Does the heat equation violate causality?

Question

I've ran across the idea that, besides simply writing partial differential equations in covariant form, they need to be hyperbolic with all characteristic speeds less than the speed of light. A straightforward generalization of the equations for a dissipative fluid to the relativistic case supposedly runs into trouble because of the presence of the heat equation:

$$\frac{\partial T}{\partial t} = \kappa \nabla^2 T.$$

In the actual relativistic theory this is generalized to something covariant like

$$u^\mu\nabla_\mu T = \kappa(g^{\mu\nu}-u^\mu u^\nu)\nabla_\mu\nabla_\nu T, $$

where $ u $ is a timelike vector (this is only schematic, there are other terms). But the point is there is still a problem with this theory because this is a parabolic equation.

I'm wondering is there a way to see something clearly pathological like superluminal signals in the heat equation? This is a little unclear to me since the equation is not wave-like. If suppose you can't send signals faster than light what would be the problem with non-hyperbolic equations?

Answer 1

What follows is certainly not a comprehensive answer addressing all of your concerns. It is an answer to the question

is there a way to see something clearly pathological like superluminal signals in the heat equation?

I would argue that yes, there is.

The general solution to the initial value problem $T(x,0) = T_0(x)$ for the heat equation on the real line is \begin{align} T(x,t) = \int_{-\infty}^{\infty}dx'\, T_0(x')G(x-x', t). \end{align} where $G$ is the Green's function for the heat equation and is explicitly given by \begin{align} G(x,t) = \frac{e^{-x^2/(4\kappa t)}}{\sqrt{\pi (4\kappa t)}}. \end{align} Notice that if $T_0(x) = \delta(x)$, namely if one were to provide a localized unit impulse of heat at the origin at $t=0$, then for all later times, the temperature distribution would equal the Green's function (which is why it is often called the impulse-response and/or source function): \begin{align} T(x,t) = \frac{e^{-x^2/(4\kappa t)}}{\sqrt{\pi (4\kappa t)}}, \qquad t>0. \end{align} But this is a Gaussian distribution over the real line which is nonzero everywhere for all $t>0$. In other words, if you heat up a point on the real line at time $t=0$, then for any $t>0$, no matter how small, the entire line will have some nonzero temperature, even though it started at zero temperature everywhere except at the origin.

This behavior allows for superluminal signaling. To see how this is so, notice that if you have a long rod from here to Proxima Centauri made of a material that precisely obeys the heat equation, and if you want to warn your ally stationed near Proxima Centauri of an impending alien attack, you just need to keep the rod cold until the moment you hear intel of an attack. At this moment, you can simply heat up the portion of the rod next to you, and she will instantly measure the portion of the rod next to her as being hotter. She can then immediately begin to prepare to defend her station.