Soft Degrees of Freedom, Gibbons–Hawking Contribution and Entropy from. . .
371
Before turning to these issues, we will first discuss the thermodynamics of a
capacitor by standard methods. In the context of general relativity, the Euclidean
approach of Gibbons and Hawking [17] consists in deriving the thermodynamics of
Kerr–Newman black holes or of de Sitter space by evaluating on-shell the Euclidean
action improved through suitable boundary terms. What these boundary terms are
in the electromagnetic sector has been discussed, for instance, in [18, 19]. That
the construction and interpretation of such boundary terms is very transparent
in the first Hamiltonian formulation is discussed, for instance, in the derivation
of the thermodynamics of the BTZ black hole [20]. We then review how the
thermodynamics of the capacitor can easily be reproduced from the Euclidean
approach [16].
2 Capacitor Thermodynamics: Textbook Approach
Consider a capacitor made of two conductors with charges +q and −q and area A.
Its capacity C =
q
V in Lorentz–Heaviside units is
C S =
4πR 1 R 2
R 2 − R 1
(1)
for two concentric spheres of radii R 1 < R 2 and
C P =
A
d
(2)
for two parallel planes of area A separated by a distance d (see, e.g. [21, chapter 2]).
The capacitor begins with 0 charge, energy, and entropy. Charges ±dq are added
on both side until one reaches ±q.
In order to charge the capacitor, one may use a circuit without any resistance so
that no heat would be produced in the process. One then would quickly arrive at the
standard results (see, e.g., chapter 14.2 of [22] in the absence of the system and its
electric polarization). A better understanding of the absence of entropy can however
be gained by considering a set-up with a resistor [23].
If a potential difference V is applied on the capacitor, there will be a current
I (t) =
q
RC e
−
t
RC . The heat lost by the system through the resistor is
Q =
∞
0
RI
2 (t)dt =
1
2
CV
2 .
(3)
Instead of a single step, the capacitor can be charged in N steps, each increasing
the voltage by
V
N . At each step, the circuit relaxes until the current vanishes. The
heat lost in all N steps is then
371
Before turning to these issues, we will first discuss the thermodynamics of a
capacitor by standard methods. In the context of general relativity, the Euclidean
approach of Gibbons and Hawking [17] consists in deriving the thermodynamics of
Kerr–Newman black holes or of de Sitter space by evaluating on-shell the Euclidean
action improved through suitable boundary terms. What these boundary terms are
in the electromagnetic sector has been discussed, for instance, in [18, 19]. That
the construction and interpretation of such boundary terms is very transparent
in the first Hamiltonian formulation is discussed, for instance, in the derivation
of the thermodynamics of the BTZ black hole [20]. We then review how the
thermodynamics of the capacitor can easily be reproduced from the Euclidean
approach [16].
2 Capacitor Thermodynamics: Textbook Approach
Consider a capacitor made of two conductors with charges +q and −q and area A.
Its capacity C =
q
V in Lorentz–Heaviside units is
C S =
4πR 1 R 2
R 2 − R 1
(1)
for two concentric spheres of radii R 1 < R 2 and
C P =
A
d
(2)
for two parallel planes of area A separated by a distance d (see, e.g. [21, chapter 2]).
The capacitor begins with 0 charge, energy, and entropy. Charges ±dq are added
on both side until one reaches ±q.
In order to charge the capacitor, one may use a circuit without any resistance so
that no heat would be produced in the process. One then would quickly arrive at the
standard results (see, e.g., chapter 14.2 of [22] in the absence of the system and its
electric polarization). A better understanding of the absence of entropy can however
be gained by considering a set-up with a resistor [23].
If a potential difference V is applied on the capacitor, there will be a current
I (t) =
q
RC e
−
t
RC . The heat lost by the system through the resistor is
Q =
∞
0
RI
2 (t)dt =
1
2
CV
2 .
(3)
Instead of a single step, the capacitor can be charged in N steps, each increasing
the voltage by
V
N . At each step, the circuit relaxes until the current vanishes. The
heat lost in all N steps is then