In my course, we considered the following Lagrangian :

$$\mathcal{L}(\phi,\phi^*, \partial \phi, \partial \phi^*) = \partial_\mu \phi \partial^\mu \phi^* - m^2 \phi^* \phi $$

We said that we want the action to be invariant under Poincaré transformation.

So :

$$ \int d^4x ~\mathcal{L}(\phi(x),\phi^*(x), \partial \phi(x), \partial \phi^*(x)) = \int d^4x' ~\mathcal{L}(\phi'(x'),\phi'^*(x'), \partial \phi'(x'), \partial \phi'^*(x'))$$

With $$ x \rightarrow x'=\Lambda x+a$$ $$ \phi \rightarrow \phi' $$ $$ \phi'(x')=\phi(x)$$

First question : When we have conservation of the action, the laws of movement will be the same with the fields before transformation and after, right ? And is the invariance of action (at a constant term not depending on $\phi$) a necessary condition for the invariance of the laws of movements ? Or is it just a sufficient condition ?

Explanations around the next question.

Now we consider a really general transformation such as :

$$ x'= x+\delta x$$

We assume that for this transformation we have :

(*) $$ \mathcal{L}(\phi'(x'),\phi'^*(x'), \partial \phi'(x'), \partial \phi'^*(x')) |det(J)|= \mathcal{L}(\phi(x),\phi^*(x), \partial \phi(x), \partial \phi^*(x)) + \partial_\mu V^\mu(\phi(x)) $$

(J is the jacobian of the transformation)

I agree that if (*) is true I will have conservation of the action and thus invariance of the laws of motion. By Noether theorem, I know that to such transformations I will have conserved quantities.

Second question :

So, in practice, when we have a Lagrangian, one of the "main thing" to do is to look for infinitesimal transformations that conserv the action (not the Lagrangian but the action), because for such quantities we have the Noether theorem that will give us usefull informations about some quantities that will be conserved. But we could imagine other quantities conserved (the Noether theorem just give us an easy way to find some conserved quantities, but not all of them).

Did I understood the philosophy around conserved quantities and invariance of Lagrangian/Action well ???

  • 1
    $\begingroup$ First question:yes.Second question: just sufficient. I am too busy to write an answer. $\endgroup$ – Valter Moretti Mar 25 '17 at 16:49
  • $\begingroup$ I could expand on the 2nd question. In the Lagrangian formulation, as you said, it is not at all obvious that the presence of conserved quantities implies the existence of continuous symmetries. However, in the Hamiltonian formalism the Noether theorem clearly shows the one-to-one connection between symmetries and conservation law. $\endgroup$ – user47224 Mar 25 '17 at 20:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.