Timeline for Maxwell's Correction to Ampere's Law
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Jan 7, 2013 at 23:40 | comment | added | Zo the Relativist | I should also add that, at this level, talking about $\mathbf{A}$ is probably going to just create more confusion than usefulness. There is a way to write down maxwell's equations that completely removes any reference to $\mathbf{E}$ and $\mathbf{B}$ and instead talks only about $\mathbf{A}$ (or, if you are not doing special relativity, $\vec A$ and $\phi$), but that is way beyond the level of this question. | |
| Dec 9, 2012 at 20:44 | comment | added | Art Brown | @ZAC: Re 2) I want to add that I am relying on superposition of sources: a) a set of fixed charges produces $\boldsymbol{E_l}$ (and crucially produces no magnetic fields), and b) a set of time-varying currents (with 0 charge density) somewhere off-stage produces the time varying $\boldsymbol{B}$ fields you specify, and hence $\boldsymbol{E_t}$. By superposition, you can add the two $\boldsymbol{E}$ fields to get the result. Superposition works because the equations are linear in the sources. | |
| Dec 9, 2012 at 18:20 | comment | added | Art Brown | @ZAC: Ah, I think I understand your question a bit better now. 1) Yes, there's only one $\boldsymbol{E}$ in the equations; it's not different from one to the next. 2) You can decompose that field into "divergence-less" ($\boldsymbol{E_t}$, aka transverse) and "curl-less" ($\boldsymbol{E_l}$ aka longitudinal or irrotational) components, $\boldsymbol{E=E_t+E_l}$; the transverse element is zeroed by the divergence operator of Gauss law, and the irrotational element is zeroed out by the curl in Faraday's law. 3) Yes, it's possible to have exactly the same V function in the two cases. | |
| Dec 9, 2012 at 13:50 | comment | added | ZAC | So just to confirm: the Electric Field in Gauss' Law and in Faraday's Law is the total electric field due to both changing magnetic fields and static charges? It is not that the E in Faraday's Law has a different meaning to the E in Gauss' Law and that you then have to add the two resulting quantities together? And it is possible to make the V in E=-grad(V)-dA/dt, when the dA/dt term and a changing magnetic field are present, to be exactly the same function V as when the changing magnetic field is not present and only the static charges are responsible for E? | |
| Dec 8, 2012 at 21:51 | history | answered | Art Brown | CC BY-SA 3.0 |