If we have an action with a scalar field non-minimally coupled to gravity, $$\int dx^4 \sqrt{-g}(-\frac{1}{2}\partial_\mu\phi\partial^\mu\phi-\frac{1}{2}\zeta R\phi^2-V(\phi)), \tag{1}$$ varying respect to the scalar field, we can get a Klein-Gordon like equation: $$\nabla^\mu\nabla_\mu \phi=\zeta R \phi+V'(\phi) \tag{2}.$$

arXiv: 1506.03545 [gr-qc] says

The equation (2) is warranted from the conservation of the energy-momentum tensor of the scalar field due to the diffeomorphism-invariance of action. $$\nabla^\mu T_{\mu\nu}=0$$

Why is this happening?


1 Answer 1


The demonstration compounds of two pieces: first you define the energy-momentum tensor of the matter fields in the nonvacuum Einstein's field equations, secondly apply the diffeomorphism invariance to the complete action.

Define the energy-momentum tensor of the matter fields
$S = \frac{1}{16 \pi} S_H + S_M$
$c = G = 1$ natural units
$S$ complete action for gravity coupled to a set of matter fields
$S_H$ Hilbert action
$S_H = \int \sqrt{-g} R d^n x$
$S_M$ action for matter
Applying the principle of least action by varying the complete action with respect to the metric, we get:
$\frac{1}{\sqrt{-g}} \frac{\delta S}{\delta g^{\mu \nu}} = \frac{1}{16 \pi} (R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu}) + \frac{1}{\sqrt{-g}} \frac{\delta S_M}{\delta g^{\mu \nu}} = 0$
$T_{\mu \nu} = -2 \frac{1}{\sqrt{-g}} \frac{\delta S_M}{\delta g^{\mu \nu}}$ energy-momentum tensor (definition)
$R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} = 8 \pi T_{\mu \nu}$ Einstein's field equations

Apply the diffeomorphism invariance to the complete action
$S = \frac{1}{16 \pi} S_H [g_{\mu \nu}] + S_M [g_{\mu \nu}, \psi^i]$
$g_{\mu \nu}$ metric tensor
$\psi^i$ matter fields
The Hilbert action $S_H$ is diffeomorphism invariant, that is coordinate invariant, when considered in isolation. As the complete action is to be diffeomorphism invariant as well, the matter action $S_M$ must also be.
Under a diffeomorphism:
$\delta S_M = \int d^n x \frac{\delta S_M}{\delta g_{\mu \nu}} \delta g_{\mu \nu} + \int d^n x \frac{\delta S_M}{\delta \psi^i} \delta \psi^i$
Since the gravitational part of the complete action does not involve the matter fields, the variation of $S_M$ with respect to $\psi^i$ will vanish. Therefore also the variation of $S_M$ with respect to $g_{\mu \nu}$ must vanish.
If the diffeomorphism is generated by an infinitesimal vector field $V^\mu (x)$, the infinitesimal change in the metric is given by the Lie derivative along $V^\mu$.
$\delta g_{\mu \nu} = \mathcal L_V g_{\mu \nu} = 2 \nabla _{( \mu} V_{\nu )}$
Setting $\delta S_M = 0$ implies:
$0 = \int d^n x \frac{\delta S_M}{\delta g_{\mu \nu}} \nabla_\mu V_\nu = - \int d^n x \sqrt{-g} V_\nu \nabla_\mu (\frac{1}{\sqrt{-g}} \frac{\delta S_M}{\delta g_{\mu \nu}})$
The symmetrization of $\nabla _{( \mu} V_{\nu )}$ was dropped as $\frac{\delta S_M}{\delta g_{\mu \nu}}$ is already symmetric.

Law of energy-momentum conservation
Demanding that the last equation holds for diffeomorphisms generated by arbitrary vector fields $V^\mu$ and using the definition of the energy-momentum tensor, we get the law of energy-momentum conservation:
$\nabla_\mu T^{\mu \nu} = 0$
Note: The derivation assumes that no matter fields appear in the gravitational action.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.