# General relativistic heat conduction equation

I read this wikipedia article https://en.wikipedia.org/wiki/Relativistic_heat_conduction It mentions the relativistic heat conduction equation: $$\frac{\partial \theta}{\partial t}=\Box\, \theta$$ Where $$\theta$$ is the Temperature and $$\Box$$ is the D'Alembert Operator How can one generalize this to situations in GR? As far as I know, it is clear that the D'Alembert Operator is invariant under Lorentz transformations and so one just has to replace the derivatives with the covariant derivatives. However what happens to the term on the LHS? Do I just replace it with the covariant derivative as well?

• checked wikipedia? Commented Jul 28, 2019 at 23:54
• Possible duplicate of Does the heat equation violate causality? Commented Jul 28, 2019 at 23:55
• It's not a duplicate, OP is wondering how to make the time derivative covariant Commented Jul 29, 2019 at 0:19

The problem you are running into is that time and space appear in fundamentally different ways in the equation. To represent this you need to incorporate some kind of four-vector $$u^\mu$$ to keep track of the rest frame. The heat equation naturally appears in relativistic descriptions of fluids, and this four-vector $$u^\mu$$ will have a natural interpretation as the velocity of the fluid. (There is a subtlety that the energy and charge in the fluid might flow with slightly different velocities, but let's ignore that and just choose a $$u$$.)
Then if you are in the rest frame of the fluid at a given point, a time derivative can be written as $$\partial_0 \rightarrow u^\mu \nabla_\mu$$
To get a space derivative you need to project out the time component $$\partial_i\rightarrow (\delta^{\mu}_{\nu}-u^\mu u_\nu)\nabla_\mu.$$ In the rest frame, this vanishes if the index $$\nu=0$$ (I am using +--- signature). You can contract this projected covariant derivative with itself to make a covariant version of the spatial Laplacian.