 with, 3, justlikethem relates to in the metric formulation.

16 Evaluating the action: Angular quantization Consider the angle as time : horizon ρ=0 t ρ φ The foliation is regular everywhere I = Z (p q i constraints) + B 1 = 0 + B 1 I Spherical symmetry (@ q = 0) plus the constraints make the bulk part equal to zero. I We are left with a term at infinity, easily calculated.

17 Chern-Simons action in angular quantization. A +1 angular decomposition: x µ =, x. I CS = k Z Tr A A A A = k Z Tr A + A F ) 4 = 0 k Tr(A ta ). Z k dtd Tr(A t A ) 4 1 The last touch is to add a boundary term to get the right Legendre transformation, W (, 3) =I CS k 3 Q 3 W = kq + kq 3 3

18 W (, 3) = k Z AdA A3 k 3 Q 3 = k 3 3 Q Q + ( 1)Q 3 Q 3. I Consistency check (a fantastic partial 3 = kq 3 These equations are the analog of M = in the metric formulation.

19 Tomorrow... Phase transitions in black hole thermodynamics I Hawking - Page I First order and second order (higher spin fields)