Matter Lagrangian for perfect fluid

[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.

[PeterDonis]

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. :frown:

[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!

[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.

[ramparts]

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.

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?

[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!