From Noether's Theorem to Stress-Energy Tensor

1. May 26, 2009

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!

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: May 26, 2009
2. May 26, 2009

RedX

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.

3. May 26, 2009

maverick280857

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

Whats going on here?

4. May 26, 2009

RedX

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.

5. May 27, 2009

maverick280857

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

6. May 27, 2009

RedX

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.