What is the velocity $u^mu$ in the stress-energy-tensor of a perfect fluid?

Question

I am currently learning about fluid dynamics in special relativity. We defined the stress-energy-tensor of a perfect fluid to be
begin{equation}
T^{mu nu} = (rho + P) u^mu u^nu + P g^{mu nu}.
end{equation}
We said that $u^mu$ is the 4-velocity of the "MCRF" of the fluid, that is the frame in which the bulk velocity is zero, i.e.
begin{equation} 
sum_a u_a^text{MCRF} = 0.
end{equation}
Where the summation goes over all particles.
It is however possible to derive the Euler-equation (in the non-relativistic limit) from the conservation of energy-momentum $T^{munu}_{quad,nu}= 0$,
begin{equation} 
rho (partial_t mathbf{v} + (mathbf{v} cdot nabla) mathbf{v})) = -  nabla P.
end{equation}
I have trouble imagining that the bulk velocity should depend on space-time coordinates. I guess you could in principle define the bulk velocity to be zero at spaces other than the fluid and $u^mu$ at the point (or volume) of the fluid. But then the Euler equation seems to be strange. The Euler equation usually treats the velocity as a vector field of the fluid and gives insight into the internal dynamics of the fluid. How can that happen with the bulk velocity which contains no information about the internal flows?
Could someone explain to me how I can interpret $u$ or $mathbf{v}$ as a vector field in the sense of how the Euler equation treats them?

mike stone · Answer

The velocity four-vector $u= gamma(c, {bf v})$ in the energy momentum tensor is a vector field just like the ${bf v}$ in the Euler equation. Indeed the space parts of the relativistic energy-momentum conservation equation
$$
nabla_mu T^{munu}=0
$$
reduce to the momentum conservation form
$$
partial_t(rho v^i)+partial_l(rho v^i v^j+delta^{ij}P)=0
$$ of the Euler equation
when $|{bf v}|/c$ is small.
Of course, just as in the Euler equations, the velocity  ${bf v}$ is the smoothed average of the velocity of the individial particles over  regions of the size of the mean-free-path. I do not know what the acronym MCRF means, or what "all particles" mean, but if one moves at the local value of $u$ then, on average, there is no local streaming of particles past you.
There  are subtleties in the relativistic definition of $u$ however.  In a relativistic system the number of particles is not necessarily conserved, but instead some quantity like Baryon number is conserved. So one can define $u$ to be the position-depenedent velocity of the local  Eckart frame in which there is no net flux of Baryon number. Alternativey one can define it to be the velcoity of the local Landau frame in which there is no net energy flux, or even a (nameless?) frame in which there is  no net entropy flux. These "$u$"s are generally different. The simple formula  you cite for $T^{munu}$ is ignoring all this.
Here is a contempory  article discussing these issues.

Javier · Answer

The MCRF is not a single frame: there is a different frame at each spacetime point. The summation for the bulk velocity doesn't run over all particles: it runs over a small region surrounding the chosen point, much smaller than the typical distances we will be considering in our study of the fluid, but much larger than the distance between molecules or the mean free path or whatever. It's a local bulk velocity.
There's nothing relativistic about this: it's exactly the same definition of fluid velocity as the one we use in classical fluid dynamics.

What is the velocity $u^mu$ in the stress-energy-tensor of a perfect fluid?

2 Answers

Add your own answers!

Ask a Question