Timeline for Properties of the dot product
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 1, 2020 at 14:37 | history | edited | Charles Hudgins | CC BY-SA 4.0 |
added 176 characters in body
|
| Nov 1, 2020 at 0:51 | history | edited | Charles Hudgins | CC BY-SA 4.0 |
deleted 3 characters in body
|
| Oct 31, 2020 at 20:32 | history | edited | Charles Hudgins | CC BY-SA 4.0 |
deleted 38 characters in body
|
| Oct 31, 2020 at 20:25 | history | edited | Charles Hudgins | CC BY-SA 4.0 |
deleted 1 character in body
|
| Oct 31, 2020 at 20:20 | comment | added | Charles Hudgins | @Toba I added a rigorous proof for posterity. Give it a look if you're curious. | |
| Oct 31, 2020 at 20:19 | history | edited | Charles Hudgins | CC BY-SA 4.0 |
added 969 characters in body
|
| Oct 31, 2020 at 20:10 | comment | added | Toba | Yes I understand that E and grad V are only approximately constant but we assumed that the path length is so small that the changes are insignificant at the level of accuracy at which we are working. I don't think a rigorous proof will be necessary since most of the derivations in Griffiths use such heuristic assumptions. | |
| Oct 31, 2020 at 19:37 | comment | added | Charles Hudgins | The path would have to be infinitesimal for $\vec{E}$ and $\nabla V$ to be constant along its length. Rather, we choose the path short enough that the deviations from being constant are not large enough to make the integrand nonpositive. Thus the integrand is strictly positive on a set of strictly positive measure, which means its integral is strictly positive, contradicting our initial assumption. If I'm not making sense, I can make the proof fully rigorous if you'd like. | |
| Oct 31, 2020 at 19:33 | comment | added | Toba | The short path is not infinitesimal in the argument I gave. It only has to be small enough that E and grad V can be considered constant along its length. | |
| Oct 31, 2020 at 19:28 | comment | added | Charles Hudgins | Because of the smoothness of $\vec{F}$ (which we assume), there must be some path of finite length along which $\vec{F} \cdot d\vec{l} > 0$. So we really are talking about an integral and not merely $\vec{F} \cdot d\vec{l}$. | |
| Oct 31, 2020 at 19:24 | comment | added | Charles Hudgins | The difference is that the "short" path used in this argument isn't infinitesimal. It has finite length. I didn't want to spell it out here, but this is basically the same argument as the one presented here: math.stackexchange.com/questions/274702/… | |
| Oct 31, 2020 at 19:18 | comment | added | Toba | There is still an implicit assumption that E.dl=-(grad V).dl . This is because in the proof of the statement you gave, a very short path was considered. If we take a very short path in the equation given in Griffiths, the integral signs disappear (since there's nothing to sum along such a path) and E.dl=-(grad V).dl results. Pardon my clumsy notation, I'm unable to use Mathjax on my smartphone. | |
| Oct 31, 2020 at 19:06 | history | edited | Charles Hudgins | CC BY-SA 4.0 |
added 16 characters in body
|
| Oct 31, 2020 at 18:54 | history | answered | Charles Hudgins | CC BY-SA 4.0 |