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Magnetic field of charge moving at constant velocity

by Shaw1950
Tags: charge, constant, field, magnetic, moving, velocity
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vanhees71
#19
Jul31-11, 04:09 AM
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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.
DaleSpam
#20
Jul31-11, 07:25 AM
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Quote Quote by Shaw1950 View Post
So two different sets of equations give you the same answer.
This is not correct, as shown by the links posted above.
RedX
#21
Jul31-11, 01:36 PM
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Quote Quote by Shaw1950 View Post
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?
I'm not sure I know the answer at the level you want it. This is kind of like the example you see in textbooks of charging a capacitor. If you form a loop around one of the wires leading up to the capacitor, and calculate the line integral of the magnetic field, then you get an answer if you use the current flowing through the loop to help you calculate this integral. But you can also make your surface go between the plates, and instead of current you'd have a changing electric flux. Both give you the same answers. Coincidence? That might be too strong of a word. It's reassuring - we expected that it'd give the same answer (and it did), and it is delightful because you don't think it'll give the same answer because electric fields are different from currents. If you want to think about it deeper that's cool. One thing though: there is such a thing as a polarization current, given by the partial derivative of the polarization vector with respect to time. The polarization vector is proportional to the electric field, for linear materials. So the partial derivative of the electric field with respect to time is like a polarization current, except not in a dielectric, but in a vacuum or ether. So in a way you can view the displacement current as a true current. I think physics might even say that the vacuum is a legitimate medium, as it contains stuff like electron/positron pairs that pop in and out of existence. and they can screen stuff, just as a dielectric screens stuff.
Shaw1950
#22
Aug1-11, 03:19 AM
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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)?
Shaw1950
#23
Aug1-11, 03:23 AM
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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.
DaleSpam
#24
Aug1-11, 06:22 AM
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Quote Quote by Shaw1950 View Post
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.
Shaw1950
#25
Aug1-11, 12:22 PM
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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!
DaleSpam
#26
Aug1-11, 12:32 PM
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Quote Quote by Shaw1950 View Post
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.
RedX
#27
Aug1-11, 12:54 PM
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Quote Quote by Shaw1950 View Post
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, [itex]\int E \cdot d\vec{l}=-\frac{d \Phi}{dt}=0 [/itex] since phi is zero.
Per Oni
#28
Aug3-11, 02:21 PM
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Quote Quote by vanhees71 View Post
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.
clem
#29
Aug3-11, 05:19 PM
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Quote Quote by Per Oni View Post
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
[tex]{\bf E}= \frac{q{\bf r}}
{\gamma^2[{\bf r}^2-({\bf v\times r})^2]^{\frac{3}{2}}}[/tex]
and
[tex]{\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}}}[/tex]
Per Oni
#30
Aug4-11, 02:13 PM
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And exactly what are you doing here?
Copy & paste, how clever!
Why couldn’t I think of that?
jtbell
#31
Aug4-11, 04:57 PM
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Quote Quote by vanhees71 View Post
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 Quote by Per Oni View Post
Sorry to be a bore about this point, but can anybody demonstrate that?
See here:

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

(two methods, take your choice)
Per Oni
#32
Aug5-11, 08:58 AM
P: 262
Jtbell that paper your referring to doesn’t look like “ a Lorentz boosted coulomb field” to me. It looks a lot more complicated including a lot of maths and Greek. Oh well, someday someone will come up with something a bit more consumer friendly.
vanhees71
#33
Aug6-11, 02:48 AM
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I find the manuscript by Fitzpatrick excellent. It's also awailable as a textbook. I can only recommend to read his chapter on the fully relativistic treatment of electromagnetics. I never understood, why there is no textbook on classical electromagnetism that from the very beginning strictly uses the relativistic framework. Instead all authors copy more or less the classical textbooks of the first half of the 20th century (although there are excellent books among them, first of all Sommerfeld's Lectures on Theoretical Physics and Becker's book). The only exception is Landau-Lifgarbages in his vol. II, but in vol. VIII he treats the constitutive equations non-relativistic as usual. But that lamento becomes off-topic now...
DaleSpam
#34
Aug6-11, 06:36 AM
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Quote Quote by Per Oni View Post
Jtbell that paper your referring to doesn’t look like “ a Lorentz boosted coulomb field” to me.
But that is exactly what it is.
Shaw1950
#35
Aug9-11, 01:08 PM
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Thank you all for the comments and feedback it was most helpful and has spurred me on to do some of the maths for myself. One of my difficulties is that many textbooks "fudge" the solution by ignoring the fact that a point charge is essentially a delta distribution and they skip essential steps in the calculation. However, you all encouraged me to make the effort, after 40 years of abstinence. Thanks!


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