Analogies between electrostatics and steady state heat equation?

Question

In electrostatics we have

$$\nabla \cdot E = \rho/\varepsilon$$ and using the divergence theorem we get

$$\int_{\partial\Omega} E \cdot \hat{n} dS = \int_\Omega \rho/\varepsilon dV.$$

This states that the electric flux out of the domain $\Omega$ is equal to the total charge inside $\Omega$. I think of this as the total 'force' that can be felt (by a charge) pushing outwards at the boundary.

Can the same thought process by applied to the steady state heat equation (I have no experience with thermodynamics). We have

$$\nabla \cdot (\nabla T) = f$$ and using the divergence theorem we get

$$\int_{\partial\Omega} \nabla T \cdot \hat{n} dS = \int_\Omega f dV.$$

Is the temperature gradient completely analagous to the electric static field? Is it like a force pushing outwards? In electrostatics the flux is out of $\Omega$ is always due to the charge density $\rho$. Is $f$ some kind of density in thermostatics? Charge density can be thought of as a contiuum of charges, but what is $f$ a continuum of?

Answer 1

Even though there is a certain similarity between the two equations, the origins are quite different. Maxwell's equations can be derived from a gauge theory, while the heat equation is a special case of a diffusion phenomenon (and hence a special case of the more general diffusion equation). Also note that the more general expression of the heat equation is

$$\rho c_p\frac{\partial T}{\partial t} - \nabla\cdot(\kappa\nabla T) = \dot q$$

where $\rho$ is the mass density, $c_p$ is the specific heat capacity, $\dot q$ is the volumetric heat source, and $\kappa$ is, in general, a tensor. If we look for stationary solutions (i.e. $\partial_tT = 0$), for a homogeneous and isotropic medium (i.e. $\kappa$ is a space-wise constant scalar) then the equation becomes

$$\nabla^2 T = -\frac1\kappa\dot q.$$

The RHS can still be interpreted as a density of heat source in space, much like $\rho$ in the electrostatic equation is a density of electric charge (divided by $\varepsilon$). However, for the heat equation, we have a time derivative, $\dot q$, that alludes to a (heat) flux, which is not the case for the charge density $\rho$. Indeed, the electrostatic equation is not describing a diffusion phenomenon.