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Is the concept of incompressible fluid valid in special theory of relativity? Can anybody answer this question without going into speed of sound and fluid dynamics ?

atom
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  • If it was invalid, there would be a lot of simulations of, e.g. gamma ray burst jets, that would be wrong. – Kyle Kanos Feb 14 '17 at 12:22
  • @KyleKanos Not sure what you mean. A GRB involves a relativistic QED plasma, with EOS $\rho\sim T^4$ and $p\sim T^4$ as well as $c_s^2\sim c^2/3$. – Thomas Feb 14 '17 at 19:52
  • @Thomas: If you're doing QED for a relativistic fireball, you're probably looking at something entirely different than what I'm talking about. – Kyle Kanos Feb 15 '17 at 02:35
  • Your edit doesn't make sense. How can somebody answer a question about whether an incompressible fluid exists without discussing A) fluid dynamics and B) the speed of sound? Incompressible flows occur because the dynamics of the flow are much slower than the speed of sound. So any discussion of incompressibility requires discussion of the speed of sound. – tpg2114 Feb 20 '17 at 07:33

2 Answers2

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In an incompressible fluid the density does not change in response to changes in the pressure. This means that the speed of sound is infinite, $$ c_s^2= c^2\left.\frac{\partial P}{\partial \rho}\right|_s=\infty . $$ Here, $P$ is the pressure and $\rho$ is is the energy density. In the non-relativistic limit $\rho=mnc^2$, where $m$ is the mass of the particles and $n$ is the particle density. This is obviously incompatible with relativity, disturbances in the fluid propagate faster than the speed of light.

Of course, non-relativistic fluids are not truly incompressible either, but the approximation is useful if the fluid velocity is much smaller than the speed of sound, $u\ll c_s$. In a relativistic fluid in which $u$ is comparable to $c$ this cannot be true.

Postscript: Note that in a non-relativistic fluid incompressibility means that $n=const$. Then $\rho=mnc^2$ is also constant. In a relativistic fluid we could either mean $\rho=const$, or $n=const$. Note that $\rho=const$ is the more natural generalization, and it is incompatible with $c_s^2<c^2$ as explained above. Constant particle density also not allowed, because $\partial P/\partial n \sim \chi/n$, where $\chi$ is the susceptibility,

Thomas
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  • Where does the sound or sound speed come in the picture? – atom Feb 14 '17 at 04:02
  • @atom Added a brief comment. – Thomas Feb 14 '17 at 05:29
  • Others, including myself, have said in other answers on this site that $p=\rho c^2$, with $c$ the speed of light, is used in relativistic fluids. – Kyle Kanos Feb 14 '17 at 12:20
  • See for instance this post & its answers – Kyle Kanos Feb 14 '17 at 13:06
  • There are no systems in nature (that we have observed) in which the speed of sound reaches the speed of light, so this is a rather extreme equation of state. For nearly ideal relativistic plasmas $c_s^2=c^2/3$. In the center of a neutron star, the speed of sound probably exceeds $c/\sqrt{3}$, as has been dicussed on this site (but not $c_s\sim c)$. – Thomas Feb 14 '17 at 19:46
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    The comments on one of my questions (http://physics.stackexchange.com/questions/213119/how-are-the-turbulent-spectra-determined-in-relativistic-turbulence) suggest that relativistic fluids are not just high speed fluids, but it is any fluid in which rest mass is not conserved. The particle-antiparticle fluid is given as an example. In which case, the flow would be both low Mach and relativistic. – tpg2114 Feb 15 '17 at 04:06
  • But an ultra- relativistic fluid like that is always compressible. In the extreme case (massless particles), the density is proportional to $T^3$. – Thomas Feb 16 '17 at 02:16
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The relativistic fluids I am familiar with are of the astrophysical context (e.g., relativistic fireballs), and in these cases one uses (or can use) the relativistic Euler equations: $$ \nabla_\mu T^{\mu\nu}=0 $$ where $T^{\mu\nu}$ is the stress-energy tensor. These still lead to conservation equations such as, $$ \frac{\partial \rho}{\partial t}+\frac{\partial}{\partial x_i}\rho u^i=0 $$ (cf. this paper by Zhang & Macfayden). The Euler equations can be used for either compressible and incompressible fluids alike (though most situations I've seen are compressible).

This paper by Moritz Reintjes (2016) explicitly derives incompressibility as an option for relativistic fluids, so I would think the answer to OPs question is a clear yes, incompressible fluids exist in relativity.

Kyle Kanos
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  • Nice to see you back and answering things -- I always learn something new. And there seems to be a whole bunch of fluids questions lately, good to have some help! – tpg2114 Feb 15 '17 at 04:07
  • @tpg2114: Yeah, the latest addition to the family has given me some extra time at night to answering. I have noticed a large influx of fluid questions, so glad I can answer some of them. – Kyle Kanos Feb 15 '17 at 11:07
  • I'm also really trying to help out with what I can in the [tag:computational-physics] tag, but it seems to get little love from the community (i.e., low vote count for Q&A) – Kyle Kanos Feb 15 '17 at 11:10
  • I'm puzzled by this paper. If both the density and the energy density are constant, then (by thermodynamic identities) so is the pressure. Then there are no non-trivial solutions left. – Thomas Feb 16 '17 at 02:13