# Analogies between electrostatics and steady state heat equation?

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?

• My guess, and I would hope you get s proper answer, is that because temperature is related to entropy, and therefore probabilistic, that even in an ideal (never actually exists) gas model, you cannot expect anything like the precision, either in theory or practice, of the divergence involved on electrostatics – user171879 Oct 14 '17 at 19:18
• There is some analogy, but the big difference is that 1.) Temperature is only positive unlike the E field which is positive or negative, and 2.) Steady state implies that f=0, not that f=constant – KF Gauss Oct 14 '17 at 20:32

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