0
$\begingroup$

Is the Stress-Energy tensor invariant in all RFs? If not (which is highly probable) how does it change?

EDIT: does the Einstein equation help? Since (without $\Lambda$) $$ R_{\alpha\beta} -\frac{1}{2}Rg_{\alpha\beta}= kT_{\alpha\beta} $$

and since the metric depends on the coordinate system, is this the connection?

$\endgroup$
  • 1
    $\begingroup$ A tensor is by definition covariant. Stress-energy tensor of what? $\endgroup$ – Qmechanic Jul 27 at 11:46
  • 1
    $\begingroup$ A tensor is "invariant" but the components of a particular representation of it, in a particular coordinate system, depend on the coordinate system. Or to put it a different way, The "laws of physics" don't depend on what coordinate system you choose to describe them. $\endgroup$ – alephzero Jul 27 at 12:03
  • $\begingroup$ @Qmechanic I actually do not know of which object I would calculate the stress energy tensor, but in general a cosmological one, like a star either resting or moving $\endgroup$ – chiara iannetta Jul 27 at 12:48
  • 1
    $\begingroup$ Are you asking how tensor components transform under coordinate transformations? $\endgroup$ – Qmechanic Jul 27 at 15:40
  • $\begingroup$ $R_{\alpha\beta}$, $g_{\alpha\beta}$, and $T_{\alpha\beta}$ all transform in exactly the same way, according to the transformation rule for components of a rank-2 covariant tensor. That is basically the point of tensor equations. If all terms transform similarly, the equation remains valid in all frames. But the fact that $T_{\alpha\beta}$ transforms like $g_{\alpha\beta}$ doesn’t mean that one is directly related to the other. $\endgroup$ – G. Smith Jul 27 at 16:23
1
$\begingroup$

The components of the stress-energy tensor are different in different reference frames. The components of any rank-2 covariant tensor transform according to the following rule:

$$T_{\alpha’\beta’}=\frac{\partial x^{\alpha}}{\partial x^{\alpha’}}\frac{\partial x^{\beta}}{\partial x^{\beta’}}T_{\alpha\beta}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.