Expert: Daniel Mazur - 10/5/2009

Question
Hi Daniel,

I have one little doubt that I'd like to clarify befor I move on to the main question of the relation between electricity and mangnetism. Here it is:
I know that the electromagnetic field of which one part is the electric field propogates with a speed 'c'.

Now suppose if we have two charges, A and B seperated by some distance in vaccum.
If I move one of the charges, say A, by a very small amount, towards the charge B, the force on the charge B will not increase instantaneously but after a time c/r where r is the initial distance between the two charges. Am I right when I say the above?
If the above stataement is correct, then will the force "on A" "due to B", also increase by the time t=c/r since we can position oursleves on B and can equally say that it is B that has moved towards A.
  
RELATION B/W ELECTRICITY AND MAGNETISM

I know that they are relativistic aspects of each other. I tried to read it from feynman, where two cases are considered, which are as follows:

We have a current carrying wire and a negative charge 'q' located outside the wire which we through along the length of the wire opposite to the direction of the current so that it is atrracted towards the wire.
The situation is considered from two frames. One is attached to the stationary wire (S frame) and the other to the moving conduction electrons which are moving with a speed 'v' (S' frame). To make things easy, we can consider the velocity of our negative charge, that we through, to be exactly eual to that of the conduction electrons so that this charge appears to be at complete rest in the second frame S'.

Now it is shown that as a consequence of length contraction, the densities of positive and negative chares in the wire don't equal each other in the case of S' frame.
And in the case of S frame, the densities of both positive and negative have been taken to be equal.
This is where my question lies that how have they been taken to be equal in the S frame, since in this frame the positive atoms are at rest and the negative electrons are going with a speed v and thus the density of the negatice charges should be D_0/[root(1 - v^2/c^2)]
where D_0 is the density of the positive charges which would also be the density of the electrons if they wee at rest.
So basically I have taken D_0 as the density of both the positive and negative charges to be equal when they are at rest with respect to each other whereas in feynman, the density has been taken to be equal in the S frame where the atoms are at rest and the electrons are moving with v.

Why isn't it that the density of positive and negative charges are equal when the conduction electrons are at rest with respect to the atoms (though they never are at "complete rest") or in other words, when they don't constitute a current in the wire.

Now of course we can "see" that there are no electric fields that are produced by current carrying wires and that's why the density of the conduction electrons must be equal to that of positive charges when they constitute current.
So by this logic, the feynman statement seems to be correct.
And then if we assume the feynman statement to be correct, then though there are no electric fileds in the current situation since we have taken the denisties to be equal in that case but then if we stop the electric current and if the electrons are brought to rest with respect to the atoms, they should have a density of
D_0 x [root(1 - v^2/c^2)] and then we should observe an electric field in this case, but that's also not so.

Please solve the confusion.  



Thanks,
Shikhin

Answer
Hi Shikhin,

I apologize for the delay, I have been on holidays without posting it up on AllExperts site.
[Q]
If I move one of the charges, say A, by a very small amount, towards the charge B, the force on the charge B will not increase instantaneously but after a time c/r where r is the initial distance between the two charges. Am I right when I say the above?
If the above stataement is correct, then will the force "on A" "due to B", also increase by the time t=c/r since we can position oursleves on B and can equally say that it is B that has moved towards A.[/Q]
Yes, you are correct on both counts. Just remember that when you switch the inertial system (IS), you need to use the coordinates of the new system, so that r/c, for example, becomes r'/c (r are distances measured in the IS of A, r' is measured in the IS of B).

[Q]
Why isn't it that the density of positive and negative charges are equal when the conduction electrons are at rest with respect to the atoms (though they never are at "complete rest") or in other words, when they don't constitute a current in the wire.[/Q]
You are trying to apply the relativistic length contraction, where it doesn't belong. Space is contracted to the moving conduction electrons, i.e. in the S' frame, not to the stationary wire. Charge densities D = Q/V stay equal in the S frame, because both positive and negative charges are counted over the same volume V = pi*rw^2*L (rw is wire radius, L is unit length measured in S).
You need to understand that relativistic effects are NOT calculated as properties of the OBJECTS moving at a certain speed in frame S! They are calculated as properties of EVERYTHING ELSE (everything related to space-time, I mean) measured in the IS tied to a moving object! This is where the source of your confusion lies. It takes some thinking, but there is a big difference between the two approaches and only the latter is correct.
If you try to calculate the charge density as D_0/[root(1 - v^2/c^2)] - I assume the expression correct, I didn't verify it - you are not using L measured in S here, you are using the L' measured in S', which is tied to the moving conduction electrons and, hence, not in S.

[Q] And then if we assume the feynman statement to be correct, then though there are no electric fileds in the current situation since we have taken the denisties to be equal in that case but then if we stop the electric current and if the electrons are brought to rest with respect to the atoms, they should have a density of D_0 x [root(1 - v^2/c^2)] and then we should observe an electric field in this case, but that's also not so.[/Q]
This is just an inverse demonstration of the error made above. Charge densities measured in S are independent of the motion status of the charges.

I think I addressed all points and gave the answer. I will welcome your feedback, as usually.

Cheers,
Daniel