Timeline for Why is $(\partial_\mu F_{\alpha\beta})F^{\alpha\beta}=F_{\alpha\beta}\partial_\mu(F^{\alpha\beta})$?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Nov 18 at 23:14 | vote | accept | user410662 | ||
| Nov 18 at 23:14 | vote | accept | user410662 | ||
| S Nov 18 at 23:14 | |||||
| S Nov 18 at 23:13 | vote | accept | user410662 | ||
| Nov 18 at 23:13 | |||||
| Jul 19 at 3:07 | history | edited | Qmechanic♦ |
edited tags; edited tags
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| Jul 19 at 2:23 | answer | added | user410662 | timeline score: 1 | |
| Jul 19 at 1:58 | comment | added | user410662 | I have realized the error in my line of thinking, see comment on other reply. | |
| Jul 19 at 1:57 | vote | accept | user410662 | ||
| S Nov 18 at 23:13 | |||||
| Jul 19 at 1:57 | vote | accept | user410662 | ||
| Jul 19 at 1:57 | |||||
| Jul 19 at 1:51 | answer | added | hft | timeline score: 1 | |
| Jul 19 at 1:41 | comment | added | naturallyInconsistent | It will be a million times more obvious that they are the same if you realise that the two terms are both $F_{\alpha\beta}g^{\alpha\gamma}g^{\beta\delta}\partial_\mu F_{\gamma\delta}$ | |
| S Jul 19 at 1:38 | review | First questions | |||
| Jul 19 at 1:55 | |||||
| S Jul 19 at 1:38 | history | asked | user410662 | CC BY-SA 4.0 |