I am having a problem with the relativistic interpretation of magnetic fields. In some textbooks one can find corresponding treatments which seem to show that relativistic length contraction of charge carriers produces the same result as Classical Electrodynamics (e.g. Berkeley Physics Course (Electricity and Magnetism) or Feynman (The Electromagnetic Field); for an online treatment see for instance http://physics.weber.edu/schroeder/mrr/MRRtalk.html (http://physics.weber.edu/schroeder/mrr/MRRtalk.html) ) , but the proof is always only done for an infinitely long wire. I tried to show now that the explanation also holds for the far-field of a finite wire but I can't make the result agree with Classical Electrodynamics:

It is straightforward to show that the field of a negative charge distribution of length L at a vertical distance z for L/z<<1 is given by

E~-1/z*arctan(L/z)

and expanding the arctan-function up to the second term

E~ -L/z^2 +L^3/(3*z^4)

Now the first term of this is simply the field of a point charge and it cancels therefore against the corresponding term for the positive charges because of overall charge invariance. This leaves just the second term to produce any field due to different length contraction, but this field would obviously decrease ~1/z^4. However, in Classical Electrodynamics the magnetic field of short wire decrease like 1/z^2 (see any derivation of the Biot-Savart law for instance), so the relativistic explanation does not appear to give the correct answer for the far field of a wire.

Any suggestions how to explain this circumstance?

Thomas

Hi tsmid,

First, I don't think that you got your expression for E right as it shouldn't involve any arctangents. Anyway, this is kind of beside the point.

Second, following the situation in the link you provided, if the wire is uncharged in the lab frame, then in the rest frame of the test charge there is a net charge on the wire. The leading contribution to the electric field is therefore 1/r^2 (since the system isn't charge neutral), so the relativstic transform of the electric field will give a 1/r^2 magnetic field to leading order just as you would expect. In other words, the positive charge just cancels the negative charge in the lab frame, so the positive charge can't cancel the negative charge in the rest frame of the test charge.

As an aside, there are some issues with current flow in a wire that simply terminates, but I don't think this matters here (you can also avoid these issues by clever use of coaxial cables to feed your system, for instance). However, this is part of the reason why an infinite wire is usually considered.

Second, following the situation in the link you provided, if the wire is uncharged in the lab frame, then in the rest frame of the test charge there is a net charge on the wire. The leading contribution to the electric field is therefore 1/r^2 (since the system isn't charge neutral), so the relativstic transform of the electric field will give a 1/r^2 magnetic field to leading order just as you would expect. In other words, the positive charge just cancels the negative charge in the lab frame, so the positive charge can't cancel the negative charge in the rest frame of the test charge.

Charge invariance (i.e. overall neutrality in this case) must hold always, even relativistically. Therefore, if the charge densities of the negative and positive charges differ, they must be distributed over correspondingly different lengths. The infinite wire tends to hide this fact. That's why I considered the finite wire here.


I don't think that you got your expression for E right as it shouldn't involve any arctangents. Anyway, this is kind of beside the point.

The electrostatic force at a vertical distance z from the wire due to a charge located at the wire's length coordinate x is (for x/z<<1)

dE ~ 1/(x^2+z^2)

If you integrate this over x from 0 to L it gives

E~ 1/z*arctan(L/z)


Thomas

Hi again Thomas,

You are of course correct that the total charge is invariant, and the argument I gave is flawed. That's why I told you there are subtleties here. For example, how exactly would you setup a situation where a finite length of wire carries a current that just begins and ends at two points. Or another issue: in the test charge rest frame the whole finite wire is now moving, so time delay effects can be significant in the far field. My handwaving explanation essentially mixed aspects of the finite and infinite wire, so you are correct to doubt it (though the essential conclusion, that a 1/r^2 electric field with give rise to a 1/r^2 magnetic field, is correct). However, if you want a real explanation then you must be more specific about your setup, and it must be physical, before I can answer.

Also, you forgot to include the directional information in the evaluation of the field strength. Please see here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html

Also, you forgot to include the directional information in the evaluation of the field strength. Please see here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html

Yes, strictly speaking I should have included the factor z/r, but if z is much larger than the length of the wire (which is the case I am considering here) one has approximately z=r i.e. z/r=1, so it doesn't make any difference for my analysis (if you expand the square root in the result given in the hyperphysics link for large values of z, you will in fact get the same result i.e. a first order field decreasing like 1/z^2 (which corresponds to the total charge) and a second order field ~1/z^4 due to the linear structure of the charge distribution).
(P.S.: I have re-edit my previous posts now in this respect. I changed there my variable r to z as well in order to avoid any confusion)


You are of course correct that the total charge is invariant, and the argument I gave is flawed. That's why I told you there are subtleties here. For example, how exactly would you setup a situation where a finite length of wire carries a current that just begins and ends at two points. Or another issue: in the test charge rest frame the whole finite wire is now moving, so time delay effects can be significant in the far field. My handwaving explanation essentially mixed aspects of the finite and infinite wire, so you are correct to doubt it (though the essential conclusion, that a 1/r^2 electric field with give rise to a 1/r^2 magnetic field, is correct). However, if you want a real explanation then you must be more specific about your setup, and it must be physical, before I can answer.

I don't see the practical realization being of relevance here. We are merely comparing one theoretical prediction with another.

But you can just assume for instance that the wire connects two differently charged objects which are sufficiently shielded so that their field does not interfere with the wire's field.

Thomas

I am having a problem with the relativistic interpretation of magnetic fields. In some textbooks one can find corresponding treatments which seem to show that relativistic length contraction of charge carriers produces the same result as Classical Electrodynamics (e.g. Berkeley Physics Course (Electricity and Magnetism) or Feynman (The Electromagnetic Field); for an online treatment see for instance http://physics.weber.edu/schroeder/mrr/MRRtalk.html (http://physics.weber.edu/schroeder/mrr/MRRtalk.html) ) , but the proof is always only done for an infinitely long wire. I tried to show now that the explanation also holds for the far-field of a finite wire but I can't make the result agree with Classical Electrodynamics:

It is straightforward to show that the field of a negative charge distribution of length L at a vertical distance z for L/z<<1 is given by

E~-1/z*arctan(L/z)

and expanding the arctan-function up to the second term

E~ -L/z^2 +L^3/(3*z^4)

Now the first term of this is simply the field of a point charge and it cancels therefore against the corresponding term for the positive charges because of overall charge invariance. This leaves just the second term to produce any field due to different length contraction, but this field would obviously decrease ~1/z^4. However, in Classical Electrodynamics the magnetic field of short wire decrease like 1/z^2 (see any derivation of the Biot-Savart law for instance), so the relativistic explanation does not appear to give the correct answer for the far field of a wire.

Any suggestions how to explain this circumstance?

Thomas
That example of how length contraction plays into the transformation is only a small part of the story. More generally the electric and magnetic fields are each without each other incomplete components of the full electromagnetic field tensor. This tensor Lorentz transforms. This is what results in being able to equate a magnetic field for an infinite length contracted line current to an electric field from the stationary line of source charges according to a different frame. You may either write the tensor containing the expression for the electric field from your case of source and Lorentz transform it and pick off the resulting magnetic field components, or you may just use the known equations for the transformation between the electric and magnetic fields that can be derived from this.

There is a 1/r^2 (monopole) field and a 1/r^4 (quadrupole) field, but you do get the coeffecient of the quadrupole term wrong by making your approximation in the integral.

As for the question of physical realization, it isn't a trivial question. One cannot just magically shield charges. The ideal way to shield the charge buildup would be to have the wire enter a charged body surrounded by a conductor. If the conductor was grounded then it could draw up charge from ground to screen the charge deposited by the current. However, now you have more wires and conductors and the problem is rather more complicated. These extra wires will carry equal and opposite current and can be regarded as infinite so that in fact their magnetic fields would just cancel at the symmetry point of the finite wire. Lets proceed with such a picture and see where it gets us. If you have a finite wire then the wire will contract in the frame where it is moving (the test charge rest frame). However, you also want to have the positive charge density get smaller because the positive charge is now not moving. This can't happen unless some positive charge on the wire has been drawn into the region shielded by our conductors in the moving frame. Now we are back to my original answer: the wire is now charged in the test charge rest frame, hence a 1/r^2 electric field exists. Like I said, my answer is essentially correct and it becomes completely correct if you state the problem physically.

There is a 1/r^2 (monopole) field and a 1/r^4 (quadrupole) field, but you do get the coeffecient of the quadrupole term wrong by making your approximation in the integral.

As for the question of physical realization, it isn't a trivial question. One cannot just magically shield charges. The ideal way to shield the charge buildup would be to have the wire enter a charged body surrounded by a conductor. If the conductor was grounded then it could draw up charge from ground to screen the charge deposited by the current. However, now you have more wires and conductors and the problem is rather more complicated. These extra wires will carry equal and opposite current and can be regarded as infinite so that in fact their magnetic fields would just cancel at the symmetry point of the finite wire. Lets proceed with such a picture and see where it gets us. If you have a finite wire then the wire will contract in the frame where it is moving (the test charge rest frame). However, you also want to have the positive charge density get smaller because the positive charge is now not moving. This can't happen unless some positive charge on the wire has been drawn into the region shielded by our conductors in the moving frame. Now we are back to my original answer: the wire is now charged in the test charge rest frame, hence a 1/r^2 electric field exists. Like I said, my answer is essentially correct and it becomes completely correct if you state the problem physically.And here I was thinking that instead of going with the current in the wire from the example he was doing a calculation for a similar but different problem of a free charge segment like a pulse from an ionised plasma beam. Yes a wire segment would be more problematic.

That example of how length contraction plays into the transformation is only a small part of the story. More generally the electric and magnetic fields are each without each other incomplete components of the full electromagnetic field tensor. This tensor Lorentz transforms. This is what results in being able to equate a magnetic field for an infinite length contracted line current to an electric field from the stationary line of source charges according to a different frame. You may either write the tensor containing the expression for the electric field from your case of source and Lorentz transform it and pick off the resulting magnetic field components, or you may just use the known equations for the transformation between the electric and magnetic fields that can be derived from this.

The length contraction should in this case be the whole story with regard to the Lorentz transformation. After all the resulting electric force in the reference frame of the test charge is exactly the same as the magnetic force in the wire's frame according to the derivations in the sources I cited above. Although these were done for an infinite wire, this shouldn't make any difference for the Lorentz transformation as the latter transforms the electric and magnetic fields locally i.e. it should be irrelevant how these fields are produced. The point is however that, as indicated above, I can't see the length contraction giving the correct result if charge invariance is properly being taken into account.

However, this brings me to another point: the argument here usually is that the magnetic part of the Lorentz force F=qvxB (where x is the cross product here) is zero in the reference frame of the test charge (as v=0 here) and thus the frame invariance of forces requires an electric force of the same magnitude. Yet I don't think this argument is correct as the rest frame of the test charge is not an inertial one but it is accelerated because the charge describes a Larmor circle around a magnetic field line in the wire's frame. So in fact there should not be any additional force required in the test charge frame.
As a comparison, consider another velocity dependent force e.g. friction: if an object slides over a surface, you wouldn't suggest either that there must be a different force acting in the object's frame because the velocity v=0 here. The point is that the velocity here is not frame dependent but referred to the surface the object is sliding over, so it does not change if you change reference frames. Similarly, the v in the Lorentz force should actually be referred to the wire producing the magnetic field, so it should be independent of the reference frame as well and thus there should not be any need for an electric field in the frame of the test charge (in general, what you should have in inertial frames other then the wire's frame is simply a constant velocity superposed on the Larmor circle i.e. a cycloid motion).

Thomas

There is a 1/r^2 (monopole) field and a 1/r^4 (quadrupole) field, but you do get the coeffecient of the quadrupole term wrong by making your approximation in the integral.

Yes, the coefficients of the two approximations may be different, but the important point is that the only non-vanishing term overall would have a 1/r^4 dependence whereas classical electrodynamics would require a 1/r^2 dependence.

Just for clarification again: the actual electric field observed would have a dependence like (neglecting any constant numerical coefficients)

E ~ n*A*L/r^2 + n*A*L^3/r^4

where L ist the length of the wire, A its cross section and and n the charge density. Now charge invariance requires that n*A*L = const= Q (total charge) i.e.

E ~ Q/r^2 + Q*L^2/r^4.

Hence the first term is not affected by any contraction of the length L but only the second term which decreases however like 1/r^4 with distance.


However, you also want to have the positive charge density get smaller because the positive charge is now not moving. This can't happen unless some positive charge on the wire has been drawn into the region shielded by our conductors in the moving frame. Now we are back to my original answer: the wire is now charged in the test charge rest frame, hence a 1/r^2 electric field exists. Like I said, my answer is essentially correct and it becomes completely correct if you state the problem physically.

But there is nothing drawn into the shielded region. On the contrary, if you consider the positive charges as resting and the negative moving, Lorentz contraction would not affect the former, and the latter would be contracted to a shorter length. The overall charge visible for the observer would thus be the same i.e. =0 (the bulk of the wire would now be negatively charged, but this would be exactly offset by positive surplus charges at both ends of the wire (and thus the total potential drop along the wire would also be unchanged i.e. the current through the wire should remain the same))

Thomas

Incorrect, if it was the negative charges moving, they would have to be moving in the opposite direction to produce the same current. You would now have to boost in the opposite direction to bring the negative charges to rest. In order for this boost to coincide with the test charge being at rest, it must now move in the opposite direction as well. Excess charge of the moving type (as judged in the lab frame) will always be drawn into the shielded region and the wire will become charged.

The test charge now sees a wire with excess charge of the positive type (excess charge was negative in the previous case) in its rest frame and so feels the opposite electric force it would feel in the other situation. This is no contradiction since you had to reverse the velocity of the test charge with the magnetic field being the same (same current), and thus the magnetic force is also reversed. Don't you see how it all comes together consistently?

Your conclusion must be incorrect: there is a 1/r^2 magnetic field in the lab frame, so there will be a 1/r^2 electric field in the test charge rest frame. I have repeatedly told you that you need to be more careful with the physical setup, and you have ignored these admonitions. You are of course free to ignore me, and I am only trying to help you out. Nevertheless, the problem is not with electromagnetism.

This thread is a little over my head, but looks like it will end soon in an agreement to disagree, so I enter to asked about some thoughts tsmid provoked with his (her?) brief reference to friction and velocity dependent forces.

Imagine a bullet traveling faster than sound along the axis of a very long tube of water. (To keep it with simple constant v, assume the energy it is losing is supplied by an electromagnetic "rail gun.") In the rest frame of the water there is a shock wave, but in the frame where the water and bullet both have essentially speed c I have may a problem with "physics is the same in all frames" and would appreciate some help. Is the speed of sound in fast moving water nearly zero? If there is length contraction of the gap between two points on the tube marked A & B, would this not compensate for the fact that the speed of sound may appear slow? How does time dilation fit in? The time for the conical shock wave hitting the tube wall to move from mark A to mark B. etc.

I.e. anyone of you want to help me understand what a bullet going thru water and making a shock wave cone scrap the tube wall "looks like" in a frame where water, tube and bullet all have essentially the same speed?

Is this an interesting question, or just confusion in my mind?

Your conclusion must be incorrect: there is a 1/r^2 magnetic field in the lab frame, so there will be a 1/r^2 electric field in the test charge rest frame.

If you could derive theoretically the 1/r^2 electric field (or point me to corresponding references) then I would be happy to acknowledge this. As it stands, my theoretical consideration yields just a 1/r^4 electric field due to the length contraction effect for the far field of a finite wire.

Thomas

As I already explained in some detail, the wire will be charged in the moving frame. A charged wire will have a monopole field as the leading term in the far field. Alternatively, since the Lorentz transformation is linear, a 1/r^2 magnetic field will give rise to 1/r^2 electric field in the moving frame. Look up any discussion of the relativistic transformation of the fields.

BillyT: Either you or my slow brain jumped out of gear: you spoke of the tube, the water, and the bullet having the same velocity, although that statement was a quantum leap from the prior gist of your post. That circumstance is a really interesting mind-bender, but before I waste a good bend, I want you to acknowledge or clarify, please. :bugeye:

BillyT: Either you or my slow brain jumped out of gear: you spoke of the tube, the water, and the bullet having the same velocity, although that statement was a quantum leap from the prior gist of your post. That circumstance is a really interesting mind-bender, but before I waste a good bend, I want you to acknowledge or clarify, please. :bugeye:No, not the same. I said "essentially the same velocity" (in a frame moving at nearly C wrt the frame where the water is stationary). I did not do the math, but I am quite confident that the difference between the speed of the bullet and the water in this frame where both have essentially speed of light (slightly less of course) is much less than the speed difference in the rest frame of the water.

I will admit that there may be a little hidden sadism or vengeance in my question/problem.

I suspect that the answer to how that (even with the bullet having very little speed wrt to the water in the frame where both are moving at nearly speed of light) bullet is still producing the shock wave that is "scrapping the tube wall" is due to the fact that the speed of sound (I think) in Deuterium water ("heavy water") is less than regular water (at the same temperature).

In this frame where the water is "extra heavy" due to what I like to call "relativistic mass," I would thus predict that the speed of sound in "extra heavy water" is very slow and the bullet is still going faster than the speed of sound in the "extra heavy water." I have been raked over the coals, especially by Aer, but even by others, active here in this thread, who certainly know more than me (or Aer) and told that my "relativistic mass" is non-sense.

I am curious to see how they explain that the bullet going very slowly thru the water (relative to the water) is still making a shock wave, as it must if "physics is the same" in all inertial frames, without essentially admitting that: Yes, that water does have extra "relativistic mass." - I have been repeatedly told "relativistic mass" is wrong, false, useless, misleading etc. So if the rescue of "physics is the same" is that the speed of sound is much less in the frame where the water, bullet & tube are moving at almost c, I want to see how they justify that low sound speed without essentially admitting "relativistic mass" is a "valid, true, useful, non-misleading" concept instead of "wrong, false, useless, misleading" one.

They are very smart and well informed. I am especially impressed by PM, so my money is on them being able to continue their view that "relativistic mass" is a "wrong, false, useless, misleading" concept, but very curious to see how they will do this.

As I already explained in some detail, the wire will be charged in the moving frame. A charged wire will have a monopole field as the leading term in the far field.
As I have shown above, the monopole term depends only on the total charge Q which must be invariant. Thus, if Q is identical for the positive and negative charges (apart from the sign), this must also hold in any other frame. Hence only the quadrupole term can yield a charging effect due to a length contraction of L (after all, you obviously can't length-contract a point charge, and the monopole field is the field of a point charge).


Alternatively, since the Lorentz transformation is linear, a 1/r^2 magnetic field will give rise to 1/r^2 electric field in the moving frame. Look up any discussion of the relativistic transformation of the fields.
Well, if the Lorentz transformation is physically consistent in this respect, then you should be able to derive the 1/r^2 electric field also without transforming the field (just assume you can't determine the magnetic field in the other frame but are stuck to the frame where there is only an E field according to relativity).

Thomas

The length contraction should in this case be the whole story with regard to the Lorentz transformation. ...
Thats not true. Proof is as simple as considering the magnetic field of a single moving point charge. It has no length to contract.

BillyT: Thank you for clarification.

As a recovering relativityist, I am presently trying to stay on the wagon and not consume any relativity, so I am not going to to try to tackle this one.

I want to compliment your question, it is an interesting one.

But, I am much less optimistic than you that the relativitiacs will discover a satisfactory solution.

BillyT: Thank you for clarification.

As a recovering relativityist, I am presently trying to stay on the wagon and not consume any relativity, so I am not going to to try to tackle this one.

I want to compliment your question, it is an interesting one.

But, I am much less optimistic than you that the relativitiacs will discover a satisfactory solution.
A relativist already gave you the solution. You ignored it.

A relativist already gave you the solution.Are you stating that the problem/or my confusion I described 25 Nov has been addressed? If yes, where?

Briefly it relates to:

Imagine a bullet traveling faster than sound along the axis of a very long tube of water. (To keep it with simple constant v, assume the energy it is losing is supplied by an electromagnetic "rail gun.") In the rest frame of the water there is a shock wave, but in the frame where the water and bullet both have essentially speed c I have may a problem with "physics is the same in all frames" and would appreciate some help. Is the speed of sound in fast moving water nearly zero? If there is length contraction of the gap between two points on the tube marked A & B, would this not compensate for the fact that the speed of sound may appear slow? How does time dilation fit in? The time for the conical shock wave hitting the tube wall to move from mark A to mark B. etc.

Well, if the Lorentz transformation is physically consistent in this respect, then you should be able to derive the 1/r^2 electric field also without transforming the field (just assume you can't determine the magnetic field in the other frame but are stuck to the frame where there is only an E field according to relativity).

The Lorentz transformations are physically consistent, and I did demonstrate that a 1/r^2 field exists without transforming the fields. I told you how your naive "charge is invariant" argument breaks down when you consider the situation carefully. You simply ignored me. The complete solution is available to you in my previous comments.

Billy T, perhaps I'm misunderstanding, but if a shockwave exists in one frame then it exists in all frames. The coordinates of the spacetime event of the shockwave passing a certain point on the tube can be easily transformed from one frame to the other. What am I missing? You might want to start a new thread.

Billy T, perhaps I'm misunderstanding, but if a shockwave exists in one frame then it exists in all frames.That is both what is troubling me and also required for physics to be the same in all frames. The coordinates of the spacetime event of the schockwave passing a certain point on the tube can be easily transformed from one frame to the other....Yes. I will not do the actual numbers, but assume that in some frame with "enough 9s" the difference between the speed of the bullet and the water is less than the speed of sound (no shock wave?) in the frame water is at rest.

For example:
Vb = 0.999,999,999,999,999,9999c
Vw =0.999,999,999,999,999,9998c
Vb - Vw =0.000,000,000,000,000,0001c may be less than the speed of sound in water's rest frame.

I expect you (or someone) will explain to me that in the frame where both have essentially the speed of light, that the speed of sound is nearly zero as the water molecule is very massive. I want to see how you can give this answer (if that is indeed it) without reference to what I call "relativestic mass".

PS put in this thread as question was stimulated by post here and I thought thread was at a standoff -which it still seems to be- anyway.

I did not ignore the solution.

I never saw it in the first place. Where are they hiding it?

I have encountered great amounts of arguement regarding whether the Lorentz / Special Relativity contraction is a real physical event or whether it an optical illusion kind of thing due to transferring information from one frame of reference to another.

These arguements have not been uttered by disgruntled ex-relativists, but by different sects of devout Relativityists.

Will it be another century before the Relativityists settle the matter among themselves?