# Classical Physics is coming back, RELOADED!!!

Discussion in 'Physics & Math' started by martillo, Jun 18, 2006.

1. ### 2inquisitiveThe Devil is in the detailsRegistered Senior Member

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3,181
No, the frequency the waves are recieved would measure as different, if the measuring apparatus is moving relative to the already-emitted beam of light. As I said before, determine the Doppler shift of the beam of light, and then one can determine what the wavelength was when it was emitted. The emitted wavelength doesn't change because of the detector's motion, the frequency does.

3. ### martilloRegistered Senior Member

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877
1100f,
As 2inquisitive says, in Classical Physics:
But I must recognize that this only happen within the "Emission Theory" of light where light is emitted at constant velocity from the source and so travels at velocity ç = c + v.
Within the "wave" model, light travels at a constant velocity c in the medium ("aether"?) and here if frequency varies the wave-lenght must also vary mantaining constant the product: (wave-lenght)*(frequency)=c.

Last edited: Jul 4, 2006

5. ### martilloRegistered Senior Member

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877
przyk,
No one needed because didn't study the problem deep enough as I did.

I repeat with mathematical rigourosity:

Last edited: Jul 4, 2006

7. ### przyksquishyValued Senior Member

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3,171
They studied it a lot deeper than you did - you've only dealt with half the so-called problem. You still haven't shown that applying the same de Broglie formula in different frames leads to any sort of contradiction - just that the wavelength is a frame-dependent quantity.

If these were real, physical compression waves propagating through some medium that could be measured directly, then yes, I'd expect the waves to transform from one frame to another according to Lorentz. This constraint doesn't apply to particle-waves. Particles are said to have wave-like properties, which itself is just inferred from diffraction patterns. Physicists just found that the mathematics they developed for waves could also be applied to particles under some circumstances - that's it.

8. ### martilloRegistered Senior Member

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877
przyk,
You don't want to apply it because you don't want an an error here!

As I said before:

It is a necessary condition essential for the formulas to be invariant in a change of coordinates that both results must be the same: the calculation of a variable directly from the formula and its calculation by the coordinate transform.

It's just rigouros mathematics.

9. ### DaleSpamTANSTAAFLRegistered Senior Member

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What are you talking about? The "values obtained applying Lorentz Transform and that of applying De Broglie formula" are exactly the same:

Both frames agree on each other's λ, whether obtained through direct calculation using p=hk or through Lorentz transforming their own k. The fact that λ does not Lorentz transform only shows that p=h/λ is not frame invariant, which everyone already agrees. It does not show that p=hk is frame varying.

The generalized deBroglie relation (p=hk) is obviously frame invariant. The classical deBroglie relation ( p=h/λ ) is obviously not. Your insistence on demonstrating one and claiming the other just shows your bias against SR, not any failing of SR itself.

-Dale

10. ### przyksquishyValued Senior Member

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Only if the formulas predict (measurable) coordinates for spacetime events - the classical de Broglie formula doesn't do this. If you insist on the wavelength transforming according to the length contraction formula, you have a pseudo-SR theory that's constructed for the sole purpose of being inconsistent with de Broglie's formula. Why would anyone want to do that?

11. ### martilloRegistered Senior Member

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877
Dalespam,
Here is the point you seem to not get properly!
With you words:
The fact that λ does not Lorentz transform not only shows that p=h/λ is not frame invariant, which everyone already agrees but it also show that p=hk is frame varying!
In other words:
If you try to get λ directly from the formula in the other referential the value obtained is different from the obtained making a Lorentz transform to this referential.
This demonstrates that the formulation does not work.
This demonstrates that the formulation is NOT INVARIANT.

12. ### przyksquishyValued Senior Member

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3,171
No. That doesn't follow.

13. ### DaleSpamTANSTAAFLRegistered Senior Member

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How so? I clearly demonstrated that p=hk is frame invariant. If you think that it is frame variant then please point out the step where I made a mistake. If not, then stop insisting that it is variant.

This is wrong. I already showed that the values are the same (see the red highlights in my previous post) and you know it.

It is fine if you wish to deliberately persist in your ignorance and illogic, but don't then try to claim that you are somehow searching for "truth" when you are plainly willing to sacrifice truth and logic to your bias against SR.

-Dale

14. ### CANGASRegistered Senior Member

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1,612
Mr. Manners asks Dale to consider that some people have a ( strong ) bias against SR because they honestly believe that they see important flaws in it.

15. ### martilloRegistered Senior Member

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877
Dalespam,
You insist in something ignoring my point of view.
I will just tell you once more time to see if you can understand. Other way it's useless to continue discussing.

What you did is just to convert the De Broglie formulation from one referential to another and then make it back to the original frame. You didn't make the calculation of λ by Lorentz transforming LENGHT in the change of coordinates. This is what I do and it is needed to do to verify if the formula is INVARIANT under the Lorentz transform of coordinates and the result show it is NOT.

I have no more ways to explain it to you.
I'm sorry but I cannot go on with the discussion this way.

The problem is not on my analisis or its exposition, the problem is that you don't WANT to see Relativity failing and you will make anything possible to show it still runs even against the evidence. In the deep you think it is right and if so much time have passed without any great physicist discovering it has flaws you think you can find the ways to "bypass the stones". I understand and I can do nothing more, at the end you just don't believe I could be right, just that.
But you should consider that may be someones although knowing some problems in Relativity just didn't find the proper answer and because of the lack of a right alternative they just continued ignoring them.

Now I will just wait for the opinion of someone else: the future!
Let time pass...

Last edited: Jul 5, 2006
16. ### imaplanck.BannedBanned

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2,237
Maybe you need to go to elsewhere then, no matter how much nonsense you explanation is some idiot is bound to agree with you somewhere, but if your theory doesn't even stand up to current measurement and observation it's never going to even get validaty as a speculative alternative here.

Last edited: Jul 5, 2006
17. ### przyksquishyValued Senior Member

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3,171
We all know that &lambda; is not invariant. You have ignored all the explanations as to why it doesn't need to be.
All the evidence I'm aware of is in favour of relativity. It has stood the test of time. After a century of experimentation, if relativity turns out to be inaccurate it'll be under obscure circumstances that are rarely encountered.
It's not a question of "bypassing the stones." Special relativity is a physical theory, and as such its primary purpose is to summarize observed data (which it does rather elegantly) and make measurable predictions. Insisting that even abstract quantities that have the SI unit of length should transform according to Lorentz is like saying physics should never use imaginary numbers - it's an unnecessary constraint.

When you think you've found a flaw in an existing theory it's a good bet you are assuming something you shouldn't, or breaking some implicit rule or other. This is the case with your criticism of F = dp/dt, for example. Try approaching it the other way around: assume that physicists aren't incompetent fools, then see if you can find rules for applying F = dp/dt such that the formula is perfectly accurate (I've already thought of a "disproof" of F = dp/dt that uses the same constraints you did, and that's even simpler than your rocket example).

18. ### martilloRegistered Senior Member

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877
przyk,
That seems interesting, show it.

19. ### przyksquishyValued Senior Member

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I'll describe it semi-rigourously like this:

Consider a large group of particles (say, neutrons), all with the same mass m and all moving in the same direction at the same velocity v, and let's say we've numbered the particles (neutron #1, neutron #2, etc).

Now we define the "object" S = {all the particles from neutron #1 to neutron #kt}

Where k is some positive constant and t is the time parameter (so if k=2, then at t=3 seconds S = {neutrons #1, 2, 3, 4, 5, 6}, with the set growing over time).

You can easily see that:

p<sub>S</sub> = mkvt, and that dp<sub>S</sub>/dt = mkv.

Since this example assumes no force on S: F<sub>S</sub> &ne; dp<sub>S</sub>/dt. QED.

It succeeds in "disproving" F = dp/dt for the same reason your rocket example does: you are allowing particles to move in and out of the set you are considering the momentum of. If you disallow this, or go down to the level of individial, indivisible particles, then dp/dt is strictly equivalent to ma in classical physics.

In your example, the "missing" term v(dm/dt) in the rocket's rate of momentum change is exactly the rate of momentum loss due to the expulsion of fuel.

20. ### DaleSpamTANSTAAFLRegistered Senior Member

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It is indeed becoming pretty useless to continue, but you are the only one ignoring anyone. Your point of view has already been refuted by me, pzryk, and even your own calculation. You have ignored all three rebuttals and also ignored all of the other excellent counter arguments. It is fine if you wish to run away from the debate, but don't try to pretend that you are the one being ignored when it is you who is stalling the discussion!

Of course that is what I did, after all we are discussing the generalized deBroglie relation! Where is "LENGHT" in p=hk? You are not even discussing the generalized deBroglie relation if you are insisting that something other than p, h, or k must be invariant. What you have actually been doing here is discussing p=h/λ and then pretending that you have shown something about p=hk.

I will restate pzryk's excellent point in a more concrete manner to see if you will ignore it again. If the deBroglie relation were not frame invariant here is what would happen: if two identical, inertial, spaceships (moving relative to each other and equipped with deBroglie labs) each did identical deBroglie experiments they would obtain some different measurement. This is what frame-varying means. You are insisting that it is frame-variant, so what specific measurement could they use to detect their motion?

Yeah, right :bugeye:. Someone like you, whose debate strategy is to simply ignore any rebuttals and continue asserting their debunked idea, is the unbiased one. I had high hopes for you when you actually did a Lortentz transform, but then you ignored some of your own results since they didn't support your case. (specifically, the loss of simultaneity for your transformed "wavelength measurement")

I agree, it is pointless to continue the discussion until you address at least some of the following points that you have carefully ignored:

1) How do you experimentally measure wavelength, and what experiment are you claiming that different frames will disagree on?

2) Why should a wavelength Lorentz transform when, according to your own calculation, the transformed results are not simultaneous in other frames?

3) How do you claim variance of p=hk (as opposed to p=h/λ ) when, as I showed, each frame agrees on the wavelength measured in the other frame using p=hk (as opposed to p=h/λ )?

Other rebuttals:
4) Since h is a constant, k was shown to transform as a four-vector, and p is also a four-vector then p=hk is manifestly invariant. How can it possibly be frame variant when each individual part is frame invariant? In which term does the frame variance enter?

5) How could your supposed frame-variant deBroglie relation be used to experimentally detect absolute motion (or at least motion relative to some frame where p=hk holds), as in the spaceship example above?

6) Since the measurable quantity is the phase of a wave at a given event, which is the frame invariant quantity r.k, how can any experiments disagree?

Until you can address some of these points that you have avoided I am not surprised that you are too intimidated or embarassed to continue even this farce of a debate.

-Dale

21. ### 1100fBannedRegistered Senior Member

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807
Wrong again.
According to classical physics, the period of oscillation of the source remains the same, but since the source now moves, the velocity of the waves changes and so does the wavelength.
This is unless you disagree with classical physics (classical as opposite to relativistic) that states that time is absolute.
BTW, the frequency change is the frequency at which the wave is detected, not the frequency of the moving source.
Or maybe you also disagree with this and state that both frequency and wavelength remain unchanged and so does the velocity of the waves which remains the same in all reference frames. I am not talking of light waves, but of sound waves or even of transversal waves in a rope. Or maybe you disagree with the fact that in a rope, there are travelling waves.

And maybe the solution is that it all depends on what exactly you are measuring, and not that if you have a wavelength, it will remain the same in all reference frames and that the transformation of the wave length is not like the transformation of a rigid ruler but that you should transform it to another frame according to its definition.

I don't understand why you are talking about light waves since we are talking about de Broglie waves.

22. ### 1100fBannedRegistered Senior Member

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807
If we summarize what you have said during this discussion:
You said that in the MM experiment there is no destructive interference , however experiment shows that there is an interference pattern.

You said that in your theory, at low velocity the equation of your electric and magnetic field are like maxwell's equation even though you don't accept their solution (that you even don't understand since you think that the solutions are only plane wave, no matter what are your initial an boundary conditions, you even said that the waves are the solutions of div.E=0).

You said that neutrino may have any spin, and that spin is noy quantized and that you do not believe in angular momentum conservation.

You have shown that you don't really understand what is a wavelength and that you believe that it should transform as rod and not as a wavwelength

And this is a partial list

Question: do you really believe that we should take your theory seriously?

I suggest that you should study real physics, it is much more interesting than your wishful thinking and it describes reality much better

Last edited: Jul 7, 2006
23. ### martilloRegistered Senior Member

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877
przyk,
I was out of the city and comming back now.
I think your example is interesting and right disproving F=dp/dt but are you conscient about it consequence?
If F=ma and not dp/dt this means Relativity is wrong! You know, the famous equation E=mc2 is derived from this equation and allowing the mass vary as m = m0/s where s=root(1-v2/c2)!
E is obtained integrating Fdx = F*vdt and substituing F by dp/dt = m(dv/dt + v(dm/dt) and making some calculations.
You can see this derivation here: Dr. Rick D. Field lecture notes
You can go to the direct link: Relativistic Energy derivation
But you can also download the complete (big) pdf file. It is very interesting, it have all the backround on Electric and Magnetic fields formulas, Maxwell Equations and more; and it comes from an established and recognized Dr.

Since you were a Relativity defener I believe you haven't realized what you have done...

Last edited: Jul 7, 2006