# Magnetic field of charge moving at constant velocity

by Shaw1950
Tags: charge, constant, field, magnetic, moving, velocity
 Sci Advisor Thanks P: 2,154 I don't understand what you mean by different approaches to solve for the full Maxwell equations. If you mean the Biot-Savart Law from magneto statics, of course it's not applicable to the magnetic field of accelerated charges since it works only in the special case of magnetostatics, i.e., stationary currents. In any case, you get the right answer by using the retarded Green's function (Lienard-Wiechert potentials for an arbitrarily moving point charge). In the case of a uniformly moving charge you get the fields also from Lorentz boosting the Coulomb field for the charge at rest. Of course, both solutions must be identical since Maxwell's electromagnetics is a relativistically covariant classical field theory.
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P: 16,489
 Quote by Shaw1950 So two different sets of equations give you the same answer.
This is not correct, as shown by the links posted above.
P: 969
 Quote by Shaw1950 I agree. Now if you solve the multiple equations of Maxwell, you will also get Biot Savart. So two different sets of equations give you the same answer. What I'm asking is not about solving equations (for the result could be a coincidence) but what is the deeper insight into why the two approaches give the same result. They don't for oscillating charges. So what is the general insight?
 P: 18 Dale, thanks for all your feedback, much appreciated. An issue remains for the low speed approximation. At low speeds there are still electrodynamic and magnetodynamic terms dE/dt and dB/dt in Maxwell and yet the solution is a magnetostatic. So why is the low speed solution magnetostatic? (1) are these dynamic terms simply both small at low speeds, or (2) do the effects of dE/dt and dB/dt cancel (and if they cancel, is there some insight about that)?
 P: 18 RedX, in terms of deeper insight, my suspicion is that the equations can be separated, perhaps Helmholtz decomposition, into parts that behave in different ways, and such a decomposition will explain the result. But I haven't figured out what that decomposition is yet.
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P: 16,489
 Quote by Shaw1950 At low speeds there are still electrodynamic and magnetodynamic terms dE/dt and dB/dt in Maxwell and yet the solution is a magnetostatic. So why is the low speed solution magnetostatic? (1) are these dynamic terms simply both small at low speeds, or (2) do the effects of dE/dt and dB/dt cancel (and if they cancel, is there some insight about that)?
It is (1). They are small at low speeds so you neglect them.
 P: 18 How would you think this specific result was viewed by Maxwell and his peers, if it was known? It surely would have seemed odd that in moving from a static particle to a frame at constant speed, a magnetic field mysteriously appeared yet the electric field was essentially the same. Did this get commented on? Of course everyone believed in the Ether and absolute frames but I'm surprised that nobody noticed or commented on this odd effect. If they had and had investigated, maybe they would have scooped Einstein by 50 years!
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P: 16,489
 Quote by Shaw1950 It surely would have seemed odd that in moving from a static particle to a frame at constant speed, a magnetic field mysteriously appeared yet the electric field was essentially the same.
Why? This kind of thing happens pretty often. For example, if you expand sin(x) and cos(x) to first order about x=0 then you get x and 1 respectively. The cos is "essentially the same" but the sin has "mysteriously appeared". I think you are trying to read more into this approximation than it warrants.
P: 969
 Quote by Shaw1950 RedX, in terms of deeper insight, my suspicion is that the equations can be separated, perhaps Helmholtz decomposition, into parts that behave in different ways, and such a decomposition will explain the result. But I haven't figured out what that decomposition is yet.
I'm not sure why you would need to decompose it. Assuming a Biot-Savart field, the flux through a circle whose center goes through the current carrying wire is zero, since the magnetic fields circle around the wire, so that they are always tangential to planes perpendicular to the direction of the current, hence no flux through such planes. From this you can show that a Biot-Savart field never creates an electric field that has a curl.

That is, $\int E \cdot d\vec{l}=-\frac{d \Phi}{dt}=0$ since phi is zero.
P: 262
 Quote by vanhees71 In the case of a uniformly moving charge you get the fields also from Lorentz boosting the Coulomb field for the charge at rest.
Sorry to be a bore about this point, but can anybody demonstrate that? I’ve seen it done for currents in conductors but it never satisfies me.
(If needed, use an online calculator to get the maths out of the way that’s more than fine by me.)

Just some of my thoughts: I can see that you will be able to derive the relativistic equation for the electric field of a point charge, which is useful for high speed particles but I’m not sure about deriving its magnetic field and especially not its relativistic magnetic field. This last one always seems a “double up” effect to me since a magnetic field is in the first place understood as a result of relativity wrt moving charges.
P: 1,250
 Quote by Per Oni Sorry to be a bore about this point, but can anybody demonstrate that? I’ve seen it done for currents in conductors but it never satisfies me. (If needed, use an online calculator to get the maths out of the way that’s more than fine by me.) Just some of my thoughts: I can see that you will be able to derive the relativistic equation for the electric field of a point charge, which is useful for high speed particles but I’m not sure about deriving its magnetic field and especially not its relativistic magnetic field. This last one always seems a “double up” effect to me since a magnetic field is in the first place understood as a result of relativity wrt moving charges.
This might have been given earlier in this thread, but the fields (from several textbooks) are given by
$${\bf E}= \frac{q{\bf r}} {\gamma^2[{\bf r}^2-({\bf v\times r})^2]^{\frac{3}{2}}}$$
and
$${\bf B}={\bf v\times E} =\frac{q{\bf v\times r}} {\gamma^2[{\bf r}^2-({\bf v\times r})^2]^{\frac{3}{2}}}$$
 P: 262 And exactly what are you doing here? Copy & paste, how clever! Why couldn’t I think of that?
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P: 11,255
 Quote by vanhees71 In the case of a uniformly moving charge you get the fields also from Lorentz boosting the Coulomb field for the charge at rest.
 Quote by Per Oni Sorry to be a bore about this point, but can anybody demonstrate that?
See here:

http://farside.ph.utexas.edu/teachin...s/node125.html