# How do you determine if a stress-energy tensor describes a fluid?

I understand that if a stress-energy tensor can be written in a frame as $$T_{\mu\nu}=\text{diag}(\rho,p,p,p)$$ then it describes a perfect fluid, but if you are given a stress-energy tensor in an arbitrary coordinate system, how do you determine if the tensor describes a perfect fluid?

• Possibly related: Stress energy tensor of perfect fluid in general frame
– Amit
Jun 18, 2023 at 23:56
• $\partial T_{\mu}^{\nu}/\partial x^{\nu}=0\;\;$ ,see :archive.org/details/… " ....The energy-momentum tensor (133.2) does not take account of any dissipative processes (including viscosity and thermal conduction), and therefore the equations relate to an ideal fluid" Jun 21, 2023 at 15:41
• Jun 27, 2023 at 17:10

… if you are given a stress-energy tensor in an arbitrary coordinate system, how do you determine if the tensor describes a perfect fluid?

One way to achieve this is to consider eigenvalue/eigenvector problem at a given point: $$(T^a{}_b-\lambda \delta^a{}_{b})V^b=0.$$

If there is a timelike eigenvector with one eigenvalue ($$\lambda_1=-\rho$$, in “mostly plus” signature) and three linearly independent spacelike eigenvectors with another common eigenvalue ($$\lambda_2=p_1=p_2=p_3=p$$) then such stress-energy tensor can belong to a perfect fluid and the timelike eigenvector ($$u^a$$, after normalization) would be the 4-velocity of its rest frame. The orthonormal set of eigenvectors at a given point will form a basis for coordinate system where the stress-energy tensor would assume canonical diagonal form. The problem of finding eigenvectors/eigenvalues is usually covered in linear algebra courses and there are many tools for it (algorithms, numerical methods, software etc.).

Another approach to the same problem is to evaluate certain Rainich-type conditions, a set of tensor equations written in terms of stress-energy tensor and the metric. And since stress-energy tensor is expressible via Einstein equations through the curvature tensor, such conditions could be seen as a purely geometric criterion for a metric (taken not just at a point but in some region) to constitute a perfect fluid spacetime.

(The original Rainich conditions provided geometric criteria for the metric to define an electrovacuum spacetime).

The paper

• Krongos, D. S., and C. G. Torre. Geometrization conditions for perfect fluids, scalar fields, and electromagnetic fields. Journal of Mathematical Physics 56.7 (2015): 072503, doi:10.1063/1.4926952, arXiv:1503.06311.

provides the following criteria for a spacetime to have a perfect fluid stress-energy tensor (specialized here for $$d=4$$).

Define trace-free stress-energy tensor (alternatively, one can instead use trace-free Ricci tensor) $$S_{ab}=T_{ab}-\frac14 T^c{}_c\,g_{ab}$$

and a scalar

$$α = − \left[\frac83 S^b_ aS^c_ bS^a_ c\right]^{1/3} .$$

The metric $$g$$ defines a perfect fluid spacetime if and only if the following conditions are met $$\begin{eqnarray}α &\ne &0 \\ K_{a[b}K_{c]d} &= &0, \\ K_{ab}v^av^b &> &0, \qquad \text{for some }v^a,\end{eqnarray}$$ where $$K_{ab} = \frac1 α S_{ab} −\frac 1 4 g_{ab}.$$

Those expressions may seem somewhat arcane, but basically those encode in the language of tensor expressions of the criteria for the necessary eigenvalue/eigenvector structure.

There is a great deal of overlap in the suggested approaches and which would be better depends on specific problem details and personal preferences.