# Relativistic Euler equation in spherical coordinates

1. Nov 18, 2015

### Geofleur

I just wanted to check that I am thinking about the coordinate transition correctly. The relativistic generalization of Euler's equation is (from Landau & Lifshitz vol. 6)

$hu^\nu \frac{\partial u_\mu}{\partial x^\nu} - \frac{\partial P}{\partial x^\mu} + u_\mu u^\nu \frac{\partial P}{\partial x^\nu} = 0$.

where $h$ is the enthalpy, $P$ is the fluid pressure, and $u^\mu$ are components of the four-velocity. The Greek indices range from 0 to 4. To get this same equation in spherical coordinates (or any new, barred coordinates), I can just bar all of the indices and replace the derivative of $u_{\overline{\mu}}$ with a covariant derivative, yes? In other words, can I simply write

$hu^{\overline{\nu}} u_{\overline{\mu}; \overline{\nu}} - \frac{\partial P}{\partial x^{\overline{\mu}}} + u_{\overline{\mu}} u^{\overline{\nu}} \frac{\partial P}{\partial x^{\overline{\nu}}} = 0$,

where the semicolon represents covariant differentiation?

The reason I am a little worried is that, to get to the relativistic Euler equation in the first place, the four-divergence of the energy-momentum tensor was used. I am wondering whether I have to go back and take the covariant four-divergence of the energy-momentum tensor expressed in barred coordinates.

Last edited: Nov 18, 2015
2. Nov 18, 2015

### PWiz

A 5 dimensional manifold?

3. Nov 18, 2015

### Geofleur

Oops, I meant 0 to 3 (I keep doing that!).

4. Nov 18, 2015

### PWiz

I don't think there's a problem with what you've done. An easy way to check is to use the unbarred indices for spherical coordinates/any non-Cartersian coordinates (i.e., write the 2nd equation as it is but use unbarred indices), then transform your 2nd equation to Cartesian coordinates (the barred indices). The derivatives of the fluid pressure transform like a tensor (since the derivative of a scalar field is a tensor) and so do the four velocities, and so you can just replace all their indices to barred ones. The covariant derivative will also transform like a tensor, but since the Christoffel symbols vanish in Cartesian coordinates, it just reduces to the regular partial derivative. You then get the first equation (but with barred indices, although that doesn't really matter). I don't think you need to bother with where the original equation comes from to determine it's form in different coordinates. (I mean you just need the tensor transformation law to see how the tensor equation changes under diffeomorphisms)

Last edited: Nov 18, 2015