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Time-dependent electric field |
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| Aug19-12, 06:10 AM | #1 |
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Time-dependent electric field
Hello! This is my doubt:
I have a particle with charge q and mass m with a relativistic velocity v·u_x in a region of electric field E = E_0 cos (wt)·u_z. I want to calculate the time evolution of the velocity. I would use the 2nd law of Newton, so that dp / dt = F where p = gamma·m·v·u_x and F = q (E + v·u_x ^ B). The thing is that as the electric field is variable, it is assumed that there will be an induced magnetic field acting on the particle (or not?). The first thing is that I don't know if I have to consider a field B acting on the particle and the second is that, in that case, when I try to calculate B, I use Maxwell's equations but I get results that do not fit. So is there only electric force or no electric and magnetic force? Thank you. |
| Aug19-12, 07:14 AM | #2 |
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Mentor
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There should be a spot without magnetic fields - in a symmetric setup, this is the center. Everywhere else, it can get tricky.
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| Aug19-12, 08:43 AM | #3 |
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In his book, The Lightning Discharge, Uman considers models of varying levels of sophistication for computing time dependent electric and magnetic fields from Maxwell's equations. Buy it, it's cheap.
Here is a link to a variety of publications and patents dealing in transient electrical events, Helmholtz equations, etc. http://www.ee.psu.edu/Directory/Facu...lications.aspx Respectfully, Steve |
| Aug19-12, 10:00 AM | #4 |
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Time-dependent electric field
lailola - your problem is not specified terribly well, but a few general observations can be made. in the rest frame of said particle, only electric forces qE operate. In any other Lorentzian frame, the relativistically exact Lorentz force law expression F = q(E+vxB) applies, where E, B, and v are all evaluated in that frame. Sticking with that one frame, integration over time using F = q(E+vxB) = dp/dt = d(γmv)/dt then gives the velocity evolution, where γ = 1/√(1-(v/c)2). This assumes no implied back-reaction coupling to whatever source produces the applied field, and further assumes radiative back reaction is negligible. You have correctly identified the relativistic energy part of all this re coupling to velocity - the remainder seems straightforward to me (well ok it gets to be a messy integro-differential equation likely needing numerical evaluation - but then what's in the detail
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| Aug19-12, 10:51 AM | #5 |
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But..in the 'laboratory' rest frame, where the particle is moving with velocity v, is there an external magnetic field B (due to time-dependent E) that I have to consider? I don't know how could it be calculated..
thank you for your answers! |
| Aug19-12, 11:15 AM | #6 |
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| Aug19-12, 11:17 AM | #7 |
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Mentor
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In the lab frame, you can have a magnetic field.
And to evaluate this in the electron rest frame, you have to convert your fields - where the conversion (given by the velocity of the electron) changes in time. Not sure whether this helps or not... |
| Aug19-12, 11:34 AM | #8 |
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| Aug19-12, 11:43 AM | #9 |
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| Aug19-12, 11:50 AM | #10 |
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| Aug19-12, 12:00 PM | #11 |
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