# Matter Lagrangian for perfect fluid

1. Dec 6, 2011

### ramparts

The stress-energy tensor is usually defined in standard GR treatments as

$$T_{\mu\nu} = -\frac{2}{\sqrt{-g}}\frac{\delta(\sqrt{g}L_m)}{\delta g^{\mu\nu}})$$

with the Lm the matter Lagrangian.

I'm curious what Lm is for a perfect fluid with density ρ and pressure P that would lead to the standard stress-energy tensor

$$T_{\mu\nu} = (\rho+P)u^\mu u^\nu + Pg_{\mu\nu}$$

in an FRW metric.

2. Dec 6, 2011

### Staff: Mentor

This paper on arxiv seems to have the basics of what you're looking for:

http://arxiv.org/abs/gr-qc/9304026

Google finds a number of other papers on the subject but they're all behind paywalls.

3. Dec 6, 2011

### ramparts

Paywall is fine by me, I'm at a university. I'll take a look through that paper, thanks - most of the actions have more information than I need (since it should just be a function of density and pressure) but I'll read through in more depth soon. I just had a go at doing something very simple which seems to work to me, which is to write the ansatz

$$L_m = (a \rho + bP) g^{\mu \nu} u_\mu u_\nu + (c\rho + dP) g^{\mu\nu}g_{\mu\nu}$$

so you can differentiate Lm with respect to the inverse metric and compare to the usual perfect fluid stress-energy tensor, using the fact that the definition of the stress-energy tensor is equivalent to

$$T_{\mu\nu} = -2\frac{\delta L_m}{\delta g^{\mu\nu}} + L_m g_{\mu \nu}.$$

Doing this gives a=b=-1/2, c=-d=-1/4 so that you have

$$L_m = \frac{1}{2} (-\rho + 3P)$$

which is nice, since it's just 1/2 the trace of the stress-energy tensor.

If there's some gross problem in doing this hopefully someone will point it out!

4. Dec 6, 2011

### dextercioby

Section 25 and 27 of the 69-page book on GR by the great Dirac has an answer to your question. You may also want to check L&L books (Fields, vol. 2, or Fluid Mechanics) or the old book by Tolman.

5. Dec 6, 2011

### ramparts

Thanks. It looks like there's a copy in my department's library which I'll look at, but for the benefit of people reading who don't have that, is there an executive summary you could give us?

6. Dec 8, 2011

### ramparts

So I've checked a couple of books and the paper PeterDonis posted (thanks!) and it seems the perfect fluid Lagrangian is actually just -ρ. Frankly that makes sense because the thing I thought it was vanishes for a radiation fluid, which is clearly wrong. L=-ρ fits the usual form for the electromagnetic Lagrangian and also a scalar Lagrangian so it makes sense to me.

The way I worked it out about must have been somewhat sloppy!