2

If I understand correctly, Sean Carroll's Spacetime and Gravity says that the energy-momentum tensor for a perfect fluid

$$T^{\mu\nu} = (\rho + p)U^\mu U^\nu + pg^{\mu\nu}$$

can be obtained as the variation of the action for a scalar field with respect to the metric

\begin{align} T^{(\phi)}_{\mu\nu} &:= \frac {-2}{\sqrt{-g}}\frac{\delta S_\phi}{\delta g^{\mu\nu}}\\ &=\nabla_\mu \phi \nabla_\nu\phi - \frac 12 g_{\mu\nu}g^{\rho\sigma}\nabla_\rho \phi\nabla_\sigma \phi - g_{\mu\nu} V(\phi).\end{align}

Is this true? How so?

Also, is it possible to rewrite this as a "canonical energy-momentum tensor"?

$$S^{\mu\nu} = \frac{\delta \mathcal L}{\delta(\partial_\mu \Phi^i)}\partial^\nu\Phi^i - \eta^{\mu\nu}\mathcal L$$

Rodrigo
  • 659
  • Could you specify where exactly this claim is made? – Javier Mar 16 '19 at 15:52
  • p. 164 "In flat spacetime this [the second expression] reduces to what we had asserted, in Chapter 1, was the correct energy-momentum tensor for a scalar field" – Rodrigo Mar 16 '19 at 17:45
  • He never says that it's the EM tensor of a perfect fluid, though. I'm not saying it can't be (it's an interesting problem), only that Carroll claims no such thing. – Javier Mar 16 '19 at 18:46
  • Does this help? http://www.damtp.cam.ac.uk/user/tong/string/string.pdf -- 4.1.1, p64. – kkm -still wary of SE promises Mar 17 '19 at 07:58
  • And perhaps this: https://physics.stackexchange.com/a/306635/115253, for a better elaboration (and the linked paper, which in turn cites the same Tong's Lectures). And since we're flat, $\eta_{\mu\nu}\equiv g_{\mu\nu}$. The pieces seem to come together. – kkm -still wary of SE promises Mar 17 '19 at 08:28

0 Answers0