Timeline for Are all Lagrangians translationally invariant?

5 events

when toggle format	what		by	license	comment
May 1, 2020 at 19:54	vote	accept	Hermitian_hermit
Apr 28, 2020 at 4:15	comment	added	Peter Kravchuk		@Matt0410 Re your last comment: yes, what you say is true, but this by itself does not imply anything. You still need to use Noether's theorem, and in the proof of that theorem equation (1.36) would fail if there were an explicit $x$-dependence in the Lagrangian.
Apr 28, 2020 at 0:39	comment	added	Hermitian_hermit		Surely if I transform my fields via any infinitesimal space-dependent transformation, I would have, by Taylor’s theorem, that $\mathcal{L}(x-\epsilon^\mu(x)) = \mathcal{L} -\epsilon^\mu \partial_\mu \mathcal{L}$. This doesn’t care about the exact functional form of the Lagrangian, just that it is some function of x? That is a function of x when a field has been substituted in
Apr 28, 2020 at 0:32	comment	added	Hermitian_hermit		Thank you for your answer. In equation (1.53) of the notes, it is stated that under a Lorentz transformation the field varies as $ \delta \phi = - \omega^\mu_\nu x^\nu \partial_\mu \phi $. Tong then states that the Lagrangian transforms the same way too as $ \delta \mathcal{L} = - \omega^\mu_\nu x^\nu \partial_\mu \mathcal{L}$ which can be written as a total derivative. Surely no assumption of Lorentz invariance has been made?
Apr 27, 2020 at 23:27	history	answered	MannyC	CC BY-SA 4.0