1. Jul 20, 2004

### alfraser

This problem has gotten me 3 different answers from 3 different profs.

1. A moving charge produces a magnetic field, and this could be measured by a stationary observer relative to the moving charge.

2. If an observer is moving parallel to the moving charge with the same velocity so that the observer does not see any movement in the charge, does this observer measure a magnetic field? I thought no.

And if 2 is correct, then what happens if as the moving charge passes the stationary observer, the observer moving parallel to the charge passes through the same location as the stationary charge. At this instant, the moving charge would be on its closest approach to the stationary observer.

I would think the stationary observer would measure a B field, but the moving observer would not measure a B field. But they are in the same location at the same time...

What is going on??

This should occur at slow ie non relativistic speeds i think.

3 profs were stumped, are you?

2. Jul 21, 2004

### zefram_c

So the charge is moving relative to the observer? In that case, he will measure a magnetic field.
Now the observer is not moving relative to the charge. No magnetic field for him.
That's exactly right: the magnetic field you measure depends on the relative velocity of the charge wrt the observer. As magnetism can be regarded as a relativistic correction to the fields of moving charges, the relative velocity enters the calculations.

One last thing: You refer a lot to stationary/moving observers and/or charges. This is not the proper way to do it. There is no point to consider a charge as "moving" relative to some absolute or "stationary" background, which in case #2 seems to have no observers. Case #2 should be simply stated "the observer sees the charge at rest wrt himself". It is only relative velocity that has any physical meaning, in pretty much all of physics.

3. Jul 21, 2004

### alpha_wolf

As zefram_c said - the measured magnetic field is a function of the measured relative velocity of the charge. If the charge is not moving in an observer's frame of reference, that observer would measure a relative velocity of zero, and therefore he would measure a magnetic field of zero. The measured releative velocity has nothing to do with position, and the question has nothing to do with relativity.

Imagine three cars. Cars A & B are moving side by side at velocity v << c in the same direction, along a road. Car C is standing at the side of that road. To a passenger in car B, car A's velocity is 0 regardless of their location. To a passenger in car C, car A's velocity is v regardless of its location. At some point, cars A & B will pass by car C, but the relative velocities each of the passengers would measure would not change at that instance, since they do not depend on location. Now charge car A with an electric charge q....

Last edited: Jul 21, 2004
4. Jul 21, 2004

### chroot

Staff Emeritus
I'm moving this post out of Theory Development. (Boy, is that a first?!) Good post, Alfraser.

- Warren

5. Jul 22, 2004

### Hurkyl

Staff Emeritus
Einstein's 1905 paper, On the Electrodynamics of Moving Bodies, starts off with this very question, and Einstein derives a transformation of electromagnetic fields from one reference frame to another.

6. Jul 22, 2004

### Njorl

You mentioned that this should be done at non-relativistic speeds. By definition, this really means that the magnetic fields are so small that they are negligible, since the magnetic field is the relativistic component of the e-field. If you can ignore time dilation etc, you can also ignore the magnetic field. It is best to just leave that caveat off.

Njorl

7. Jul 22, 2004

### Broken

What did they say and what university do you go to? I find that hard to swallow.

8. Jul 22, 2004

### JohnDubYa

Ho! Let's leave the universities out of this, okay?

9. Jul 23, 2004

### alfraser

Thanks for the replies, it makes a lot more sense now. Hehe yes I really did get three different answers.

1.) 1st thought you would still measure a magnetic field if you were moving parallel to the moving charge at the same velocity. (That was my Electricity and Magnetism prof!!. But he's really an astronomer so we'll cut him some slack. )

2.) 2nd prof thought two observers would probably measure two different magnetisms, but wasn't sure why.

3.) 3rd prof I asked understood that there would be two fields, but his explanation was wrong I'm sure, didn't use relativity and had some inconsistencies in it.

In their defence none study EM, and they are all good in their respective fields.

And yes I'm leaving which University out of it!!

Thanks,
ALF

PS sorry about the posting location, I thought I posted it in the general physics area!

10. Jul 24, 2004

### pmb_phy

As mentioned above, the answer lies in relativity. As Einstein said
That was from a letter from Einstein to the Michelson Commemorative Meeting of the Cleveland Physics Society as quoted by R.S. Shankland, Am. J. Phys., 32, 16 (1964), page 35.

All this basically means that a magnetic field in one frame can be a combination of electric and magnetic field in another frame and an electric field in one frame can be a combination of electric and magnetic field in another frame. In modern notation this can all be expressed in terms of the Faraday tensor (Electromagnetic field tensor) represented by the letter "F". For the definition of this tensor please see

Some people like to phrase all this by saying that the E-field and B-field are all just aspects of one thing, the EM field. Before relativity the E-field and B-field were spatial vectors (aka 3-vectors). In SR the EM field, represented by the faraday tensor F, is a 4-tensor. The E-field and B-field only take on meaning when one introduces a frame of reference, or, equivlently, an observer equipped with a 4-vector which defines his frame of reference. That 4-vector is called his 4-velocity U. In SR the E-field and B-field are then 4-vectors defined in terms of F and U. E.g. the E-field E 4-vector is given by the tensor product of F and U, i.e.

E = cF*U

is the electric field 4-vector measured by the observer whose 4-velocity is U. In component form

Eu = cFuv Uv

A similar relation holds for the B-field 4-vector (for details see General Relativity, Robert Wald, page 64, Eq. (4.2.21) for E-field and Eq. (4.2.22) for B-field).

I highly dissagree. This question has everything to do with relativity. Why do you think it doesn't?

That is incorrect. Even in Classical EM the B-field of a moving charge, q, is given by

B = q(u0/4*pi)vxr/r3

Given a large enough charge (or E-field) there can be a non-neglegible B-field. For example: If the E-field 3-vector in S has components (Ex, Ey, Ez) then in S' (moving in the +x direction with respect to S with speed v) the z-component of the B-field 3-vector in S' is given by

B'z = gamma[Bz - (v/c2)Ey]

if v << c and Bz = 0 then

B'z ~ - (v/c2)Ey

This can't be ignored if Ey is large, even though v << c.

I hope this helps ALF

Pete

Last edited: Jul 24, 2004
11. Jul 24, 2004

### alpha_wolf

Because it can be correctly answered without resorting to relativity, by analizing the relative velocities, which are a pre-relativistic concept (see Lorentz and Galilei transformations), and deriving the magnetic field from that via the classic formula. Relativity simply gives a more complete picture, as it does to many questions that can be answered classicaly.

Last edited: Jul 24, 2004
12. Jul 24, 2004

### pmb_phy

Pre-relativistic concepts did not properly define the electric and magnetic fields. That was what relativitiy did and that is what classical EM now incorporates into it. It only appears that relativity is not used but any such appearance is incorrect. One of the purposes of relativity was to clarify this issue of the relative existance of the electric and magnetic forces between different inertial frames etc. That is one of the reasons why Einstein's 1905 paper was called "On the Electrodynamics of Moving Bodies".

You claim "can be correctly answered without resorting to relativity, by analizing the relative velocities, which are a pre-relativistic concept (see Lorentz and Galilei transformations), and deriving the magnetic field from that via the classic formula" - To do this you have to provide a definition of "magnetic field" and "electric field" and that was what relativity provided.

Let me give you a clear cut example of a problem that existed before 1905 if we tried to use your methods. Consider a long wire carrying a current I which lies at rest on the x-axis in the inertial frame S. Let the wire be uncharged as well. Now transform from frame S to frame S' which is moving in the +x direction. If you use your method and find the force per unit charge on a stationary particle then how do we know that this is a "real" electric field? If it were really an electric field then it'd have to behave accoording to Maxwell's equations and that means Gauss's law would have to hold true. Application of Gauss's law in S' would yield a finite linear charge density on the wire. This means that charge density can be zero in one frame and not zero in another frame. Recall that we started off by dictating that the wire is uncharged as measured in S. What reason do you claim that classical pre-relativity gives for saying that there charge density has a relative esistance?

Thanks

Pete

Last edited: Jul 24, 2004