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Is Lorentz transformation absolute

  1. May 12, 2008 #1
    An observer at the mid point between two frames A and B that are moving toward her, measures each of their speeds to be 0.75c. According to SR, the Lorentz transformation will determine the speed of A and B measured by the other to be less than 1.5c. (in fact less than c)

    Being aware of the effects of SR, the observer at the mid point will also be aware that their measurement of the speed of A and B is subject to the Lorentz transformation and therefore is a measure of some speed greater than 0.75c contracted to 0.75c by the Lorentz effect. After accounting for the Lorentz factor that gave their measurement for A and B they find the speed of each is greater than 0.75c.
    Since the observer at the mid point is not necessary in the measurements made by A and B, should the observer calculate the speed of A and B as measured by each as a Lorentz transformation of 0.75c, or the greater, untransformed speed?
    Why?
     
    Last edited: May 12, 2008
  2. jcsd
  3. May 12, 2008 #2

    Doc Al

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    I thought you just said that A and B are moving with a speed of 0.75c with respect to that midpoint observer? Where do you get the speed of 0.375c?
     
  4. May 12, 2008 #3
    Sorry Doc Al, I left that line in from a previous version of the question. Thanks for mentioning it.
     
  5. May 12, 2008 #4

    Doc Al

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    :confused: What do you mean here? Let's call that midpoint observer M. Both A and B are moving at speed 0.75c with respect to M.
    :confused: What do you mean by "after accounting for the Lorentz factor"?

    All measurements of the speed of A with respect to M are made using distances (and times) as measured by M. So there's no need for a "Lorentz factor".
     
  6. May 12, 2008 #5
    Let me restate the question using your suggestion of M as my frame.
    If I (M) calculate the speed of A as measured by B, I must invoke the Lorentz transformation as it affects measurements made by B. I must also invoke it to calculate the speed of B as measured by A as it affects the measurements made by A. If I calculate the speed of M (my frame) as measured by B, I invoke the Lorentz transformation as it affects the measurements made by B. If B calculates the speed of B as measured by M, they invoke the Lorentz transformation as it affects the measurements made by M (me).
    Now, being fully aware of the effects of SR on all measurements of constant linear motion, how do I consider the initial v of B or A, input into the Lorentz factor and applied to the Lorentz transformation of the measurements made by M(me) on the axis of motion and rate of time, is "valid" if I by the definition of relative motion, my(M's) measurement of v must already be affected by the Lorentz transformation?
     
  7. May 12, 2008 #6
    The statement of the problem doesn't make sense to me. The role of the observer is to make the measurements. If the observer is not measuring anything, why is that observer there? Furthermore, the observer doesn't measure the speed of reference frames, the observer defines the reference frame.

    The correct language should be in the form of observer A in reference frame A measures position and clock settings of these events as follows... observer B in reference frame B measures position and clock settings of these events as follows...

    I wouldn't start jumping into Lorentz factors until you identify the invariant spacetime interval. And also the simplest way to clear up a paradox in SR is to draw a spacetime diagram, the picture itself if drawn with care will usually illustrate clearly the resolution to the problem.
     
  8. May 12, 2008 #7

    Doc Al

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    The reason why you need to use the Lorentz transformations (the relativistic addition of velocity formula, in this case) is because the only data you have to work with are measurements made in frame M. In order to transform the speed measurement of A with respect to M into a speed measurement of A with respect to B, you need to use the relativistic addition of velocity formula. It doesn't "affect the measurements"--it allows you to predict them. If B directly measured the speed of A, he'd of course get the same answer--without the need to use any Lorentz transformations.
    The same "addition of velocity" formula applies here.
    No. Since we are given the speed of B with respect to M, no calculation is needed to deduce the speed of M with respect to B. It's the same! (Opposite direction, of course.)
    You lost me on that last sentence. But read my comments above and see if you still have a question.
     
  9. May 12, 2008 #8
    Yes, if B measures the speed of A, und you want to now the speed of A in a third reference frame like M, you must apply the Lorentz-Transformation (or better the "relativistic velocity addition rule").

    No. That is another situation which is completely different, because the relative speed between two observers does not change. So if B measures that the speed of M is -0.75c, than M measures that the speed of B is 0.75c. Only the sign +- of the velocity is changing and no Lorentz-transform is needed.

    There is no problem - see the explanation above.
     
  10. May 12, 2008 #9
    The principle of relativity removes the property absolute rest from the laws of mechanics. If M and B have a relative speed of 0.75c, neither can say they are at rest or in motion. Such a claim by either would be a nonsensical statement without an absolute frame of rest from which to make a measurement. The property rest is then a relative measure. If the equations of mechanics are upheld in a frame, the coordinates of the frame hold the property rest. This is the key, the frame is not at rest in the classical sense, the frame holds the property rest with respect to the equations of mechanics being upheld in (accurately predicting) the measured motion (mechanics) of bodies.
    To say a frame is at rest or in constant linear motion is nonsensical except with respect to another frame. There is no test of mechanics that will distinguish rest from constant linear motion, because there "is" no such distinction.
    So with this line of reasoning, let me restate the question.
    I am in the frame M speeding toward B at 0.75c. A is also speeding toward M in the same direction at 0.75c and therefore speeding toward B at 1.5c. But B does not measure the speed of A to be 1.5c, since B's measurements are contracted and time is dilated with respect to M, B will always measure the speed of A to be less than c.
    If your scratching your head because I said M was speeding toward B, but then I attributed the contraction and dilation to the measurements made by B, you will see my point.
    How does M, between A and B, both speeding toward M at 0.75c, reason and predict the measurement of the speed of A as made by B will be less that c due to the contraction and dilation suffered at B, when the same reasoning would require the contraction and dilation should have been suffered at M and not B in the previous example?
    If you're thinking it is simply a matter of relative motion, you are stating my original question. Why does M not prime the measurement of the speed of B with the Lorentz transformation before considering the speed of A as measured by B?
     
  11. May 13, 2008 #10
    It seems that you have got problems to understand the relativity principle and the definition of reference frames. Of course the observer in M can say he's at rest an B is in motion; and also B can say he's at rest and M is in motion. And because both can say this, the relativity principle is fulfilled. This is as true for the Galilei-transformation (without electrodynamics) as it is true for the Lorentz-transformation. So, when we say A or B or M are at rest, we mean that we measure those conditions with instruments (rods and clocks) which are at rest in specific inertial reference frames. So let us discuss your example more precisely:

    To define the inertial reference frames, let us state: A is resting in M1; B is resting in M2; C is resting in M3.

    a) Using rods and clocks which are at rest in M3 we measure that the speed of C is 0c, that the speed of A is -0.75c, and that the speed of B is 0.75c. Therefore in M3 the relative speed between A and B is 1.5c. (Of course, 1.5c is not a signal velocity and therefore there is no violation of special relativity).

    b) Using rods and clocks which are at rest in M1 we measure that the speed of C is 0.75c, that the speed of A is 0c, and that the speed of B (using relativistic velocity addition) is 0.96c. Therefore in M1 the relative speed between A and B is 0.96c.

    c) Using rods and clocks which are at rest in M2 we measure that that the speed of C is -0.75c, that the speed of A (using relativistic velocity addition) is -0.96c, and that the speed of B is 0c. Therefore in M2 the relative speed between A and B is 0.96c.

    That's all.
     
  12. May 13, 2008 #11

    Doc Al

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    If you are in the frame M, then B is speeding towards you. (You are at rest with respect to yourself.)
    No, A is not speeding toward B at 1.5c--meaning: The speed of A as measured by B (and vice versa) is not 1.5c.
    Since M is moving with respect to B, B will have to use the Lorentz transformations in order to interpret any measurements of A made by M.
    Not yet.
    I don't see your reasoning at all. M does not require any "transformations" to interpret data that he has directly measured: The speeds of A and B with respect to M are given in the problem! Of course, anyone (in any frame, not just M's) who wishes to calculate--based on the given data--the speed of B as measured by A will have to use the "addition of velocities" formula. So?
    I don't see your point at all. If A moves at 0.75c with respect to M, then M moves at 0.75c with respect to A.
     
  13. May 13, 2008 #12

    russ_watters

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    Yes.
    You have that precisely backwards. Knowing and accepting that there is no absolute rest or motion means either can be said to be at rest or in motion.

    It seems like you make the statement that there is no absolute rest/motion, but still won't let the idea go.
    Yes. As is motion.
     
  14. May 13, 2008 #13
    I think we're getting down to the point of confusion or disagreement.
    Russ, you claim I have it "precisely backwards" and yet you claim motion and rest are relative measures.
    Let me state what I think you are saying in that it may help make more clear what I am saying.
    You say "Knowing and accepting that there is no absolute rest or motion means either (frame) can be said to be at rest or in motion".
    I agree with that statement as far as it goes, but by itself it clarifies nothing. But take it as a true statement about a specific frame (which you seem unwilling to do) and you will find - I (M) am in motion and B is at rest. This is a valid observation, it does not change the quantitative values of dimension expressed by the Lorentz transformation between me (M) and B.
    Doc Al, you claim "You are at rest with respect to yourself". By this I'm guessing you mean that relative motion must be a quantitative property of change you assign to the observed frame instead of your own frame. If that is what you mean, I don't know why you would make this distinction. If you could prove this distinction, you would be proving the existence of an "absolute" frame of rest. If you want me to disprove it I would only suggest that if you are at rest with respect to yourself then you must according to you and Russ, be in motion with respect to yourself. It becomes clear that "with respect to yourself" is "self-referential" and of no meaning at all.
    When I state the properties of a frame, I make the assumption the frame is, or extends from an internal (note that is internal, not inertial) observer. This means there is no concept of "with respect to yourself" that is not "with respect to your frame of reference". Your frame cannot "claim" to be at rest or motion except with respect to another frame, but it "always" possesses the property rest (or constant linear motion) with respect to the equations of mechanics.
    Once another frame is identified the property rest and motion only serve to establish a quantitative transformation of the dimensions from one frame to the other.
    Before I go on to the specific question again, can you verify or clarify what I've said in terms of what you are arguing against.
     
  15. May 13, 2008 #14

    Doc Al

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    If you think you can have a non-zero speed with respect to yourself, I don't think there's much to talk about and little point in continuing.
     
  16. May 13, 2008 #15
    Doc Al, I don't.
    It is now clear I don't even know what you mean by the phrase "I am at rest with respect to myself".
    Please explain.
     
  17. May 13, 2008 #16
    Hello Chrisc.

    From #10 by Histspec.

    ------------------------------------------------------------------------------------

    To define the inertial reference frames, let us state: A is resting in M1; B is resting in M2; C is resting in M3.

    a) Using rods and clocks which are at rest in M3 we measure that the speed of C is 0c, that the speed of A is -0.75c, and that the speed of B is 0.75c. Therefore in M3 the relative speed between A and B is 1.5c. (Of course, 1.5c is not a signal velocity and therefore there is no violation of special relativity).

    b) Using rods and clocks which are at rest in M1 we measure that the speed of C is 0.75c, that the speed of A is 0c, and that the speed of B (using relativistic velocity addition) is 0.96c. Therefore in M1 the relative speed between A and B is 0.96c.

    c) Using rods and clocks which are at rest in M2 we measure that that the speed of C is -0.75c, that the speed of A (using relativistic velocity addition) is -0.96c, and that the speed of B is 0c. Therefore in M2 the relative speed between A and B is 0.96c.

    That's all.

    --------------------------------------------------------------------------------------

    I don't think you could wish for a more logically presented explanation.

    Matheinste.
     
  18. May 13, 2008 #17

    Doc Al

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    You don't what?
     
  19. May 13, 2008 #18
    I do not "think I can have non-zero speed with respect to myself."
    As I said, I don't know what you mean by this.
     
  20. May 13, 2008 #19
    matheinste, I think you are missing the point.
    You and Histspec begin your "logic" with:
    "a) Using rods and clocks which are at rest in M3 we measure that the speed of C is 0c."
    With respect to what?
     
  21. May 13, 2008 #20
    Hello Chrisc.

    Are there more than one of you ?

    Matheinste.
     
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