Timeline for Can ideal dipoles be associated to a covariant four-current?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 16, 2023 at 18:44 | vote | accept | Woe | ||
| Jun 16, 2023 at 18:44 | history | bounty ended | Woe | ||
| Jun 16, 2023 at 18:44 | comment | added | Woe | I understand. I will accept the answer now, thank you very much. | |
| Jun 16, 2023 at 9:24 | comment | added | LPZ | Well your formula for delta and the partial derivatives are false to start with. Furthermore, even your integration is not done carefully. You need to integrate over a space like slice, which changes due to the LR (relativity of simultaneity). | |
| Jun 16, 2023 at 9:17 | comment | added | LPZ | Just to be clear, you want to know were was your mistake in your original derivation? | |
| Jun 15, 2023 at 23:40 | comment | added | Woe | Correct, thank you. I was troubled by these deltas. On the other hand, generally, I see from your answer that if $\mathbf{p}$ and $\mathbf{m}$ transform the way you showed, the electromagnetic sources are indeed covariant (provided we consider the extra term from the continuity equation). Therefore, it would remain only to show these transformations explicitly. It should, of course, be possible to do that from the integral definitions of $\mathbf{p}$ and $\mathbf{m}$ as I tried, but my results are incorrect for $\mathbf{p}$, and this is not due to the deltas. Do you have any clues on that? | |
| Jun 15, 2023 at 22:57 | comment | added | LPZ | Actually I did put the intermediate step with the chain rule. They are equal simply due to the general fact: $$\partial_tf(x+ut) = u\partial_xf(x+ut)$$ | |
| Jun 15, 2023 at 22:52 | comment | added | Woe | In these new added steps, you wrote $(\partial_x'-u \partial_t')\delta(x'+u t') = (1-u^2)\partial_x' \delta(x'+u t')$. Taking one extra intermediate step for this calculation (the chain rule), one would have $(\partial_x'-u \partial_t')\delta(x'+u t') = \delta'(x'+u t')-u^2\delta'(x'+u t')$. These $\delta'$ are derivatives w.r.t. different variables ($x'$ and $t'$), no? Why in your result you took them to be equal? | |
| Jun 15, 2023 at 8:00 | history | edited | LPZ | CC BY-SA 4.0 |
added 171 characters in body
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| Jun 15, 2023 at 7:59 | comment | added | LPZ | I don't think that I made a mistake. I've added intermediate steps for clarity. | |
| Jun 14, 2023 at 23:59 | comment | added | Woe | You wrote $\mathbf{p}\cdot\nabla = \gamma^{-1}p_x\partial_x'+\mathbf{p}_{\perp}\cdot\nabla_{\perp}$. Should not it be $\mathbf{p}\cdot\nabla = \gamma p_x(\partial_x'-u\partial_t')+\mathbf{p}_{\perp}\cdot\nabla_{\perp}$? Even if the time derivative term is zero, your version has $\gamma^{-1} p_x$ while mine has $\gamma p_x$. I think this is the only difference in the $p_x$ transformation, even though I obtained it by integration. | |
| Jun 13, 2023 at 12:49 | history | answered | LPZ | CC BY-SA 4.0 |