The stress-energy tensor for a perfect fluid is given by $$T_{\mu\nu}=(\rho+p)U_{\mu}U_{\nu}-pg_{\mu\nu}$$ where U is the 4-velocity. The matrix components of the SEM are written as $$T_{\mu\nu}=diag(\rho,p,p,p).$$ Will this always be the components of the stress energy tensor despite having different components for the metric in each case or do they change according to the formula?

  • $\begingroup$ Hello! I do not understand the second equation: do not mix index notation and matrix one, it can lead to misleading ideas. For example: T00 is rho, ok.. but in which frame? What you wrote is valid only in the local rest frame of the local fluid element. $\endgroup$
    – Quillo
    Feb 14, 2022 at 16:39
  • $\begingroup$ The first equation is valid in any frame correct? or is it just the MCRF $\endgroup$
    – aygx
    Feb 14, 2022 at 17:33
  • $\begingroup$ You cannot identify any arbitrary stress energy tensor with stress energy tensor for an ideal fluid. For instance, you cannot explain the inhomogeneities in general fluids (like shear, vorticity, divergence)using only perfect fluid stress energy tensor. Check this article : arxiv.org/abs/gr-qc/9812046v5 $\endgroup$
    – KP99
    Feb 14, 2022 at 18:01
  • $\begingroup$ @KP99 Thank you for the article link . It cleared up some misconceptions I had regarding the SEM tensor. Is it true that the perfect fluid definition only holds in the MCRF, or can there still be a perfect fluid where it has different 4-velocity components? $\endgroup$
    – aygx
    Feb 14, 2022 at 18:18
  • 1
    $\begingroup$ It's a tensor like you said yourself. The components transform as those of a tensor. $\endgroup$ Feb 14, 2022 at 22:38

1 Answer 1


Consider your stress-energy tensor: $$ T_{\mu\nu}=(\rho+p)U_{\mu}U_{\nu}-p g_{\mu\nu} $$ This is a legit tensor written in terms of other tensors (the scalar fields $\rho$ and $p$, the metric and the velocity field): the RHS is the constitutive relation for $T_{\mu\nu}$ in terms of your macroscopic fields ($U_{\mu}$, $\rho$...). You can use more complex constitutive relations (possibly involving more macroscopic fields) for dissipative fluids, see e.g. this and this questions and links here.

Note that $\rho$ and $p$ are scalars by construction, because are the physical properties of the matter in the local reference frame comoving with the local matter element (more precisely: this is the rest frame defined by the conserved current in your system, whatever it is, typically the baryon current).

If you want the energy in the local comoving reference frame, you just have to calculate $T_{\mu\nu} U^\mu U^\nu$.

This representation of $T_{\mu\nu}$ is tensorial, meaning that it's always valid (for any spacetime and fluid configuration). The second representation that you're proposing is valid only in the reference frame of the fluid (aka, a frame that is instantaneously comoving with a local fluid element). Locally, in this frame, the velocity is represented by $U^\mu \sim (1,0,0,0)$. Therefore, $T^{00}= \rho$, and so on... (you find the diagonal matrix representation). Again, this is valid for every metric, but in a specific "tetrad" of the fluid element.

See also: Wiki article, Stress-energy tensor for a perfect fluid in GR, Stress-energy tensor. Why this general form?, What is a "perfect fluid", really?.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.