# How are the turbulent spectra determined in relativistic turbulence?

In a non-relativistic compressible fluid, the turbulent energy spectra are well-understood and appear to follow the Kolmogorov hypothesis. It would also appear that relativistic turbulence also follows the expected -5/3 slope in the inertial range of the spectra.

However, one point (of many) that I am unsure about, is how one could actually compute either a frequency spectrum or a wavenumber spectrum in a relativistic turbulent flow. What does a frequency or a wavenumber mean when the Lorentz factor leads to time dilation and length contraction?

Do frequencies appear smaller (time dilated) depending on the velocity of a particular eddy? Do wavenumbers appear larger depending on the velocity? It would seem that if there are "eddies" with velocity $u_1 \ll c$ and a wavenumber $\kappa_1$ and other "eddies" with velocity $u_2$ where $u_1 \ll u_2 < c$ with a wavenumber $\kappa_2$, it is possible that $\kappa_1 = \kappa_2$ with the Lorentz factor but $\kappa_1 \neq \kappa_2$ without it.

It looks like it must not change because studies show that turbulence still follows the $k^{-5/3}$ slope in the inertial range and if the Lorentz factor lead to shifts in the frequency or wave number, it seems extremely unlikely that the slope would match traditional turbulence. But on the other hand, relativistic speeds change both time and lengths, so I don't know how that doesn't influence the frequency or wavenumber depending on the velocity.

If I had my giant hotwire probe in space as some relativistic grid turbulence passed by, and my probe measures the velocity every 1 second, what am I really measuring in the fluid when the fluid has a different "clock"? And how do I use that signal to compute a frequency spectrum?

• Should the fourth word be incompressible instead of compressible? Do you have a reference for the compressible case? Note also relativistic may mean two different things. It may mean the velocities are close to $c$. It may also mean that there is no conserved mass -- think a gas of particles and antiparticles -- but fluid velocities can remain small. Commented Oct 18, 2015 at 3:57
• @user2309840 No, it does mean compressible (although, incompressible also works). The spectra in compressible turbulence is also well understood, although several mechanisms are not quite as well understood as incompressible turbulence. I could dig out references about it if you would like. I also was unaware of the different meanings of relativistic and was thinking of the case where velocities (maybe not all velocities though) are close to $c$. Commented Oct 18, 2015 at 15:09
• fwiw My personal terminology has "relativistic" and "no conservation of rest mass" as orthogonal properties. But there is yet another meaning of "relativistic": internal energy much greater than rest mass energy. That is, the random thermal velocities that lead to pressure are relativistic, even if the bulk motions aren't. A nice result is that any fluid that approaches relativistic in this thermal sense becomes characterized by a $\Gamma = 4/3$ adiabatic index, and that's pretty much all there is to them.
– user10851
Commented Oct 19, 2015 at 0:18
• I assume you are asking how to derive the wavenumber/frequency spectrum from a spectrum of flow and/or density fluctuations/variations, some of which may have speeds reaching significant fractions of c? Commented Jan 21, 2020 at 15:14
• @honeste_vivere Correct -- what does the spectrum look like and how do we compute it when the velocity fluctuations may approach large fractions of the speed of light locally. Commented Jan 21, 2020 at 15:23

## 1 Answer

An incomplete answer:

In the non-relativistic case, one often sees correlation functions that involve three velocities $v$. In the relativistic case, the velocities are typically all replaced with more Lorentz covariant objects, for example the stress tensor $T^{\mu\nu}$. (See for example this paper by Westernacher-Schneider et al.) While the stress tensor and three velocities of course transform under the Lorentz group, the transformation rules for the stress tensor, since it's a tensor, are a bit a nicer.

Pace tpg2114, I've frankly always been a little disappointed by the numerical evidence I've seen for the 5/3 scaling law. Figure 3 (left) in the paper I cited above for example looks like a huge effort by a serious numerical relativist in the ideal case, and yet it's still noisy. Figure 2 in this very impressive paper also purportedly supports 5/3 scaling but now with some non-negligible viscosity, and yet it's not what I would call a smoking gun. There was even this paper which studies ideal, incompressible, non-relativistic hydro and involved an enormous simulation on graphics cards and yet produced what I thought at the end of the day was an underwhelming plot. It may just be extremely different to get enough separation of scales in a numerical simulation to see 5/3 over more than a decade in $k$. But the theorist in me wonders if maybe there is also more to the story. (I've also only given 2+1 dimensional examples above where there is actually more going on -- an enstrophy cascade. I'm not familiar with the numerical story in 3+1 dimensions.)

Anyway, the way these simulations are usually set up, I understand, is to pump in energy at one scale and pump it out at another. In the 2+1 dimensional case, one pumps in at small scale because of the inverse cascade and pumps out at large scale. In the 3+1 dimensional case, I assume one pumps in at large scale and out at small scale. Having done that, there is then automatically a preferred frame -- from having a box, from deciding how to pump energy in and out. One is presumably supposed to compute the correlation functions in that frame. Then from these correlation functions one can extract the 5/3 scaling.

Now if all the components of the stress-tensor two-point function satisfy the 5/3 scaling, it's possible one would see the scaling independent of frame because of the linearity of the transformation law for the stress-tensor under boosts. I don't know if anyone has looked at this in detail. I suspect the story is more complicated.