# From Noether's Theorem to Stress-Energy Tensor

## Main Question or Discussion Point

Hi

The following is a standard application of Noether's Theorem given in most books on QFT, in a preliminary section on classical field theory. Reproduced below are steps from the QFT book by Palash and Pal, which I am referring to, having read the same from other books. I have some trouble with a step which I have highlighted in bold. I would be grateful if someone could clarify my doubts. Thanks in advance!

Consider infinitesimal transformations of the coordinate system

$$x^{\mu} \rightarrow x'^{\mu} = x^{\mu} + \delta x^{\mu}$$

under which the fields transform as

$$\Phi^{A}(x) \rightarrow \Phi'^{A}(x) = \Phi^{A}(x) + \delta \Phi^{A}(x)$$

The change in the action resulting from these transformations is given by

$$\delta \mathcal{A} = \int_{\Omega'} d^{4}x' \mathcal{L}(\Phi'^{A}(x'),\partial_{\mu}'\Phi'^{A}(x')) - \int_{\Omega}d^{4}x \mathcal{L}(\Phi^{A}(x),\partial_{\mu}(\Phi^{A}(x)))$$

where $\Omega'$ is the transform of $\Omega$ under the coordinate change. This can be rewritten as

$$\delta \mathcal{A} = \int_{\Omega}d^{4}x \left[\mathcal{L}(\Phi'^{A}(x),\partial_{\mu}'\Phi'^{A}(x)) - \mathcal{L}(\Phi^{A}(x),\partial_{\mu}(\Phi^{A}(x))\right] + \int_{\Omega'-\Omega}d^{4}x \mathcal{L}(\Phi'^{A}(x),\partial_{\mu}'\Phi'^{A}(x))$$

The last term is an integral over the infinitesimal volume $\Omega'-\Omega$, so we can replace it by an integral over the boundary $\partial\Omega$,

$$\int_{\Omega'-\Omega}d^{4}x \mathcal{L}(\Phi'^{A},\partial_{\mu}'\Phi'^{A}) = \int_{\partial\Omega}dS_{\lambda} \delta x^{\lambda}\mathcal{L}(\Phi^{A},\partial_{\mu}\Phi^{A}) = \int_{\Omega}d^{4}x \partial_{\lambda}(\delta x^{\lambda}\mathcal{L}(\Phi^{A},\partial_{\mu}\Phi^{A}))$$
While I understand how the last integral has been written, by employing Gauss's Law, I don't understand how

$$\int_{\Omega'-\Omega}d^{4}x \mathcal{L}(\Phi'^{A},\partial_{\mu}'\Phi'^{A}) = \int_{\partial\Omega}dS_{\lambda} \delta x^{\lambda}\mathcal{L}(\Phi^{A},\partial_{\mu}\Phi^{A})$$

Can someone please explain this step? Where did the $\delta x^{\lambda}$ come from?

Thanks.

EDIT - I think I got it: $d^{4}x = dS_{\lambda}x^{\lambda}$

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Yeah, you pretty much got it with your edit. This entire derivation looks mathematically funny, but I think it's correct.

I don't think it's trivial to say that d^4x=dS dot delta x; you have to imagine the geometry in your head. Say your coordinate transformation is a stretch in the z-direction only. Then why is the new surface only created with its normal to z? You also create surfaces parallel to the z-axis when you stretch the box in the z-direction, but these are only infinitismal and can be neglected. The z-plane that used to be the boundary before the stretch is the non-infinitismal surface that's being changed by your stretch in the z-direction.

Thanks RedX. Some more help is needed in interpreting whats going on in the next couple of steps:

At this point, it is convenient to define the variation for fixed $x$, which gives the change in the functional forms only. For any function $f(x)$ whose functional form changes to $f'(x)$, we can write

$$\bar{\delta}f(x) \equiv f'(x) - f(x) = [f'(x')-f(x)]-[f'(x')-f'(x)] = \delta f(x) - \partial_{\mu}f(x)\delta x^{\mu}$$

where we have again ignored terms of higher order in $\delta x^{\mu}$ in writing the last equality.
Whats going on here?

Take the function f(x)=x^2. You can change it by adding x^3 to it, and the new function is:

g(x)=x^2+x^3

If you label g(x) as f'(x), then: f'(x)-f(x)=$$\overline{\delta}}$$=x^3

But you can also change the function by messing with the argument. So change the argument to x+1:

h(x)=(x+1)^2

If you label h(x) as f(x'), then f(x')-f(x)=2x+1

If you add these two changes, you get the total change $$\delta f$$.

So they are just classifying the different types of changes, except they are using infintinismal changes.

Thanks RedX. Some more help is needed in interpreting whats going on in the next couple of steps:

Whats going on here?
Shouldn't we have $\partial_{\mu}f'(x)\delta x^{\mu}$ in the last equality?

Shouldn't we have $\partial_{\mu}f'(x)\delta x^{\mu}$ in the last equality?
The last quotation you have: "where we have ignored higher terms ... in writing the last equality" explains why. You're talking about infinitismal changes. So $$\partial_{\mu}f'(x)\delta x^{\mu}$$ is already first order because of the small delta x^mu. So if f'(x)=f(x)+$$\overline{\delta}f(x)$$, then $$\partial_{\mu}f'(x)\delta x^{\mu}= \partial_{\mu}(f(x)+\overline{\delta}f(x))\delta x^{\mu}$$, and having two deltas is too small to care about - one is enough. So f(x) can be used instead of f'(x) for that term.