A
16 π
4
This is exactly what is required to make the laws of black holes the same as the laws of
thermodynamics.
Why does one get this intrinsic gravitational entropy which has no parallel in other
quantum field theories. The reason is gravity allows different topologies for the spacetime
manifold.
Boundary at infinity
IDENTIFY
S 1
S 2
In the case we are considering the Euclidean-Schwarzschild solution has a boundary at
infinity that has topology S 2 × S 1. The S 2 is a large space like two sphere at infinity and 31
the S 1 corresponds to the imaginary time direction which is identified periodically. One can fill in this boundary with metrics of at least two different topologies. One of course
is the Euclidean-Schwarzschild metric. This has topology R 2 × S 2, that is the Euclidean two plane times a two sphere. The other is R 3 × S 1, the topology of Euclidean flat space periodically identified in the imaginary time direction. These two topologies have different
Euler numbers. The Euler number of periodically identified flat space is zero, while that
of the Euclidean-Schwarzschild solution is two.
τ2
surface term
= 1 M (τ _ τ
2
2
1 )
volume term
τ1
= 1 M (τ _ τ
2
2
1 )
Total action = M ( τ 2 − τ 1)
The significance of this is as follows: on the topology of periodically identified flat space one can find a periodic time function τ whose gradient is no where zero and which agrees with the imaginary time coordinate on the boundary at infinity. One can then work out
the action of the region between two surfaces τ 1 and τ 2. There will be two contributions to the action, a volume integral over the matter Lagrangian, plus the Einstein-Hilbert
Lagrangian and a surface term. If the solution is time independent the surface term over
τ = τ 1 will cancel with the surface term over τ = τ 2. Thus the only net contribution to the surface term comes from the boundary at infinity. This gives half the mass times
the imaginary time interval ( τ 2 − τ 1). If the mass is non-zero there must be non-zero matter fields to create the mass. One can show that the volume integral over the matter
Lagrangian plus the Einstein-Hilbert Lagrangian also gives 1 M ( τ
2
2 − τ 1). Thus the total
action is M ( τ 2 − τ 1). If one puts this contribution to the log of the partition function into the thermodynamic formulae one finds the expectation value of the energy to be the mass,
32
as one would expect. However, the entropy contributed by the background field will be
zero.
The situation is different however with the Euclidean-Schwarzschild solution.
τ = τ2
surface term
volume term = 0
= 1 M (τ _ τ
2
2
1 )
fixed two
sphere
τ = τ1
r = 2M
surface term from corner
= 1 M (τ _ τ
2
2
1 )
Total action including corner contribution = M ( τ 2 − τ 1)
1
Total action without corner contribution =
M ( τ 2 − τ 1)
2
Because the Euler number is two rather than zero one can’t find a time function τ whose gradient is everywhere non-zero. The best one can do is choose the imaginary time coordinate of the Schwarzschild solution. This has a fixed two sphere at the horizon where τ
behaves like an angular coordinate. If one now works out the action between two surfaces
of constant τ the volume integral vanishes because there are no matter fields and the scalar curvature is zero. The trace K surface term at infinity again gives 1 M ( τ
2
2 − τ 1). However
there is now another surface term at the horizon where the τ 1 and τ 2 surfaces meet in a corner. One can evaluate this surface term and find that it also is equal to 1 M ( τ
2
2 − τ 1).
Thus the total action for the region between τ 1 and τ 2 is M( τ 2 −τ 1). If one used this action with τ 2 − τ 1 = β one would find that the entropy was zero. However, when one looks at the action of the Euclidean Schwarzschild solution from a four dimensional point of view