# Magnetic Field Caused by Relativistic Effect

1. Feb 28, 2007

### farmboy

I'm a newbie, so please correct me if this is not a good post.

Question:

1
The absolute speed of current-carrying electrons of a steady current in a wire are very slow, with respect to the wire.

2
So how does the magnetic field arise from the relativistic imbalance between the moving electrons and the sum of the stationary charges of the valence electrons and metallic atomic cores?

3
Is this because there is a very high density of conduction electrons in the wire (~10^23/cm^3), so even with the electrons moving at only a few m/sec or less, still aggregate in total to a large coulombic effect because there are so very many?

Thanks!
farmboy

2. Mar 1, 2007

### daniel_i_l

Welcome!
The magnetic force around a wire as a result of relativity works like this:
Lets say you have a wire going from left to right with a positive charge ontop of it moving to the right with the same speed as the E's in the wire. In this frame, the only force on the P (positive charge) is the magnetic field of th E (electrons) which pulls the P down.
But suppose we look in the E's frame - here the E's and the P are at rest so why should it be pulled down (the magnetic force is 0)? The trick is in the first frame, there was an equal density of positive and negative charges in the wire - and the negative charge was moving to the right - so in the electrons frame (which is the second frame) the electrons are denser due to length contraction giving a net negative charge - this makes an electric field around the wire which (when taking time dialation into account) accelerates the P exactly like the magnetic force in the first frame.
So you can see that the magnetic force is just the electric force looked at in a different frame - if you look in the frame of the moving charge then all the forces on it that were magnetic in the frame were it was moving can now be described in terms of the electric field.

3. Mar 1, 2007

### farmboy

But the speed of the electrons is very slow...

Thanks Daniel.

This is my understanding -- that a magnetic field is a relativistic view of an electric field -- and that the single, unified, underlying process is that of photon exchange between charges. So a magnetic field "is just really an electric field." The speed of the observer with respect to the speed of the current-carrying charges makes the difference.

My question is: If in fact the current-carrying charges are moving slowly (below m/s, showable from the wire cross section, the total current, etc) -- then how does a relativistic correction involving a term like (v^2/c^2) come into play? It must be that although a single charge will have a fantastically small relativistic correction because the above term is so very small -- that in aggregate, the ~10^23/cm^3 in the wire add up to a sizable net relativistic effect.

It looks to me like it is nature's great gift that we are given such a high density of "neutral charge" in a metallic wire, yet are able to generate an "electric field by another name" with this relativistic imbalance between the moving electrons at ~m/s, and the underlying core positive ions and underlying negative valance electrons.

fb

4. Mar 1, 2007

### gabee

"But the speed of the electrons is very slow..."

Actually the electrons in a conductor move with speeds on the order of 106 meters per second, the NET charge is what is drifting very slowly along the wire. You can think of the electrons in a conductor as wildly bouncing around everywhere at high speeds with the whole group slowly drifting to the left. Does that help?

5. Mar 1, 2007

### jostpuur

I must point out one other point of view on this matter, although length contraction was already mentioned.

If there is a test particle flying close to a wire, the force that all electrons in the wire exert on the test particle is huge. I don't bother calcute how many zeros there are in such force, but you can estimate it if you want. On the other hand, the protons exert also a huge force, which cancels the force from the electrons. However, if the electrons are moving, the force that they exert is altered due to relativistic effects. Since the velocity of the electrons is small, the force is altered relatively very little. If the force from electrons is something like $$F(1-v^2/c^2)$$, and then the stationary protons exert a force $$-F$$, you are left with $$-Fv^2/c^2$$. You might think that is a small number, because velocity was small, but if the $$F$$ was big in the first place, there could as well be a detectable effect, and this is where the magnetism comes from. Okey? I never heard of this in lectures, but I think that is an important point of view. I noticed this myself, so feel free to be skeptical...

Your idea that magnetic fields are only electric fields is one the right track, but you shouldn't really use those words, because electric and magnetic fields are still different. Try to find something on electromagnetic field tensor $$F^{\mu\nu}$$. The idea is that there is some absolute quantity, and $$E$$ and $$B$$ are merely different ways of dealing with this quantity. (Analogously like the coordinates of a point are only some way dealing with a point.)

6. Mar 1, 2007

### farmboy

Gabee, Jostpurr, Daniel --

Thanks for the great responses!

fb

7. Mar 1, 2007

### rbj

from Wikipedia:

A thought experiment one can do to show this is with two identical infinite and parallel lines of charge having no motion relative to each other but moving together relative to an observer. Another observer is moving alongside the two lines of charge (at the same velocity) and observes only electrostatic repulsive force and acceleration. The first or "stationary" observer seeing the two lines (and second observer) moving past with some known velocity also observes that the "moving" observer's clock is ticking more slowly (due to time dilation) and thus observes the repulsive acceleration of the lines more slowly than that which the "moving" observer sees. The reduction of repulsive acceleration can be thought of as an attractive force, in a classical physics context, that reduces the electrostatic repulsive force and also that is increasing with increasing velocity. This pseudo-force is precisely the same as the electromagnetic force in a classical context.

here's the quantitative version (from the talk page):

The classical electromagnetic effect is perfectly consistent with the lone electrostatic effect but with special relativity taken into consideration.

The simplest hypothetical experiment would be two identical parallel infinite lines of charge (with charge per unit length of $\lambda \$ and some non-zero mass per unit length of $\rho \$ separated by some distance $R \$. If the lineal mass density is small enough that gravitational forces can be neglected in comparison to the electrostatic forces, the static non-relativistic repulsive (outward) acceleration (at the instance of time that the lines of charge are separated by distance $R \$) for each infinite parallel line of charge would be:

$$a = \frac{F_e}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho}$$

If the lines of charge are moving together past the observer at some velocity, $v \$, the non-relativistic electrostatic force would appear to be unchanged and that would be the acceleration an observer traveling along with the lines of charge would observe.

Now, if special relativity is considered, the in-motion observer's clock would be ticking at a relative *rate* (ticks per unit time or 1/time) of $\sqrt{1 - v^2/c^2}$ from the point-of-view of the stationary observer because of time dilation. Since acceleration is proportional to (1/time)2, the at-rest observer would observe an acceleration scaled by the square of that rate, or by ${1 - v^2/c^2} \$, compared to what the moving observer sees. Then the observed outward acceleration of the two infinite lines as viewed by the stationary observer would be:

$$a = \left(1 - v^2 / c^2 \right) \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho}$$

or

$$a = \frac{F}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} - \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} = \frac{ F_e - F_m }{\rho}$$

The first term in the numerator, $F_e \$, is the electrostatic force (per unit length) outward and is reduced by the second term, $F_m \$, which with a little manipulation, can be shown to be the classical magnetic force between two lines of charge (or conductors).

The electric current, $i_0 \$, in each conductor is

$$i_0 = v \lambda \$$

and $\frac{1}{\epsilon_0 c^2}$ is the magnetic permeability

$$\mu_0 = \frac{1}{\epsilon_0 c^2}$$

because $c^2 = \frac{1}{ \mu_0 \epsilon_0 }$

so you get for the 2nd force term:

$$F_m = \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} = \frac{\mu_0}{4 \pi} \frac{2 i_0^2}{R}$$

which is precisely what the classical E&M textbooks say is the magnetic force (per unit length) between two parallel conductors, separated by $R \$, with identical current $i_0 \$.

Last edited: Mar 1, 2007
8. Mar 1, 2007

### Hans de Vries

9. Apr 13, 2007