From Noether's theorem to canonical Energy-Momentum tensor using translations In this text that I am reading it says that the transformation $\delta \phi(x)$ is a symmetry if the Lagrangian changes by a total derivative:
$$\delta \mathcal{L}= \partial_{\mu}F^{\mu} .  $$
From Noether's theorem we know that the current is conserved:
$$j^{\mu}=\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\delta\phi-F^{\mu}.$$
Here the author uses this equation and "translations" to derive the Energy-Momentum tensor. But I cannot follow the intermediate steps of computation.
Suppose this:
$$x^\nu \to x^\nu-\epsilon^\nu; \phi(x) \to \phi(x)+\epsilon^\nu\partial_{\nu}\phi(x).$$
The Lagrangian also transform as
$$\mathcal{L}(x) \to \mathcal{L}(x)+\epsilon^\nu\partial_{\nu}\mathcal{L}(x).$$
If we apply the above equation for the current we can write:
$$j^{\mu}=\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}
\epsilon^\nu\partial_{\nu}\phi(x)
-F^{\mu}.$$
The text here jumps to four conserved currents given below:
$$(j^{\mu})_{\nu}=\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}
\partial_{\nu}\phi(x)
-\delta^\mu_{\nu}\mathcal{L}=T^\mu_{\nu}.$$
The question is indeed about the intermediate steps or lines to obtain the final result, in particular, how $F^{\mu}$ gives $\delta^\mu_{\nu}\mathcal{L}$ in the last line when we get the currents for all $\nu$? And where goes $\epsilon^\nu$?
 A: May I ask what text you are reading? My understanding of the stress energy tensor is as follows. The Noether condition is written as,\begin{equation}
\partial _\mu \bigg[\frac{\partial \mathcal L}{\partial (\partial _\mu  \phi )}\delta \phi +\mathcal L \delta x^\mu\bigg]=0
\end{equation}
In the discrete case we can imagine separate infinitesimal time and space translations. Field theory kind of smooshes (technical term) space and time together into a spacetime manifold. If we consider an active infinitesimal transformation $x^\nu \rightarrow x^\nu -\lambda ^\nu$, hence $\phi(x)\rightarrow \phi(x+\lambda)=\phi(x)+\lambda^\nu \partial _{\nu}\phi(x)$. If the Lagrange density is not an explicit function $x^\nu$ then we expect the following to be conserved,
\begin{equation}
\frac{\partial \mathcal L}{\partial (\partial _\mu \phi)}\partial _\nu \phi -\delta ^\mu_\nu \mathcal L\nonumber
\end{equation}
In such a transformation the form variation $\delta \phi$ is only dependent of the derivatives of the field so,
\begin{equation}
\delta \phi \longrightarrow  \lambda ^\nu \partial _\nu \phi
\end{equation}
The 4-vector variation is $-\lambda^\nu$,
\begin{equation}
\delta x^\mu \longrightarrow -\lambda ^\nu
\end{equation}
Pull out the $\lambda ^\nu$. Then $\partial _{\mu}$ is non-zero if $\delta ^{\mu}_{\nu}$.
Please let me know if you don't agree! 
