Timeline for Maxwell equations: exterior derivative vs partial derivative
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 22, 2022 at 17:03 | comment | added | Cameron Gibson | I cannot think of an obvious way to do this (of course something that is identically zero can be rewritten in a trivial way). I do not know why you would want to do this though. Your equation is just equivalent to the fact that partial derivatives commute, which results in zero when contracted with the Levi-Civita symbol. | |
| Feb 22, 2022 at 10:01 | comment | added | gammadragon | I would like to write the equation in my last comment in terms of an exterior derivative if possible. | |
| Feb 22, 2022 at 0:38 | comment | added | Cameron Gibson | It’s already written in terms of differential forms: $F$ is a differential form. What precisely are you after? | |
| Feb 21, 2022 at 19:00 | comment | added | gammadragon | I don't mean the one-form current $J$ from the Maxwell equation, I mean the conserved two-form current which satisfies $$\partial_\mu (\star F)^{\mu\nu}=\partial_\mu J^{\mu\nu}=0.$$ I want to know how I should write the equation for this current in differential forms. | |
| Feb 21, 2022 at 17:57 | comment | added | Cameron Gibson | $J=d \star F$, so $dJ=0$ since $d^2=0$. Is this what you are after? Why are you saying $J=\star F$? | |
| Feb 21, 2022 at 17:50 | comment | added | gammadragon | Thanks for your answer! What then is the differential form version of my equation? If I set $J^{\mu\nu}=\star F$ Would it be something like $$d(\star J)=0,$$ because that seems odd to me as it would just return $dF=0$ right? | |
| Feb 18, 2022 at 19:12 | history | edited | Cameron Gibson | CC BY-SA 4.0 |
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| Feb 18, 2022 at 18:59 | history | edited | Cameron Gibson | CC BY-SA 4.0 |
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| Feb 18, 2022 at 18:54 | history | answered | Cameron Gibson | CC BY-SA 4.0 |