# Relativistic Coulomb Force

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

1. ### Q-reeusValued Senior Member

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Can you handle advanced tensor maths? If not, a waste of time pointing you to 'proofs' based on Lorentz invariance of ME's, which proofs inevitably can always be made to have a circular reasoning element to them. Maybe this brief Wiki article gives the honest answer that is easily appreciated but maybe not satisfying to someone intent on pursuing an endless 'but then why that' regression:
https://en.wikipedia.org/wiki/Charge_invariance

3. ### The GodValued Senior Member

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......The property of charge invariance follows from the vanishing divergence of the charge-current four-vector{\displaystyle j^{\mu }=(c\rho ,{\vec {j}})}

, with {\displaystyle \partial _{\mu }j^{\mu }=0}

......

That shd put many off.

As Dan says, maths is maths.....But in case of relativity, both SR and GR, I have my reservations. You (or anyone else) are at liberty to call me names, but something is really amiss here.

Last edited: May 20, 2017

5. ### Q-reeusValued Senior Member

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Seems you missed reading the part I thought most relevant for you:
"The origin of charge invariance (indeed, all relativistic invariants) is under speculation presently. There may be some hints proposed by string/M-theory. It is possible the concept of charge invariance may provide a key to unlocking the mystery of unification in physics (the theoretical unification of gravity, electromagnetism, the strong, and weak nuclear forces)"

In other words, it's at rock bottom an experimental fact of nature. Given the invariance of c, then ME's are also Lorentz invariant and from that charge invariance follows. But as I hinted last post, you can then ask 'but why do the ME's have the form they do? Coz that's how it is with nature. What is amiss? Do you really want to keep kicking a can down the street here, after OP question has been settled?

7. ### danshawenValued Senior Member

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You run afoul of the Banuch-Tarski paradox whenever you try and define a differential element of volume for something like a relativistic electric charge in order to calculate charge density for a particle defined as having zero volume, that's why.

Any sort of density modeling will have this problem because the calculation of a density involves a volume at some point, and volume is neither a Lorentz invariant, nor anything that makes sense in a single dimension when time dilation always affects all three dimensions for bound particles of inertial mass/energy equally, regardless of the direction of relative motion.

Gaussian integration was modeled for the rest frame relative to charges moving in more or less straight trajectories with respect to our scale, not for atomic structure. On our scale, changing direction means acceleration. Not so on a quantum scale for an electron bound in the electron cloud of an atom.

The way an electron changes direction is fundamentally different on an atomic level, as is the volume element its localized form of propagation defines.

Last edited: May 20, 2017
8. ### tsmidRegistered Senior Member

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As I said, the length contraction in Relativity only depends on v^2, that is the speed (squared.) A rod moving with speed v in the +x direction has the same length contraction factor as one moving with speed v in the -x direction (or any other direction).

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9. ### tsmidRegistered Senior Member

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OK, yes, I see now what you mean, I forgot the further factor sin(θ) from the Jacobian for the spherical coordinate system. But that would merely change it to sin^3(θ) from sin^2(θ), which means the average (from 0 to π) becomes 4/3π (≈ 0.42) instead of 1/2. That means the relatistic correction factor for the Coulomb force is now

f_av ≈ 1- v^2/c^2*(1-2/π)

Well, it still doesn't seem to be consistent.

How do you want to compensate a discrepancy of the order v^2/c^2 with terms v^4/c^4 and higher (which are the remaining terms in the expansion)

Last edited: May 20, 2017
10. ### tsmidRegistered Senior Member

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I am not implying changing velocities, only an isotropic probability distribution for the direction of the velocity vector at a given point (as you would expect it in a random medium).

11. ### przyksquishyValued Senior Member

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You gave an example involving electrons moving around in a glass of water. Unless the electrons are escaping out of the glass in all directions they have to be constantly changing direction (accelerating), in which case the formula you started with doesn't apply. If you want to calculate the electric field the way you are doing it (adding contributions from individual charges) then the correct formula to use for your case is Jefimenko's equation (https://en.wikipedia.org/wiki/Jefimenko's_equations), which explicitly accounts for the time-delay effect.

12. ### tsmidRegistered Senior Member

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Again, I am not assuming the different directions for the velocity vector are associated with a change of direction for one electron, in as much assuming e.g. a Maxwell velocity distribution is not associated with a change of velocity of one electron. It is a time averaged probability of having a certain velocity at a given point in space given a sufficiently large number of electrons. And since it is a point (in practice of course only a sufficiently small volume), there is no difference in the time delay to the test charge. All the different probability states contribute the same to the electric field at the test charge, the only difference being the θ-dependence in the relativistic Coulomb law.

13. ### danshawenValued Senior Member

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THAT'S a factor of v^2/c^2, a term inside the radical for gamma. That particular term is dimensionless.

Directed Lorentz contracted lengths, which are vectors multiplied by relativity's gamma, very much require a direction of relative motion (just like a relative velocity does). Do they not?

Mathematicians seem to prefer the idea that length contraction occurs with respect to the centers of whatever is contracting, BUT THIS IS ONLY A MATHEMATICAL CONVENTION, NOT A PHYSICAL MEASUREMENT, and it is for this reason, I no longer care how they define such things, or even whether or not thid convention agrees with the convention(s) which apply to the Lorentz inertial roadbed, assumed to be at rest relative to any moving coordinate systems. You would not believe how many theoretical teeth I needed to pull one by one, even to get a straight answer about this being only a mathematical convention. Since when does anyone do solid geometry on anything that does not have a well defined origin, other than my means of mathematical convention? All the time, apparently.

I give up. There are just too many realtivistic misconceptions here to deal with. Reasoning like this for atomic structure is simply not going to work.

My advice is to completely work out problems like the pole and barn paradox, the twin paradox, and the Ehrenfest paradox. Don't depend too heavily on boost matrices for turning the mathematical relativity crank until you understand relativistic dynamics for velocities <=c, and consider all of the invariant quantities you can depend on not to change.

14. ### Q-reeusValued Senior Member

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Evidently you missed the sense of what I wrote and you quoted in #66:
"...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..."
So why go on to just have the weighting factor sin θ operate on only one term in the expression for E'? Makes no physical or mathematical sense. A correct worked example was linked to here: http://www.sciforums.com/threads/relativistic-coulomb-force.159363/page-2#post-3455551.
Which admittedly jumped a bit from the last integral expression to get the final result Q' = Q. Do you dispute the correctness of that derivation?

Just when you think it's all wrapped up, off it all goes again. Sigh.

15. ### Q-reeusValued Senior Member

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When will you cease to pollute this thread with irrelevant nonsense Dan? Banuch-Tarski paradox is mathematical tomfoolery having precisely ZERO relevance to this thread topic. And afaik no-one thinks it has any practical relevance to the physical world. 'Density' issues in general have zero relevance here. Gauss's law works completely independent of the size of a bounding surface of integration. Whatever charge exists within can have any exotic 'ultimate form' you like - all that matters is the magnitude of such charge.

Your bizarre claim that at the 'quantum scale/atomic level' changing direction doesn't involve acceleration is utter nonsense. I'll go with Lubos on that one:
https://physics.stackexchange.com/questions/20187/how-fast-do-electrons-travel-in-an-atomic-orbital
Actually do give this thread a miss Dan.

16. ### tsmidRegistered Senior Member

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Sorry, I didn't see your post with that reference. Still, that just gives the result for the integral without mathematical proof. But I noticed now that in my post #66 above I still made a mistake for the normalization factor for the θ- integral. It should have been 1/2 rather than 1/π. With that the v^2/c^2 dependence indeed disappears even with only the first term of the series expansion.

17. ### Q-reeusValued Senior Member

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Nothing really close in any online list of integrals, and I can't think of a variable substitution that would make it easier. Also don't have anything like Mathematica installed to attempt a machine solution. Given no-one protested the result in that Q & A article, one assumes it works out as claimed. If you want to be sure, maybe contact Mike who gave the derivation:
https://physics.stackexchange.com/users/8007/mike
Or put in a fresh query that thread.
That you got a correct result with only the severely truncated Taylor series terms is interesting and not something that should generally apply. Good that there has been a gain despite all the fluff here.

18. ### danshawenValued Senior Member

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And someone teaching you relativity by claiming that the convention is to measure lengths or length contraction from the center mark of the meter stick, that somehow isn't utter nonsense? The origins of the rest vs. moving frames never need to be lined up in order to calculate how they are different, either, I suppose? The question of which end contracts is impertinent, but the question of which simultaneous event occurs first from a moving frame, that's perfectly fine? Stranger and stranger.

I don't know. Relativity's clock postulate at least tries to instruct people in the right concepts. Acceleration on atomic scales is different from experience on larger scales because it is occuring at a scale in which it is simply not possible to define something like acceleration for a wave function. There's no such thing as "position", and here you go, doing geometry on wave functions for all you are worth. Newton's acceleration and Euclid's solid geometry both need to die and not be resurrected to explain anything.

You really don't get that concept, Q-reeus? Think harder.

I'll just watch and shut up now. By all means, please continue.

Last edited: May 21, 2017
19. ### Q-reeusValued Senior Member

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Nobody is going back to the original Bohr model. As Lubos made clear enough imo, despite having to deal with a momentum probability distribution, there still must be the concept of centripetal acceleration balancing against electrostatic attraction. In a non-radiating ground state orbital.
I can only hope there is no further cause for any here to continue, Dan.

20. ### danshawenValued Senior Member

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I noticed. Carry on the good work. Pay no attention.

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21. ### tsmidRegistered Senior Member

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If you put the definite integral into Wolfram Alpha it comes up with 'Standard Computation Time Exceeded' which made me believe first that this does not have a trivial solution. However, I tried now the indefinite integral and it strangely comes up pretty quickly with

-sqrt(2)*cos(θ) / sqrt[ v^2/c^2*(cos(2θ)-1) +2 ]

and obviously v^2/c^2 drops out both for θ=0 and θ=π as cos(2θ)-1 = 0 then. The definite integral over this range thus evaluates to 2.

Well, the point is that the nominator of the expression for the electric field is exactly 1-v^2/c^2 , and since the first term of the Taylor expansion of the denominator is 1, this means that the sum of all the remaining terms must exactly evaluate to +v^2/c^2 in order for the velocity dependence to disappear. But there is only one term in v^2/c^2 left, which is the second term +3/2*v^2/c^2*sin^2(θ) and which indeed exactly evaluates to v^2/c^2 when averaged over θ ( average of sin^2(θ) = 2/3 over the sphere). This means that all the higher orders of v/c must cancel mutually. We can check: the next order would be v^4/c^4 with the terms -3/2*v^4/c^4*sin^2(θ) and +15/8*v^4/c^4*sin^4(θ). The first term gives -v^4/c^4 (again, average of sin^2(θ) = 2/3 over the sphere) and the second evaluates to +v^4/c^4 (average of sin^4(θ) = 8/15 over the sphere.

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22. ### Q-reeusValued Senior Member

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Great. So hopefully we are now all on the same page and agree that relativistic velocities have no effect on net charge i.e. charge invariance is completely compatible with SR. Phew - threads don't generally end that way!

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23. ### The GodValued Senior Member

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But you see due to length contraction etc, the charge density at least is not invariant. Some kind of charge redistribution happens.

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