Wikipedia writes that

"if the action has an infinite-dimensional Lie algebra of infinitesimal symmetries parameterized linearly by $k$ arbitrary functions and their derivatives up to order $m$, then the functional derivatives of $L$ satisfy a system of $k$ differential equations."

Such a transformation may look like $$\delta\varphi = \text{i} \omega_a(x)T^a F[\varphi] + \text{i}\sum\limits_{j=1}^{m} F^{\mu_1..\mu_j}[\varphi](\partial_{\mu_1}..\partial_{\mu_j})\omega_a(x)T^a \quad a=1,..,k$$

They specify the parameters $k$ and $m$, however only state how $k$ is interpreted. What about $m$? How does the choice of $m$ affect Noether's second theorem?


1 Answer 1


To quote a recent paper, arXiv:1510.07038,

Noether’s second theorem gives us strong identities, which constrain the form of the equations of motion and the current $j^\mu$.

In particular, for the case $m=1$, they show that

$$ E^k F^a_k[\varphi] - \partial_\mu\big( E^k F^{a\mu}_k[\varphi] \big) \stackrel{!}= 0 \tag{1}$$

where $E^k$ are the equations of motion for the field $\varphi_k$ and the functionals $F$ as specified in the question. Different choices of $m$ will alter the explicit shape of (1). However,

the authors know of no physically interesting examples.


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