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I am currently trying to better understand Noether's theorem, so I picked up Itzykson and Zuber's QFT book. I am struggling a bit to understand the computation that leads to the variation of the action.

On page 22 of their book they proceed in the following manner.

Under a space-time translation we have

$$\mathcal{L}(\phi_i(x),\partial_{\mu} \phi_i(x)) \rightarrow \mathcal{L}(\phi_i(x+a),\partial_{\mu} \phi_i(x+a)).\tag{1.94}$$

They also define $$\delta \phi_i(x) = \delta a^{\mu} \partial_{\mu} \phi_i (x)\tag{1.95a}$$ $$\delta \partial_{\mu} \phi_i(x) = \delta a^{\nu} \partial_{\nu} \partial_{\mu} \phi_i(x) + \partial_{\mu} [\delta a^{\nu}] \partial_{\nu} \phi_i (x) \tag{1.95b}$$ (this part I understood, I got them easily by Taylor expanding the term on the right in the first transformation equation)

However, my problem is how to get to the final form of the variation of the action defined as

$$ \delta I= \int{d^4x \left(\partial_{\nu} \mathcal{L} - \partial_{\mu} \left[\frac {\partial \mathcal{L}}{\partial (\partial_{\mu}\phi_i(x)} \partial_{\nu} \phi_i(x)\right]\right) \delta a^{\nu}}.\tag{1.96}$$

They mention that they arrive at this form using integration by parts, and I am unable to see how to do so.

I have seen alternative derivations in Peskin and Schroeder's book and in Schwartz's book, however I want to be able to go through the derivation in Itzykson and Zuber's book as it is the method that seemed the most natural to me.

My approaches:

I use the general $\delta I$ form $$\int{d^4x \bigg(\frac{\partial\mathcal{L}}{\partial \phi_i} \delta \phi_i + \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi_i) } \delta \partial_{\mu} \phi_i} \bigg)$$

Which leads to (after partial differentiation)

$$\int{d^4x \bigg(\frac{\partial\mathcal{L}}{\partial \phi_i} \delta \phi_i - \partial_{\mu}(\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi_i)}) \delta \phi_i + \partial_{\mu}(\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi_i)}\delta \phi_i)}\bigg)$$

I take the 3rd term as being 0(This is where I believe I am wrong), and thus I am left with the first terms from which I am unable to get to the form mentioned in the book.

(I am using the $A\partial_{\mu}B = -\partial_{\mu}AB$ trick to go on with my integration by parts).

I am doing this purely from interest, it is not homework or so, and I would be pleased if someone could help me with better understanding the computation needed.

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  • $\begingroup$ P&S assume that the field variation $\delta\phi_i$ vanishes on the boundary of the region of integration. The integral of the last term is then indeed zero because it can be converted into a surface integral over that region (Gauss theorem). Please check if I&Z are making similar assumptions. $\endgroup$
    – Kurt G.
    Jul 27 at 14:59
  • $\begingroup$ They just hop from equ (1.95b) to (1.96). However, I do believe it is the case. $\endgroup$ Jul 27 at 15:26
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So, just proceed, plugging in (19.5), $$\int{d^4x \bigg(\frac{\partial\mathcal{L}}{\partial \phi_i} \delta \phi_i + \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi_i) } \delta \partial_{\mu} \phi_i} \bigg)\\ = \int\!\!{d^4x \bigg(\frac{\partial\mathcal{L}}{\partial \phi_i} \partial_{\nu} \phi_i (x) \delta a^{\nu} + \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi_i) }\bigl ( \partial_{\nu} \partial_{\mu} \phi_i(x) ~~\delta a^{\nu}+\partial_{\nu} \phi_i (x) ~~\partial_{\mu} [\delta a^{\nu}] \bigr )\bigg) ~~ }\\ = \int\!\! {d^4x \bigg( \partial_\nu \mathcal{L} ~~\delta a^{\nu} - \partial_{\mu} \left[\frac {\partial \mathcal{L}}{\partial (\partial_{\mu}\phi_i)} \partial_{\nu} \phi_i(x)\right] \delta a^{\nu} \bigg) }, $$ where the first term is from the evident chain rule, and last term in the middle line (square brackets) has been integrated by parts, throwing away the surface term which vanishes at infinity, as all fields, etc.. are supposed to vanish there.

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    $\begingroup$ Thank you a lot ! I was stuck on the chain rule of the first term and now I see it clearer ! $\endgroup$ Jul 28 at 8:46

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