Timeline for Do divergence and curl of Lorentz force have some physical meaning?
Current License: CC BY-SA 3.0
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| Mar 15, 2023 at 17:05 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Nov 2, 2016 at 15:15 | history | tweeted | twitter.com/StackPhysics/status/793833822483800064 | ||
| Nov 2, 2016 at 7:45 | history | edited | Les Adieux | CC BY-SA 3.0 |
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| Nov 2, 2016 at 7:40 | history | edited | Les Adieux | CC BY-SA 3.0 |
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| Nov 2, 2016 at 7:38 | comment | added | Les Adieux | @RobJeffries Uhh Thank you so much for that page! Going to edit the edit lol. What about the divergence instead? | |
| Nov 2, 2016 at 7:27 | comment | added | ProfRob | As you have perhaps seen, the curl of the Lorentz force is only zero for static fields. So your edit is wrong. physics.stackexchange.com/questions/118498/… | |
| Nov 2, 2016 at 7:14 | history | edited | Les Adieux | CC BY-SA 3.0 |
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| Feb 3, 2016 at 8:04 | comment | added | FraSchelle | The physical meaning of $\nabla\times F$ and $\nabla\cdot F$ directly, as far as I can see, are the usual ones : the longitudinal and transverse component of the force field. But be warn that the transverse and longitudinal components of the force is not related directly to the transverse and longitudinal component of the gauge fields, because of the cross product $v\times B$ in the Lorentz force. Using the Maxwell equations (as you tried) it must be possible to express the longitudinal and transverse component of the force using only one component of the field. | |
| Feb 3, 2016 at 8:00 | comment | added | FraSchelle | All these expressions make perfect sense inside some integrals, and using the Stokes theorem. Then you can relate the work -- being $\int F\cdot dr=\iint\nabla\times F\cdot dS$ in terms of the voltage drop $\int E\cdot dr=\int\nabla\varphi\cdot dr=\varphi_{2}-\varphi_{1}$ for instance. For the magnetic field, it may be more transparent to use the gauge potential. In integrals there is no difference of course. | |
| Jan 27, 2016 at 15:09 | history | edited | Les Adieux | CC BY-SA 3.0 |
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| Jan 27, 2016 at 15:02 | comment | added | honeste_vivere | I think you are missing a dot product symbol in the first boxed equation between the $\mathbf{v}$ and ( )'s. | |
| Jan 27, 2016 at 14:52 | history | asked | Les Adieux | CC BY-SA 3.0 |