7. Entropy and Potential

 

In classical thermodynamics, internal energy has the following relationship with entropy.

 

$dU=Tds-PdV$

 

If $dV=0$ regardless of temperature $T$ like solid,

 

$dU=TdS=dQ=>dS=dU/T=dQ/T$.

 

If internal energy($U$) is mostly composed of potential($\Phi$) and kinetic energy($E_k$) of electrons, nuclei, and atoms, $TdS=dQ=d\Phi+dE_k \approx dE_k$ is reasonable.

 

If $E_k/Q = -1/\beta - \ln\gamma(1+\beta))$, then

$$dS \approx -\frac{dQ}{T\beta}-\frac{dQ}{T}\ln\gamma(1+\beta)$$

$$S \approx -\frac{Q}{T\beta}-\frac{Q}{T}\ln\gamma(1+\beta)+C$$

If $Q/T\approx mC_v, 1/\beta, \ln\gamma(1+\beta)$ is proportional to the logarithmic potential, we can define entropy like this.

 

$S = S_0-k_B((\ln\phi)^2+ (\ln\phi_e)^2)$

($S_0$: initial entropy, $k_B$: Boltzmann constant, $\phi, \phi_e$: density of gravitational, electomagnetic potential)

 

In a quantized space like an atom, the potential is discontinuous, which is the source of specific heat($C_v$), but in a place where subatoms or molecules move, such as a gas or plasma, the potential will act like pressure.

 

[Below is not accurate]

Assuming that the potential density $\Phi/c^2 = \phi_e\phi \approx \rho{e^\rho}$,

$$\frac{Q}{T}=\ln\Phi=\rho + \ln\rho$$

$r$ is smaller than the certain value, $Q/T \approx \rho = GM/2rc^2$.

Here, we can guess that $r$ is proportional to temperature T in case that $Q, M$ are constant and $r$ is small enough. But $r$ is large, $\ln\rho$ is more important and heat absorption/release changes.