Timeline for Is the equation $[\nabla_{\mu},\nabla_{\nu}]=F_{\mu\nu}$ correct? If yes, how does it have to be interpreted?
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 21, 2021 at 6:41 | comment | added | oliver | @Filippo: I was once like you :-) But I noticed that I had to decide for one side because it is exhausting to express everything in two slightly different languages, and most of all, making the tools and using them at the same time. And so I decided for the evil side ;-) But there are many physicists more capable than me, who see nothing special in mastering physics and mathematics at the same time. | |
| Mar 20, 2021 at 15:44 | comment | added | Filippo | @oliver "Street language" :D I am currently a physics student (maybe I'll change to mathematics after my bachelor degree), but indeed, I have a tendency towards rigorous mathematics and I'm not very good at understanding the street language of physicists ^^ | |
| Mar 19, 2021 at 20:28 | comment | added | oliver | @Filippo: I see that you are a little tending to the mathematical side of things (do you happen to be a mathematician?), and from this point of view it is understandable that you probably expected to get a declaration of $\psi$ as a function. For me as a physicist it is just natural to assume that $\psi$ in the context of a field theory is a function. After all, physicists are often a little lazy with notation, but it helps getting quicker to the point, at least if those with whom one communicates speak the same kind of "street language". | |
| Mar 19, 2021 at 20:21 | comment | added | Filippo | @oliver I thought that the $\psi$ was just a decoration and that the calculation could be performed without it. Thus, I didn't get that you were using the product rule. This is one of the situations were naively doing the calculation without the $\psi$ leads to a wrong result. Thank you for not only doing the calculation, but also talking about the physical interpretation! | |
| Mar 19, 2021 at 19:49 | comment | added | Filippo | @Jbag1212 Now I understand, thank you! | |
| Mar 19, 2021 at 19:46 | comment | added | Jbag1212 | @Filippo What is the issue with the linked picture? The pink terms come from applying the product rule to the corresponding blue and orange terms. They then cancel with green and red term. In other words, they were not "added" they came about from properly applying the product rule | |
| Mar 19, 2021 at 19:36 | comment | added | Filippo | @Jbag1212 I have the feeling that there has been a misconception: The picture doesn't show my calculations, it's a screenshot of oliver's answer. | |
| Mar 19, 2021 at 19:32 | comment | added | Jbag1212 | @Filippo You haven't properly applied the product rule to the orange term or the blue term $\partial_\nu (A_\mu \psi) \neq (\partial_\nu A_\mu) \psi $ | |
| Mar 19, 2021 at 18:26 | comment | added | Filippo | Can you explain, please? | |
| Mar 19, 2021 at 18:23 | comment | added | oliver | @Filippo: you are using highly complicated notation from differential geometry and don't know how to apply the product rule of differentiation...? | |
| Mar 19, 2021 at 18:21 | comment | added | Filippo | The picture I linked to shows that the rosa terms don't appear before the fourth equal sign. | |
| Mar 19, 2021 at 18:19 | comment | added | oliver | What do you mean by "After adding the rosa terms"? I'm just using algebra and the rules of differentiation. No assumptions. Nothing to add. | |
| Mar 19, 2021 at 18:14 | comment | added | Filippo | Well, that's what you do to obtain the desired result, don't you? $\textbf{After}$ adding the rosa terms. | |
| Mar 19, 2021 at 18:11 | comment | added | oliver | @Filippo: what about cancelling the red term with the second pink term, and the first pink term with the green term? | |
| Mar 19, 2021 at 18:03 | comment | added | Filippo | Here's how I came to my conclusion. The pink terms are the ones I was talking about. | |
| Mar 19, 2021 at 17:52 | comment | added | oliver | @Filippo: not at all, after the fourth equal sign, both $A_\mu\partial_\nu$ as well as $A_\nu\partial_\mu$ appear each once with a "+" and once with a "-" sign. Hence they both cancel. | |
| Mar 19, 2021 at 17:37 | comment | added | Filippo | It seems like you are assuming $A_{\mu}\partial_{\nu}=A_{\nu}\partial_{\mu}$: After the fourth equal sign you add $A_{\nu}\partial_{\mu}$ and subtract $A_{\mu}\partial_{\nu}$. Otherwise, you would have ended up with the equation $[\nabla_{\mu},\nabla_{\nu}]=F_{\mu\nu}+A_{\mu}\partial_{\nu}-A_{\nu}\partial_{\mu}$ like I did (see my question). | |
| Mar 19, 2021 at 17:16 | history | edited | oliver | CC BY-SA 4.0 |
added 12 characters in body
|
| Mar 19, 2021 at 17:10 | history | edited | oliver | CC BY-SA 4.0 |
added 12 characters in body
|
| Mar 19, 2021 at 17:04 | history | edited | oliver | CC BY-SA 4.0 |
added 359 characters in body
|
| Mar 19, 2021 at 16:55 | history | answered | oliver | CC BY-SA 4.0 |