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I have derived by varying the following action

$$ \mathcal{S}_{free} = \int d^4x\sqrt{-g} \left( -g^{\mu\nu}\frac{1}{2}\nabla_\mu\phi\nabla_\nu \phi - V(\phi)\right) $$ the following stress-energy tensor $$ T_{\mu\nu}=\nabla_\mu\phi\nabla_\nu\phi-\frac{1}{2}g_{\mu\nu}g^{\rho\sigma}\nabla_\rho\phi\nabla_\sigma \phi-g_{\mu\nu}V(\phi) $$ with the equation of motion as $$ \nabla_\mu\nabla^\mu\phi-\frac{\partial V}{\partial \phi}=0 $$

I'm trying to explicitly show that $\nabla_\mu T^{\mu\nu} = 0$ using the e.o.m but I just can't seem to figure this out! My problem is getting $\nabla_\mu T^{\mu\nu}$ to have an expression which is the e.o.m multiplied with something so that it is shown to be zero.

What I've tried: $$ \nabla_\mu T^{\mu\nu} = \nabla_\mu\left(\nabla^\mu\phi\nabla^\nu\phi+\frac{1}{2}g^{\mu\nu}g^{\rho\sigma}\nabla_\rho\phi\nabla_\sigma\phi+\frac{1}{2}g^{\mu\nu}V(\phi)\right) $$ $$ =(\nabla_\mu\nabla^\mu\phi)\nabla^\nu\phi+(\nabla_\mu\nabla^\nu\phi)\nabla^\mu\phi+\frac{1}{2}g^{\mu\nu}g^{\rho\sigma}(\nabla_\mu\nabla_\rho\phi)\nabla_\sigma\phi+\frac{1}{2}g^{\mu\nu}g^{\rho\sigma}(\nabla_\mu\nabla_\sigma\phi)\nabla_\rho\phi+\frac{1}{2}g^{\mu\nu}\nabla_\mu( V (\phi)) $$

Edit:

I've arrived using the hints below to $$\nabla_\mu T^{\mu\nu}= \nabla_\mu\nabla^\nu\phi\nabla^\mu\phi-\nabla^\nu\nabla_\mu\phi\nabla^\mu\phi $$

Is it correct to assume $\nabla_\mu\nabla^\nu\phi = \nabla^\nu\nabla_\mu\phi$?

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  • $\begingroup$ What have you tried? $\endgroup$
    – Prahar
    Nov 7, 2019 at 4:02
  • $\begingroup$ I have added the bit that I have tried on the question. @Prahar $\endgroup$ Nov 7, 2019 at 11:15
  • $\begingroup$ You can use the chain rule for the $\nabla_\mu(V(\phi))$ term and the symmetry of $g^{\sigma \rho}$. $\endgroup$
    – hof_a
    Nov 7, 2019 at 12:12
  • $\begingroup$ Your algebra is correct. For the last point - you know the explicit definition of covariant derivative right? Why don’t you use that to check whether what you need is correct? $\endgroup$
    – Prahar
    Nov 7, 2019 at 14:10
  • $\begingroup$ Oh thanks, yeah I checked it and it turns out to be correct since $\Gamma^\lambda_{\alpha\mu} = \Gamma^\lambda_{\mu\alpha}$! $\endgroup$ Nov 7, 2019 at 14:14

1 Answer 1

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$\nabla^{α}(\nabla_{α}φ\nabla_{β}φ) -\cfrac{1}{2}g_{αβ}g^{μν}\nabla^{α}(\nabla_{μ}φ\nabla_{ν}φ) - \nabla^{α}g_{αβ}V(φ) = 0 \Rightarrow $ $\nabla^{α}\nabla_{α}φ\nabla_{β}φ +\nabla_{α}φ \nabla^{α}\nabla_{β}φ - \cfrac{1}{2}g_{αβ}g^{μν}\nabla^{α}(\nabla_{μ}φ)\nabla_{ν}φ - \cfrac{1}{2}g_{αβ}g^{μν}\nabla_{μ}φ\nabla^{α}\nabla_{ν}φ - \nabla^{α}g_{αβ}V(φ) = 0 \Rightarrow$ $\nabla^{α}\nabla_{α}φ\nabla_{β}φ +\nabla_{α}φ \nabla^{α}\nabla_{β}φ - \cfrac{1}{2}\nabla_{β}(\nabla^{ν}φ)\nabla_{ν}φ - \cfrac{1}{2}\nabla_{β}(\nabla^{μ}φ)\nabla_{μ}φ - \nabla_{β}φ\cfrac{dV}{dφ} = 0 \Rightarrow $ $\nabla^{α}\nabla_{α}φ\nabla_{β}φ - \nabla_{β}φ\cfrac{dV}{dφ} = 0\Rightarrow $ $\nabla_{β}φ ( \nabla^{α}\nabla_{α}φ - \cfrac{dV}{dφ}) = 0 $

Since $\nabla_{β}φ = 0 \rightarrow φ = $ constant , $\nabla^{α}\nabla_{α}φ - \cfrac{dV}{dφ} =0$

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