# Absolute Reference Frame

Discussion in 'Physics & Math' started by Prosoothus, Mar 27, 2006.

1. ### Physics MonkeySnow Monkey and PhysicistRegistered Senior Member

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Dale,

The solution to the puzzle I mentioned earlier lies in the fact that, in the third link, the magnetic force was calculated to order v^2 but the electric force was not. This is inconsistent. If you look back at my post, you will see that the transverse (to the motion) electric field picks up a factor of &gamma; so the electric force is not simply e^2 / 4 D^2 i but rather &gamma; e^2 / 4 D^2 i . This force is clearly corrected from the v = 0 result by a term of order v^2. To be precise, &gamma; e^2 / 4 D^2 i ~ (1 + 1/2 v^2) e^2 / 4 D^2 i . The magnetic force is - v^2 &gamma; e^2 / 4 D^2 i (factor of &gamma; here too!), but to order v^2 we can simply ignore the &gamma;. The result when you add everything together is (1 + 1/2 v^2 - v^2) e^2 / 4 D^2 i = (1 - 1/2 v^2) e^2 / 4 D^2 i , thus resolving the problem. Note that the ratio of the two forces as calcaluted in the third link is indeed correct for small v, and actually it holds for arbitrary v as you can see above.

3. ### DaleSpamTANSTAAFLRegistered Senior Member

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Hi PM,

I am looking at it and I don't see it. The formula for the electric force, Coulomb's law is exact. The magnetic force is Bqv which is also exact. I don't see the approximation to order v^2 anywhere. Are you saying that the Biot-Savart law is an approximation or that I dL = qv is?

-Dale

5. ### Physics MonkeySnow Monkey and PhysicistRegistered Senior Member

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Dale,

Ah, there's your problem. Coulomb's law is not exact, it neglects terms of order v^2. Look either at your fourth link or at my post to see that the transverse electric field picks up a factor of &gamma; as does the transverse magnetic field. This means that the electric force between two moving charges is increased from the predictions of Coulomb's law.

7. ### Tom2Registered Senior Member

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726
That's not the claim I am talking about. I am talking about your claim that SR predicts that the two charges can simultaneously attract, repel, and be neutral with respect to each other, depending on the frame of reference. I have given references that show explicitly that this is a prediction of Galilean relativity, which allows observers to take on any speed one likes. In SR the force between the charges is always repulsive, just as it should be.

So all your thought experiment shows is that Maxwell's electrodynamics is incompatible with Galiliean relativity, which is a fact that is well known to physics undergraduates the world over. Congratulations.

8. ### DaleSpamTANSTAAFLRegistered Senior Member

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Doh! Of course. Coulomb's law is only an exact solution of Maxwell's equations in a purely electrostatic situation. Unfortunately, in your corrected derivation of the EM forces, as soon as you introduce γ you are already using time-dilation and length-contraction. So you cannot afterwards use the EM force equations to derive time-dilation as I did. My equations are correct, but circular: γ falls out because γ was put in earlier.

So that begs the question: can the γ factor be derived directly from Maxwell's equations, or do they require the wave solution and the usual "light-clock" derivation.

-Dale

9. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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but as you note later, there may be no length contraction in the perpendicular to velocity direction. Time does not have any vectory property so do you really need two or more clocks? I.e. is it possible to get the SR expresion for time dilation from my "electron clock" slowing?

10. ### DaleSpamTANSTAAFLRegistered Senior Member

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The Lorentz transform is a transform in which c is invariant. However, the Lorentz transform is not the only transform under which c is invariant. The Voigt Transform also has an invariant c. Where the Lorentz transform has length contraction in the direction parallel to motion the Voight transform has length expansion in the direction perpendicular to the motion (both by a factor of γ). Also, the Voigt transform has a time dilation factor of γ² compared to the Lorentz factor of γ. They both have relativity of simultaneity.

So with light clocks to derive the whole transform you need a perpendicular one and a parallel one. You start with the perpendicular one and make the Lorentz assumption that the additional time is due purely to time dilation or you make the Voigt assumption that the additional time is due purely to length expansion. You then go to the parallel clock to get the other property because the first is not sufficient.

I think the same thing would happen with your electron clocks. You would need one perpendicular to the motion to derive the time dilation and then one parallel to the motion to derive length contraction. My point was just that you need two clocks and that even when you do it you are left with both the Lorentz and the Voigt transforms as potential canidates with no theoretical justification for picking one over the other. You don't need the two clocks because time dilation is a vector, just to get both the time and space effects.

-Dale

11. ### Physics MonkeySnow Monkey and PhysicistRegistered Senior Member

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Dale,

I can't resist telling you about this fascinating and deep connection. You have brought up the Voigt transformation, and it's easy to see that this transformation is actually a multiple of Lorentz transformation. That this is so is no accident. In fact, you can easily check that Maxwell's equations are invariant under an arbitrary conformal transformation defined as x -> a x, y -> a y, z -> a z, t -> a t (a > 0) provided there are no sources. This transformation is clearly not a Lorentz transformation, and it doesn't preserve the invariant interval. However, the paths of light rays are not affected. What does this mean? Well, ask yourself, what does the conformal transformation does keep fixed? The answer is the causal structure of spacetime. In other words, events that are spacelike separated remain spacelike separated and so forth. The size of the interval between two events changes, but the sign of the interval remains fixed. This is simply the deep statement that photons have no knowledge of distance and time, only of before and after! This in turn is a consequence of the masslessness of photons. Besides being cool, the relevance of this to the present situation is simply to note that electromagnetism by itself will never be able to decide between the Lorentz and Voigt transformations. You have to add matter.

Now, the cute thing about your ratio of accelerations equation is that you don't actually have to know much about the electric field. To be precise, if you want to compute the ratio of the accelerations to order v^2, it suffices to know the electric force to order 1 since the magnetic force is already of order v^2. You are forced to add the extra caveat that your equation is only valid to order v^2, but this isn't too big a deal. You see, the relativity group of nature should be continuous corresponding to the continuous range of relative velocities. This means, though it is perhaps not obvious, that you only need to know the infinitesimal transformations to get the whole group (with some important caveats that we don't need to worry about right now). Everything you've done is therefore valid, up to order v^2. I'm getting tired so I will come back later to talk more about the Lorentz vs. Voigt issue.

Last edited: Apr 3, 2006
12. ### Physics MonkeySnow Monkey and PhysicistRegistered Senior Member

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Ok, so I left off with the observation that pure electromagnetism is invariant under the Lorentz transformation but also under any change of scale. This invariance is deeply connected to the massless nature of the photon. Pure electromagnetism cannot therefore decide between the Voigt transformation and the Lorentz transformation.

So what about matter? Massive particles have, well, mass, and this mass defines a length scale which is nothing but the Compton wavelength of the particle. Because they come with an invariant length scale, we don't expect massive particles to be invariant under conformal transformations. With that in mind, the first thing to notice about the Voigt transformations is that the transformations don't form a group! For example, try transforming to S' moving at v with respect to S and then to S'' moving at -v with respect to S'. You won't get x'' = x, t'' = t, y'' = y, z'' = z, S is different from S''! What you find is that x'' = &gamma;<SUP>-2</SUP> x, t' = &gamma;<SUP>-2</SUP> t, etc. So the result of doing v then -v isn't the identity transformation, it's a conformal transformation. This isn't surprising since the Voigt transform was really a combination of the Lorentz transformation and a conformal transformation. The Lorentz part clearly inverts when successively doing L(v) then L(-v), but the conformal part builds up. I would have to do one more conformal transformation to really get back to the same coordinate system. Photons don't care about the difference, but massive particles do.

But the real transformation that connects coordinates in S and S'' must be the identity, right? All this means is that we aren't allowed to apply the Voigt transformation to every frame. In other words, the Voigt transformation implies a preferred frame relative to which it is valid, and we can measure our speed relative to that frame with simple experiments. Thus if we believe the principle of relativity, Voigt can't be right. Of course, Nature could have turned out that way, although it almost hurts me to think about such an ugly possibility.

Last edited: Apr 9, 2006
13. ### DaleSpamTANSTAAFLRegistered Senior Member

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That's pretty neat. I had played around with the Voigt transformation a bit as far as deriving it, but hadn't actually applied it to anything. It didn't even occur to me to check wether or not Voigt(v).Voigt(-v)=I as you would expect. Even if photons wouldn't mind something like that it would certainly make the math more cumbersome.

-Dale

14. ### CANGASRegistered Senior Member

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We sure don't want to hurt THE MATH.

15. ### DaleSpamTANSTAAFLRegistered Senior Member

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Hehe. Point taken CANGAS

I guess it is just "icing on the cake" that nature is so cooperative and considerate of my rather limited math abilities!

-Dale

16. ### CANGASRegistered Senior Member

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From an early time in my formal education, I quickly discovered that I loved mathematics, although it insisted on resisting my advances and consistently remained aloof to my advances ( I was not a quick math learner ). As time went by, I eventually learned that the real world is a harsh mistress, and sometimes pure mathematical analysis did not give me the answer needed to stay on top of the curve.

In friendly contravention to your recent statement, it has sometimes been my personal discovery, in everyday life and in scientific musing, that nature can and will and has been uncooperative and has sometimes forced me to forsake math and embrace a workable strategy.

Hence, my thinly disguised sarcasm that "we must not hurt the math".

Unfortunately, there are times we must hurt the one we love, and HURT THE MATH.

17. ### martilloRegistered Senior Member

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864
Prosoothus wrote:
Good question Prosoothus, here Relativity fails. Physics Monkey's reasoning fails because he takes the force in one frame and just calculate the Lorentz transform of it but this have been demonstrated to work for neutral conductors (the electric and magnetic "interchange" of forces when changing frames...) but does not work for beams of charged particles.
This problem shows the real necessity of the existency of an absolute frame in Physics.
You can also see: "Considerations against Relativity"

A New Physics is rising...

Last edited: Apr 14, 2006
18. ### DaleSpamTANSTAAFLRegistered Senior Member

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So then point out exactly where PM made his error. Which equation is wrong and why?

-Dale

19. ### martilloRegistered Senior Member

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864
Dalespam,
The point is that PM have derived the relativistic transformation of a force between referential frames applying Lorentz transforms. This have been already proven to have some sense with electric and magnetic forces (although discusible) on neutral conductors only. In relativistic transformation for neutral conductors the electric and magnetic forces of both carriers of force varies in such a way that what one augments the other diminishes and the total effect of the force remains the same.
This was very well presented at the site: http://www.phys.ufl.edu/~rfield/classes/fall03/ but unfortunately the sub-links aren't working now. I don't have other links but may be you can find others at the web.
The problem is that for that to work is necessary the participation of the electrons (as carriers of the moving charges) and the protons (as the neutralizing charges). Both contribute in different ways to the final force in the two considered referentials.
For pure electrons beams there are no neutralizing charges to contribute on the force.
This way The Lorentz transformation of force doesn't work for the electric and magnetic forces of pure beams of electrons.
Relativistic predictions really fail in this case presented by Prosoothus. They show different behaviours for the same phenomenon just under a change of referential of observation. This means inconsistency of the theory.

Last edited: Apr 14, 2006
20. ### Tom2Registered Senior Member

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726
That's not true, the Lorentz transformations work fine on individual charged particles, and on beams of charged particles. All you have to do to see this is perform a Lorentz transformation on the field of a line of charge. You'll obtain the electric and magnetic fields for a steady current that you would expect from Maxwell's theory.

This shouldn't be surprising in the least because Einstein showed that SR was buried in classical electrodynamics all along. So when you apply SR to EM fields, of course you recover the classical electrodynamic prediction for moving charged objects.

21. ### DaleSpamTANSTAAFLRegistered Senior Member

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Your assertion makes no logical sense whatsoever. You agree that the theory can handle the more complicated and more general case where there are both positive and negative charge carriers. How then is it possible that the theory cannot handle the simpler and more specific case where the number of positive charge carriers is zero? This is logically equivalent to working a problem where you calculate the net force on an object sliding down ramp due to gravity and friction and then claiming that you can't use the same method of analysis for the simpler case when the ramp is frictionless.

If you think PM made a mistake you need to be specific. Which step is wrong? Which equation has an error due to the fact that there are no positive charge carriers? Your general dismissal of his approach is logically unsound and completely insufficient, so be specific.

-Dale

22. ### martilloRegistered Senior Member

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864
Dalespam, Tom2 and Physics Monkey,

I must apologize, I made a mistake. I have studied the subject deeper and there's no problem with the relativistic predictions in this cases. PM derivation is right.

I will make a correction in my page.

Last edited: Apr 15, 2006
23. ### DaleSpamTANSTAAFLRegistered Senior Member

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No problem. It took me years to study the subject deep enough and finally come to the conclusion that each of my objections were mistaken. I certainly don't mind others doing that as well.

-Dale