# From Noether's Theorem to Stress-Energy Tensor

• maverick280857
In summary, the infinitesimal transformation of the coordinate system x^{\mu} \rightarrow x'^{\mu} under which the fields transform as\Phi^{A}(x) \rightarrow \Phi'^{A}(x) = \Phi^{A}(x) + \delta \Phi^{A}(x)produces a change in the action which is given by\delta \mathcal{A} = \int_{\Omega'} d^{4}x' \mathcal{L}(\Phi'^{A}(x'),\partial_{\mu}'\Phi'^{A}(

#### maverick280857

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: $d^{4}x = dS_{\lambda}x^{\lambda}$

Last edited:
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 what's 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.

maverick280857 said:
Thanks RedX. Some more help is needed in interpreting what's 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?

maverick280857 said:
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.