For a homogeneous and the isotropic universe, the spacetime metric $ds^2$ is given by the FRW form in comoving coordinates: $$ds^2=dt^2-a^2(t)\Big[\frac{dr^2}{1-kr^2}+r^2(d\theta^2+\sin^2\theta d\phi^2)\Big].$$ It determines the LHS of the Einstein's field equation.

In Kolb and Turner's Cosmology book, it is said that to be compatible with the symmetries of the metric $ds^2$, the stress-energy tensor $T_{\mu\nu}$ should be diagonal and by isotropy, the spatial components must be equal.


  • How can I see that the symmetries of $g_{\mu\nu}$ must be present in that of $T_{\mu\nu}$? It is not obvious to me.

  • How do the symmetries dictate that the tensor $T_{\mu\nu}$ is diagonal? All I know is that if a Cartesian tensor $T_{ij}$ is invariant under rotation it will satisfy $$T^\prime_{ij}=R_{ik}T_{kl}(R^{-1})_{lj}=T_{ij}$$ where $R$ is the $3\times 3$ rotation matrix.

  • 2
    $\begingroup$ Recall that $T_{\mu \nu} \sim \frac{\delta S}{\delta g^{\mu \nu}}$ $\endgroup$
    – Avantgarde
    Mar 13, 2018 at 14:19
  • $\begingroup$ @Avantgarde Nice point. $\endgroup$
    – SRS
    Mar 13, 2018 at 20:08

1 Answer 1


By means of Einstein field equations $T_{\mu \nu}$ could be expressed through geometric quantities, namely the Einstein tensor, and so it must have the same symmetries as Einstein or Ricci tensor.

The FLRW metric possesses a large group of isometries which is one of $SO(4)$, $ISO(3)$ or $SO(3,1)$ for the values of $k=1,0,-1$. The orbits of those groups are slices $t=\rm const$. Under those symmetries $R_{00}$ (and hence $T_{00}$) must be a constant scalar, $R_{0i}$ (and $T_{0i}$) must be an invariant 3-vector field and $R_{ij}$ ($T_{ij}$) must be a invariant symmetric rank-2 3-tensor field.

There are no invariant 3-vector fields under isometries that include full $SO(3)$ rotations (since rotating a nonzero vector around an axis that is not parallel to it would change it), so $T_{0i}\equiv 0 $. And the invariant symmetric tensor $T_{ij}$ must be proportional to the 3-metric (if it's not, than at a given point there is (at least) one eigenvector with eigenvalue different from the rest and so a $SO(3)$ rotation which changes this eigenvector would also alters the tensor).

The structure of stress-energy tensor could be simplified further if we write it with one upper and one lower index (since $T^i_j\sim \delta ^i_j$): $$T^{\mu}_{\nu}=\mathrm{diag}(A,B,B,B),$$ where $A$ and $B$ could depend only on time $t$.


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