For a given scalar field $\phi$, the stress energy tensor is \begin{equation} T_{\mu\nu}=\partial_{\mu}\phi\partial_{\nu}\phi - \frac{1}{2} g_{\mu\nu}(\partial_{\alpha}\phi\partial^{\alpha}\phi). \end{equation} How can I get a second order differential equation for $\phi$, assuming that the metric $g_{\mu\nu}$ satisfies Einstein's equation? I tried to use the identity $\nabla^{\mu}G_{\mu\nu}=0$, getting the following equation \begin{equation} 0=\nabla^{\mu}T_{\mu\nu}=g^{\mu\rho}\nabla_{\rho}T_{\mu\nu}=g^{\mu\rho}(\partial_{\rho}T_{\mu\nu}-\Gamma^{\alpha}_{\rho\mu}T_{\alpha\nu}-\Gamma^{\alpha}_{\nu\rho}T_{\mu\alpha}). \end{equation} But turns out that this is not a second order differential equation. How can I get such equation?
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1$\begingroup$ Are you sure this is the stress tensor you want to use? Usually the stress tensor for a scalar field is quadratic in the field. I would have expected the first term in your stress tensor to be $$\partial_\mu \phi \partial_\nu\phi,$$ instead of $$\partial_\mu \partial_\nu \phi.$$ In that case, you could get the field equation by just requiring that the stress tensor is divergence-free. $\endgroup$ – asperanz Dec 30 '17 at 22:05
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$\begingroup$ You're right, it's $\partial_{\mu} \phi \partial_{\nu} \phi$. That's what I've tried to do without any success. $\endgroup$ – Elias Dec 31 '17 at 18:31
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Note that what you did does not provide a second order differential equation for $\phi$, since $T_{\mu\nu}$ is already second order in the derivatives.
The stress-energy tensor of a matter sector is given by: $$ T_{\mu\nu}=\frac{1}{\sqrt{g}}\frac{\delta S_m}{\delta g^{\mu\nu}} $$
so we need to find an action $S_m=\int d^4x\,\sqrt{g}\,\mathcal{L}[\phi,\partial_\mu\phi]$ such that $$ T_{\mu\nu}=\partial_{\mu}\phi\partial_{\nu}\phi - \frac{1}{2} g_{\mu\nu}(\partial_{\alpha}\phi\partial^{\alpha}\phi) $$ and since $\phi$ is scalar, and $T_{\mu\nu}$ is second order in the derivatives, we can easily conclude that: $$ \mathcal{L}=g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi $$
Now, you calculate the equations of motion for $\phi$:
$$ \frac{\delta S_m}{\delta \phi}=0 $$ that is: $$ \partial_{\mu}(\sqrt{g}\,\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)})=0 $$ a second order differential equation for $\phi$, for a given $g_{\mu\nu}$ determined.
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$\begingroup$ I'm not familiar with the Lagrangian formulation of RG. I've been told that one way to obtain the equation is by means of the Bianchi identities $\nabla_{[\mu}R_{\nu\gamma\delta]}=0$ and Einstein's equation. $\endgroup$ – Elias Dec 30 '17 at 2:08
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$\begingroup$ I simply use the definition of $T_{\mu\nu}$. There is no mention to the lagragian of $g_{\mu\nu} $. Your proposal, of using $\nabla_{[\mu}R_{\nu\gamma]\delta\lambda}=0$ will not work. You are going to find the conservation of $T_{\mu\nu}$ again. See this in the part of 'Features'/Comservation of energy and momentum $\endgroup$ – Nogueira Dec 30 '17 at 2:22
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$\begingroup$ After the OP's edit, $T_{\mu \nu}$ only contains first derivatives of the scalar field, which means that requiring its divergence to be zero will result in a second-order differential equation in $\phi$. $\endgroup$ – Michael Seifert Jan 10 '18 at 19:21
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HINT: It'll be easiest to just write the stress-energy tensor in terms of the covariant derivative (note that these are equivalent for scalar fields): $$ T_{\mu \nu} = \nabla_\mu \phi \nabla_\nu \phi - \frac{1}{2} g_{\mu \nu} \nabla^\rho \phi \nabla_\rho \phi $$ Now take the divergence of $T_{\mu \nu}$ without writing down any Christoffel symbols; leave all your derivatives in terms of $\nabla_\mu$, rather than $\partial_\mu$. Use the facts that $\nabla_\mu$ obeys the product rule, and that by definition $\nabla_\mu g_{\rho \sigma} = 0$.
BTW, the assumption that $g_{\mu \nu}$ satisfies Einstein's equations is completely unnecessary for this derivation.
answered Jan 10 '18 at 19:25
