Thermodynamics of photon gas: Carnot cycle Masatsugu Sei Suzuki Department of Physics (Date: October 12, 2016)

Here we discuss the thermodynamics properties of photon gas including the Carnot cycle. ______Part-1

Apply the thermodynamic relation dE  TdS  PdV to a photon gas. Here one can write that E  Vu T )( , where u(T), the mean energy density of the radiation field, is independent of the volume V. The radiation pressure P  u T 3/)( . (a) Considering S as a function of T and V, express d S in terms of d T and d V. Find

(S / T)V and (S / V)T . (b) Show that the mathematical identity (2S / VT)  (2S / TV) gives immediately a differential equation for u(T) which can be integrated to yield the Stefan Boltzmann law 4 u T)(  T 4 . In fact, it is well known that u T )(  B T 4 , where the dispersion of c photon is given by   cp and the Stefan-Boltzmann constant is defined by  2k 4   B . B 60ℏ c 23

Part-2 Carnot cycle

Consider a Carnot engine that uses as the working substance a photon gas (two isothermal stages 1-2 and 3-4, and two isentropic stages 2-3 and 4-1. The schematic diagram for the four processes in the P-T diagram for the photon gas is shown in Fig.

Note that the P-V diagram for the photon gas is rather different from that for the ideal gas. (c) Calculate the entropy S and the pressure P of the photon gas by using the relations given in (a) and (b).

(d) Given T1 and T2 (T1>T2) as well as V1 and V2 (V1>V2), determine V3 and V4. (e) Discuss the work, the heat and the internal energy in the isothermal process (1→2). What is

Q1 transferred as heat from the high temperature reservoir at T1 to the photon gas during this process? (f) Discuss the work, the heat and the internal energy in the isentropic process (2→3). (g) Discuss the work, the heat and the internal energy in the isothermal process (3→4).

(h) Discuss the work, the heat and the internal energy in the isentropic process (4→1). What is Q2

as heat transferred from the photon gas to the low temperature reservoir at T2 during this process ? (i) Show the total work done by the photon gas during one cycle is given by

4 W  B T 3(T T )( V V ) 3c 1 1 2 2 1

(j) Show that the energy conversion efficiency for the photon gas is the Carnot efficiency for the W T ideal gas which is given by    1 2 . Q1 T1

(( Solution )) (a)

dU TdS  PdV where

1 U VuT( ) , P  u(T ) 3

Note that u(T) is the mean energy density of the radiation field (independent of V).

TdS dU  PdV 1 d[ Vu ( T )]  u ( T ) dV 3 4 du udV  V dT 3 dT or

4 V du dS  udV  dT 3T T dT

 S   S     dV    dT  V T  T V since S is assumed to be dependent only on V and T. Then we have

 S  4u  S  V du    ,     V T 3T  T V T dT

(b)

2 S   S  1 du  [   ]T  (Maxwell’s relation) VT V  T V T dT

 2S   S  d  4u   [   ]V    TV T  V T dT  3T 

Then we have

 du  T  u 1 du d  4u  4        dT  T dT dT 3T 3 T 2       or

du du 3T  4T  4u dT dT or

du du 3T  4T  4u dT dT or

du T  4u dT

Then we have

1 dT du  4  u  T

lnu  4lnT + const.

In conclusion, u  T 4 . Note that in fact, u(T) can be expressed by

4 u(T )  B T 4 c

(c) TdS  dE  PdV 4 V B  4 B 4  c 4  d VT   T dV  c  3V

4 1  B d([ VT 4 )  T 4dV ] c 3 4 1  B 4[ T 3VdT  T 4dV  T 4dV ] c 3

Then we get

4 2 4 3 4  4 3  dS  B 4( VT dT  T dV )  B d VT  c 3 c  3 

The entropy S is given by

16 S  B VT 3  const 3c

The state of equation is obtained as

4 B 4 V T 4 E 4 T P   c  B 3V 3V 3c

So that the pressure P is independent of V, and depends only on T.

(d)

Process 1→2 (at T = T1),: isothermal process

4 P  P  B T 4 (which is independent of V) 1 3c 1

Process 3→4 (at T = T2),: isothermal process

4 P  P  B T 4 3 3c 2

Process 2→3, and process 4→1; isentropics process. S = const, which means that VT 3  const , 4 T 4 or PV 3/4  const because of P  B . Here we have 3c

3 3 3 3 V2T 1 V T23 , V4T 2  V T11

(e)

Process 1→2

Q12  dU 12  PdV 4 4  BTVV4()  B TVV 4 ()  c1213 c 121 16  B T4 ( V  V ) 3c 1 2 1

The heat Q1 taken up is

16 Q  Q  B T 4 (V V ) 1 12 3c 1 2 1

The work done by the photon gas is

4  W  PdV  B T 4 (V V ) 12 3c 1 2 1

(f) Process 2→3

Q23  0

U23  Q 23  W 23

  W23 4 4  BVT4  B VT 4 c32 c 21 3 4BT1 4 4  B 4 [3 VT22  VT 21 ] c T2 c 4 B VT3 ( T  T ) c 21 2 1

(g) Process 3→4

Q34  dU 34  PdV 4 4  BTVV4()  B TVV 4 ()  c2433 c 243 16  B T4 ( V  V ) 3c 2 4 3 16  B TT3 ( V  V ) 3c 12 2 1

The heat Q2 is

16 Q  Q  B T 3T (V V ) 2 34 3c 1 2 2 1

The work done by the photon gas is

 W34  PdV 4  B T 4 (V V ) 3c 2 4 3 4  B T T 3 (V V ) 3c 12 1 2

(h) Process 4→1

Q23  0

U41  Q 41  W 41

  W41 4 B (VT4  VT 4 ) c 11 42 3 4 B 4T1 4 (VT11  3 VT 12 ) c T 2 4 B VT3 ( T  T ) c 11 1 2

(i) The total work done by the photon gas is

 W W

 (W12  W23  W34  W41 ) 4  B [T 4 (V V )  3V T 3 (T  T )  T 3T (V V )  3 TV 3 (T T )] 3c 1 2 1 12 2 1 1 2 1 2 11 1 2 4  B T 3 (T  T )(V V ) 3c 1 1 2 2 1

(j)

Q1 16 B 3  T1 (V2 V1 ) T1 3c

Q2 16 B 3  T1 (V2 V1 ) T2 3c

Then we have the relation

Q Q 1  2 T1 T2

4 Q Q  B T 3 (T T )(V V )  W 1 2 3c 1 1 2 2 1

The efficiency ;

W Q Q T    1 2  1 2 , Q1 Q1 T1 which is equal to the Carnot efficiency.

Fig. Carnot cycle. TH  T1 . TL  T2 . QH  Q1 . QL  Q2 . W  Q1  Q2 . ______APPENDIX 2. (Point 40) Photon gas : (a) Ignoring the zero-point energy, show that the partition function Z for a gas of photons in volume V is given by V  lnZ 2 ln(1  eℏ  ) d  , 2 3   c 0 and hence, by integrating by parts, that V 2( k T ) 3 ln Z  B 45 ℏ3c 3  x3  4  dx  . 0 exp( x) 1 15 (b) Show that 4 VT 4 16  VT 3 4 VT 4 4 T 4 F   SB , S  SB , U  SB , P  SB , 3c 3c c 3c and hence that U 4U U  3 F , PV  , and S  , 3 3T  2k 4 where   B is the Stefan-Boltzmann constant. SB 60 ℏ c23

((Solution)) Helmholtz free energy F;

Using the one particle partition function for the mode with k

 1 Zexp(  n ℏ  )  1k k ℏ n0 1 exp(    k ) we have the partition function for the total system (canonical ensemble)

ZC   Z 1k k or

ℏ  k lnZC   ln(1  e ) k

2V ℏ  4 k2 dk ln(1  e   k ) (2 ) 3 

2V 2  4d  ln(1  e ℏ  ) (2 ) 3 c 3 V  2d  ln(1  e ℏ  )  2c 3 

where the factor 2 comes from the two kinds of the polarization vectors, and k  ck . Taking the integral,

2V 4  1 ln Z ℏ 3 d  C 3 3  ℏ  (2)3 c0 e  1 2V 4 1  x 3  dx 3 3 ℏ 3  x (2)3 c  0 e  1 V1  x 3  dx 2 3ℏ 3  x 3c ()  0 e  1

F  kTBln Z C V1  x 3   kT dx B 2 3ℏ 3  x 3c ()  0 e  1

The integration by part leads to

F k 4  x3dx 1  2k 4 4   B T 4   B T 4    T 4 ℏ 332  x ℏ 33 SB V 3 c 0 e 1 45 c 3c or

4 V F  SB T 4 3c where

 x3dx  4   x 0 e 1 15 or 1 F  U 3 using the value of U.

We note that

4 SBV 3 ln ZC  T 3ckB

The entropy S:

F 16 V 4U S  ( )  SB T 3  T V 3c 3T

The total energy:

4 V 16 V 4 V U  F  ST  SB T 4  SB T 4  SB T 4  3F 3c 3c c

The pressure:

F 4 U P  ( )  SB T 4  V T 3c 3V

The Gibbs free energy:

1 1 G  N  F  PV  U  U  0 3 3 leading to

  0

In other words, the chemical potential is zero for the photon gas.

The Stefan Boltzmann constant:

c 24k  24 k  [B ]  B =5.670367 x 10 -8 W/(m2K4) SB 415(ℏc )3 60 ℏ 3 c 2