Perfect and viscous fluids in general relativity

Question

After reading a lot of papers on beyond standard cosmology recently, I'm becoming confused and a bit nervous about my own knowledge on physics! By definition, a perfect fluid is described by an energy-momentum tensor that has the general following form (I'm using the $\eta = \mathrm{diag}(1, -1, -1, -1)$ convention, and $c = 1$): $$\tag{1} T_{ab} = (\rho + p) \, u_a \, u_b - p \, \eta_{ab}. $$ In principle, the isotropic pressure $p$ could be any non-linear function of the energy density $\rho$ (the perfect fluid may be self-interacting). A scalar field is a good example of this.

But then, fluids with bulk viscosity $\zeta$ (no shear viscosity, no heath flux) have the same energy-momentum expression as (1), but add a new term to the isotropic pressure: $$\tag{2} p \quad \Rightarrow \quad \tilde{p} = p - \zeta \Theta, $$ where $\Theta = \nabla_{\mu} u^{\mu}$ is the expansion scalar. In isotropic/homogeneous FLRW cosmology, we have $\Theta = 3 \dot{a} / a$.

Fluids with bulk viscosity are generally described as non-perfect, since they generate entropy, while perfect fluid have constant entropy. But then, this is where I'm getting confused! What is a "perfect fluid" in general relativity? Is it simply a fluid that has (1) as energy-momentum tensor (so this would include all the fluids with bulk viscosity as "perfect fluids"), or do we have to add a constraint to (1)? If so, how can we translate that constraint, mathematically, to get a "perfect fluid"?

And is it true that all perfect fluids have constant entropy?

This question is related (but not the same) as this old one from mine:

What is a "perfect fluid", really?

Answer 1

GR books that I have read have always taken 'perfect/ideal fluid' as a synonymous with eq (1) for $T_{ab}$ with no further comment. You are right to point out that this does not correspond to the usage in fluid mechanics, so probably the GR community should be persuaded to care about this and together with the rest of fluid mechanics come up with an agreed common terminology.

Of course the ideal gas is something else again, since it has a given equation of state. But this is relevant here because the ideal gas is not a perfect fluid! I mean by this that an ideal gas, in the thermodynamic sense, can be viscous, and it usually is, since we invite students to derive an expression for its viscosity using kinetic theory.

You might like to note that the thermodynamic definition of ideal gas is just Boyle's law and Joule's law. That is:

(1) pressure inversely proportional to volume at fixed temperature

(2) internal energy a function of temperature alone.

The equation of state ($p = n k T$) can be derived from these conditions, but not the heat capacity and indeed the heat capacity need not be independent of temperature. Therefore the adiabatic index can also be temperature-dependent (as it is for most real gases.) When an ideal diatomic gas is heated through the rotational and vibrational temperatures, and on up to relativistic behaviour, the equation of state cares not a jot and carries on being $p = n k T$ all the way.

Finally, I don't think a way out of this terminology quagmire is to define 'perfect fluid' as different from 'ideal fluid'. I think that policy would only maintain the confusion.

Answer 2

The coefficients $\rho$ and $p$ are not allowed to be arbitrary functions in hydrodynamics. They must depend on $x$ only through a slowly varrying local temperature $T(x)$ and velocity $u_\mu(x)$. In dimensionless terms, their derivatives will be small when multiplied by the mean free path. This makes the notion of a derivative expansion well defined.

It is the zeroth order term in this expansion that corresponds to the perfect fluid. The viscosity term $\nabla_\mu u^\mu$ that you wrote is indeed a correction to the perfect fluid because it comes in at first order. In fact, (1) is not general enough at first order because then we can also have terms in the energy momentum tensor like $\nabla_{(\mu} T_{\nu)}$. The best source on this I can recommend is the set of lectures https://arxiv.org/abs/1205.5040 by Pavel Kovtun.

Answer 3

There may be some variations between authors, but the definition of a perfect fluid in GR that I know most readily is that it is a fluid that has no heat conduction and no viscosity in its momentarily comoving reference frame. This is a generalization of an ideal gas from thermodynamics and you can derive the stress-energy tensor that you showed from there. (Schutz, for example, does this in Chapter 4.) No heat conduction gets $T^{0i} = 0$ and no viscosity gets you that $T^{ij}$ is proportional to $\delta^{ij}$.

I don’t recall ever seeing discussion of bulk viscosity, so that part is curious. I looked it up, and I see what you're saying about it "modifying" the pressure formally. It still has the viscosity term in there though, so I don't understand that to be a perfect fluid anymore, appearances notwithstanding.

A key feature of the perfect fluid, is, as you said, that entropy is conserved. This falls out as a consequence of the conservation law $T^{\alpha\beta}_{\ ;\beta} = 0$ for the perfect fluid without the bulk viscosity. I'm not sure that it would follow with the $\nabla_\mu u^\mu$ terms in there, but I didn't check.