# Classical Electromagnetism vs. 3rd Law? (second round: Relativity)

1. May 13, 2010

### nonequilibrium

Ouch, had just written a big post when my university internet connection timed out on posting, so I'll keep this one short but powerful:

I came to the understanding E and B fields are modified when changing reference frames. But what about the following situation:

Charge q (named 1) is fixed, charge q (named 2) is moving with uniform velocity v. 1 exerts an electrical force on 2, and 2 does the same on 1, but the electrical forces are different due to relativistic modifications (and the magnetic field generated by 2 does not affect 1, being fixed).

Even more so: switch reference frames (keep 2 fixed) and now the forces are switched, meaning it also ignores the principle of relativity.

Surely I must have ignored something elementary (and possibly obvious). Am I right in assuming Lorentz Force gives the effective force? (as E and B are defined that way...) Maybe I have to account for Faraday's Law, but then again, I'm not sure if that is already included in the modificated E-field.

Curious as to what the answer may be,
mr. vodka

NB: As a side, I'm interested in finding out where the statement "charge is invariant" comes from. If there is no theoretical reason for this and purely an experimental one: how can it be experimentally deduced? My problem is that q is defined in terms of E, and we say E is variant, so how do you know maybe q is variant, but we put all the variance in the E-factor? (of course I'm not implying all physicists are wrong and I'm right, I'm just trying to grasp (and thus ask) how we can know it)

2. May 14, 2010

### mikeph

Your 3rd paragraph is very non-relativistic in its description: "1 is still and 2 is moving". From this you derive an asymmetry in the interaction between the two. Well, relativity says that you can't do this- each observer is equally valid* and the mutual interaction is therefore identical (and opposite - which is where your N3 comes in).

*This is Einstein's 1st postulate.

If you re-organise your thoughts in paragraph 3 with the extra consideration that 2 effectively sees 1 as the moving charge which causes a magnetic field, everything should fall into place.

3. May 14, 2010

### Phrak

Are you sure the subject was electric charge rather than the other things that are called charge in an alalogous maner? In the case of electric charge, there is something called the charge continuity equation that is a direct consequence of Maxwell's equations. It just says that the negative change of the charge within a region will equal the current out.

By the way, charge is not invariant to the relative motion of an observer. You should use the word "invariant" within a context such as motion, or over time. It doesn't mean anything without context.

E is not invariant with respect to motion. You said this originally, remember? "E and B fields are modified when changing reference frames." This is true.

4. May 14, 2010

### nonequilibrium

Hm wow, both posts seemed to misinterpret what I was saying. Of course it is my fault for apparently leaving so much room for false interpretation. Let me try to be clear:

Of course I meant "stand in an inertial reference frame in which charge 1 is standing still and 2 is moving with a uniform (non-zero) velocity." Now in this frame, the electric field produced by 2 in the point where 1 is fixed, is different in value from the one created by 1 in the point where 2 is passing (assuming a certain time instant t). Does this not contradict Newton's 3rd law? I don't know what you mean by "everything should fall into place".

As for Phrak, there I have to correct two things:
- I have read charge is invariant to the relative motion of an observer in more than one book. As for the word "invariant", I am using it to talk about the fact its value does not change in different (inertial) reference frames (I don't know of any other uses of the word in physics, but indeed, I could have been more specific).
- Indeed I never tried to claim E is invariant. My point was the following: if the invariance of charge is purely experimental, how can we actually test it? I mean: if we conduct an experiment with moving charges and then calculate the forces present after having done the experiment (as charge is defined in terms of forces, right?), how do we know the charge is invariant? Surely the forces are not invariant, and nor would you ever expect that of course, because as I said: E is invariant and E causes the force, but how do you know "all the variance" comes from the E-field's variance, leaving the charge invariant. To put it foolishly (but maybe more clearly): how do you know not both the charge and E are variant, both changing with half the factor that is supposed to explain the variance of E (of course again: it is not ever that simple, but I'm trying to get the gist of my question across).

Thank you.

5. May 14, 2010

### mikeph

The electric field is independent of velocity of the charge; it doesn't matter if charge 2 is moving or stationary, all that matters is the distance between them.

6. May 14, 2010

### nonequilibrium

The electric field E of a point charge q moving with uniform velocity v in empty space is given by

$$\overline{E} = \frac{q(1-\beta^2)}{4\pi \epsilon_0 r^3 (1-\beta^2 \sin \theta^2)^{\frac{3}{2}}} \overline{r}$$

(for a point situated with respect to q by polar coordiantes (theta,r) with theta the angle between the velocity vector and the position vector, and with $$\beta = \frac{v}{c}$$)

(ref.: Classical Electromagnetism via Relativity, Rosser, p. 114)

7. May 14, 2010

### mikeph

Ok, if you use this formula, beta is a scalar and the same as measured in either charge's rest frame. The only difference is r bar points in the opposite direction, and so E bar does as well.

8. May 14, 2010

### nonequilibrium

How does that resolve the matter?

9. May 14, 2010

### mikeph

You are saying the electric fields as experienced by each charge are different, but that equation shows they are exactly the same.

10. May 14, 2010

### nonequilibrium

Why? Because you just state it? I don't see any argument.

Here is mine: if you're in an inertial reference frame where 1 is still and 2 is moving with a velocity v, you see an electric field induced by 2 on 1 described by the lengthy formula I posted above. On the other hand, because 1 is standing still according my reference frame, I see it induces and electric field on 2 described by Coulomb's Law. Clearly these two quantities are not of the same size.

Now please, I'm not trying to argue I'm definitely right and you are wrong, but I'd appreciate a clear-cut argument from your end.

11. May 15, 2010

### Phrak

Very good. I misspoke. Please, excuse me for being misleading. Electric charge is conserved. Specifically, the current into a volume is equal to the time rate change of contained charge. This is true in any inertial frame.

What is not invariant with respect to motion is charge density. As you can imagine, the dimensions of a box change due to Lorentz contraction. The rate change of current in and out of a region is due to both contraction of the box and time dilation of the current.

12. May 15, 2010

### Staff: Mentor

First, I believe that Newton's 3rd law does hold in this specific case since the charges are not accelerating, you would have to work it out to be sure.

However, in general you are completely correct that Newton's 3rd law does not always hold between two charges in EM. But, Newton's 3rd law is a special case of a more general law which does always hold: conservation of momentum. The key to understand is that the fields themselves carry momentum. Here is my favorite page on the subject:
http://farside.ph.utexas.edu/teaching/em/lectures/node91.html

13. May 16, 2010

### nonequilibrium

Phrak - Aha, I see what you meant. But indeed, I have heard charge is invariant, but how can this be checked experimentally (cf. my initial question)?

DaleSpam - How does Newton's 3rd law hold? If E1 = kq/r en E2 = the length expression in post 6, then surely q*E1 does not equal q*E2. Is this not a breach of Newton's 3rd law? Am I overlooking something?
And I see, apparently there are cases where Newton's law effectively don't work together with Maxwell's, interesting... I must say not all mathematics on that page is familiar to me yet, but I'm getting there. It is interesting this arises due to the fields having momentum. It does strike me as odd that this should lead to a problem, as force can be generally defined as dp/dt. But I'll take your word for it (until I know all the math).
Also, could you also take a look at my second question? How does my experiment hold up if you switch reference frames? In the case of a breaking of Newton's 3rd law, in a different reference frame the forces would be switched, breaking the principle of relativity :s Surely I'm wrong on this one, but I can't see it.

14. May 16, 2010

### Staff: Mentor

For a moving charge you need to use the http://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential" [Broken], not Coulomb's law. I think when you use that you will find that Newton's 3rd law holds here, but I haven't worked it out myself so I am not sure.

Maxwell's equations are relativistic (in fact, they were relativistic before special relativity was developed). So if you switch reference frames they still hold exactly the same. Again, you would need to use the Lienard Wiechert fields, which are the solution to Maxwell's equations for an arbitrarily moving point charge.

Last edited by a moderator: May 4, 2017
15. May 17, 2010

### nonequilibrium

DaleSpam, I appreciate you trying to help, but it seems you've completely ignored me on what I'm saying twice in a row now... Did you truly read what I said?

I gave the correct expressions for E1 and E2 (as E1 is the Coulomb Law "version" because 1 is standing still, and for E2 I referenced to an earlier post where I stated the electric field for a moving charge). Note: with E1 I mean the electric field generated by 1 on 2. Now on sight you see E1 is not equal to E2, yet Newton's third law says q*E1 = q*E2. Can you respond directly to this? Also, if you switch reference frames, then if 1 experienced the smaller force in the first reference frame, then now 2 experiences the smaller force. Is this not a direct contradiction with the theory of relativity? Thank you for your time.

16. May 17, 2010

### Count Iblis

I don't understand the problem here. You have two charges that are moving relative to each other. In each rest frame the other charge is moving. This looks very symmetrical to me....

17. May 17, 2010

### nonequilibrium

It is in the forces that there lies the (presumed...) assymetry. Let me try to make it very clear what I mean, because I seem to be very vague:

Situation: inertial reference frame where charge q (named 1) is still and charge q (named 2) is moving with constant velocity v. If E1 is the electric field created by 1, then E1 = kq/r. For E2, since it is moving, we have:

$$\overline{E_2} = \frac{q(1-\beta^2)}{4\pi \epsilon_0 r^3 (1-\beta^2 \sin \theta^2)^{\frac{3}{2}}} \overline{r}$$

If F1 is the force by 1 on 2, then F1 = qE1. Also, F2 = qE2. Obviously, F1 is not equal to F2. Conclusion: Newton's third law doesn't hold.

EXTRA: If it doesn't hold, one is experiencing a greater force than the other (you can calculate which, but it's not important for the concept). Say F1 < F2. Now view the whole ordeal from a different reference frame where 1 is moving and 2 is still. Now 1 is generating the adjusted electric field, and by complete analogy F1 > F2. Is this not a contradiction with the theory of relativity?

I hope I have finally made my question clear! I await your responses :)

18. May 17, 2010

### Staff: Mentor

Sorry, I am not ignoring you I just have not checked your work myself.

The correct fields for an arbitrarily moving point charge are given by the Lienard Wiechert potential. This reduces to Coulomb's law for a stationary particle, but I am not sure if it reduces to the expression you gave for E2.

Yes. My direct response is that it doesn't matter. There are definitely scenarios in EM where Newton's 3rd law fails, regardless of whether or not this is one. In those cases momentum is still conserved because the fields carry momentum.

No. The theory of relativity says that the laws of physics are the same in all reference frames. It does not say that all reference frames will agree on all quantities. In this case Maxwell's equations hold in both reference frames.

19. May 17, 2010

### Count Iblis

Ok, I see your point now.

The way you've written the formula for E2 obscures that it is time dependent; you are focussing on some particular moment. If you think about that, you can more easily understand what is going on.

If the third law were to hold in some reference frame, that would mean that the momentum changes of the charges would be equal in magnitude and opposite in direction at each moment during some time interval dt.

If you now transform to some other inertial frame, then the two oppsite momentum changes happening between t and t + dt transform to two different but also opposite momentum changes, but they no longer happen at the same times in the new inertial frame (in general). This would not be a problem if the forces were equal as a function of time, but that's not the case. So, the conclusion is that if the third law holds in one inertial frame, it won't hold in another in general.

We know for sure that the third law does hold in the center of mass inertial frame by symmetry. It is easy to see that the Lorentz boost from that frame to the frame you consider will lead to simultaneous events in the CoM frame to become non simultaneous in your frame.

20. May 19, 2010

### nonequilibrium

Aha! Thank you both a lot for replying; I think I get it, and I must say it's truly interesting.

So generally (and in my example) the third law of Newton doesn't hold up in its rudimentary form and this is due to relativity. The fact I assumed there would also follow a contradiction in the theory of relativity (for my example), was because I assumed (necessary) simultaneity in all frames. The law for conservation of momentum still holds though, due to there being a momentum associated with E,B-fields.

Does this summary look fitting?

Thanks once again :)