Timeline for Why do we have redundant degrees of freedom?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 19, 2019 at 22:34 | comment | added | my2cts | In an electromagnetic wave there are 5 degrees of freedom. The amplitude, the polarisation and the three components of $\vec k$. The sixth degree of freedom is the Coulomb potential. | |
| Jul 19, 2019 at 20:08 | comment | added | InertialObserver | I see.. so we start with 4 DOFs, which for free EM waves is 2 redundant ones. To get the wave equations for E B we need to fix a gauge, which in turn removes 2 DOFs. It’s not obvious to me, however, that fixing a gauge will always lead to the correct physical DOFs | |
| Jul 19, 2019 at 20:02 | comment | added | knzhou | @InertialObserver I wouldn't phrase it that way -- it's more like the $A_\mu$ are the true DOFs the whole time, and the counting is $4 - 2 = 2$. The $E$ and $B$ fields are secondary things that you can define in terms of them that happen to be useful, but they aren't the primary objects. Nonetheless, you can still do the counting (as $6 - 4 = 2$), just not as naturally, since $E$ and $B$ aren't the DOFs in the Lagrangian. | |
| Jul 19, 2019 at 19:55 | comment | added | InertialObserver | i see what you're saying in the second paragraph.. For free EM, we start with 6 and then maxwells equations give us 4 more constraints leaving us with 2 DOFs. This doesn't still help me figure out how to count the gauge DOFs added by introducing $A_\mu$ | |
| Jul 19, 2019 at 19:28 | history | answered | knzhou | CC BY-SA 4.0 |