= Solution
The metric dependence of the <minimally coupled scalar field> <Lagrangian> is $\partial L/\partial g^{\alpha\beta}=-\tfrac12\nabla_\alpha\phi\nabla_\beta\phi$. Part (ii) therefore gives \b[the stress-energy tensor]
$$
\boxed{T_{\alpha\beta}=\nabla_\alpha\phi\nabla_\beta\phi-g_{\alpha\beta}\left[\frac12\nabla_\rho\phi\nabla^\rho\phi+V(\phi)\right].}
$$
Use <metric compatibility> and the <symmetric Hessian of a scalar field> to compute
$$
\begin{aligned}
\nabla_\mu T^\mu{}_{\alpha}
&=(\Box\phi)\nabla_\alpha\phi+\nabla^\mu\phi\,\nabla_\mu\nabla_\alpha\phi
-\nabla^\mu\phi\,\nabla_\alpha\nabla_\mu\phi-V'(\phi)\nabla_\alpha\phi\\
&=(\Box\phi-V'(\phi))\nabla_\alpha\phi.
\end{aligned}
$$
This is the <scalar stress-energy divergence identity>, an instance of the <diffeomorphism Noether identity for a scalar field>. On the scalar equation of motion it gives \b[stress-energy conservation], $\boxed{\nabla_\mu T^\mu{}_{\alpha}=0}$.
Back to article page
