- #1
sam_bell
- 67
- 0
Hi,
I keep running my brain in circles while trying to get a solid grip on Noether's theorem. (In Peskin and Schroeder they present this as a one-liner.) But I'm having trouble seeing the equivalence between "equations of motion are invariant" and "action is invariant (up to boundary term)". Now I know that when the equations of motion are satisfied then there is no change in the action for infinitesimal variations. More exactly for variations which are zero on the boundary. Thus, there is a solution field [itex]\phi_0[/itex], and neighboring fields [itex]\phi_0 + \delta\phi[/itex], all of which have the same boundary values. If I apply a symmetry transform [itex]U[/itex] on all these fields, then their boundary values need not all transform the same way (right?). If they don't have the same boundary values, then it doesn't feel like we should be comparing their action anymore. Or at least that we're comparing the wrong set of fields. And if that's the case, then who's to say that the transformed field [itex]U(\phi_0)[/itex] continues to be an extrema of the action (i.e. solution of equations of motion)?
Sam
I keep running my brain in circles while trying to get a solid grip on Noether's theorem. (In Peskin and Schroeder they present this as a one-liner.) But I'm having trouble seeing the equivalence between "equations of motion are invariant" and "action is invariant (up to boundary term)". Now I know that when the equations of motion are satisfied then there is no change in the action for infinitesimal variations. More exactly for variations which are zero on the boundary. Thus, there is a solution field [itex]\phi_0[/itex], and neighboring fields [itex]\phi_0 + \delta\phi[/itex], all of which have the same boundary values. If I apply a symmetry transform [itex]U[/itex] on all these fields, then their boundary values need not all transform the same way (right?). If they don't have the same boundary values, then it doesn't feel like we should be comparing their action anymore. Or at least that we're comparing the wrong set of fields. And if that's the case, then who's to say that the transformed field [itex]U(\phi_0)[/itex] continues to be an extrema of the action (i.e. solution of equations of motion)?
Sam