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. $$x^{\nu }\to x^{\nu }-{ϵ}^{\nu }$$ Then our fields transform to first order Taylor's expansion as $$\varphi (x)\to \varphi (x+ϵ)\approx \varphi (x)+{ϵ}^{\nu }{\partial }_{\nu }\varphi (x)$$

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 ${\tilde{x}}^{\nu }=(0,0,0,0)$. 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 $\tilde{\varphi }$ such that it's value you at a point $x$ will be the same as the untransformed field at $x+ϵ$. In other words $$\tilde{\varphi }(x-ϵ)=\varphi (x)$$ or $$\tilde{\varphi }(x)=\varphi (x+ϵ)$$ And again to first order, $$\tilde{\varphi }(x)=\varphi (x)+{ϵ}^{\nu }{\partial }_{\nu }\varphi (x)$$

We are interested in the variation in our action $$S[\varphi ]=\int d{x}^4\mathcal{L}(\varphi ,{\partial }_{\nu }\varphi )$$ as we undergo such a space-time translation we will be interested in the principal linear variation of $$\mathrm{\Delta }S=S[\tilde{\varphi }]-S[\varphi ]$$ which we will call $\delta S$. Specifically, $$\mathrm{\Delta }S=\int d{x}^4\mathcal{L}(\tilde{\varphi },{\partial }_{\nu }\tilde{\varphi })-\int d{x}^4\mathcal{L}(\varphi ,{\partial }_{\nu }\varphi )$$ or $$\mathrm{\Delta }S=\int d{x}^4\mathcal{L}(\varphi +{ϵ}^{\nu }{\partial }_{\nu }\varphi ,{\partial }_{\mu }\varphi +{ϵ}^{\nu }{\partial }_{\nu }{\partial }_{\mu }\varphi )-\int d{x}^4\mathcal{L}(\varphi ,{\partial }_{\nu }\varphi )$$

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

Now for solutions of the equations of motions, $$\delta S=\int d{x}^4{\partial }_{\mu }\left[{ϵ}^{\nu }{\partial }_{\nu }\varphi \frac{\partial \mathcal{L}(\varphi ,{\partial }_{\mu }\varphi )}{\partial ({\partial }_{\mu }\varphi )}\right]$$

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, $$\mathrm{\Delta }\mathcal{L}=\mathcal{L}(x+ϵ)-\mathcal{L}(x)$$ and again to first order $$\delta \mathcal{L}=\mathcal{L}(x)+{ϵ}^{\nu }{\partial }_{\nu }\mathcal{L}(x)-\mathcal{L}(x)={ϵ}^{\nu }{\partial }_{\nu }\mathcal{L}(x)$$ 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. $${ϵ}^{\nu }{\partial }_{\nu }\mathcal{L}(x)={\partial }_{\mu }\left[{ϵ}^{\nu }{\partial }_{\nu }\varphi \frac{\partial \mathcal{L}(\varphi ,{\partial }_{\mu }\varphi )}{\partial ({\partial }_{\mu }\varphi )}\right]$$ We should be careful, however, because the right-hand side holds only for $\varphi$ which are solutions to our equations of motion (Euler-Lagrange) whereas the left-hand side holds generally, i.e., for any $\varphi$, 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 $\varphi$ and ${\partial }_{\nu }\varphi$ 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 $\varphi$, we can replace it with it's functional form in terms of $x$ since now we are really only varying $ϵ$ and not the functional form of $\varphi$.

The last few steps are fairly straightforward. $$0={\partial }_{\mu }\left[{ϵ}^{\nu }{\partial }_{\nu }\varphi \frac{\partial \mathcal{L}(\varphi ,{\partial }_{\mu }\varphi )}{\partial ({\partial }_{\mu }\varphi )}\right]-{ϵ}^{\nu }{\partial }_{\nu }\mathcal{L}$$ Now we rewrite the last part as, $${ϵ}^{\nu }{\partial }_{\nu }\mathcal{L}={ϵ}^{\nu }{\partial }_{\mu }{\delta }_{\nu }^{\mu }\mathcal{L}$$ You should make sure the above is clear. Then we get, $$0={ϵ}^{\nu }{\partial }_{\mu }[{\partial }_{\nu }\varphi \frac{\partial \mathcal{L}(\varphi ,{\partial }_{\mu }\varphi )}{\partial ({\partial }_{\mu }\varphi )}-{\delta }_{\nu }^{\mu }\mathcal{L}]$$ 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, ${ϵ}^{\nu }$. $$(j^{\mu }{)}_{\nu }={\partial }_{\nu }\varphi \frac{\partial \mathcal{L}(\varphi ,{\partial }_{\mu }\varphi )}{\partial ({\partial }_{\mu }\varphi )}-{\delta }_{\nu }^{\mu }\mathcal{L}=T_{\nu }^{\mu }$$ 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.