# Relativistic Coulomb Force

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

1. ### danshawenValued Senior Member

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Yes and no. On the one hand, relativistic time dilation is path independent, but I do not believe it is necessarily quantum spin independent, because quantum spin / entanglement itself is at once the fastest process (not a velocity) in the universe, and also the basis of time.

The twin paradox works as well for a long ride on the rim of a relativistic merry-go-round (or more accurately, a centrifuge), as it does with orbiting spacecraft. A relativistic merry-go-round is much larger than atomic structure, but it is true that even radioactive decay of unstable atoms would proceed more slowly on the rim of that relativistic merry-go-round than they would at its center. It could be argued that average v=0 on the relativistic merry-go-round as well, or for a plane with an atomic clock on board, like the experiment my former mentor Carroll O. Alley performed. His experiment was an early confirmation of relativity, well before we had GPS satellites using it every day.

I would argue the case for v=0 differently for atoms only because the relativistic merry-go-round is usually not treated as a wave function. Bound energy that is moving <c or at rest persists in time because of entanglement. Unbound energy propagating at c persists in time because of entanglement also. Energy is conserved because both forms of energy persist in time and are related by E=mc^2 and entanglement. What is really needed here is an equivalent of the Schrödinger wave equation for bound energy. Hint: it won't be propagating at c, and this is the REAL reason that relativity does not apply to atoms the same way it does to a relativistic merry-go-round.

After a radius of appropriate length, it appears to an observer on the rim a good deal longer for the merry-go-round to complete one revolution than it does to an observer at its center. Even the stationary estimation of the constant pi is different than it appears to an observer moving with the rim. The meter sticks closest to the observer appear unaffected, of course, but looking at meter sticks on the far side of the outer rim, half of them are Lorentz compressed because a component of v is in the opposite direction of relative motion.

This is the case with the Milky Way galaxy as a relativistic merry-go-round as well. The stars we view 180,000 light years away at the most distant rim have already completed a greater amount of rotation than we are observing now, and also they are experiencing a much greater amount of time dilation than the stars, nebulae and black holes nearer the center. The stars on the opposite side of the Milky way also will appear to be much closer together (more densely packed) on the far rim than they actually are.

Atomic structure may not be a relativistic merry-go-round, and I don't think it would be a "structure" in any sense of the word without entanglement. Entanglement inteacts most strongly with the basis of time and direction of propagation here, or else atomic structure itself would not be possible. It's not all about electric charge, because electrons, if they are moving, still exist in free space. What force or mechanism holds them together?

This was a very, very good follow-up question exchemist.

Last edited: May 18, 2017

3. ### exchemistValued Senior Member

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No Dan, entanglement is not relevant to this problem. Let's keep it simple. It is just an issue of how the Lorentz contraction affects how the electric field from a moving charge is perceived.

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5. ### danshawenValued Senior Member

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You can't have it both ways. Is the electron cloud in atomic structure like an Ehrenfest relativistic merry-go-round, or is it not?

7. ### exchemistValued Senior Member

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I suspect it is, but that, like most of quantum physics, is unrelated to entanglement.

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8. ### danshawenValued Senior Member

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What holds an electron (itself) together? All of it is negatively charged. It should fly apart. In quantum mechanics, if it stops moving, it does (cease to exist, waveform collapses). What is it supposed to be moving relative to? Itself? Mass/energy is conserved. Where exactly does the energy go when its waveform collapses? Its entangled twin would be my guess.

Relativity is not compatible with quantum mechanics unless you add entanglement.

Last edited: May 18, 2017
9. ### exchemistValued Senior Member

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That is not the question at issue. This thread, so far as I can see, is asking about why, in a neutral atom, the nucleus does not see a net -ve charge from the electrons that appears, from its frame of reference, to be less than its own +ve charge.

Nobody is asking about any supposed constituent pieces of an electron - which, according to the prevailing model, there are not. If you try to work that in, you are derailing the thread.

10. ### Q-reeusValued Senior Member

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Instead of continuing on to engage in page after page of rambling nonsense that completely misses the issue, try going back to the basic assumptions in #1 and actually rethink them carefully. You wrote:
"If we have a random distribution for the direction of v, we have to average over sin^2(θ), which yields a constant factor 1/2....."

Firstly, it should be perfectly obvious the approximate expression you obtained: ≈ (1- v^2/c^2*(1-1.5*sin^2(θ)) ), will not at all yield an overall factor of 0.5.
But further, what makes you think the problem is to be evaluated in 2D? Because that is what your logic is assuming - averaging the normal component field as a 2D slice, around a 1D circle.
But the evaluation must be over a 2D spheroidal surface embedded in 3D space. That's what Gauss's law implies be done. I suggest going back to square one, work out the appropriate integral expression over a surface not around a ring. Actually, way back, leading to continuity eq'n (5.9), it's already a given that charge invariance applies. Proving it holds in the specific case of a moving point charge is still worthwhile though.
Best also imo to pay no heed to futile rambling diversions elsewhere this thread.

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11. ### danshawenValued Senior Member

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The nucleus assuredly sees a balanced negative electron cloud, and the bigger the nucleus, the more electrons are needed to balance its own positve charge. The Periodic Table, among other things, depends on equal number of protons vs electrons.

Last edited: May 18, 2017
12. ### danshawenValued Senior Member

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For once, someone here understands an appropriate application of geometry. The nucleus is a good thing to designate as an origin for a coordinate system for doing geometry.

If the electron cloud were flat (possible, with relative motion), would charge in any direction be affected? My answer would be 'no'. What do you think?

13. ### tsmidRegistered Senior Member

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I am not sure (anymore) how you understand 'net velocity' but by the general definition it refers to the vectorial sum of the individual velocities. The latter don't have to be zero in order for the net velocity to be zero. The net velocity of the air molecules in your car tyre may be zero, but still you have pressure in your tyres. This is because p=nkT and the temperature T depends on v^2 rather than v (so effectively the speed rather than the velocity). The length contraction effect in Special Relativity (which is essentially responsible for the velocity dependence of the Coulomb force) also depends only on the speed rather than the velocity, so is not overall zero for a zero net velocity.

14. ### tsmidRegistered Senior Member

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The average of sin^2(θ) is 1/2. Inserting this gives f_av ≈ 1 - v^2/4c^2 .

2D? But the problem is only 1D. The only variable is the angle θ the between velocity vector v and the radius vector r. Otherwise the situation is rotationally symmetric for an isotropic distribution of velocity vectors.[/QUOTE]

15. ### exchemistValued Senior Member

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Yes we all know that.

The the issue therefore is where tsmid's reasoning has gone off the track. That's what he is asking, effectively.

Maybe Q-reeus is onto it. I have not followed the maths myself.

16. ### danshawenValued Senior Member

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Velocity is a vector (direction, magnitude). Speed is not (magnitude only).

Relativity does not in any way suggest that the relative motion responsible for time dilation, length contraction, is a speed and not a velocity. Quite the opposite; the rest frame can actually be defined as the vector sum of ± c, for any inertial frame you wish to measure it in. There is no time dilation or length contraction associated with something at rest relative to any measurement.

I don't know where you got the impression, relativity was based on relative speeds. It most certainly is not, unless the speed to which you are referring is c, which is a speed reserved in relativity only for the case of massless photons, not electrons. A vector sum of ± c is still zero, even in the case of photons, for all inertial reference frames at rest relative to the measurement of c in them..

17. ### Q-reeusValued Senior Member

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First part true - if applied to a simple line integal situation. Namely, the summed normal component of E field (not force f!) crossing perimeter of a circle.
To repeat: the actual problem is that of computing the flux of E crossing a closed 2D surface. Typically, for convenience, one uses a sphere of area A = 4πr².

Which implies a θ dependent weighting function must operate on the net expression for E, not just your simplistic sin^2(θ), when integrating normal component of E over a surface. Not hard to figure what that weighting function is for case of a spherical surface! Think about it some more. I could point to worked examples available on the web. Better if you can grasp the actual situation for yourself, work it through afresh, and then check it against a standard derivation.

I previously mentioned the continuity eq'n (5.9) in that article linked to in #1 as implying charge invariance (not just conservation of charge). As well, fig's 46 & 47 make it obvious the net number of field lines are conserved - merely rearranged. So we know the answer to your apparent conundrum must be consistent with Lorentz invariance for charge.
See above. NOTE: Only the exact expression for E, not your approximate first term of Taylor series one, will yield charge invariance even when the appropriate integral is used.

Last edited: May 19, 2017
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18. ### danshawenValued Senior Member

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Fancy. Sure you don't need to affirm the identity 1=1 first?

I acknowledge a lot of correct math here to confirm something plainly obvious to anyone in touch with a simple physical reality that atoms have neutral electric charge on any scale in which a Coulomb force can be observed.

The tail does not wag the dog. Math is modeled after the physical world, not the other way around. Don't need line integrals. Don't need Green's theorem. Relativity all by itself is enough. The original problem was a misapplication of relativity, velocities in particular, not Maxwell's equations. Those are consistent with relativity or else you have made a mistake.

Last edited: May 19, 2017
19. ### Q-reeusValued Senior Member

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As usual Dan, your commentary is, well, not particularly illuminating. So, anticipating further contention from tsmid and yourself, might as well short-circuit things and provide that worked example:
https://physics.stackexchange.com/questions/63547/gausss-law-with-moving-charges
End of story - one hopes!

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20. ### danshawenValued Senior Member

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I see what you mean.

I suppose there is sometimes a case for wanting to be completely general about solving a problem like that.

As the answer with the least number of votes on the physics stack exchange commented, it should be obvious to the casual observer who is also an experimentalist that this calculation isn't necessary on the same grounds I objected. It models a real situation where the outcome is already known and then proceeds in the most convoluted manner possible to arrive at precisely the same conclusion. Just like doubting an identity like 1=1, and waste time a lot of time proving it for the general case.

A bit too "axiomatic" for my taste, that's all. I can imagine situations in which that approach might have an advantage. Thanks, Q-reeus.

21. ### Q-reeusValued Senior Member

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There was only one official Answer Dan, and it got 4 +ve votes. Not axiomatic either - just a straightforward precise calculation done right. So tsmid hopefully now knows and knows why that weighting factor is just sinθ.

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22. ### exchemistValued Senior Member

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Exactly! The number of field lines is conserved: they are merely rearranged.

That is the point. Very well put and clear. Thanks.

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23. ### The GodValued Senior Member

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How?

The above is of course incorrect conclusion but even otherwise you have made a mess of your 5.3.4.

1. The taking of average over theta is bad, we are talking about Electrical Field which is a vector quantity. It has to be seen at every instant for each point.

2. The section talks of a single charge movement, you are talking about some random distribution of charged particles.

3. The additional factor on account of SR, gives a value quite close to unity (as usual) even for decent speeds, so question of abnormal forces does not arise in normal low speed set ups.

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