# Relativistic Coulomb Force

Discussion in 'Physics & Math' started by tsmid, May 16, 2017.

1. ### Q-reeusValued Senior Member

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And how would one determine charge density for, in standard QED, a notional point charge? Maybe a metal globe with a uniform surface charge is relevant as an academic exercise. Remember the OP issue is one of net charge in one frame vs another.

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3. ### The GodValued Senior Member

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Same way as one determines the curvature of spacetime at singularity.

But for non point charges (beyond OP but reasonably relevant), the change in charge density is present.

5. ### Q-reeusValued Senior Member

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For those who believe in such absurdities maybe such a 'determination' makes sense.
Sure. Suppose a spherical shell in it's rest frame S has uniform surface charge density ρ. In frame S', a Lorentz boost γ = 1/√(1-v²/c²) turns the sphere into an oblate ellipsoid. Applying Pythagoras's theorem, I obtain the surface charge density ratio, as a function of polar angle θ in S, to be
ρ'/ρ = 1/√((1/γ²)sin²θ + cos²θ)
Do you agree?

7. ### The GodValued Senior Member

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The fact is that charge density changes, exact distribution notwithstanding.

Once the charge density changes, can we not have a solid of some tricky shape (say volume density) part whose charge is different for rest and otherwise condition. The overall charge under motion for the entire object can still be the same.

I know you can show me by integrating over the full volume that charge is invariant but what about charge of part(s) of the volume.

8. ### Q-reeusValued Senior Member

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A moving slab of electrically polarized matter has an induced orthogonal magnetic polarization, and vice versa. All thanks to ME's and SR.

9. ### The GodValued Senior Member

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Thanks, whole...Inclusive one.....is ok. It is the part which is troublesome. You have also agreed on regions of low and high charge densities. A simpler intuitive stuff could be that charge density remains uniform from spherical to ellipsoidal distribution.

Ideally speaking the relativistic contraction, and thus the deformation of original object, should ensure same total charge in rest and otherwise condition for all or any given part of the original object.

10. ### tsmidRegistered Senior Member

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OK, so the electric field from an ensemble of charges with an isotropic velocity distribution is independent of the velocity i.e. given by the classical Coulomb law. But what about an anisotropic distribution? Assume you have an equal number of positive and negative charges homogeneously distributed inside a box, both initially at rest, so the electric field is zero everywhere outside the box. Now you accelerate the negative charges to a certain velocity so that they bounce back and forth between two facing sides of the box (ignoring collisions with other charges). With the positive charges still at rest, this means a stationary test charge situated outside the box facing one of the other sides of the box would see a net electric field (γ-1) times the field that would be produced by all the negative charges alone. This also means the test charge suddenly has an electrostatic potential energy. Where does this energy come from, considering that the work done on the negative charges in the box has been completely converted into their kinetic energy?

11. ### Q-reeusValued Senior Member

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Taking a breather at that point, would you agree that admission is itself good news? I think so.
You are almost describing the case of a fundamental mode cavity oscillator filled with a dielectric. Conventional microwave theory predicts precisely zero external field. Well actually for finite skin depth there will be a subtle modification of the internal cavity fields resulting in very weak leakage to the outside. And for 'pure' dielectric cavity oscillators, some leakage always occurs. Beyond that, I discovered a long while ago that the fully relativistic field eqns do predict a net exterior field, even for a perfectly conducting cavity oscillator, but I won't elaborate here.

At any rate, suppose the distribution of moving charges corresponds roughly to a leaky dielectric cavity oscillator. What you have is a coupled system. Lattice charges + conduction charges + exterior charge. That system will obey the ME's and no mysterious energy excess or deficit is predicted. Some RF energy source must continue to supply energy to compensate for losses. The usual case in RF/MW systems.

12. ### tsmidRegistered Senior Member

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I was not implying any net oscillatory behaviour of the charges. The overall charge distribution in the box is still random and overall neutral at any time, only that, in contrast to the 3D-case, the charges are moving just in one dimension inside the box (say the x-axis). So there isn't even a net current at any time. Yet still, if the positive and negative charges have different speeds, a remote test charge located on a line perpendicular to the x-axis should see a net electric field because of the different γ-factor with which the field of each individual charge is amplified.

13. ### Q-reeusValued Senior Member

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OK that wasn't clear before. What you are really talking about is a collection of dipole oscillators having random phases but strung out along a line and with their axes all along that line. The fields of a dipole oscillator are well known to antenna folks:
http://www.waves.utoronto.ca/prof/svhum/ece422/notes/05-dipole.pdf see eqn. (19)
Along the oscillation axis, E field further out drops off as 1/r² and is technically non-propagating. The propagating i.e. 1/r radiation field is zero there and a maximum in the equatorial plane where θ = π/2.
I think you can figure what that will give as an overall emission from the line of linear oscillators.

Last edited: May 26, 2017
14. ### przyksquishyValued Senior Member

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I told you earlier: you're only getting an apparent problem because you're starting with the wrong formula for the electric field.

First the obvious: the general way to determine the electric and magnetic fields produced by a system of charges and currents is to solve Maxwell's equations. Maxwell's equations are linear and its solutions obey the superposition principle, so for working out the electromagnetic field the only thing that matters is the net charge and current densities throughout the source producing it. In your case the problem is easy: you've described a situation in which the net charge is zero and there are no net currents throughout the source, so the electric field both inside and outside your box is zero, regardless of what individual charges are doing at the microscopic level.

So electromagnetic theory predicts there will be no electric field around your box. The only possible remaining question is how this reconciles with the formula you posted in the opening post of this thread from the lecture notes you linked to. The answer is what I already told you: that formula is a special case for a charge moving in rectilinear motion. It formally only applies for a charge moving in a straight line with constant velocity from $t = -\infty$ to $t = +\infty$. In your problem the charges are bouncing back and forth inside the box, i.e., accelerating, so the formula does not directly apply.

For a charge moving with a non-constant velocity, the electric field at a particular location (the generalisation of the Coulomb field) is given by [1]
$\displaystyle \boldsymbol{E} = \frac{q}{4 \pi \epsilon_{0}} \biggl[ \frac{\boldsymbol{e}_{r'}}{r'^2} + \frac{r'}{c} \, \frac{\mathrm{d}}{\mathrm{d}t} \biggl( \frac{\boldsymbol{e}_{r'}}{r'^{2}} \biggr) + \frac{1}{c^{2}} \, \frac{\mathrm{d}^2}{\mathrm{d}t^{2}} \, \boldsymbol{e}_{r'} \biggr] \,.$​
This is expressed in terms of the "retarded" distance $r'$ -- the distance relative to where the charge was when it produced the electric field, accounting for the limited speed of propagation of electromagnetic influences. (See [1] for more explanation.) If you apply this to your example with charges bouncing in a box you should find: 1) if the charges are moving in such a way that there is no net current then contributions from the terms in $\frac{\mathrm{d}}{\mathrm{d}t} \bigl( \dotsm \bigr)$ and $\frac{\mathrm{d}^2}{\mathrm{d}t^{2}} \bigl( \dotsm \bigr)$ should cancel out, so you're essentially left adding Coulomb electric fields, in which case 2) if the charges cancel then the resulting electric field will be zero.

This should deal with your charges in the box. Finally, if you apply the formula above to the special case of a charge moving with constant velocity, and reexpress it in terms of the position relative to the charge's current location (instead of the retarded location) then you should (I haven't personally done the calculation) recover the formula from your opening post.

[1] The Feynman Lectures on Physics, Vol II, Ch. 21: Solutions of Maxwell's Equations with Currents and Charges, available online at http://www.feynmanlectures.caltech.edu/II_21.html.

Last edited: May 26, 2017
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15. ### exchemistValued Senior Member

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Just thought I would say what a lot of good physics there is on this thread now. I am enjoying reading it and learning.

16. ### Q-reeusValued Senior Member

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No static field, but....Any material body above 0 K is constantly emitting (and equally absorbing from the environment if in thermal equilibrium) thermal radiation, having roughly a Planck distribution of intensity vs wavelength or frequency. Yet there is no net current anywhere. But there is a non-zero rms current density. As you stated earlier, linearity applies so emission from all accelerating charges must be added according to the superposition principle. Randomness of phase and polarization does not lead to total cancellation of the fields from individual radiant collision events:

17. ### przyksquishyValued Senior Member

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Yes there is. The very meaning of heat is that individual atoms and molecules in matter move around in a random fashion, and these have nonzero and nonuniform charge densities (e.g. the positive charge is concentrated in the nuclei) and they interact.

If I need to clarify: by "no net density/current" I meant $\rho(\boldsymbol{x, t}) = 0$ and $\boldsymbol{j}(\boldsymbol{x}, t) = 0$ for all $\boldsymbol{x}$ and $t$. If you put that in classical electromagnetic theory then the result is that there is no (non-background) electromagnetic field. Obviously that is an approximation and matter in the real world is not so simple (or classical, for that matter), but this isn't pertinent to the question tsmid was asking.

18. ### tsmidRegistered Senior Member

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The period during which each charge is accelerated is negligible here compared to the period during which it has constant velocity. The reflection region at the walls is about the size of atom i.e. ≈ 10^-8 cm. For a total length of the box of let's say 10 cm this is a fraction 10^-9. So if you have 10^9 charges in the box, on average only one of those would be seen as accelerated at any time. That one charge can not possibly cancel the charge of the 10^9 -1 other charges in uniform motion.

Besides, you could also simply assume that the charge leaves the volume and is just replaced by a charge entering the volume at the same place travelling in the opposite direction. This would yield exactly the same situation for the charge configuration in the volume but without any acceleration involved at all.

Physics is all about applying the appropriate equation to the situation. Your equation is clearly inappropriate here.

Last edited: May 26, 2017
19. ### tsmidRegistered Senior Member

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I was not thinking of dipole oscillations either. The charges are supposed to be all moving independently of each other, just constrained by the walls of the box

20. ### przyksquishyValued Senior Member

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What makes you think it's the "time spent accelerating" that determines the magnitude of the effect? Did you actually try to find or work this out or are you guessing?

This is what I keep trying to tell you. The equation you are using is a special case and is not appropriate to the situation you are considering.

False. It is strictly more general than the equation you posted. It applies for any charge moving in any arbitrary way, constant velocity or not, with no need for handwaving or guessing about what the effect of changing velocity will be.

The same is true for the Jefimenko equation I linked to earlier, which is basically the same thing except expressed as an integral over the (arbitrarily time dependent) local charge and current densities.

21. ### Q-reeusValued Senior Member

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Nitpicking. You should have further quoted the very next sentence of mine: "But there is a non-zero rms current density."
In summary, you translated tsmid's query as that of a perfectly static ensemble of randomly mixed +ve & -ve charges? OK. I didn't get that picture at all.

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22. ### Q-reeusValued Senior Member

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This is the problem with your descriptions - vague and contradictory. Mention box and naturally a 3D situation is implied. Yet you want it to be just 1D motions. And want within that constraint that +ve & -ve charges freely pass each other without strongly interacting i.e. scattering off each other. Unrealistic.
I could describe a realistic and tightly specified scenario that covers what you are grasping about for. It suggests a trivially null outcome but is in fact extremely interesting when properly worked through and gets back to comment made in #88. But I won't.