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Imagine some kind of thermoionic device, where electrons jump off a metal in vacuum due to a fixed temperature bath and in the presence of a bias voltage.

In a steady state regime (and quasi-equlibrium regime), there will be a net particle current. Thanks to the first law of thermodynamics,

$$\Delta U=T\Delta S+\mu\Delta N$$, we can define the energy and entropy currents, $$\dot{U}=T\dot{S}+\mu J$$, where $U$ is the energy, $S$ is the entropy, $T$ the temperature, $N$ the number of particles, $\mu $ the chemical potential of the metal, $\dot{U}$ the energy current, $\dot{S}$ the entropy current and $J$ the particle current.

In a microscopic picture, $J\propto \int f(E)\tau(E) \mathrm{d}E$ and $\dot U\propto \int E f(E)\tau(E) \mathrm{d}E$, where $\tau(E)$ takes into account the probability of escaping the metal and $f$ is the Fermi-Dirac distribution.

By construction, $\dot{U}$ will be positive (as long as the metal is not receiving electrons). But $T\dot{S}=\dot{U}-\mu J$ can be negative or positive.

What is the physical interpretation of $T\dot{S}$? Why it can take positive and negative values?

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we can define the energy and entropy currents

What you defined are the rates of change ($\dot U$, $\dot S$). In the case of energy, which is conserved, the rate of change must correspond to the net flow. But entropy is different; it is not conserved.

Your example can be generalized to

$$T\dot S=\dot U-\sum_i F_i\frac{\partial x_i}{\partial t},$$

where $\dot S$ is the rate of system entropy increase and $F_j$ and $x_j$ are respectively generalized forces and displacements; examples include

  • negative pressure and volume
  • surface tension and surface area
  • electric field and polarization
  • voltage and charge
  • magnetic field and magnetic moment
  • force and distance
  • stress and volumetric strain
  • chemical potential and amount of matter

Along with energy, the generalized displacements above are conserved except for the electrical polarization and magnetic moment; thus, we ignore those forms of work for simplicity. By the divergence theorem expressed as $\frac{dA}{dt}=-\nabla\cdot\boldsymbol{J_A}+\dot\sigma =-\nabla\cdot\boldsymbol{J_A}$, where $\boldsymbol{J_A}$ is the flux of A and $\dot\sigma$ represents generation (absent for a conserved quantity), we obtain the continuity relation

$$T\dot S=-\nabla\cdot\boldsymbol{J_U}+\sum_iF_i\nabla\cdot\boldsymbol{J_i}.$$

From the chain rule, $A\nabla\cdot\boldsymbol{B}=-\boldsymbol{B}\nabla\cdot A+\nabla\cdot (A\boldsymbol{B})$; thus, letting $A=\frac{1}{T}$,

\begin{align}T\dot S&=T\left(\boldsymbol{J_U}\cdot\nabla\frac{1}{T}-\nabla\cdot\frac{\boldsymbol{J_U}}{T}-\sum_i\boldsymbol{J_i}\cdot\frac{F_i}{T}+\nabla\cdot\frac{\sum_iF_i\boldsymbol{J_i}}{T}\right)\\&=T\left[\underbrace{\boldsymbol{J_U}\cdot\nabla\frac{1}{T}-\sum_i\boldsymbol{J_i}\cdot\frac{F_i}{T}}_{\substack{\text{this looks like}\\\text{a generation term}}}\underbrace{-\nabla\cdot\left(\frac{\boldsymbol{J_U}-\sum_iF_i\boldsymbol{J_i}}{T}\right)}_{\substack{\text{this looks like}\\\text{a flux term}}}\right].\end{align}

Looking again at the divergence theorem, we infer a relation

$$T\dot S=T(\dot \sigma-\nabla\cdot \boldsymbol{J_S}),$$

where we identify

$$\dot\sigma =\boldsymbol{J_U}\cdot\nabla\frac{1}{T}-\sum_i\boldsymbol{J_i}\cdot\nabla\frac{F_i}{T}$$

as the entropy generation rate and $$\boldsymbol{J_S}=\frac{\boldsymbol{J_U}-\sum_iF_i\boldsymbol{J_i}}{T}$$

as the conserved entropy flux or current.

As derived here, for example, $\boldsymbol{J_U}=\boldsymbol{J_Q}+\boldsymbol{J_W}+\boldsymbol{J_H}$, where $\boldsymbol{J_Q}$ is the heat flux, $\boldsymbol{J_W}$ is the work flux (excluding flow work) and $\boldsymbol{J_H}$ is the enthalpy flux related to mass transfer (recall that the molar enthalpy $h=\mu+Ts$, where $\mu$ is the chemical potential). The subtraction of $\sum_iF_i\boldsymbol{J_i}$ from $\boldsymbol{J_U}$ above effectively removes $\boldsymbol{J_W}$ and $\sum_i\mu_i\boldsymbol{J_i}$ for mass flow $\boldsymbol{J_i}$. This allows us ultimately to write the above relation as

$$\boldsymbol{J_S}=\frac{\boldsymbol{J_Q}+T\sum_is_i\boldsymbol{J_i}}{T},$$

where $s_i$ is the entropy, if any, carried by generalized displacement $i$ (e.g., the flow of charge carriers).

In addition, applying the chain rule to the entropy generation rate gives

$$\dot\sigma=\frac{1}{T^2}\left(-\boldsymbol{J_U}\cdot\nabla T+\sum_iF_i\boldsymbol{J_i}\cdot \nabla T-T\sum_i\boldsymbol{J_i}\cdot\nabla F_i\right);$$ thus, $$T\dot\sigma=\underbrace{-\boldsymbol{J_S}\cdot\nabla T}_{\substack{\text{contributes to entropy}\\\text{generation from flow down}\\\text{a temperature gradient}}}+\underbrace{\sum_i-\boldsymbol{J_i}\cdot\nabla F_i.}_{\substack{\text{contributes to entropy}\\\text{generation from flow down}\\\text{all other gradients}}}$$

What this tells us is that entropy can be transferred in and out of a system through heating and through the entropy carried by matter, for example. This is $\boldsymbol{J_S}$.

In addition, entropy is generated whenever flow of a generalized displacement (all $\boldsymbol{J_i}$ plus the entropy flow $\boldsymbol{J_S}$ itself) is driven by a gradient in any internal property (all $\boldsymbol{\nabla F_i}$ plus temperature $\boldsymbol{\nabla T}$). The direction of spontaneous flow is opposite the direction of the gradient in the intensive property; in this way, Nature smooths out disturbances. This is $\boldsymbol{\dot\sigma}$.

Depending on how much entropy you transfer in and out and how much entropy you generate, $\boldsymbol{\dot S}$ can be positive or negative.

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  • $\begingroup$ Thanks for explaining the equations better. Could you give some insight on why the normal particle current $J$ can sometimes go in the opposite direction of entropy current $J_S$? What would this mean? that electrons getting out provide more entropy to the system? $\endgroup$
    – Mauricio
    Commented Jun 8, 2022 at 22:21
  • $\begingroup$ $\boldsymbol{J_U}$ also depends on $\boldsymbol{J_i}$ (in the electron example, electrons carry energy), so we cannot conclude that the entropy current is opposite the particle current just because a single minus sign appears in the numerator. I've edited the answer to clarify. In fact, the entropy current direction ends up matching the electron current direction (in the absence of other effects). $\endgroup$ Commented Jun 8, 2022 at 22:48
  • $\begingroup$ @Mauricio If you found this answer useful, please consider upvoting it; otherwise, you may wish to indicate if anything remains unclear. $\endgroup$ Commented Jul 10, 2022 at 19:39
  • $\begingroup$ Is there a way to straightforwardly apply this to a black body? i.e. fixed temperature etc., and the only flux of anything is related to the flow of photons in/out of the body. I'm reading 'Physics of Solar Cells' by Wurfel and he says Boltzmann derived a simple relation between the emitted entropy current and energy current of a black body: I_s=4/3* I_e/T $\endgroup$ Commented Jun 23, 2023 at 0:46
  • $\begingroup$ *(maybe not fixed temperature actually, just defined as absorbing all incident photons) $\endgroup$ Commented Jun 23, 2023 at 0:59

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