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Nonstandard relativistic dynamics

  1. Aug 10, 2008 #1
    Consider please that we know selleri's transformation equations for the space-time coordinates of the same event or other such nonstandard transformations.
    How could we use them in order to approach rfelativistic dynamics or electrodynamics.
    Thanks in advance
     
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  3. Aug 11, 2008 #2

    DrGreg

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    For the benefit of reader's not familiar with "Selleri's transformation equations" they are

    [tex]x' = \gamma(x - v t)[/tex]
    [tex]t' = \frac{t}{\gamma}[/tex] ..........(1)​

    as described in this post

    Here, [itex](ct, x)[/itex] are standard Einstein-sychronised coordinates in one "stationary" frame, and [itex](ct', x')[/itex] are non-standard coordinates in another "moving" frame. The Selleri transform is effectively the Lorentz transform expressed in terms of a non-standard coordinate chart.

    Selleri coordinates are related to Einstein-synced coordinates [itex](ct'', x'')[/itex] for the same "moving" observer by

    [tex]x' = x''[/tex]
    [tex]t' = t'' + \frac{v x''}{c^2}[/tex] ..........(2)​

    They have a metric

    [tex]ds^2 = \frac{dx'^2}{\gamma^2} - c^2 \, dt'^2 + 2 \, v \, dx' \, dt'[/tex] ..........(3)​
    _______

    There are two ways to answer the question.

    One is to make use of our knowledge of SR in standard coordinates. As we know energy-momentum [itex](E/c, p)[/itex] is a 4-vector, we can express that 4-vector in any coordinate chart we wish, so, for example, we will get

    [tex]p' = \gamma \left( p - \frac{v E} {c^2} \right) [/tex]
    [tex]E' = \frac{E}{\gamma}[/tex] ..........(4)

    [tex]p' = p''[/tex]
    [tex]E' = E'' + p'' v[/tex] ..........(5)​

    (from (1) and (2) respectively).

    For those who favour the Lorentz Ether approach and who don't want to use Einstein's postulates you will need to think up some extra postulates which give rise to the energy-momentum equations (4) above.

    A similar approach applies to electromagnetic tensors, etc.
     
  4. Aug 11, 2008 #3
    Thanks Dr.Greg. Presenting equations (4) you have taken into account that momentum is space like and energy is time like?
     
  5. Aug 12, 2008 #4

    DrGreg

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    Yes.

    If you know about "4-vectors" you will know that the energy-momentum 4-vector [itex](E/c, \textbf{p})[/itex] behaves like the position 4-vector [itex](ct, \textbf{x})[/itex] -- they both transform via the Lorentz transform. If this is true in standard Einstein coords, then these 4-vectors must also transform the same as each other in other coordinate systems too. If this is unclear, just remember that

    [tex]\textbf{p} = m \frac {d\textbf{x}}{d\tau}[/tex]
    [tex]E = m c^2 \frac {dt}{d\tau}[/tex] ​

    [itex]m[/itex] and [itex]\tau[/itex] are invariants, i.e. coordinate-independent.

    The physics is that we know that the Einstein-synced definitions of energy and momentum are conserved, so any linear combination of these two quantities are also conserved.

    The same general principle of change-of-coordinates will apply to Maxwell's equations, but that gets a bit more complicated (and a bit out of my depth).
     
  6. Aug 14, 2008 #5
     
  7. Aug 14, 2008 #6

    DrGreg

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    I haven't had time to read the paper you quote, yet, but you might find some of the posts (including mine) in this long thread interesting.
     
  8. Aug 15, 2008 #7
    Independent of the paper I have quoted consider please that we have derived all the nonstandard Lorentz transformations from the Lorentz-Einstein ones using a "physically" correct synchronization procedure.
    Under such conditions is it necessary to ask if the relativity principle is respected, if there is something in contradiction with Einstein's special relativity theory, with transitivity or it is correct as you stated that they are no more then the LET with a shift in time (space?).
     
  9. Aug 15, 2008 #8

    DrGreg

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    I'm not quite sure what exactly you mean by '"physically" correct synchronization procedure'. But in all the cases of different sychronisations we have ever considered in these threads, there is always a formula that converts between our coodinates and Einstein-synced coordinates. So I see these coordinates as just a mathematical re-expression of SR in an alternative coordinate system; the physics is the same but the maths looks different. It's no different than converting between cartesian xyz coordinates and polar [itex]r\phi\theta[/itex] coordinates. Equations may look very different in the two different systems but they represent the same physical reality.


    It depends on how you interpret the phrase "relativity principle". Rizzi et al suggest the principle should be worded as

    "Under the same conditions" means that both observers would have to make their measurements in operationally the same way, without relying on any external reference. One of the problems with Selleri synchronisation is that you need to know your frame's velocity relative to the "ether" frame. If you were inside an inertial sealed box, with no access to the outside world, you could not set up Selleri coordinates.

    Having now skimmed through Rizzi et al's paper, without reading all the details and none of the proofs, it looks like a well-written and well-researched paper that agrees with my point of view.
     
  10. Aug 16, 2008 #9
    I have in mind Selleri and Abreu and Guerra (on arxiv). They do not involve in their derivations the Lorentz-Einstein transformations, performing the so called "external clock synchronization and I suppose that is why you state "One of the problems with Selleri's synchronization is that you need to know your frame's velocity relative to the ether frame.
    In my oppinion external clock synchronization could be avoided. Consider that the I' inertial reference frame is filled with a transparent ideal transparent dielectric characterized by a refraction index n. Light signal propagates in the positive direction of the O'X' axis with speed
    c'n=c/n (1)
    A clock C'1(x') synchronized using such a light signal will display a time
    t'n=nx'/c (2)
    A clock C'2(x') located at the same point but Einstein synchronized will display a time
    t'E=x'/c (3)
    Combining (2) and (3) the result is
    t'E=t'n+x'(1-n)/c (4)
    and that is what I call "physically correct" relationship between the readings of two differently synchronized clocks located at the same point in space. Would you call it "synchronization gauge"?
    Starting with the Lorentz-Einstein transformations for x and tE and imposing the condition that the time trasformation do not depend on the space coordinate (n=1+V/c) we recover Selleri's transformations.
    Do you consider that the approach is correct.
    In what concerns Rizzi's definition
    "Under the same conditions" means that both observers would have to make their measurements in operationally the same way, without relying on any external reference.
    I consider that using the Selleri transformations they help me to find out the x' space coordinate as a function of x and tE measured in I and to find out t'n as a function of tE. Under such conditions do we perform the same experiment? Have we violated so far the relativity principle?
    Thanks in advance for the patience with which you answer my questions helping me to understand the interesting subject.
     
  11. Aug 20, 2008 #10

    DrGreg

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    Let's see if I understand this.

    You are saying an observer O' emits a flash of light at time zero, and it is received at location P' (a fixed distance x' from O') by two methods. One method travels through a pipe through an optically dense medium (refractive index n) to arrive at clock C'1 and the other method travels outside the pipe, through vacuum, to arrive at co-located clock C'2. On reception of the flash, each clock is set according to your equations (2) and (3).

    Am I right so far?

    The problem with this is that the reception of the two signals are two different events which occur at two different times. Your calculations seem to assume they are the same event.

    The procedure I described above (where c is the empirically determined two-way speed of light in vacuum, and c/n is the empirically determined two-way speed of light through the stationary medium), would result in the co-located clocks C'1 and C'2 being synchronised with each other.

    Your calculation, is really just taking the approach that the one-way speed of light in vacuum is postulated to be c/n (in one direction) and then deriving the Selleri transform from that. Introducing a medium is just confusing two different physical situations which happen to be described by the same equation (with different meanings).

    So I think the maths is right but the physical interpretation is wrong. Unless I've misunderstood what you are saying.
     
  12. Aug 20, 2008 #11
    Thanks for your help. I would be happy having you next door to me. That is the big service the Forum is offering.
    My answer to your fist question is yes.
    The receptions of the two of the two signals are two different events that take place at the same point in space (x') at different times, displayed by two different clocks located at that point, synchronized with the clock located at the origin, O' using synchronizing signals which propagate with c and with the one way speed c/n. What I try to do is to use the Lorentz-Einstein transformations and to express t'E by t'n via a physically correct relationship
    t'E=x'/c (1)
    t'n=nx'/c (2)
    t'E=t'n+(x'/c)(1-n) (3)
    Doing so I recover all the results obtained by Selleri and Abreu and Guerra (arXiv) without using the Lorentz-Einstein transformations. As far as I know such transition from the reading of one clock to the reading of another one is a current procedure.
    Consider the photographic detection of a source of light S'(x') by an observer located at O' The source of light emits a light signal at a time t'e the observer located at O' taking the snappshot at a time t'r the two times being related by
    t'r=t'e+x'c. In practice we express the coordinates of the event associated with the emission by the time coordinate associated with the event of reception.

    Your calculation, is really just taking the approach that the one-way speed of light in vacuum is postulated to be c/n (in one direction) and then deriving the Selleri transform from that. Introducing a medium is just confusing two different physical situations which happen to be described by the same equation (with different meanings).
    Please be more explicit. As I have mentioned above c/n is an one way outgoing speed of light. Reicnebach's equation enables us to find out the corresponding one way speed of the incomong signal.
    What would be the correct interpretation if there is one?
     
  13. Aug 20, 2008 #12

    DrGreg

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    Everything you say in post #11 is OK, considered in isolation.

    You can consider that the one-way speed of light O' to C' through vacuum is c/n (where n is just some number to be determined later) and then your whole argument makes sense.

    What I have difficulty with is bringing in a medium of refractive index n. To assume that the one-way speed of light through a medium of refractive index n is c/n, when we already know that the two-way speed of light is c/n, you are saying the one-way and two-way speeds are the same, so the synchronisation is Einsteinian. If you respond to this by saying we don't already know that the two-way speed of light is c/n, then what is n and what is the point of mentioning a medium at all?

    Or another way of putting it, under the "light through a medium" interpretation, equations (1) and (2) in post #11 refer to two different reception events, so you can't just combine them in equation (3). It is not the difference between emission and reception events that I object to, it's the difference between receiving a signal through a medium in (2) and receiving a signal through a vacuum in (1). In (3) and your argument that follows, it is implicit that t'E and t'n are different coordinates for the same event.
     
  14. Sep 8, 2008 #13
    Thanks for your answer. My question is now:
    If we present the Selleri transformations as
    x'=[tex]\gamma[/tex](x-VtE)
    t'a=tE/[tex]\gamma[/tex]
    in order to underline the clocks which display the involved times, then it would be correct to mention the same facts in the expression of the energy and momentum as
    E'a=EE/[tex]\gamma[/tex]
    p'a=pE/[tex]\gamma[/tex]
    with subscript E for standard synchronized and a for nonstandard synchronized?
     
  15. Sep 8, 2008 #14
    Is this about anything more than defining new variables? If not, no new physics is involved.

    If it is proposed that measurable quantities such as Einstrein's x" are actually Selleri's x', this would constitute a differet theory. Is there a thing whos measure disagrees between theories?
     
  16. Sep 9, 2008 #15

    DrGreg

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    Sorry, I changed notation between post #2 and the notation used in some previous threads.

    My [itex](t, x)[/itex] is the same as your [itex](t_E, x_E)[/itex].
    My [itex](E, p)[/itex] is the same as your [itex](E_E, p_E)[/itex].
    My [itex](t', x')[/itex] is the same as your [itex](t'_a, x'_a)[/itex].
    My [itex](E', p')[/itex] is the same as your [itex](E'_a, p'_a)[/itex].

    So in your notation

    [tex]E'_a = \frac{E_E}{\gamma}[/tex]
    [tex]p'_a = \gamma \left( p_E - \frac{V E_E} {c^2} \right) [/tex]​
     
  17. Sep 9, 2008 #16

    DrGreg

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    You are right, this is nothing more or less than a change of variables and there's no new physics. Some authors such as Selleri seem to think that Selleri coordinates are somehow more "correct" than Einstein-synced coordinates but I don't understand that logic.

    If you double-check my equations, x' and x'' are the same anyway, but t' and t'' differ. But you can't measure either t' or t'' directly, except in the special case when x''=0 and both values agree. Coordinate time values at a distance from the observer carrying the clock are assigned by convention and the whole Selleri/Einstein business is an argument over which convention to use. In the end it doesn't really matter as it's just a change of coordinates. Einstein sync is mathematically easier because it gives you an orthogonal coordinate system.
     
  18. Sep 10, 2008 #17
    The test, would be to compare the two delta t's, then, wouldn't it?
     
  19. Sep 11, 2008 #18
    when does anisotropy fade away?

    Consider the Selleri transformation for the space coordinate
    x'=[tex]\gamma[/tex](x-VtE) (1)
    to which we can add
    y'=y (2)
    taking into account the invariance of distances meaured perpendicular to the direction of relative motion. By definition
    x=rcos[tex]\theta[/tex] (3)
    y=rsin[tex]\theta[/tex] (4)
    the formula that accounts for the aberration becoming using Selleri coordinates
    tan[tex]\theta'[/tex]=([tex]\gamma-1[sin\theta]/(x-VtE) (5)
    If the synchronizing light signal is emitted at t=0 then tE=r/c and (5) becomes
    tan\theta'=\gamma-1sin\theta/(cos\theta-V/c) (6)
    a result which is usually obtained via the Lorentz transformations.
    Could we say that under such conditions anisotropy fades away!
     
  20. Sep 11, 2008 #19

    DrGreg

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    For constant x'', [itex]dt' = dt'' = d\tau[/itex], where [itex]\tau[/itex] is proper time which you can measure. So measurements still can't distinguish between the two alternative coordinate systems. One way of looking at this is when you integrate this, the two coordinate systems assume different constants of integration, but the constant of integration is arbitrary.

    The fact is that Einstein's primed coordinates and Selleri's double-primed coordinates (in my notation) are both equally valid coordinates systems, both compatible with the ruler length of any observer stationary in the frame, and both compatible with the proper time of any observer stationary in the frame. The only difference is when two stationary observers compare each other's clocks. In Einstein (i.e. Minkowski) coordinates, the speed of light [itex]dx'/dt'[/itex] is c in all directions. In Selleri coordinates, the speed [itex]dx''/dt''[/itex] varies by direction, although the round-trip there-and-back average is still c.
     
  21. Sep 11, 2008 #20

    DrGreg

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    Re: when does anisotropy fade away?

    If I understand correctly, you are essentially considering space only and not time. Selleri space coordinates are identical to Einstein space coordinates (in magnitude), it's only the time coordinates that differ. So no real surprise you get the same answer.

    Or I may have misunderstood. It might help to explicitly spell out exactly what event it is you are measuring.
     
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