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Light sphere question

  1. Nov 27, 2009 #1
    Assume there is a stationary observer O and a moving observer O' at v in collinear relative motion.

    When origined with O, O' emits a spherical light pulse.

    Now, the light pulse is described by O as x^2 + y^2 + z^2 = (ct)^2 and by O' as x'^2 + y^2 + z^2= (ct')^2.

    By considering only the x-axis points of the light sphere, it is the case that x'^2= (ct')^2. Thus, ct' = ± x'.

    Here is my question. What are the equations strictly from the coordinates and proper time of O to describe ct' = ± x'. This means, what are the x points in O and what are the times in O for the light sphere of O'.
     
  2. jcsd
  3. Nov 27, 2009 #2

    Dale

    Staff: Mentor

    ct = ± x
     
  4. Nov 27, 2009 #3
    This violates R of S.
     
  5. Nov 27, 2009 #4

    Dale

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    No it doesn't
     
  6. Nov 27, 2009 #5
    How descriptive. Perhaps you are right.

    The light sphere in O' is located at ct' = ± x'.

    t' = ( t - vx/c^2 )λ

    x' = ( x - vt )λ

    The only way your equation would work for the light sphere of O' is if v = 0.
     
  7. Nov 27, 2009 #6

    atyy

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    Substitute x=ct into your equation for t'

    t' = (t - vct/c2
    t' = λ(1-v/c)t

    Solve for t

    t = t'/(λ(1-v/c))

    Substitute x=ct into your equation for x', followed by substituting above equation for t

    x' = (ct - vt)λ
    x' = λ(c-v)t
    x' = λ(c-v)t'/(λ(1-v/c))
    x' = (c-v)t'/(1-v/c)
    x' = c(1-v/c)t'/(1-v/c)
    x' = ct'
     
  8. Nov 27, 2009 #7
    Solve for t

    t = t'/(λ(1-v/c))


    How can you do this?

    Events simultaneous for O' will not be simultaneous for O.

    Thus, there wil be a t1 and t2 in O, not a common t.
     
  9. Nov 27, 2009 #8

    atyy

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    Simultaneous means there are two events at two different spatial coordinates at the same coordinate time in one frame.

    But there's only one event here - described as the photon being at (x,t) in one frame, oe equivalently as the photon being at (x',t') in another frame.

    That's what your equation t' = (t - vx/c2)λ means. I just plugged and chugged.
     
    Last edited: Nov 27, 2009
  10. Nov 27, 2009 #9
    I just wrote the transformation equations down.

    You cannot use the same t without violating the R of S.

    Keep in mind, O' is moving.

    Also, there are two events, ± x'.

    Also, if the two events are simultaneous in O and O', I can run this into a contradiction of R of S with collinear relative motion.

    Something is wrong here.

    Since O' emitted the light, there exists points that are equidistant in O' and occur at the same time in O' but are not simultaneous on O.

    These are the points I am looking for.
     
  11. Nov 27, 2009 #10

    atyy

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    Event A is (xa=ct, ta=t)
    Event B is (xb=-ct, tb=t)

    ta=tb=t, so event A and B are simultaneous in O.

    Use your equations to find ta' and tb'.
     
    Last edited: Nov 27, 2009
  12. Nov 27, 2009 #11

    Dale

    Staff: Mentor

    cfrogue, atyy already presented some pretty convincing math, but let me try two other ways:

    1) Lorentz transform approach:

    We have the standard form of the Lorentz transform
    1a) t' = ( t - vx/c^2 )γ
    1b) x' = ( x - vt )γ

    And we have any arbitrary equation in the primed frame
    1c) x' = ct'

    To obtain the corresponding equation in the unprimed frame we simply substitute 1a) and 1b) into 1c)

    1d) ( x - vt )γ = c(( t - vx/c^2 )γ)

    Which simplifies to
    1e) x = ct


    2) First principles approach:

    We know that the second postulate is that c is the same in all reference frames, so we can immediately write that the speed of the light pulse is c. This implies
    2a) x = ct + B

    Since we know that the origins coincided with the flash we know that x=0 and t=0 is a point on the light pulse, so we can use that to solve for B
    2b) 0 = c0 + B
    2c) B = 0

    Substituting 2c into 2a gives
    2d) x = ct

    Note that approach 1 is a general approach that will work for any equation that you care to write. Approach 2 is specific to this problem since we are dealing with light pulses and will not work in general. I would typically recommend approach 1.
     
  13. Nov 27, 2009 #12
    Event A is (xa=ct, ta=t)
    Event B is (xb=-ct, tb=t)


    This assumes the events in O will be simultaneous with the events in O' given ta = t = tb.

    You should already know events like this cannot be simultaneous in both frames with v ≠ 0.

    So we see that we cannot attach any absolute signification to the concept of simultaneity, but that two events which, viewed from a system of co-ordinates, are simultaneous, can no longer be looked upon as simultaneous events when envisaged from a system which is in motion relatively to that system.
    http://www.fourmilab.ch/etexts/einstein/specrel/www/
     
  14. Nov 27, 2009 #13

    atyy

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    What? OK, I don't understand your scenario. What events are simultaneous in which frame?
     
  15. Nov 27, 2009 #14
    You do not have vt as the origin of the light sphere in O' with this approach. O' is moving with relative motion v. Recall, the light postulate proclaims light emits from the emission point in the frame spherically in all directions. Thus, since O' emitted the light pulse, these conditions must be met. We must have simultaneity in O' that is not simultaneous in O.

    I have worked on this for days and cannot make it happen.

    I will take you down the road of my failures.

    This math is not convincing and is not correct.
     
  16. Nov 27, 2009 #15
    I am clearer in my last post.

    I cannot do the math and have tried.

    I get junk I keep throwing it out.
     
  17. Nov 27, 2009 #16

    Dale

    Staff: Mentor

    Please post your work, I am sure I can point out the place you went wrong.
    I accept that it is not convincing to you, but it is correct. It might help if you identified which equation you think is wrong, why you think it is wrong, and what you think it should be instead. That might be easier than posting your work if there is a lot of material.
     
  18. Nov 27, 2009 #17
    No this ain't about me, it is about your equations.

    1d) ( x - vt )γ = c(( t - vx/c^2 )γ)

    You cannot use t in the context of O.

    There are two t's t1 and t2 since,

    So we see that we cannot attach any absolute signification to the concept of simultaneity, but that two events which, viewed from a system of co-ordinates, are simultaneous, can no longer be looked upon as simultaneous events when envisaged from a system which is in motion relatively to that system.

    http://www.fourmilab.ch/etexts/einstein/specrel/www/

    I have the following


    x = (v( t1 + t2))/2

    for the x point and all this is useless.

    Remember, if we fail to realize that what is simultaneous in O' is NOT simultaneous in O, then we are committing a serious act of crack pottery.
     
  19. Nov 27, 2009 #18

    Dale

    Staff: Mentor

    OK, so what do you think is wrong about 1d)? It is simply the substitution of 1a) and 1b) into 1c). You gave 1c) so I assume that either you think I wrote the Lorentz transform wrong in 1a) and 1b) or you think that for some reason substitution is no longer a valid algebraic operation in relativity. If you think that I wrote it wrong, then what do you think is the right formula for the Lorentz transform, and if you think that substitution is no longer a valid algebraic operation can you express why you believe that?
     
  20. Nov 27, 2009 #19

    DrGreg

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    There is no need to worry about the simultaneity, or not, of A and B. Just consider A on its own and do the transformation.

    Ignore B and consider A. Show that when xa = cta then x'a = ct'a. (The details of this have already been given in several posts.) Do you disagree with this?

    Then, as an entirely separate argument, ignore A and consider B. Show that when xb = −ctb then x'b = −ct'b.
     
  21. Nov 27, 2009 #20

    Let me think about this a while.

    I see some sense here.
     
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