abskebabs Posted May 30, 2007 Share Posted May 30, 2007 Hi everybody. This may have been mentioned before, and I apologise if it had, but this has been puzzling me just now. I know that a moving charge generates a magentic field, but all motion is relative, so if I was moving at the speed of the charge parallel to it, would I still detect a magnetic field? In the framework of relativity how can we answer this question? Link to comment Share on other sites More sharing options...
timo Posted May 30, 2007 Share Posted May 30, 2007 - No. If the charge was at rest relative to you, then you wouldn't detect a magnetic field. - You get the answer either directly from the Maxwell equations or, if you prefer it more "relativity-like" by transforming the field-strength tensor [math]F^{\mu \nu} = \left( \begin{array}{cccc} 0 & E_x & E_y & E_z \\ -E_x & 0 & B_z & -B_y \\ -E_y & -B_z & 0 & B_x \\ -E_z & B_y & -B_x & 0 \end{array} \right) [/math] according to the normal transformation rules for tensors. Link to comment Share on other sites More sharing options...
Klaynos Posted May 30, 2007 Share Posted May 30, 2007 I've just done an electrodynamics course (the exam was yesterday), and it really is quite interesting how moving charges give results that look so much like relativity, which is pretty much sensible when you think about it Link to comment Share on other sites More sharing options...
Meir Achuz Posted May 30, 2007 Share Posted May 30, 2007 I've just done an electrodynamics course (the exam was yesterday), and it really is quite interesting how moving charges give results that look so much like relativity, which is pretty much sensible when you think about it I am not sure what you mean by "look so much like relativity,". Only relativity can give the fields of a moving charge. There is no other way. Link to comment Share on other sites More sharing options...
Klaynos Posted May 30, 2007 Share Posted May 30, 2007 I am not sure what you mean by "look so much like relativity,".Only relativity can give the fields of a moving charge. There is no other way. Yes, but we annalysed it without being told what we where doing, so lots didn't realise that doing the t - t_c stuff in the charge density was the same as what we did in relativity the year before so when the gamma coeficient drops out it just seemed nice... Link to comment Share on other sites More sharing options...
abskebabs Posted May 30, 2007 Author Share Posted May 30, 2007 Thanks for your responses, I understand now, and feel I should have thought a little harder about the question before posting it; it would have been more rewarding to reach the conclusion myself. It's interesting how there seems to be a certain kind of symmetry here, in the case that we only need to relate observables we see in our own frame to accuarately work out other relative quantities, and again this is in line with the postulate that the laws of physics are the same in all inertial reference frames! I should have realised that if a magnetic field were to be detected moving alongside a moving charge, this would real an "absolute velocity" which shouldn't be allowed by principles of relativity. We can calculate quantities in this way with the minimium amount of available information about "observables." Wonderful:-) Link to comment Share on other sites More sharing options...
Tom Mattson Posted May 31, 2007 Share Posted May 31, 2007 Hi everybody. This may have been mentioned before, and I apologise if it had, but this has been puzzling me just now. I know that a moving charge generates a magentic field, but all motion is relative, so if I was moving at the speed of the charge parallel to it, would I still detect a magnetic field? In the framework of relativity how can we answer this question? Here's another puzzle for you, along the same lines. Say you have a pure electric field in one inertial frame. Can you perform a Lorentz transformation to some other inertial frame in which the field is purely magnetic? Hint: You can get the answer fairly quickly by considering the field strength tensor in Post #2. Link to comment Share on other sites More sharing options...
abskebabs Posted May 31, 2007 Author Share Posted May 31, 2007 Here's another puzzle for you, along the same lines. Say you have a pure electric field in one inertial frame. Can you perform a Lorentz transformation to some other inertial frame in which the field is purely magnetic? Hint: You can get the answer fairly quickly by considering the field strength tensor in Post #2. Tbh im not sure, but I could give you an answer to the question though I doubt its correctness. I do not understand fully understand what all the terms in the tensor mean, indeed I have only recently acquired a vague notion of what a tensor actually is. Nevertheless, to answer your question; I would go with no, because it doesn't matter what velocity you are travelling with respect to a charge/electric field, you always need an electric field for there to be a magnetic field, and that electric field needs to be in relative motion to whatever is measuring the magnetic field. Therefore as you cannot have the latter without the former, how could you have a purely magnetic field? Link to comment Share on other sites More sharing options...
Severian Posted May 31, 2007 Share Posted May 31, 2007 I've just done an electrodynamics course (the exam was yesterday), and it really is quite interesting how moving charges give results that look so much like relativity, which is pretty much sensible when you think about it That's because it is relativity! Maxwell's equations are relativisticly invariant (which is not bad for 1861, being 44 years before Einstein's Special relativity in 1905). Link to comment Share on other sites More sharing options...
abskebabs Posted June 1, 2007 Author Share Posted June 1, 2007 That's because it is relativity! Maxwell's equations are relativisticly invariant (which is not bad for 1861, being 44 years before Einstein's Special relativity in 1905). Interestingly how did maxwell himself interpret the speed of light predicted from the permitivity and permeabillity of free space? Did he think that the so called predicted speed was relative to the vacuum, or not at all? I guess if he lived a bit longer, he could have himself realised what this predicted speed meant and beat Einstein to being the founder of relativity by a few decades:D ! I remember my lecturer saying that one time. Link to comment Share on other sites More sharing options...
Sepiraph Posted June 1, 2007 Share Posted June 1, 2007 That's because it is relativity! Maxwell's equations are relativisticly invariant (which is not bad for 1861, being 44 years before Einstein's Special relativity in 1905). Being a E & M theory, it has to be if it was to be correct ... Although I think Einstien deserves the credit for special relativity, his insight goes beyond merely a E & M theory. Link to comment Share on other sites More sharing options...
Klaynos Posted June 1, 2007 Share Posted June 1, 2007 That's because it is relativity! Maxwell's equations are relativisticly invariant (which is not bad for 1861, being 44 years before Einstein's Special relativity in 1905). Yes I suppose that was more my point Link to comment Share on other sites More sharing options...
Meir Achuz Posted June 1, 2007 Share Posted June 1, 2007 That's because it is relativity! Maxwell's equations are relativisticly invariant (which is not bad for 1861, being 44 years before Einstein's Special relativity in 1905). But Max could not find the field of a moving charge, until Einstein and Lienard-Wiechert. Link to comment Share on other sites More sharing options...
Norman Albers Posted June 1, 2007 Share Posted June 1, 2007 We can L-transform any differential current source (locally in space) to have only an electric field representation. I went back and forth with Puthoff on this, and he agreed here but I have not yet shown that this means the vacuum needs no rotational or spin characteristics to transmit "magnetic" fields. I suspect it does not need these characteristics, and this is on my "list of things to do today". Link to comment Share on other sites More sharing options...
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