Proof Minkowski Spacetime is Poorly Conceived

Discussion in 'Alternative Theories' started by danshawen, Apr 21, 2016.

  1. James R Just this guy, you know? Staff Member

    You have r=ct and omega=(angle)/t, but those two 't's are not the same thing, are they?
    What, exactly are you saying that v is the velocity of? And what is omega the angular frequency of? You mention a particle, but it's not very clear to me what you're doing.

    Your last line appears to be dimensionally incorrect in that it equates something with units of velocity to something with units of velocity squared. What went wrong?
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  3. exchemist Valued Senior Member

    Not that it matters much, but you cannot patent a scientific theory anyway.
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  5. przyk squishy Valued Senior Member

    You seemed to be under the impression that frequency isn't accommodated in Minkowski's framework. It is, for instance as part of the four wavevector. Furthermore, you can use the component transformation rule for four vectors to derive how the frequency of a plane wave changes if you change from one reference frame to another. That's what the relativistic Doppler effect is. I honestly have no idea what the rest of your OP is supposed to have to do with Minkowski.
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  7. danshawen Valued Senior Member

    They are the same because the speed of light is what is invariant, not t.
  8. danshawen Valued Senior Member

    I was responding to Ophiolite's specific concern, which is estimable to say the least. How many patents are traceable to E=mc^2? And the only thing Einstein himself ever patented had to do with refrigeration. You will recall he was a patent clerk for a while.
  9. danshawen Valued Senior Member

    Although I admit that the math so far would suggest a plane wave, that is not the identity this expression addresses. Let it marinate a while. You will understand it.
  10. exchemist Valued Senior Member

    Very few, I should have thought. I can think of very few pieces of technology that rely for their operation on mass-energy equivalence. Whereas thousands depend on Newtonian mechanics and thousands more on Faraday's ideas about electromagnetism.

    (Einstein was a patent office examiner, not a clerk - there is a big difference.)
  11. James R Just this guy, you know? Staff Member

    Not a very helpful response, danshawen.

    Let's try again, then.

    ... in which v is the tangential velocity of some point at distance r from an axis, moving around that axis of rotation with angular speed omega.

    That is, \(r\) is defined as the radius of some unspecified particle, \(c\) is the speed of light and \(t\) is defined to be the time \(r/c\).

    Perhaps this particle is a sphere rotating on its own axis with angular speed omega. That's my guess, lacking any actual information on the matter. Then \(v\) is the tangential velocity of a point on the "equator" of the particle.

    \(\omega\) is the angular speed of rotation. In time \(t\) (defined arbitrarily as \(r/c\)) the particle rotates through an angle given by

    \(\theta = \omega t = \frac{\omega r}{c}\)

    Now \(v = \omega r\) and \(\omega = \frac{c \theta}{r}\) so

    \(v = c \theta\).

    In other words, a point on the equator of the particle has linear velocity given by the speed of light multiplied by the angle through which the particle rotates in a time taken for light to travel the distance of one particle-radius.

    But wait! What if we define \(c\) to be the speed of Fred the Tortoise (who I happen to know moves at 35 centimetres per minute).

    Then, the radius of the particle in terms of Fred-travel time is \(r=ct\), as before. Work it through and we find that

    \(v = c \theta\)

    which tells us that a point on the equator of the particle has linear velocity given by the speed of Fred the Tortoise multiplied by the angle through which the particle rotates in a time taken for Fred to travel the distance of one particle-radius.

    Here's where I get lost. How does \(v=c^2\) follow from any of the above calculations about Fred the Tortoise?

    I guess the stuff in ALL CAPS must somehow be very important, but I'm not seeing it.
  12. danshawen Valued Senior Member

    No. As I have said, c is the invariant, not t. Time dilation, time intervals, and the rate at which time passes is different everywhere. In particular, TIME DILATION IS DIFFERENT INSIDE OF PARTICLES, DEPENDENT ON r. It is c that is constant and invariant, whether the path of the photon is along the radius, or the path of the photon is tangential.

    You're doing fine. By all and every means, keep thinking about it.

    This is just a whole lot more fun since I have worked through the details myself (for a change).

    Google scholars needn't bother searching for this one. You won't find it anywhere else.
  13. James R Just this guy, you know? Staff Member

    Show me where you got \(v=c^2\) from, danshawen.

    It can't be right - whatever it is supposed to mean - because it is dimensionally incorrect.
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  14. danshawen Valued Senior Member

    It is still dimensionally correct because internal to particles of matter or antimatter, delta t=o, the same as it is for a photon propagating in linear mode. This is why matter persists.

    This is too weird for me to make up, and I am as surprised about the finite result as anyone else. I originally thought it should be faster.

    A similar dimensional argument could be made against the units of tangential velocity where I started:

    v = omega x r has the units: (angle x length)/time. If any length is interpreted as light travel time, then (time x time)/time is simply time, and not a velocity, isn't it? This is why mathematicians do not regard dimensional analysis as a rigorous mathematical check. They are more comfortable working with dimensionless constants, real or complex variables with no dimensions or ones they can more consistently define.

    As I said at the beginning of my analysis, any velocity is fundamentally a ratio of how fast something moves compared to how fast something else moves, like a stopwatch or a beam of light. This part is not rocket science, although the rest of what I am saying decidedly is.

    The fastest process here is the basis of time itself. Hint: it has less to do with the linear mode of propagation than it does with a superposition of rotational and linear propagation modes. This we all intuit from common experience. Look around you. It works. There is a reason fundamental particles of matter persist as though time has stopped for the energy that is bound within it.

    If you have better math to demonstrate this, show us.
    Last edited: Apr 23, 2016
  15. Ophiolite Valued Senior Member

    No, it's right up your street.
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  16. danshawen Valued Senior Member

    And, you don't think Minkowki was weirder only because you concentrated on geophysics. I do understand your point of view. I like that kind of skepticism a lot. We need more of it, not less.
  17. Dr_Toad It's green! Valued Senior Member

  18. exchemist Valued Senior Member

    Dimensional correctness is a necessary but not sufficient check, that's all. So if your dimensions don't balance, you have most definitely made a mistake somewhere.
  19. exchemist Valued Senior Member

    Er, quite so.

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  20. Schneibster Registered Member

    We'll see. The mods seem to disagree, but I won't let that bias me.

    Actually this is incorrect. The hyperfine transitions (between of the ground state of cesium are used to define time; one second is a defined multiple of how long one such hyperfine transition takes to occur (the multiplier is exactly 9,192,631,770). A meter is defined in terms of a second and the speed of light; it is exactly 1/299,792,458th of the distance light travels in a second. The hyperfine transition of cesium is used because it is extremely stable; this is why cesium is used in atomic clocks.

    Minkowski didn't create intervals. He discovered them in the math of SRT. Specifically, he discovered them in the Lorentz transforms. The hyperbolic trigonometric version of the Lorentz transform yields a quantity called "rapidity," which you can find out about by reading Baez' page on symmetries. <-Linkie As confirmation you can check out the Physics FAQ's page on velocity <-Linkie where you will also find discussion of rapidity.

    You will find that the hyperbolic trig version of the Lorentz transforms explicitly treats velocity as a rotation on a hyper-hyperboloid of revolution. The hyperbolic rotation angle is the rapidity. This is a very intuitive way to think about foreshortening (apparent decrease of length with velocity). It also in the same breath gives an intuitive way to think about time dilation; some distance in x (or y or z or some combination) turns into distance in t; thus, the object with velocity relative to the observer's frame of reference appears shorter, and simultaneously appears to experience time more slowly, to the observer. The equations turn out to be exactly the same transforms that are applied during ordinary rotation in 3D, but with the addition of a hyperbolic dimension, time.

    Because distance turns into time and vice versa, 1) it is apparent that although the shape of time is different from the shapes of the space dimensions, they are all "the same sort of thing" since they can be substituted for one another simply by changing velocity; and 2) since the change from distance to time and vice versa is at a fixed rate, if we properly add space and time together we can define an invariant. This invariant is called "interval." To give it its full name, it is "Minkowski spacetime interval."

    The interval is

    \(d^2 = x^2 + y^2 + z^2 - c^2t^2\)
    d is the Minkowski spacetime interval,
    x, y, and z are distances in the three spatial dimensions,
    c is the speed of light, and
    t is the time.

    Note that t must be multiplied by c to yield a distance; we can see therefore that a second of time is equal to a light-second (299,792,458 m) when transformed due to velocity, and vice versa. Note also that the negative sign before the time term makes it hyperbolic. This is apparent in GRT by the standard tensor sign convention of -+++.

    None of this has anything to do either with FTL or with entanglement. It's just bog standard SRT.

    You cannot apply these formulae to light or Doppler shifts because at the speed of light the length in the direction of travel becomes zero, and the rate of the passage of time also becomes zero. The Lorentz transform claiming the length of a material object is zero is a clear indication that it has passed beyond its region of applicability. Press further, with a velocity beyond c, and the length becomes an imaginary number; ridiculous results like these are the usual result of trying to use a theory in a domain where it no longer applies. Nobody knows what a length of zero or even worse an imaginary length means; it shows no apparent physical meaning.

    There is no omission. Photons simply do not follow the same rules as matter particles do.

    No, it's not. The conversion factor between time and space has been confirmed by a multitude of experiments. So has Minkowski's equation. So have the Lorentz transforms.

    That's possible, but you have to be really careful how you do it and what conclusions you draw. And most especially, you have to be incredibly careful how you define velocity when you do it; it's possible to define velocity as rapidity, which does not vary in the same manner as defining velocity as a fraction of the speed of light. Rapidity is not a fraction of the speed of light; it's the hyperbolic tangent of that fraction. It is very easy to become confused and draw all sorts of erroneous conclusions.

    Like everyone else who has commented on this, I don't see where you got \(c^2\). But I'll also add that I don't see where you got r. What is the size of a photon? There is no answer to this question that means anything in terms of the photon's spin.

    Note also that the rotational axis (not the spin axis! Spin (technically spin angular momentum) is a purely quantum mechanical degree of freedom, and can be measured at any angle, not just the axis of the photon's motion!) must be along the photon's line of motion; otherwise, one side of the photon would be moving faster than the speed of light and the other slower. Another way to say this is to say that light is a transverse wave, not a longitudinal wave. And this fact is confirmed by light's ability to be polarized; longitudinal waves cannot be polarized.

    So basically we can't define a photon's size, so we can't define r, and nobody understands where you got \(c^2\).

  21. Schneibster Registered Member

    OK, now I see where you got \(c^2\), but you got it from defining r; and r is undefined since the size of a photon is undefined.

    I don't agree that you've defined any sort of rotation that leads to anything exceeding the speed of light. Your definition depends on defining r, and you cannot.

    Sorry, this is a fail. Without r you cannot define anything at all, and even if you did nothing you've said here has anything to do with intervals, Minkowski, Lorentz, or even SRT.
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  22. Dr_Toad It's green! Valued Senior Member

    Schneib, I see you're in your element here. Sorry we don't have the same quality of crackpottery at my place for you to take on!
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  23. danshawen Valued Senior Member

    That is exactly correct.

    To determine that, you must understand that a pair of photons of sufficient energy are capable of creating a particle of matter and a particle of antimatter.

    Exactly how do you imagine that process takes place. Please note that I am not providing any more details of the process other than to calculate that it may only occur if there is both a tangential and a radial component of the propagating energy, and that according to the classical formula applied to a consideration of both components as light travel time, it is transparent that whatever is going on inside is doing so at a rate of c^2.

    This is not the only way to derive the same result, just the simplest one.
    Because this analysis works without any of those, and because the precepts upon which Minkowski based those things collapse of their own weight the very instant you can define a process like this one which is faster than c.

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