# A question about moving electric charges and their force

1. Sep 20, 2014

### jethomas3182

I want to find out whether I'm doing something wrong, and where to look for similar thinking.

Imagine a thin wire that carries a constant electric current. Negative charges move to the right. Positive charges are stationary. The charges balance on average, or maybe there are a few more negative charges to the left and fewer negative charges to the right. Apart from perhaps a small gradient, the distribution of positive charges and of negative chargeds are both uniform.

Choose a positive charge at the center. What forces are on that charge? My first thought is to say that there are an equal number of stationary positive charges on both sides, so those cancel. And there are nearly equal negative charges on both sides, so those mostly cancel too.

It takes time for the force from an electric charge to arrive at the center, so the forces we add up to get the total force at one time will actually reach back into the past. They are "retarded" forces. But that doesn't matter because it's a steady state and there were just as many charges in each location in the past as there are now.

But wait, it does matter. Right now the electrons are uniformly distributed along the wire. At any time in the past the were uniformly distributed. But the ones we add up to get the force right now are not uniformly distributed. The farther we go back into the past to find charges that affect the center right now, the more those charges move before they reach their uniform distribution today. If the charges travel at speed v, then a charge that was at location -t at time -t will be at location -t+vt now. The charges that have a uniform distribution now were spread out across the various times that they created the force that affects the center now.

Similarly, the charges on the right which are uniformly distributed now, were bunched together when they created the force that affects the center now. A charge that was at location t then, is at location t+vt now.

So the charges on the left affect the center less than the charges on the right. When we sum, we sum over a path that's denser with electrons on the right than on the left.

Charges that move away from each other should exert more force on each other, than forces that are getting closer.

Have I left out something? Not magnetism. There is no magnetism unless both charges are moving.

I think not relativity. Time dilation and space contraction are the same whether you are coming or going, right?

All I can think of, is that maybe when a charge is approaching, its electric force is kind of squeezed in front of it. So the force is denser even though the charges themselves are less dense, and maybe that cancels out. Or possibly that hypothetical gradient creates the average velocity, and the charge effect due to velocity is exactly balanced out by the charge effect due to gradient?

Where can I find the results the actual experts got when they figured this out?

2. Sep 20, 2014

### Staff: Mentor

Are you wanting to consider a resistive wire or a superconducting wire?

3. Sep 20, 2014

### jethomas3182

I started simply by imagining stationary positive charges and moving negative charges.

I'm not sure what makes a one-dimensional wire resistive or superconducting. I'm not clear what makes a high-temperature copper wire resistive. Something about crystal structure? Flaws in crystals, and interfaces between crystals? Thermal motion that jostles electrons into random directions?

If positive charges behind moving electrons attract more than equivalent positive charges ahead, that would create resistance in itself and then I'd need a reason why that doesn't happen in superconductors. But I'm not at all sure it happens in thin copper wires either.

4. Sep 20, 2014

### Staff: Mentor

The reason I ask is because all you need to know is the fields at the location of interest and the motion of the charges. If you are in a conductor then there is an E field and if you are in a superconductor then there is not.

The stuff about the retarded potential is only relevant in so far as it alters the field, which isn't much.

5. Sep 20, 2014

### jethomas3182

I'm pretty sure I want an E field, but if it isn't too much trouble could you do it both ways? Or point me to something that explains it?

6. Sep 20, 2014

### Staff: Mentor

I am not sure what you mean by "do it both ways".

Since the protons are at rest there is no magnetic force on them regardless of the presence or absence of a B field. If there is an E field (conductor) then there is an electric force on them. If there is not an E field (superconductor) then there is no electric force on them.

7. Sep 20, 2014

### jethomas3182

Yes, the protons are at rest so they neither create nor are affected by a B field.

I figured that each individual electron has an E field and those are summed to create a composite E field.

I imagine that in a superconductor maybe all the individual E fields sum to zero? It seems like it would be a great big change if they didn't have individual E fields because they are in a superconductor.

So my concern is about the right way to sum all the individual E fields.

Last edited: Sep 20, 2014
8. Sep 20, 2014

### Staff: Mentor

Maxwell's equations are linear, so you can indeed simply sum individual solutions to make a composite solution. However, electrons are strange things, definitely not classical point charges, so in practice this is not the way to calculate the field inside the wire. Instead, you simply use Ohm's law to determine the E field directly.

9. Sep 20, 2014

### jethomas3182

I looked at a derivation of Maxwell's equations that did not consider delays at all. That made things simpler. When I started to consider delays, the first strange result I got was that it looked like the force from charges moving away did not balance the force from the same charges approaching. At first that seemed strange because they look symmetrical. But it doesn't violate any conservation laws, all that has to balance is that any two charges have the same force on each other, not that charges that are now an equal distance from a third charge now, should have the same force on the third charge now.

It's possible that if I consider electrons as classical point charges and do it the correct way, I might get approximately correct results. And the first obvious correction is the lightspeed delay.

I'll simplify it. Say that right now, we have a plus charge at 0 and we have negative charges at -5, -3, -1, 1, 3, 5 and the negative charges are moving at velocity v where light travels at 1 distance unit per time unit.

To figure the force of the negative charges on a charge at place 0 and time 0, the charge that's now at -1 was at time -1/(1-v) and distance -1/(1-v) when it created the force that affects 0 at time 0.

The charge that's now at -2 was at time -2/(1-v) and distance -2/(1-v) when it created the force that affects 0 at time 0.

The charge that's now at 1 was at time = -1/(1+v) and distance 1/(1+v), and the charge at 2 was at time -2/(1+v) and distance 2/(1+v).

In general, a charge that's now at distance +t was at time -t/(1+v) and distance t/(1+v)) when it created the force that affects 0 at time 0. And a charge that's now at distance -t was at time -t/(1-v) and distance -t/(1-v) then.

The magnitude of the force varies by the inverse distance squared. So the charge which is now at +t has a force $$\frac{(1+v)^2}{t^2}$$. The charge that's now at -t has force $$\frac{-(1-v)^2}{t^2}$$

Add those two forces and we get $$\frac{4v}{ t^2 }$$

They don't balance.

Unless I have made a mistake.

It occurs to me that this might look like a crackpot theory that doesn't fit the way classical physics was done. My intention was to find out how this problem was actually solved, but if it's too far from mainstream I can ask elsewhere.

10. Sep 21, 2014

### Staff: Mentor

11. Sep 21, 2014

### jethomas3182

Thank you! I can work with that!

So, what I have done is predict the electric force in the direction of motion, given nothing but the stationary electric force and the geometry of the situation. Lienard Wiechert gives the real prediction.

$$E(r,t) = \frac{1}{4\pi} \frac{q(n-\beta)}{\gamma^2 (1-n.\beta)^3 |r - r_s|^2}$$

For my special case I can throw away the second term since it's constant motion.

I'll throw away various fudge factors like $\frac{q}{4\pi}$
$n$ is a unit vector, I can replace it with 1 or -1.

$$E(r,t) = \frac{1-\beta}{\gamma^2 (1-\beta)^3 |r - r_s|^2}$$

$\beta_s(t) = \frac {v_s(t)}{c}$ I called that v.

$$E(r,t) = \frac{1-v}{\gamma^2 (1-v)^3 |r - r_s|^2}$$

$$\gamma(t) = \frac{1}{\sqrt{1-|\beta(t)|^2}}$$

$$E(r,t) = \frac{(1-v)(1-v^2)}{ (1-v)^3 |r - r_s|^2}$$

And $|r-r_s|$ is what I call t, the distance between where the moving charge was then and the location its electric field affects now.

$$E(r,t) = \frac{(1-v)(1-v^2)}{(1-v)^3 |t|^2}$$

Now go back and get the signs right for n.

For -r, n=-1.

$$E(-r,t) = \frac{(-1-v)(1-v^2)}{(1+v)^3 |t|^2}$$
$$E(-r,t) = \frac{-(1-v^2)}{(1+v)^2 |t|^2}$$

For +r, n=1
$$E(r,t) = \frac{(1-v)(1-v^2)}{(1-v)^3 |t|^2}$$
$$E(r,t) = \frac{(1-v^2)}{(1-v)^2|t|^2}$$

$$E(r,t) + E(-r,t) = \frac{1-v^2}{t^2} \frac{(1+v)^2 -(1-v^2)}{(1+v)^2(1-v)^2}$$

$$= \frac{1-v^2}{t^2} \frac{4v}{(1+v)^2(1-v)^2}$$

So it looks like I'm off by a factor of $1-v^2$. I may have made additional mistakes.

I'll look for where that error came from. In the meantime it looks real. When charges are approaching you and an equal number of charges are leaving, other things equal the electric field from the departing charges is stronger than that from the arriving ones.

12. Sep 21, 2014

### jethomas3182

The 1-v^2 term is $\frac{1}{\gamma^2}$ which is something that gets stuck into relativistic equations. I'll look at why that is, but as I expected it has the same effect coming and going, so it only affects the magnitude of the effect.

I can use it to simplify.

$$E(r,t)+E(−r,t) = \frac{1-v^2}{t^2} \frac{4v}{(1+v)^2(1-v)^2}$$
$$=\frac{1-v^2}{t^2} \frac{4v}{(1-v^2)^2}$$
$$=\frac{1}{t^2} \frac{4v}{1-v^2}$$

13. Sep 22, 2014

### jethomas3182

After another leisurely break I looked at it again. I decided my notation would be confusing for someone else to read.

I want to look at the special case of electric charge with constant motion, and the force in line with the direction of motion.

Starting from first principles, I set c=1. The unit of distance is the light-second or the light-nanosecond or whatever.

Calculate the electric force on a unit charge at t=0, at the origin, from a unit charge now at distance w whose velocity is v < 1 to the right.

The charge is at w now. At some time -t in the past it was at distance d so that the force traveled distance d in time t. d=t

If w > 0, d = w - vt
If w < 0, d = |w| + vt

since d = t,

w = t(1+v) or |w| = t(1-v)

t = w/(1+v) or t = -w/(1-v)

So if the distance is now w, the force f arriving from d should be 1/d^2 by coulomb's law.

f = (1+v)^2/w^2 or f = -(1-v)^2/w^2

The sum of forces from both sides is 4v/w^2

But the Lienard/Wiechert formula is different. I think it would be the same as mine if you divided its force by gamma^2.

$$f = \frac{(n-\beta)}{\gamma^2 (1-n.\beta)^3 |r-r_s|^2}$$

$\frac{(n-\beta)}{(1-n.\beta)}$ in the linear case simplifies to

$\frac{(1-\beta)}{(1-\beta)} = 1$ when n=1 which happens when $r_s$ <0 (since r=0)

$\frac{(-1-\beta)}{(1+\beta)} = -1$ when n=-1 which happens when $r_s$ > 0

$$f = \frac{\pm 1}{\gamma^2 (1-n.\beta)^2 |r-r_s|^2}$$

$$\frac{1}{\gamma^2} = 1-\beta^2$$

$$f_r = \frac{-(1-\beta^2)}{(1+\beta)^2|r-r_s|^2}$$ or
$$f_l = \frac{(1-\beta^2)}{(1-\beta)^2|r-r_s|^2}$$

$$f_r = \frac{-(1-\beta)}{(1+\beta)|r-r_s|^2}$$ or
$$f_l = \frac{(1+\beta)}{(1-\beta)|r-r_s|^2}$$

Add our two solutions for $r_s$, and we get

$$f_r + f_l = \frac{4t}{(1-\beta^2)|r-r_s|^2}$$

This is my solution times $\gamma^2$.

Unless I have misread the notation, which seems likely. The text appeared to say that the formula does not take the current position and adjust the force to retarded time, but instead it takes the force from retarded time and adjusts it as if the force instantaneously came from the present position! That would be quite a trick! But it still leaves the forces unequal coming and going.