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How can a single moving charge generate a magnetic field relativistically |
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| Jul25-11, 01:22 PM | #1 |
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How can a single moving charge generate a magnetic field relativistically
Hey,
I just remember reading about how the magnetic force can be thought of as a relativistic effect in the sense that the moving charge will see the charges in the wire contract and so it will see a higher density of positive (or negative) charges along the wire. However if this is true how would a single charge moving generate a magnetic field? since the effect is due to the other charge seeing a greater aggregate of charges while moving. If my question isnt clear ill try to re-word it thanks -Storm |
| Jul25-11, 01:37 PM | #2 |
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Is there anything like 'charge contraction'?I am noob when it comes to relativity,but I've never heard about that before.Heard about length contraction though.
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| Jul25-11, 01:51 PM | #3 |
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Storm, I actually asked this question a while back and never got an answer for it. Maybe someone knows so we can both find out!
Shoku, length contraction is the reason that this effect happens supposedly. That's what is meant by charge contraction. |
| Jul25-11, 02:00 PM | #4 |
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How can a single moving charge generate a magnetic field relativistically
So does that mean charge actually decrease in magnitude or something like that?
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| Jul25-11, 02:09 PM | #5 |
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Look up relativity and electromagnetism on google or wikipedia for more. |
| Jul25-11, 03:32 PM | #6 |
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Even for a single charge the length contraction argument holds. The field itself is "contracted". Einstein derives the transformation of the fields themselves (without reference to the charges) in his original OEMB paper.
Remember that the EM field is a rank-2 tensor, so it transforms a little differently than a four-vector. |
| Jul25-11, 05:50 PM | #7 |
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| Jul25-11, 06:33 PM | #8 |
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An (antisymmetric) rank-2 tensor, like the EM field, just means it acts like a plane -- more specifically, a bivector.
This link has a figure visualizing an electromagnetic field from different reference frames. http://www.av8n.com/physics/magnet-r...bib-maxwell-ga Look at the red EM field (the red plane). It lines up with the t' axis, so the observer in the primed reference frame sees it as a pure electric field. (It could be caused by a charged wire... or, by a single electron!) Since it's in the (r,t') plane, the primed observer sees this electric field along the r direction. Now, consider it from the unprimed frame (green axes). It has a nonzero projection onto the rz-plane; that projection is represented as the gray rectangle. Notice that r and z are both "space" axes, so this is a magnetic field in the unprimed frame. (Since we usually represent magnetic fields by the normal to the plane, this would be the component along the third spatial direction -- probably [itex]B_\phi[/itex].) This is what DaleSpam was saying: we only need to transform the fields to see how this works. And Shoku Z, you raise an interesting question about whether charge can decrease in magnitude in different reference frames. See Melvin Schwartz's excellent Dover paperback, "Principles of Electrodynamics": http://www.amazon.com/Principles-Ele.../dp/0486654931 He points out that if that were so, we'd be in trouble! The electrons are moving much faster than the nucleus, so if charge varied with motion, their charge would no longer cancel the charge in the nucleus, and matter would fly apart. In general, Schwartz's book is an excellent and accessible way to learn about the origins of magnetism from electricity. His entire program is to begin with electrostatics, review special relativity, and derive magnetism and Maxwell's equations therefrom. I don't like a few of his conventions (e.g. using [itex]ict[/itex] for the time dimension in relativity), but they're fine for his intended audience, and I was able to look past these distractions. |
| Jul25-11, 06:37 PM | #9 |
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"How can a single moving generate a magnetic field": - Cheap answer 1: moving charge = current, current generates magnetic field. - Cheap answer 2: perform a Lorentz boost on the field of a stationary charge, end up with moving charge and a magnetic field. - More intuitive answer: see this link for a much more complete discussion: http://en.wikipedia.org/wiki/Moving_...ductor_problem -- but briefly, whether you're looking at the charge as a moving charge with an electric + magnetic field, or a stationary charge with just an electric field, the force you feel will be the same...so maybe the answer is that it's really just a crutch to talk about a "magnetic field" or an "electric field", when in fact whether it's an electric or magnetic field that is causing any given force depends entirely on how you look at it. I would be interested in hearing your thoughts on this, though, since I've long been interested in getting an intuitive understanding of magnetism (as opposed to a purely mathematical understanding that leads to answers like 1 & 2). For example, why would seeing a higher density of positive (or negative) charges make a charge "see" a magnetic field? |
| Jul26-11, 06:56 AM | #10 |
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I dont think its so much a matter of relativity explaining the existence of a magnetic field as it is a matter of relativity requiring the existence of a magnetic field.
the electric field of a single moving electron is indeed contracted. but there is also a magnetic field too. |
| Nov2-11, 04:08 PM | #11 |
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Good question and hope that I could provide some help to your concerns
Electric field and magnetic field are observables depending on the frame of reference where the observation takes place. Imagine that a single charge is moving along a straight line, in one case, an observer is moving with the single charge at the same speed and direction. Under such circumstance, the single charge appears to be at rest to the observer. By then, the observer would declare that he could only sense electric field and no magnetic field could be found. In another case where the observer is moving along with single charge in the same direction, but at different speed. Then he would notice the single charge is moving in front of him at a constant speed. When he turns on the measuring instrument, we would notice both electric and magnetic fields. Now that Einstein's relativity could come in beautifully to consolidate the differences among the different observers by saying that the electric and magnet field are observed properties rather than intrinsic properties of the single charge. The values at different frames of reference could be transformed from one to another by using Relativity formulations By now, the electric and magnetic fields are totally unified and no longer separable. We call it electromagnetic field. The electric and magnetic field in a traditional sense are only one aspect of electromagnetic field and entirely depending on the frame of reference where the observation is made. Your analogy of contracting wire is a OK one while it is still trying to explain the effect from traditional electric field point of view. The result from using such analogy would agreed with Einstein Relativity's prediction, but it is not the fundamental explanation for such effect. |
| Nov2-11, 05:51 PM | #12 |
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While what you're saying is empirically true, from a macroscopic level anyway (which is all the question concerns), I'd argue that your conclusion is not as obvious as you make it out to be...because, for the force on the charge in the moving frame to agree with that in the stationary frame, you need to assume that the particle is either a point (which produces infinities and inevitably has subtleties of its own), or that it's perfectly spherical / "rigid" -- even while in motion, which would seemingly contradict relativity (specifically, Lorentz contraction). However, with the assumption that it remains spherical, the self-force on each part of the particle is perfectly cancelled by that self-force on the rest of the particle...which you can say is obvious because its macroscopic dynamical behavior while moving "has to" agree with that while it's at rest because of relativity -- but that's just begging the question, since you're citing relativity to explain a relativistic effect...while in my opinion, the full explanation is not at all obvious -- e.g. that the Lorentz-transformed / retarded self-field would perfectly cancel for the moving particle. Does that make sense? |
| Nov5-11, 02:28 AM | #13 |
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I assume that you would prefer to explain all those effects based on Lorentz Transformation (please correct me if I am wrong). Lorentz Transformation gives the same formulations as Einstein's Relativity’s predictions. However, Lorentz Transformation was based on a false postulate of ETHER in attempt to address the failures of detecting the changes in speed of light. Also, the electrons do not have to be Rigid. We would put some electric charges on two softballs and throw them away side by side at a speed near to light. The two softballs would not change the shapes at all and all relativity predictions will still hold true. The Contraction we are talking about based on either Lorentz Transformation or Einstein’s relativity only describes our perception or in another word, measurement. It has nothing to do with the softballs themselves. BTW, my attempt of using softballs to represent electrons is not appropriate. An electron is not a ball like object. In eyes of quantum mechanics, an electron does not have a certain shape or position. It is a wave or a ghost. |
| Nov5-11, 07:22 AM | #14 |
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In the past I thought just like a lot around here: “magnetism”? Nothing but a Lorentz boost of a moving electrostatic field, end of story.
That was until I had a proper look at these formulas: (coppy/past of these formulas doesn't work so I've put them in by hand) See: [URL="http://http://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity#Joules-Bernoulli_equation_for_fields_and_forces"] E'=Y(Eo + V X Bo) and B'=Y(Bo - E X V / C^2) Would you normally give the length of say a car and then state whether or not the car was moving? Why then the need to think of a Lorentz boost in case of a moving magnetic field? If you move by hand a magnet across a conductor, an electric field will be set up in that conductor. Does the speed of your hand approach C? I think not. Include gamma to work out E? I think not. Normally, conduction electrons in a current move even slower than your hand. Include gamma to work out B? I think not. |
| Nov5-11, 07:33 AM | #15 |
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[tex]F_{\mu '\nu '}=\Lambda^{\nu}_{\nu '}\Lambda^{\mu}_{\mu '}F_{\mu\nu}[/tex] and the tensor transformation law for a rank 1 tensor is: [tex]x_{\mu '}=\Lambda^{\mu}_{\mu '}x_{\mu}[/tex] See eq 56 at http://farside.ph.utexas.edu/teachin...es/node10.html |
| Nov5-11, 09:46 AM | #16 |
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| Nov5-11, 09:58 AM | #17 |
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Or, am I misunderstanding? In any case, I'd be interested in hearing an expanded explanation. |
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