Analogies between electrostatics and steady state heat equation?

Physics Asked on January 4, 2022

In electrostatics we have

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

$$int_{partialOmega} E cdot hat{n} dS = int_Omega rho/varepsilon dV.$$

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

$$int_{partialOmega} nabla T cdot hat{n} dS = int_Omega f dV.$$

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?

One Answer

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

$$rho c_pfrac{partial T}{partial t} - nablacdot(kappanabla T) = dot q$$

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

$$nabla^2 T = -frac1kappadot q.$$

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.

Answered by Phoenix87 on January 4, 2022

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