# Magnetism seems absolute despite being relativistic effect of electrostatics

1. Feb 14, 2012

### universal_101

I know that magnetic force due to a current carrying wire on a test charge moving w.r.t the wire(along the wire), can be interpreted as the electrostatic force if we use the first order relativistic corrections for Time Dilation or Length contraction of the charges of the wire, in the frame of the the test charge.

But what I don't seem to understand is rather very simple situation.

Let's consider a simple model of a conducting wire,

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Now, let's suppose there is some current in the wire and the electrons are moving at speed 'v' w.r.t the the wire,
secondly, a stationary test charge w.r.t the wire lying around.

Naming the above scenario as (1)

Now, the test charge starts moving in the direction of electrons with the same speed 'v'.
This time in the reference frame of the test charge, electrons are stationary and nucleus(positive charge) is moving at speed 'v'.

Naming this scenario as (2)

And so the question arise, the two scenario are identical w.r.t principle of relativity. That is, in the first case only negative charges are moving, but there is no force on the charge. But in the second case when positive charges are moving there is a force on the test charge(magnetic force towards wire). Whereas, the two cases are essentially identical w.r.t principle of relativity.

2. Feb 14, 2012

### Staff: Mentor

3. Feb 14, 2012

### universal_101

The explanation is quite good, but then we have another problem if we assume this explanation to be correct.

This is, why then we don't see any force when the situation changes from NO current to some current. Since, due to the motion of charges, which are responsible for current should also go through the length contraction when compared to the stationary state of these charges, when there is NO current.

Therefore, I think, according to the above explanation, there should be a magnetic force even if we switch ON or OFF the current.

4. Feb 14, 2012

### Staff: Mentor

If there is no current then the electrons are not moving therefore there is only one single reference frame and the test charge is stationary as are all of the charges in the wire. There is no length contraction, no net charge, and no force on the test charge with the current off.

5. Feb 14, 2012

### universal_101

Yes, that is correct.

But there should be all these effects when the current is ON, that is, when electrons are moving and therefore there should be length contraction and thus Force.

6. Feb 14, 2012

### kmarinas86

From http://physics.weber.edu/schroeder/mrr/MRRtalk.html:

It is not clear to me what as to which plane would the charges actually length contract toward.

Say I have a sequence of charges like this at 0 current condition:

Code (Text):

+ + + + + + +
- - - - - - -

At a length contraction factor of 1/2, do we have:

Code (Text):

+ + + + + + + +
--------

or

Code (Text):

+ + + + + + + +
--------

or

Code (Text):

+ + + + + + + +
--------

or what?

Would we have a situation where the field along the wire appears to lack uniformity because of such contraction, with, say, the leading and trailing end of the wire being more positive? This doesn't seem very intuitive or logical if you ask me.

Last edited: Feb 14, 2012
7. Feb 14, 2012

### Staff: Mentor

Yes, when the current is ON then the electrons and the test charge are moving, all these effects are present, and there is a force.

In the lab frame the force is attributed to the magnetic force on the moving charge due to the current in the neutral wire. In the electron/test-charge frame the force is attributed to the electrostatic force on the stationary charge due to the net charge in the wire.

8. Feb 14, 2012

### Staff: Mentor

It wouldn't be as simple as that. Remember the relativity of simultaneity.

Suppose that you suddenly turn the current on at time t=0 in the lab frame. So at t=0 electrons begin leaving one end of the wire at a certain rate and entering the other end of the wire at the same rate, so there is no net charge. In the moving frame the beginning of the electrons leaving one end is not the same time as the beginning of the electrons entering the other end, so there is a net charge.

9. Feb 14, 2012

### kmarinas86

10. Feb 14, 2012

### Staff: Mentor

The same conclusion is true in the steady state situation, it is just easier to describe in the charging situation.

11. Feb 14, 2012

### kmarinas86

I am unable to assume that the lack of uniformity in the charging situation is somehow analogous to the steady state situation. They are completely different. In the charging situation, there is a changing drift velocity. Such is not the case in the steady state situation.

Let's be a little more direct here: To what plane does the electron bulk flow actually contract towards as a result of the bulk's relative velocity with respect to the rest frame of the positive charges?

For our sakes, let's assume that the observer is positioned between the 4th and 5th + charges and that the - charges are migrating to the right.

Last edited: Feb 14, 2012
12. Feb 14, 2012

### Staff: Mentor

They are not completely different, in fact, shortly after both ends have switched they are exactly the same.

If that is not good enough for you then you are welcome to pursue the math on your own, but do not (as you did in your ASCII drawings) forget the relativity of simultaneity, and do not forget the charges that are entering and leaving the ends of the wires.

13. Feb 14, 2012

### kmarinas86

So what does the un-uniformity look like?

What does your comment about the relativity of simultaneity have to do with the situation with steady state current?

Finally, to what plane do the - charges actually contract in the rest frame of the + charges?

Last edited: Feb 14, 2012
14. Feb 14, 2012

### Staff: Mentor

Uhh, uniform.

15. Feb 14, 2012

### kmarinas86

Note: The previous post was recently edited to say "un-uniformity".

Let's keep this REALLY simple. Assuming that the wire is neutral (no net charge) and that the wire is 1 meter long and that I have a length contraction of electrons, why should I get from that a uniform charge distribution when the electrons are drifting through wire (current)?

I would TOTALLY expect an un-uniform distribution, assuming length contraction applies to the bulk flow of electrons.

I STILL don't have an answer to my question as to what do the electrons actually length contract towards.

Last edited: Feb 14, 2012
16. Feb 14, 2012

### Staff: Mentor

In the steady state the four-current (density) is uniform and constant in the lab frame, therefore it is uniform and constant in the test-charge frame also.

Why? Why do you expect a gap of any kind in the steady state?

Length contraction occurs, as always, in the direction of motion. The word "towards" doesn't make any sense in this context. The word "towards" implies something changing over time. Length contraction does not change over time in an inertial frame.

Last edited: Feb 14, 2012
17. Feb 14, 2012

### kmarinas86

Ok, then let me ask it this way: From the lab frame, where is the center of contraction for the bulk of electron flow in a straight wire conductor? The contraction is only "linear", so I assume that this "center" of contraction must be a geometric plane. Where is that located in relation to the observer?

SR says that objects (read: multiple particles) will length contract. So, logically speaking, you can treat the + charges and - charges as two separate "objects" at different speeds. I assume this to mean not only the particles by themselves, but the entire bulks of the particles as a whole. For an object to contract, the distance in-between also has to contract. You don't have just the fundamental particles contracting. In the extreme case, going from 0 current to a very high current would cause the following to occur:

This

Code (Text):

+       +       +       +       +
-       -       -       -       -

into this

Code (Text):

+       +       +       +       +
-----

or

Code (Text):

+       +       +       +       +
-----

or

Code (Text):

+       +       +       +       +
-----

et cetera

Last edited: Feb 14, 2012
18. Feb 14, 2012

### Staff: Mentor

No, I already covered the non-steady state situation in post 8. None of your suggestions are correct, neither in the transient nor in the steady-state conditions.

19. Feb 14, 2012

### kmarinas86

Can you describe the nature of the length contraction?

The rest frame of the straight conductor is chosen. If the negative charge of the free electron steady current is seen uniformly spread throughout the wire, then I have no choice but to assume that the free electron charge density is inversely proportional to the factor $\sqrt{1-\left(\frac{v}{c}\right)^2}$, where v is the electron drift velocity of the current. Logically then, the amount free electrons in the same conductor must increase to the same degree as the free electron charge density does. The density and count of other charges is not affected. This would violate the neutral wire assumption. If that is not the case, prepare or find a picture of what you think actually happens.

Your "uniformity" assumption combined with a length contraction factor of $\sqrt{1-\left(\frac{v}{c}\right)^2}=1/8$, then we would have something like this:

Code (Text):

+       +       +       +       +
---------------------------------

Last edited: Feb 14, 2012
20. Feb 14, 2012

### DrGreg

I think the following comment I made 2 years ago applies here:

The application to this thread is that the electrons are not rigidly linked to each other so there's no reason for the rest-distance between them when moving to equal the rest-distance when not. In fact they will spread out to fill whatever space is available to them.

There is no "centre of contraction" because contraction-due-to-acceleration doesn't occur.