Noether's Theorem and the Stress Energy Tensor

A common derivation that you will find in QFT texts is the derivation of the Stress Energy Tensor using Noether's theorem. I am just starting to study QFT using David Tong's notes and Schwartz' book. However, I have found that in both the expositions are incomplete and the terminology and notation is sometimes confusing.

As such, I didn't feel I grokked the derivations. So to fill in the gaps, I am writing out all the gory details for my own edification and the hope that this might help someone else facing similar frustrations. After starting this, I also found these notes by Hagen Kleinart to be particularly clear.

Noether's theorem is a statement about continous symmetries of an action. With such a symmetry there is an associated conserved current. The symmetry we will be looking at is a space-time translation. To that end we consider an infinitesimal space time translation. $$x^{\nu} \rightarrow x^{\nu} - \epsilon^{\nu}$$ Then our fields transform to first order Taylor's expansion as $$\phi(x) \rightarrow \phi(x+\epsilon) \approx \phi(x) + \epsilon^{\nu}\partial_{\nu}\phi(x)$$

I want to say more about the transformations above.In particular, we are viewing these transformations as active transformations. This is something I always get confused with. But I am thinking of the above like so.

With an active transformation we are changing the field value at the same point in space-time as opposed to labeling the points of space-time. If we were talking about particles, then this would be equivalent to physically moving a particle to another location. For example, imagine I am standing at point \(x^{\nu} = (0, 1, 0, 0)\) and then take a step in the negative \(x\) direction so I am now standing at the point \(\widetilde{x}^{\nu} = (0, 0, 0, 0)\). This is an active transformation of my location.

Now with fields we are going to "move" the fields. By that I mean we are going to define a new field \(\widetilde{\phi}\) such that it's value you at a point \(x\) will be the same as the untransformed field at \(x+\epsilon\). In other words $$ \widetilde{\phi}(x -\epsilon ) = \phi(x)$$ or $$ \widetilde{\phi}(x ) = \phi(x+\epsilon)$$ And again to first order, $$ \widetilde{\phi}(x) = \phi(x) + \epsilon^{\nu}\partial_{\nu}\phi(x)$$

We are interested in the variation in our action $$ S[\phi] = \int dx^4 \mathcal{L}(\phi, \partial_{\nu} \phi)$$ as we undergo such a space-time translation we will be interested in the principal linear variation of $$\Delta S = S[\widetilde{\phi}] - S[\phi]$$ which we will call \(\delta S\). Specifically, $$\Delta S = \int dx^{4} \mathcal{L}(\widetilde{\phi}, \partial_{\nu} \widetilde{\phi}) - \int dx^4 \mathcal{L}(\phi, \partial_{\nu} \phi) $$ or $$\Delta S = \int dx^{4} \mathcal{L}(\phi + \epsilon^{\nu}\partial_{\nu} \phi, \partial_{\mu} \phi + \epsilon^{\nu}\partial_{\nu}\partial_{\mu} \phi) - \int dx^4 \mathcal{L}(\phi, \partial_{\nu} \phi) $$

Now as typically done we will do a taylor expansion and only keep first order terms and we find. $$ \delta S = \int dx^4 \left(\mathcal{L}(\phi, \partial_{\mu} \phi) + \epsilon^{\nu}\partial_{\nu} \phi \frac{\partial \mathcal{L}(\phi, \partial_{\mu} \phi)}{\partial\phi} + \epsilon^{\nu}\partial_{\nu}\partial_{\mu} \phi \frac{\partial \mathcal{L}(\phi, \partial_{\mu} \phi)}{\partial (\partial_{\mu}\phi)} - \mathcal{L}(\phi, \partial_{\nu} \phi)\right) $$
Next we replace $$\epsilon^{\nu}\partial_{\nu}\partial_{\mu} \phi \frac{\partial \mathcal{L}(\phi, \partial_{\mu} \phi)}{\partial (\partial_{\mu}\phi)}$$ by $$\partial_{\mu}\left[\epsilon^{\nu}\partial_{\nu} \phi \frac{\partial \mathcal{L}(\phi, \partial_{\mu} \phi)}{\partial (\partial_{\mu}\phi)}\right] -\partial_{\mu}\left(\frac{\partial \mathcal{L}(\phi, \partial_{\mu} \phi)}{\partial (\partial_{\mu}\phi)}\right) \epsilon^{\nu}\partial_{\nu} \phi $$ Substituting into \(\delta S\) and canceling terms we get, $$ \delta S = \int dx^4 \left[\epsilon^{\nu}\partial_{\nu}\phi \left( \frac{\partial \mathcal{L}(\phi, \partial_{\mu} \phi)}{\partial\phi} -\partial_{\mu}\left(\frac{\partial \mathcal{L}(\phi, \partial_{\mu} \phi)}{\partial (\partial_{\mu}\phi)}\right) \right) + \partial_{\mu}\left[\epsilon^{\nu}\partial_{\nu} \phi \frac{\partial \mathcal{L}(\phi, \partial_{\mu} \phi)}{\partial (\partial_{\mu}\phi)}\right] \right] $$

Now for solutions of the equations of motions, $$ \delta S = \int dx^4 \partial_{\mu}\left[\epsilon^{\nu}\partial_{\nu} \phi \frac{\partial \mathcal{L}(\phi, \partial_{\mu} \phi)}{\partial (\partial_{\mu}\phi)}\right] $$

Now we are getting close to the finish line. The last steps are as follows. First, we notice that the change in the Lagrangian Density under this symmetry transformation is, $$ \Delta \mathcal{L} = \mathcal{L}(x+\epsilon) - \mathcal{L}(x)$$ and again to first order $$ \delta \mathcal{L} = \mathcal{L}(x) + \epsilon^{\nu}\partial_{\nu} \mathcal{L}(x) - \mathcal{L}(x) = \epsilon^{\nu}\partial_{\nu} \mathcal{L}(x)$$ Now we have two different expressions for \(\delta \mathcal{L}\) as a function of our symmetry transformation and just set them equal to each other. $$ \epsilon^{\nu}\partial_{\nu} \mathcal{L}(x) = \partial_{\mu}\left[\epsilon^{\nu}\partial_{\nu} \phi \frac{\partial \mathcal{L}(\phi, \partial_{\mu} \phi)}{\partial (\partial_{\mu}\phi)}\right] $$ We should be careful, however, because the right-hand side holds only for \(\phi\) which are solutions to our equations of motion (Euler-Lagrange) whereas the left-hand side holds generally, i.e., for any \(\phi\), and infinitesimal space-time translation. One other subtle point that I struggled with for a while is that one typically thinks of \(\mathcal{L}\) as a function of \(\phi\) and \(\partial_{\nu}\phi\) so one would expect, \(\partial_{\nu} \mathcal{L} = 0\) because there is no explicit dependence on \(x\). But here we are viewing \(\mathcal{L}\) as a function of \(x\). One can think of it as if after fixing \(\phi\), we can replace it with it's functional form in terms of \(x\) since now we are really only varying \(\epsilon\) and not the functional form of \(\phi\).

The last few steps are fairly straightforward. $$ 0 = \partial_{\mu}\left[\epsilon^{\nu}\partial_{\nu} \phi \frac{\partial \mathcal{L}(\phi, \partial_{\mu} \phi)}{\partial (\partial_{\mu}\phi)}\right] - \epsilon^{\nu}\partial_{\nu} \mathcal{L} $$ Now we rewrite the last part as, $$\epsilon^{\nu}\partial_{\nu} \mathcal{L} = \epsilon^{\nu}\partial_{\mu} \delta_{\nu}^{\mu} \mathcal{L}$$ You should make sure the above is clear. Then we get, $$ 0 = \epsilon^{\nu}\partial_{\mu}\left[\partial_{\nu} \phi \frac{\partial \mathcal{L}(\phi, \partial_{\mu} \phi)}{\partial (\partial_{\mu}\phi)} - \delta_{\nu}^{\mu} \mathcal{L} \right]$$ Noether's theorem says that we get a "current", \(j^{\mu}\), such that it's four-divergence is zero, \(\partial_{\mu}j^{\mu}=0\), and in this case we have a separate current for each translation direction, \(\epsilon^{\nu}\). $$(j^{\mu})_{\nu} = \partial_{\nu} \phi \frac{\partial \mathcal{L}(\phi, \partial_{\mu} \phi)}{\partial (\partial_{\mu}\phi)} - \delta_{\nu}^{\mu} \mathcal{L} = T^{\mu}_{\nu}$$ I did this for a single scalar field for clarity, but it's relatively straightforward to generalize this. I hope this helps someone. I mainly wrote this up for my own understanding and ended up clarifying quite a lot of my own understanding in the process.