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Peskin & Schroeder's expression of the Noether current If a (quasi-)symmetry is defined as a transformation that changes the action by a surface term i.e. $$S\to S'=S+\int d^4x \partial_\mu K^\mu(\phi_a),\tag{1}$$ or equivalently, the Lagrangian changes by a 4-divergence, $$\mathscr{L}\to\mathscr{L}'=\mathscr{L}+\partial_\mu K^\mu,\tag{2}$$ then considering transformations on the fields only, the expression of the Noether current turns out to be $$j^\mu(x)=\sum\limits_a\frac{\partial \mathscr{L}}{\partial(\partial_\mu\phi_a)}\delta\phi_a -K^\mu.\tag{3}$$ P & S give the example of an internal transformation where $K^\mu=0$, and a spacetime transformation (namely, spacetime translation) under which $K^\mu\neq 0$.


Lewis Ryder's expression of the Noether current Here, the symmetry of the action is defined as a transformation that leaves the action invariant i.e. $$S\to S'=S.\tag{4}$$ Considering transformations $$x^\mu\to x^{\mu'}=x^\mu+\delta x^\mu,\\ \phi(x)\to\phi'(x)=\phi(x)+\delta\phi(x),$$ they derive the following expression for the Noether current $$j^\mu(x)=\sum\limits_a\frac{\partial \mathscr{L}}{\partial(\partial_\mu\phi_a)}\delta\phi_a -\Theta^{\mu\nu}\delta x_\nu\tag{5}$$ where $\Theta^{\mu\nu}$ is the stress-energy tensor given by $$\Theta^{\mu\nu}=\sum\limits_a\frac{\partial \mathscr{L}}{\partial(\partial_\mu\phi_a)}\partial^\nu\phi_a -\eta^{\mu\nu}\mathscr{L}.\tag{6}$$


Question $1$ Between $(3)$ and $(5)$ which expression of the Noether's current is more general?

Question $2$ By generalizing Ryder's definition of symmetry $(4)$ (to a quasi-symmetry, i.e., $(1)$), we will obtain $$j^\mu(x)=\sum\limits_a\frac{\partial \mathscr{L}}{\partial(\partial_\mu\phi_a)}\delta\phi_a -\Theta^{\mu\nu}\delta x_\nu-K^\mu.\tag{7}$$ Should $(7)$ be regarded as the most general expression of the Noether current?

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1 Answer 1

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Yes, OP's eq. (7) is the general expression for the Noether current of a single global continuous symmetry in field theory. Peskin & Schroeder (3) are only considering purely vertical transformations, cf. e.g. this related Phys.SE post.

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    $\begingroup$ Sorry to jump in so late: although Peskin & Schroeder consider purely vertical transformations, they give as an example a horizontal tranformation $x^{\mu}\rightarrow x^{\mu}-a^{\mu}$, and this transfomation generates a Lagrangian variation $\delta\mathcal{L}=\partial_{\mu}(a^{\nu}\delta^{\mu}_{\phantom{\mu}\nu}\mathcal{L})$ which they interpret as the improved current. 1. Aren't we double counting things, given that in (7) here we already accounted for horizontal variations? 2. Shouldn't we, on account of horizontal variations, force a strict symmetry instead of a quasi-symmetry? $\endgroup$
    – Condereal
    Feb 13 at 11:26
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    $\begingroup$ Hi Condereal. Thanks for the feedback. 1. It is true that quasi-symmetries comes in equivalence classes, where each equivalence class corresponds to a different conservation law, cf. e.g. my Phys.SE answer here. 2. No, it turns out that's not necessary. $\endgroup$
    – Qmechanic
    Feb 13 at 11:38
  • $\begingroup$ Thanks for the answer! What should I do with the improved current term if I want to write the Noether current using equation (7) in the OP in the case of a translation like $x^{\mu}\rightarrow x^{\mu}-a^{\mu}$? $\endgroup$
    – Condereal
    Feb 13 at 11:42
  • $\begingroup$ Have you studied how the corresponding ambiguity works in classical mechanics, cf. e.g. my Phys.SE answer here? $\endgroup$
    – Qmechanic
    Feb 13 at 11:50
  • $\begingroup$ I read and re-read your post suggestion, but it keeps confusing me. Are you saying, that using or not the improved current term depends on whether the Lagrangian depends on $x^{\mu}$ or not? $\endgroup$
    – Condereal
    Feb 13 at 13:14

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