I need to prove that the energy-momentum tensor density is defined as:
\begin{equation}
\mathcal{T}_{\mu\nu}=-\frac{\partial \phi}{\partial x_\nu}\frac{\partial\mathcal{L}}{\partial(\frac{\partial \phi}{\partial x\mu})}+\mathcal{L}\delta_{\mu\nu}
\end{equation}
satisfies the equation $\frac{\partial \mathcal{T}_{\mu\nu}}{\partial x_\mu}=0$.
My Approach:
By using chain rule of differentiation and employing the Euler-Lagrange equation for $\phi$:
\begin{equation}
\frac{\partial \mathcal{T}_{\mu\nu}}{\partial x_\mu}=
-\frac{\partial\partial \phi}{\partial x_\mu\partial x_\nu}\frac{\partial\mathcal{L}}{\partial(\frac{\partial \phi}{\partial x\mu})}-\frac{\partial \phi}{\partial x_\nu}\frac{\partial\partial\mathcal{L}}{\partial x_\mu\partial(\frac{\partial \phi}{\partial x\mu})}+\delta_{\mu\nu}\frac{\partial \mathcal{L}}{\partial x_\mu}=
-\frac{\partial\partial \phi}{\partial x_\mu\partial x_\nu}\frac{\partial\mathcal{L}}{\partial(\frac{\partial \phi}{\partial x\mu})}-\frac{\partial \phi}{\partial x_\nu}\frac{\partial \mathcal{L}}{\partial \phi}++\delta_{\mu\nu}\frac{\partial \mathcal{L}}{\partial x_\mu}
\end{equation}
Now by using:
\begin{equation}
\delta \mathcal{L}=\frac{\partial \mathcal{L}}{\partial \phi}\delta \phi+\frac{\partial \mathcal{L}}{\partial \frac{\partial \phi}{\partial x_{\mu}}}\delta{\frac{\partial \phi}{\partial x_{\mu}}}
\end{equation}
in the above equation we get the continuity equation of $\frac{\partial \mathcal{T}_{\mu\nu}}{\partial x_\mu}=0$.
I am sure of the case $\mu=\nu$ but not sure of $\mu \neq \nu$.
It would be great if this can be resolved.
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$\begingroup$ F is usually reserved for the field tensor, so I changed it to the more common T. Also your $\delta_{\,u\nu}$ should be $g{\,u\nu}$. Conventions are convenient. $\endgroup$– my2ctsCommented Jan 6 at 12:38
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