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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!

[tex]\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})[/tex]

Can someone please explain this step? Where did the [itex]\delta x^{\lambda}[/itex] come from?

Thanks.

EDIT - I think I got it: [itex]d^{4}x = dS_{\lambda}x^{\lambda}[/itex]

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 howConsider infinitesimal transformations of the coordinate system

[tex]x^{\mu} \rightarrow x'^{\mu} = x^{\mu} + \delta x^{\mu}[/tex]

under which the fields transform as

[tex]\Phi^{A}(x) \rightarrow \Phi'^{A}(x) = \Phi^{A}(x) + \delta \Phi^{A}(x)[/tex]

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

[tex]\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)))[/tex]

where [itex]\Omega'[/itex] is the transform of [itex]\Omega[/itex] under the coordinate change. This can be rewritten as

[tex]\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))[/tex]

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

[tex]\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}))[/tex]

[tex]\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})[/tex]

Can someone please explain this step? Where did the [itex]\delta x^{\lambda}[/itex] come from?

Thanks.

EDIT - I think I got it: [itex]d^{4}x = dS_{\lambda}x^{\lambda}[/itex]

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