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?

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    $\begingroup$ As far as I read about, it is pathological. That's why people don't use parabolic equations for relativistic scenarios, they tend to use corrected hyperbolic equations, in the case of heat equation that would be something like maxwell cantaneo equation. I can leave some links later if you want $\endgroup$
    – Hydro Guy
    Nov 19, 2014 at 4:26
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    $\begingroup$ Matteo Smerlak discussed this back in 2012. $\endgroup$
    – Kyle Kanos
    Nov 19, 2014 at 4:29
  • $\begingroup$ Thank you both for the references. Yes this question was in the context of the Eckart theory (and Landau-Lifschitz) discussed in that paper. $\endgroup$
    – octonion
    Nov 19, 2014 at 4:37
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    $\begingroup$ @octonion, here a reference that I was studying arxiv.org/abs/0902.3663 $\endgroup$
    – Hydro Guy
    Nov 19, 2014 at 12:58

1 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.

  • $\begingroup$ Does your example work? I would think getting the necesarry accuracy would run into problems with the uncertaincy principle. $\endgroup$
    – Taemyr
    Nov 19, 2014 at 13:53
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    $\begingroup$ @Taemyr this is a thought experiment needed to demonstrate the principle, not a recipe for an experiment. $\endgroup$
    – Ruslan
    Nov 19, 2014 at 13:54
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    $\begingroup$ Am I missing something, or does this answer only consider the classical heat equation, not the covariant one OP wants? $\endgroup$
    – Kyle Kanos
    Nov 19, 2014 at 15:39
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    $\begingroup$ @KyleKanos In the question (v3), the OP writes "...because of the presence of the heat equation: $$T_t = \kappa\nabla^2T.$$ Later, in the question I quote in my response, the OP uses the term "heat equation" again without emphasizing that it's not the one he explicitly refers to earlier. Further, based on his other language, it also seemed to me that the OP was asking somewhat generally about non-hyperbolic equations, using the equations he wrote down as examples. Finally, I included a disclaimer specifically since I knew the OP was at least asking more than the question I was answering. $\endgroup$ Nov 19, 2014 at 16:47
  • $\begingroup$ Okay, just making sure that it's not me. $\endgroup$
    – Kyle Kanos
    Nov 19, 2014 at 17:08

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