Skip to main content
10 events
when toggle format what by license comment
Sep 2, 2015 at 20:35 history edited Qmechanic CC BY-SA 3.0
added 3 characters in body; edited tags
Sep 1, 2015 at 19:41 vote accept knzhou
Sep 1, 2015 at 12:47 answer added Qmechanic timeline score: 10
Aug 26, 2015 at 17:07 history tweeted twitter.com/#!/StackPhysics/status/636585651866062848
Aug 25, 2015 at 4:02 answer added Timaeus timeline score: 7
Aug 25, 2015 at 3:07 comment added Omry The new lagrangian has two terms, the second of which is identical to the original one. The first one, however, always can be chosen to vanish (Lorenz gauge), and I don't think this is the case for its original term. ($\partial_\mu A_\nu \partial^\nu A^\mu$)
Aug 25, 2015 at 0:12 comment added or1426 I'm slightly ashamed to admit I managed to forget the swapping the derivatives trick. Thanks to your prompt I performed the calculation and obtained exactly the same results. I'm not sure why this could be occurring however given that the canonical momentum has changed it seems plausible that the canonical position is changed by the "swapping" of derivatives as well (as this procedure involves integrating over $x^\mu$). Both the momenta are gauge invariant so the only thing I can think of is some transformation happening to $x^\mu$. I'd be very interested in an answer to this question!
Aug 24, 2015 at 23:27 comment added knzhou We get terms like $\partial_\mu A_\nu \partial^\nu A^\mu$. We use integration by parts to swap the derivatives, getting $\partial^\nu A_\nu \partial_\mu A^\mu$. It's allowed because the $\mathscr{L}$ is always in a $d^4x$ integral to get $L$ for any physical applications.
Aug 24, 2015 at 23:03 comment added or1426 Could you expand a bit on your method "expanding and integrating by parts"? I'm struggling to see how you end up integrating anything.
Aug 24, 2015 at 21:59 history asked knzhou CC BY-SA 3.0