# Explicit derivation of free scalar stress-energy tensor $\nabla_\mu T^{\mu\nu}=0$

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 $$\nabla_\mu\nabla^\nu\phi = \nabla^\nu\nabla_\mu\phi$$?

• What have you tried? Nov 7, 2019 at 4:02
• I have added the bit that I have tried on the question. @Prahar Nov 7, 2019 at 11:15
• You can use the chain rule for the $\nabla_\mu(V(\phi))$ term and the symmetry of $g^{\sigma \rho}$. Nov 7, 2019 at 12:12
• 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? Nov 7, 2019 at 14:10
• Oh thanks, yeah I checked it and it turns out to be correct since $\Gamma^\lambda_{\alpha\mu} = \Gamma^\lambda_{\mu\alpha}$! Nov 7, 2019 at 14:14

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