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In general relativity, suppose as usual that we have the following action for the matter fields

\begin{equation} S_{\mathrm{matter}} = \int_M d^4 x \sqrt{-g} L_{\mathrm{matter}} , \end{equation}

where $L_{\mathrm{matter}}$ is the matter Lagrangian. Then, again as usual, we define the energy-momentum tensor of the matter fields as

\begin{equation} T^{ab} = \frac{2}{\sqrt{-g}} \frac{\delta S_{\mathrm{matter}}}{\delta g_{ab}} \end{equation}

My question is the following: Are there any (physically interesting) cases where we have different $S_{\mathrm{matter}}$, but the same corresponding stress-energy tensor?

It seems like an obvious approach here is to take e.g. the action for the scalar field

\begin{equation} S\left[ \Phi \right] = \int_M d^4 x \sqrt{-g} \left( - \frac{1}{2} g^{ab} \nabla_a \Phi \nabla_b \Phi - V\left( \Phi \right) \right) \end{equation}

and attempt to add to this a piece $S'\left[ \Phi \right]$ such that $\frac{\delta S'}{\delta g_{ab}} =0$. However, I'm struggling to find any plausible such $S'\left[ \Phi \right]$.

So: Do (interesting) cases with different $S_{\mathrm{matter}}$ but the same $T^{ab}$ actually exist? And if so, can people help me find an example? Thanks!

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  • $\begingroup$ Comment to the question (v1): Do total divergence terms count as physically interesting? $\endgroup$
    – Qmechanic ♦
    Apr 15, 2016 at 22:06
  • $\begingroup$ Also, does the action have to stay Lorentz-invariant? because terms with factors of $\frac{1}{\sqrt{g}}$ will help you out here, but it will make the integrand of S not a volume element. $\endgroup$
    – Zo the Relativist
    Apr 15, 2016 at 22:23
  • $\begingroup$ Hi! Both good questions -- ideally (a) Lorentz invariant terms, and (b) not divergence terms. What I was thinking was: Are there any cases where we have matter obeying genuinely different dynamical equations, but nevertheless we get the same stress-energy tensor? $\endgroup$
    – Jimeree
    Apr 16, 2016 at 0:49
  • $\begingroup$ The stress tensor is non-zero whenever the action depends on the metric. In order to get the same stress tensor you might add some topological terms to the action. For instance in odd dimensions you can add the Chern-Simons term $\int A \wedge dA + A \wedge A \wedge A $. In even dimensions, you can add the Gauss-Bonnet term which integrates to the Euler Characteristic of the manifold. $\endgroup$
    – Prahar
    Apr 18, 2016 at 13:56
  • $\begingroup$ Thanks! So consider $S_{GB} = \int_M d^4 x \sqrt{-g} G = \chi ( M )$. I take it that $\frac{\delta S_{GB}}{\delta g_{ab}} = 0$, so $\frac{\delta ( \sqrt{-g} G ) }{\delta g_{ab}} = 0$. Now consider $S_{GB\Phi} = \int_M d^4 x \sqrt{-g} G f (\Phi )$. We have $\frac{\delta S_{GB\Phi}}{\delta g_{ab}} = 0$ if $\frac{\delta f (\Phi )}{\delta g_{ab}} =0$, true if e.g. $f (\Phi ) \sim \Phi^n$. Even then, however, we have $\frac{\delta S_{GB\Phi}}{\delta \Phi} \neq 0$. So: $T^{ab}$ is unchanged after adding $S_{GB\Phi}$ to the action, but the equations of the $\Phi$ are different. Does that sound right? $\endgroup$
    – Jimeree
    Apr 19, 2016 at 16:17

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