From Noether's theorem to canonical Energy-Momentum tensor using translations

Question

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 $\epsilon^\nu$?

Answer 1

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!

Answer 2

My understanding is that you have to consider $\delta \phi$ as a sum of partial derivatives: $\delta\phi = \sum_{\nu=0}^3\epsilon^\nu \partial_\nu\phi$. Then $\epsilon^\mu\partial_\mu L$ (without Einstein summation, as we consider one specific $\mu$) becomes $\delta^\mu_\nu\epsilon^\nu \partial_\mu L$. Then you can take out $\epsilon^\nu$ from your conserved current equation and argue that since $\epsilon^\nu$ can be any arbitrary perturbation, you get to define $(j^\mu)_\nu$ the way you did.