I Relationship between the divergence of a field and its cause?

[fluidistic]

I am quite confused. I know about the continuity equation, for instance $\frac{\partial \rho}{\partial t}+\nabla\cdot \vec J = \sigma$. In steady state, $\frac{\partial \rho}{\partial t}=0$ which implies $\nabla\cdot \vec J = \sigma$. In words, the divergence of the vector field $\vec J$ is equal to the source of that field, $\sigma$. This means that in steady state, if I focus on an infinitesimal volume element, and I compute the net flux that goes of that volume element, I would get $\sigma$. Great.

Here comes the problem. In thermoelectricity, the total heat flux is worth $\vec J_Q = -\kappa \nabla T +ST\vec J$ where $S$ is the Seebeck coefficient. The first part of this flux is the usual thermal flux, the second part is what you may call a Peltier flux (it has no name in the literature but it is exactly this). The heat equation of a thermoelectric material has the form $$\nabla \cdot(\kappa \nabla T)+q_\text{Joule}+q_\text{thermoelectric source terms}=0$$. In other words, it is the divergence of the usual heat flux $-\kappa \nabla T$ that's worth the total heat sources. It's not the divergence of the total heat flux that's worth the sum of all heat sources. Why?! I do not understand this.

If you were to compute the divergence of the total heat flux, you'd get some terms that "aren't real" in the sense that they correspond to no measurable heat sources. These terms do not appear in the heat equation, they are not manifest in any experiment, as far as I know. Therefore I do not understand well what the divergence of a field means. What would mean the divergence of the total heat flux in the case of thermoelectricity, for instance?

 

[renormalize]

$\vec J_Q = -\kappa \nabla T +ST\vec J$

Can you clarify your notation? Is the un-subscripted vector $\vec J$ meant to be the electric current-density within the conductive thermoelectric material $\vec J_E=\sigma\vec E=-\sigma\nabla V$ or is it something different?

 

[fluidistic]

Can you clarify your notation? Is the un-subscripted vector $\vec J$ meant to be the electric current-density within the conductive thermoelectric material $\vec J_E=\sigma\vec E=-\sigma\nabla V$ or is it something different?

Sure. In the case of thermoelectricity, $\vec J = -\sigma \nabla V -\sigma S \nabla T$ where $S$ is a scalar (to keep things simple. Of course it could be more complex, and be a tensor with anisotropy involved but this doesn't change my question) and $\sigma$ is the electrical conductivity, also a scalar to keep things simple.
One therefore gets:

$$\nabla \cdot \vec J_Q = \nabla \cdot(-\kappa \nabla T)+T\vec J\nabla S + \underbrace{S\vec J\cdot \nabla T}_\text{???}$$.

And $$\nabla \cdot (V\vec J) = -\vec J^2 /\sigma - \underbrace{S\vec J \cdot \nabla T}_\text{???}$$.

I computed this term because it is present in the heat equation (which should be better renamed as "internal energy equation" IMO). Indeed, the (steady-state) heat equation is nothing but $\nabla \cdot \vec J_U =0$ where $\vec J_U=\vec J_Q +V\vec J$, where $\vec J_U$ is the flux of internal energy (U stands for internal energy as used in Callen's textbook on thermodynamics for instance, and many, many others.).

One gets $$\nabla \cdot \vec J_U = \nabla \cdot (\kappa \nabla T)+\underbrace{\vec J^2/\sigma}_{q_\text{Joule}} \underbrace{- T\vec J \nabla S}_{q_\text{thermoelectric source terms}}=0$$. This is the heat equation of an isotropic thermoelectric material. There is no trace of the (???) term: $S\vec J \cdot \nabla T$, which is a heat source term that is not manifest. The heat equation dictates how the temperature behaves and this term is not present. However it is present as a heat source term when one computes the divergence of the total heat flux. But this term is canceled out by part of the divergence of the $V\vec J$ term.

Therefore, one gets that the divergence of the total heat flux is not equal to the sum of all heat sources that appear in the heat equation. How are we supposed to understand this?

 

[fluidistic]

Ok, so after having thought more about it, I think I partially understand what's going on.
In short, the heat equation should really, really be renamed "internal energy equation".
The divergence of the total heat flux should indeed be equal to the sources of all heats. Therefore, I believe that $\nabla \cdot(-\kappa \nabla T)+T\vec J\nabla S + \underbrace{S\vec J\cdot \nabla T}_\text{???}$ accounts for all heats, even though the last term does not appear in the heat equation.

The steady-state heat equation should give the sum of all energy sources, and it turns out that part of the energy coming from the total heat flux is exactly canceled out by an energy flux that isn't thermal. That is, it comes from the $V\vec J$ part and is due to the fact that electrical charges moves and change their energy by doing so, in part proportionally to $\nabla T$ due to thermoelectricity. As a result, the heat equation does not contain all sources of heats. It contains all sources of internal energy, which includes all sources of heats plus other sources of energy.

And this reconciliate with the usual knowledge of the relation between the divergence of a field and what creates it, at least to me. Everthing makes sense.

I would be curious to understand or to know the exact origin of the (???) term though, and whether it is possible to measure it experimentally. That's the last part I feel I haven't completely understood out of this question.