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In many treatments of Noether theorem [e.g. Srednicki's book eq. (22.27), Banados' review eq. (2.65), David Tong's notes on QFT eq. (1.38), Peskin & Schroeder eq. (2.12), etc.] an almost explicit form of current is given which is

$$j^\mu=\text{canonical momenta} \times \text{variation of field}- \text{boundary term}.$$

While the explicit form of the 1st term is provided in these texts, the 2nd one is just mentioned as $K_\mu$ defined by

$$\delta S= \int d^4 x~\partial_\mu K^\mu. \tag{1}$$

These texts tend to find $K_\mu$ by performing an explicit off-shell symmetry variation of the action, whereas they find the first term by simple substitution of the Lagrangian in the explicit formula. Luckily in Francesco-Mathieu-Senechal (section 2.4.2 eqn. $(2.141)$) a remarkably simple explicit form of $K^\mu$ is given in terms of the Lagrangian density $\mathcal{L}$: $$K^\mu=-\mathcal{L}~\frac{\delta x^\mu}{\delta \omega_a} \tag{2.141'}$$ where $\omega_a$ are the parameters of transformation i.e. $$x'^\mu=x^\mu+\omega_a \frac{\delta x^\mu}{\delta \omega_a}.$$

In principle this term can be derived from (1) and the transformation equation. But I cannot think of how to proceed. Any hints towards deriving $(2.141')$ would be appreciated.

A related soft comment/question: I wonder why the above mentioned reviews/notes/books don't seem to include this explicit form for $K^\mu$. In the context of deriving EOMs, I have two sets of people in the community: those who prefer to vary their action and those who plug and chug in Euler Lagrange equations. But here, in this very similar situation, why would people even want to use different techniques for different kind of terms? This is probably a sociological question not meant for PSE, and I am not asking it: I am simply pointing out, in case it is not and there is some calculational efficiency or some subtlety hiding in one of these methods which makes people prefer one over the other.

Qmechanic
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1 Answers1

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  1. For what it's worth, eq. (2.141') is part of the bare Noether current associated with a horizontal transformation, cf. e.g. this Phys.SE post. It appears in many textbooks, e.g. Goldstein & Ramond.

  2. The full Noether current also has an improvement term in case of quasisymmetry. The explicit form of the improvement term is usually calculated case by case.

  3. Srednicki, Banados, Tong, Peskin & Schroeder, Schwartz, etc, only consider vertical transformations. Horizontal transformations are dealt with by first transcribing them as vertical transformations. Not surprisingly, the improvement term they derive reproduce eq. (2.141').

  4. Generically, improvement terms do not have to be of the form of eq. (2.141').

Qmechanic
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