In the "Cambridge Lectures on Supersymmetry and Extra Dimensions" of F.Quevedo it is written on page 59 ($T^{\mu\nu}$ stands for the energy-momentum tensor):

The metric $g_{\mu\nu}$ as gauge field couples to the "current" $T^{\mu\nu}$ via $g_{\mu\nu}T^{\mu\nu}$.

He argues apparently on the basis of an analogy since a couple of lines above it is said:

"We introduced a gauge field $A_{\mu}$ coupling to a current $J^{\mu}$ via the interaction term $A_{\mu}J^{\mu}$".

The problem I have is that the electromagnetic interaction term $A_{\mu}J^{\mu}$ appears in the Lagrangian of the Maxwell equations, however, I have never seen in a Lagrangian of General Relativity (GR) the term $g_{\mu\nu}T^{\mu\nu}$.

According to my knowledge the total Lagrangian in GR is $L_{tot}= L_{EH} +L_{matter}$. Neither in $L_{EH}$ (essentially proportional to the curvature scalar $R$) nor in $L_{matter}$
(could be for instance $L_{KG}= \frac{1}{2}((\partial_\mu \phi) (\partial_\nu\phi) g^{\mu\nu} -m^2\phi^2)$ ) a term $g_{\mu\nu}T^{\mu\nu}$ appears.

Actually if the action $S_{matter}=\int d^4x \sqrt{-g} L_{matter}$ of the matter fields is varied one gets:

$$\delta S_{matter} =\frac{1}{2} \int d^4x \sqrt{-g} T^{\mu\nu}\delta g_{\mu\nu}$$

which looks very similar, but as it is a variation is not the same as $g_{\mu\nu}T^{\mu\nu}$.

So what is meant by "The metric $g_{\mu\nu}$ as gauge field couples to the "current" $T^{\mu\nu}$ via $g_{\mu\nu}T^{\mu\nu}$" ? Does it make any sense if the coupling term does not appear in the GR-Lagrangian ? And above all, as far as I remember the trace of the energy-momentum tensor is zero for the EM-field. Could a coupling term make sense if it is under some circumstances zero ?


By "definition", the energy momentum tensor in GR is defined by the functional derivative.

$$ T^{\mu\nu} = \frac{\delta}{\delta g_{\mu\nu}}\mathcal{L}_{matter} $$

This "functional derivative equation" can be "integrated" to obtain

$$ \mathcal{L}_{matter} = g_{\mu\nu} T^{\mu\nu} + C$$,

where as $C$ is some functional of the other fields in the theory (but not $g_{\mu\nu}$.)

  • $\begingroup$ Hmm but in general, $T=T[g]$, and therefore you cannot integrate it just like that. It's not that simple, I believe. $\endgroup$ – AccidentalFourierTransform Jul 31 '18 at 15:34
  • $\begingroup$ Agreed. This only works if $T_{\mu\nu}$ does not depend on $g_{\mu\nu}$. In fact, the argument can be reversed to show that the matter Lagrangian can only be written as $g_{\mu\nu}T^{\mu\nu}$ if $T^{\mu\nu}$ is independent of $g_{\mu\nu}$ $\endgroup$ – mmeent Jul 31 '18 at 15:42
  • $\begingroup$ I think, T is to be viewed as a given external current, which isn't subject to a back reaction of the field g (this is the case when we don't write a kinetic term for the matter fields in our Lagrangian so we won't get equations of motion for them). $\endgroup$ – Photon Aug 1 '18 at 10:05

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