Relativity Theory is THEORETICALLY wrong!

Discussion in 'Pseudoscience Archive' started by martillo, Apr 21, 2007.

  1. martillo Registered Senior Member

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    I have found many new interpretations (I mean different physical phenomena to explain the same results) for well known experiments rather than the relativistic explanations!: http://www.geocities.com/anewlightinphysics/new_interpretations/Summary_of_new_interpretations.htm
    Someone will find the proper explanation on Gravity Probe B results at its time.

    What is important now is that Relativity Theory is THEORETICALLY wrong as demonstrated in: http://www.geocities.com/anewlightinphysics/sections/Section1-1_Considerations_against_Relativity.htm

    A new theory is NEEDED!
     
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  3. (Q) Encephaloid Martini Valued Senior Member

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    Needed for what?
     
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  5. przyk squishy Valued Senior Member

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    Hi martillo,
    It doesn't look like your arguments have changed since the last time I saw them (except for one which seems to have disappeared). But anyway. I'll deal with your point (C) about gyroscopes now because it's fairly trivial:
    The answer is: relative to any inertial reference frame. Relativity denies the existence of a single, unique absolute frame, but it doesn't claim the opposite extreme: that all reference frames are indistinguishable and equally valid. The fact that things behave differently in rotating frames is obvious to anyone who's ever been on a merry-go-round or in a car turning a sharp corner, so you don't need to bring up pendulums, gyroscopes, or the Sagnac effect to prove this point.

    STR claims that the laws of physics are invariant with respect to transformations in the Poincaré group. This includes boosts, translations, and fixed rotations (as opposed to rotating frames, where the angle of rotation is not constant), so you can see relativity as proposing a preferred set of reference frames if you like.

    As for (A), I don't see why this:
    should be a problem for you (I'd expect ageing rates to be relative).

    As for this:
    It may seem counter-intuitive, but it isn't actually a contradiction. In general, \(\frac{\part a}{\part b}\) doesn't necessarily equal \(\left( \frac{\part b}{\part a} \right)^{-1}\) in mathematics.

    There's nothing magical about reciprocity, so if you like I could come up with a though experiment that might make it seem more reasonable to you.
     
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  7. martillo Registered Senior Member

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    Hi przyk,
    Yes, thanks to some discussions in this forum I have realized that Relativity can be developed with constant mass. It's only necessary to redefine the momentum as p=γmv where γ=1/(root(1-v2/c2)). So the argument that mass is constant although considered true is not any proof that Relativity is wrong.
    Yes it does, General Relativity does. It even include any accelerated frame!
    Boosts, translations and fixed rotations relative to what??? You must consider a first basic reference frame to say that and which is it? That frame must not be accelerated to be an inertial frame but not accelerated relative to what?
    It must exist a basic frame or a set of frames which we could call them to be at rest! Then you after can state that any frame having a boost, translation or constant rotation relative to them is an inertial frame.
    You must consider that any frame obtained by a boost, translation or constant rotation from an accelerated frame will also be an accelerated frame.
    Then once we accept that "rest" frames in the Universe exist just one of them selected by some property of the universe like possible symmetry would be the "privileged" absolute frame of the Universe.
    Iknow it would not be easy to determine but I believe some day it will be.
    The problem is that ages are not relative! Age is directly related to all physiological phenomena that have haened to an individual in his history. You cannot say something has happened if one frame of observation is selected but that thing hasn't if another one is selected!
    Suppose that in one frame of observation twin1 aged more than twin2 and so twin1 has a long bear while twin2 has not. That situation cannot change just changing the frame of observation. The fact that the twin has or not has bear cannot depend in the reference frame.
    A consistent theory will give the same observation of that kind of things (which can be called Intrinsic properties) in any frame of observation. Otherwise is an inconsistent and a wrong theory!
     
  8. martillo Registered Senior Member

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    896
    (Q),
    Well I think nobody wants to have to believe in a wrong theory. It could give wrong predictions and could make us think or even do something wrong.
     
  9. (Q) Encephaloid Martini Valued Senior Member

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    It hasn't yet. So, what's your point?
     
  10. spuriousmonkey Banned Banned

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    It is just a theory.
     
  11. przyk squishy Valued Senior Member

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    And it accounts for the pseudo-forces in those frames with the existence of a gravitational potential gradient and space-time curvature. I don't think GR handles rotating frames, though.
    I suppose the short answer is to say that there's an absolute state of acceleration. Once you've detected an inertial reference frame (not very difficult), the Poincaré group gives you all the other reference frames with the same properties.
    This gives absolute acceleration. Relativity denies absolute velocity. This, by the way, is exactly the way it was with Galilean relativity.

    Just in case you didn't know, a "boost" is a transformation into a frame in relative motion. For example:
    \(t' = \gamma (t - \frac{v}{c^2}x)\)

    \(x' = \gamma (x - vt)\)​
    is a Lorentz boost along the x-axis.
    If the laws of physics are the same in an entire set of reference frames, you cannot attribute special properties to just one.
    All this tells me is that relativity is incompatible with your own worldview. A theory only contradicts itself if it predicts (in two different ways) that one observer will make two contradictory observations.

    In any case, the success of relativity is due to its history of making accurate predictions. If you like, you can imagine that your mother ship is in an absolute reference frame, claim that time dilation and length contraction occur relative to that frame only, and still show that the twins will observe exactly what relativity predicts they will. In this sense, you can consider relativity to be an illusion if you want.
    If an event occurs in one reference frame, it occurs in all reference frames. It's only a question of when.
    The problem here is simultaneity. Two events that occur simultaneously in one frame do not occur simultaneously in all frames. Look at the equation for the Lorentz boost I posted above. Notice that t' is also a function of x. Again, you can call relativity of simultaneity "real" or "apparent", but it is possible to show that either one of the moving twins will "naturally" map out a reference frame related to your mother ship frame by a Lorentz transformation.

    Imagine one of the twins moving in the +x direction at velocity v. Suppose he wants to place a clock in front of him, and another behind him. From the point of view of the mother ship, the clock he pushes forward will move faster than v, and so will dilate more than the twin. The clock he pushes back will move slower than v, and so will experience less time dilation. If the clocks were synchronized before our twin moved them, they won't be synchronized afterwards. But he'll think they're still in sync because of the difference in time it takes light to reach him from each of the clocks.
     
  12. przyk squishy Valued Senior Member

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    3,203
    I'll address point (B) here.
    The transformation of a wavelength (and distances in general) is not that simple. The expression of a wave (assuming unit amplitude and no phase) looks like this:
    \(\cos(kx + \omega t)\) (wavelength \(\lambda = \frac{2 \pi}{k}\), period \(T = \frac{2 \pi}{\omega}\))​

    The inverse Lorentz boost along the x axis is:
    \(t = \gamma (t' + \frac{v}{c^2}x')\)

    \(x = \gamma (x' + vt')\)​

    Substituting into the expression \(kx + \omega t\) gives:
    \(\gamma k (x' + vt') + \gamma \omega (t' + \frac{v}{c^2}x')\)​
    This can be rearranged to give:
    \(\gamma (k + \frac{v}{c^2}\omega) x' + \gamma (\omega + vk) t'\)​
    So we can rewrite our wave as:
    \(\cos(k' x' + \omega' t')\)​
    where:
    \(k' = \gamma (k + \frac{v}{c^2}\omega)\)

    \(\omega' = \gamma (\omega + vk)\)​

    You can see that it is perfectly possible for the wavenumber \(k\) to be zero (corresponding to an infinite wavelength) in one frame and non-zero in another, provided \(\omega \neq 0\). Where \(\omega = 0\), \(k' = \gamma k\), and we get the familiar length contraction formula for the wavelength:
    \(\lambda' = \frac{1}{\gamma}\lambda\)​
     
    Last edited: Apr 23, 2007
  13. martillo Registered Senior Member

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    896
    (Q),
    Well, you haven't but I believe there are some contradictions and inconsistencies and is what I'm presenting here.
     
  14. (Q) Encephaloid Martini Valued Senior Member

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    So, it is YOU personally who needs a new theory. Any particular reason why?
     
  15. martillo Registered Senior Member

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    896
    przyk,
    It does but the rotaton is treated the same as classically. If you have an object with a composed motion of linear velocity relative to an observer plus a rotation you just need to apply Lorentz Transform in the time varying plane of both, the observer and the object (that which also has the direction of the rotation) plus a classical rotation. It is a composed transform.

    But this way you are accepting there is a special set of frames and is not what General Relativity says. When I'm referring to the frame that can be determined by pendulums and gyroscopes I'm referring to GR statement of no privileged frames of reference at all.

    Yes and I believe it exist. We can just choose a frame with directions determined by gyroscopes far away from massive objects to ensure no gravitatonal effects and the center of the frame to be the center of the Universe which I believe must exist (although difficult to determine).
    These way we can define rest frames in the Universe. One of them would be the Absolute Frame of the Universe which could be determined by some property of the Universe like symmetries in it.

    I'm talking about special properties of the Universe and not of the laws of the Universe.

    A theory also contradicts itself if two different observers make contradictory observations what is present in the presented problem.


    The crossing event of the twins after the symmetrical travel gives only one event to be measured by all the frames in the instant when the center of all the frames coincide. This avoids the problem of the relativity of the simultaneity. This is one of the important features of the problem.
     
    Last edited: Apr 26, 2007
  16. martillo Registered Senior Member

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    896
    (Q),
    Yes, I just want the truth.
     
  17. martillo Registered Senior Member

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    896
    przyk,
    (copy/paste didn't work for the formula)
    First note that the right equation is λ’ = γλ since the lenght observed by the frame "at rest" (λ) must be smaller than the lenght observed by the moving frame: the contraction must be λ < λ’ .
    Second, you just arrived at the same formula as me so where is the problem???
     
  18. przyk squishy Valued Senior Member

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    3,203
    Hit "quote" to reply. By the way, I'm using \(\TeX\) (added to the forum a few months ago) to typeset formulae. There's a thread about it [THREAD=61223]here[/THREAD], in case you find it useful.
    The more general formula is:
    \(\frac{1}{\lambda'} = \gamma \left( \frac{1}{\lambda} + \frac{v}{c^2} \, \frac{1}{T} \right)\)​
    where \(T\) is the period of the wave.

    \(\lambda' = \frac{1}{\gamma} \lambda\) is true only if the period \(T\) (in the unprimed frame) is infinite.

    \(\lambda' = \gamma \lambda\) is true only if the period \(T'\) (in the primed frame) is infinite.

    The general relation between \(T\) and \(T'\) is:
    \(\frac{1}{T'} = \gamma \left( \frac{1}{T} + v \frac{1}{\lambda} \right)\)​

    I'll respond to your other post when I have time.
     
  19. (Q) Encephaloid Martini Valued Senior Member

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    That would be your own, personal, self-gratifying truth?

    Will that so-called truth be of any use to you? What do you expect to gain?
     
  20. martillo Registered Senior Member

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    896
    (Q),
    Yes, it is self-gratificating to find at least part of a truth.

    I don't have a practical use for now but it is all under development and something can surge...

    I used to dream to gain some things, it was a good incentive while having a hard work taking much of my time... but now I don't expect anything. I will receive just what others really would want to give, it all depends on the value it would have for otherones not for me.
     
    Last edited: Apr 23, 2007
  21. Singularity Banned Banned

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    i am sure this exactly what the greatest scientist were told by the orthodox community, so stop being narrow minded.
     
  22. (Q) Encephaloid Martini Valued Senior Member

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    What is under development? What will surge?

    Others already have found value in the existing theory. In fact, it works very well, with solid experimental results. What makes you think you will provide any more value?
     
  23. martillo Registered Senior Member

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    896
    (Q),
    You should take a look at the main page: http://www.geocities.com/anewlightinphysics

    Well, a right theory has more value than a wrong theory.
     

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