GR8677 #22


Problem



Electromagnetism}Lorentz Transformation
When an electric field is Lorentztransformed, afterwards, there might be both a magnetic and electric field (in transverse components). Or, more rigorously, one has,
for motion in the direction.
(A) Obviously not. Suppose initially, one has just , afterwards, there's still .
(B) True, as can be seen from the equations above.
(C) Not true in general. Suppose . Afterwards, the transverse components are off by a even if there's no B field to start with.
(D) Nope.
(E) Mmmm... no need for gauge transformations.


Comments 
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danielsw98667 20191021 07:23:51 
Obviously, the answer is A if we will apply the principle of Special Relativity  Electromagnetic Field. random numbers generator

  fredluis 20190809 04:20:09  I believe Jeremy ans was better. In the copper there is no significant gain in number of conductive electron but decrease in their sweeping velocity because of the impurities seen as obstacles. pressure cleaning   joshuaprice153 20190808 07:37:32  Hey, would you mind if I share your blog with my twitter group? There’s a lot of folks that I think would enjoy your content. Please let me know. Thank you. towing service   djh101 20140827 16:38:09  You can eliminate almost all of the answers without really knowing anything about relativity.
A. Assume observer 1 = observer 2. According to this, an identity transformation will transform an electric field into a magnetic field.
C. From basic E&M, you should know that there's a little bit of symmetry going on between electricity and magnetism. With only that in mind, you can eliminate both C and D. They can't both be true, so by symmetry, neither is true.
D. See above.   wittensdog 20090925 17:53:58  In general, what happens is that the field tensor transforms with two factors of whatever matrix corresponds to the given Lorentz transformation. The conditions on the matrices that are in the Lorentz group are not terribly strict, and certainly we know from the simple case of a boost along the xaxis that the components can mix into each other (think about the usual space and time fourvector, and how it transforms).
In general, for a Lorentz transformation, whatever fourvector or tensor you have can have its components mix together pretty generally.
Also, at least in the classical realm, fields NEVER depend on gauge in any way. The issue of gauge fixing only ever comes up when you start working with potentials, which are not measurable. So immediately throw out anything involving gauge dependence when asked about fields.   FortranMan 20081017 23:15:56  For more on this topic, look around for Relativistic Electrodynamics, specifically in Griffiths section 12.3. Imagine the case of a large parallelplate capacitor moving perpendicular to their surface areas, and imagine what the Lorentz contraction does to the charge distributions on the plates. This ends up increasing the electric field that's perpendicular to the direction of motion. Note for a parallelplate capacitor situated such their motion is parallel to their surfaces, this contraction does not occur and the field remains unchanged.   grae313 20071007 15:50:57  On choice (E), does the Lorentz transformation already assume the Lorentz guage?
rorytheherb 20081006 10:22:16 
yeah I'm looking at griffiths EM and it's not really illuminating this particular question. i knew of course that E and B mix in S.T.R., but when I saw choice E, I recalled the Lorentz gauge and got duped. Can anyone clarify why how we can say E and B mix _without_ first specifying a gauge?

dean 20081009 21:00:55 
I'm not sure there's any relevance at all. Gauge transformations talk about potentials, not fields. A (the vector potential) may change depending on what gauge you're using, but B does not, and can not, else we'd have a physically different situation. Lorentz transformations talk about fields directly in different inertial frames.

 

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