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Has ayone seen the lost simultaneity?

  1. Jun 13, 2004 #1
    The figure is of a moving frame moving to the right at velocity = 1/3c. On each end of the frame is a mirror (L and R) located at 4(1/3 + 1/3 + 1/3 + 1/3) x (1 second) from the midpoint of the frame. A pulse of light emitted at M (on either the moving or stationary frame) . We define the stationary point M when the pulse is emitted.

    Code (Text):

     L |    |____|____|____M____|____|____|    |R
    1. In the first second the photon headed left will arrive at 1c to the left of M while the left mirror has moved to a point where the photon is located during the first second. The right photon is located 1c to the right of M at the end of 1 second..

    1 sec. L |<____________M___________> |R

    2. After 2 seconds the photon on the left has reflected back to M while the right photon has just reached the right mirror:

    2 sec. ____________>M ____________>|

    3. After 3 seconds the left photon is now located at the shifted M' (3)(1/3) units to the right of M. The right photon reflecting off the R mirror is also located (3)(1/3) units to the right of M.

    3 sec.

    As the frame has moved to the right (3)(1/3) units the photons meet at the position of the moving observer who, quite naturally, concludes the photons were emitted simiultaneously.

    We conducted the above analysis from the stationary frame. From symmetry considerations conducting the analysis in the moving frame gives the same result: The moving observer concludes the photons were emitted simultaneously in the moving frame.

    This you can easily demonstrate for yourself. Where is the loss of simultaneity? If any?
  2. jcsd
  3. Jun 14, 2004 #2
    neither your observer/emitter or reflectors are moving in relation to each other.

    net result = same as stationary frame
  4. Jun 14, 2004 #3
    The point I was making (among others) is that the emitter can be moving or stationary.

    Right: the reflectors are attached to the moving frame. The Left reflector moves to the oncoming photon heading Left. The right photon begins to chase the moving reflector to the Right. Watching from the stationary platform we see the photons finally meeting where the obseserver on the moving platofrm is then located.

    The moving observer will see the lights simultaneously meeting her, after the reflections, whatever the time/space dilations may be, or not be. Pick your frame, you always get simultaneous emissions of the photons.
    The moving frame contains L reflectror, the Observer midpoint to the reflectors and the Right reflectors. The photons are onblivious to the motion of the frame and only reflect when a mirror is present. :smile:
  5. Jun 14, 2004 #4
    yep yep, according to Einstein's relativity all inertial frames will act the same way

    and function under the same physics as something "supposedly" stationary, since he didn't define an absolute rest frame
  6. Jun 14, 2004 #5
    this is sweet i learn something in another thread and get to come here and talk like i know what's going on kekeke
  7. Jun 14, 2004 #6
    You have demonstrated that the observer will always measure light pulses meeting at the middle, regardless of whether or not the observer frame is moving.
    So if light is moving at c+v and c-v the results are similar to a frame in which light travels at c, as the losses and gains in the speed of light cancel out.
    This is effectively a two-way light speed test, and it is very difficult to record the fact that the speed of light could be different from c.
  8. Jun 14, 2004 #7
    You might recognize this as the cerulean etc example of another thread that 'proved' the moving observer would conclude the photons were not emitted simultaneously. If you look closely, you are able to use either frame as a basis of calculations. Where is the loss of simultaneity? Howver, the moving observer sees only the final arrival of both photons after some time dilation. The stationary observer in the stationary case sees the emitted to return time as 2 + 2x(1/3) = 8/3 sec. The stationary observer sees the return time in the moving frame as 9/3 sec.due to the movement of the point M in 3 sec time span.

    Inany event, whatever time is determined by the moving observer for the round trip of the reflected lights, thephotons always arrive simultaneously at the moving observer.

    If I am not mistaken, the Cerulean model had the moving observer concluding the photons were not emitted simultaneously because the photon first hit the left reflector then the right reflector. The moving observer cannot make this claim as he cannot see when the photons struck the left and right mirrors. He has to wait, just like the stationary case, until the photons arrive at the midpoint M.

    I am not sure what you mean by the losses and gains in the speed of light. Each photon simply travels the same distance in the same of amount of time. Again, the cerulen example concluded othewise.

    Wespe quoting Paul Newman in movie, 'The Color of Money', "I'm back."
  9. Jun 14, 2004 #8


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    yes of course both the observer moving relative to what you have defined as the 'stationey frame' and the observer in the stationery frame agree that both photons were emitted simulataneously.

    this comes as no great suprise as simultaneity only fails at distance.
  10. Jun 14, 2004 #9
    I don't understand " . . .simultaneity only fails at distance". The example here was argued extensively where the moving observer was deemed to conclude the photons were not emitted simultaneously, by some, not me.. However, in the example discussed the lack of simultaneity was based, as far as I was able to determine, on the fact that the left mirror met the photon before the right mirror received a photon. This was the "physical state" that indictated the moving observer would conclude the photons were not emitted simultaneously. The analysis did not extend to to the expanded reflection scenario that I described. I could never understand how the moving observer could observethe photons before they reached her, do you know what I mean?

    So again, what is the distance condition you refer to?
  11. Jun 15, 2004 #10


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    Two 'events' (really they are the same event) having the same spatial and temporal location will appear to be simultaneous to all observers.
  12. Jun 19, 2004 #11
    Could you please elaborate on the meaning o "distance". Thanx
  13. Jun 19, 2004 #12

    Doc Al

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    the relativity of simultaneity

    Disagreements about the simultaneity of two events only happen when those events take place at different places (that is, when they are separated by a distance).

    For example: Let's say you and I bang our heads together. Since that "event" takes place at the same place, every frame will agree that we hit our heads together "at the same time".

    But if instead, you and I are miles (light years?) apart and we decide to slap ourselves in the head at the same time (how? we check our synchronized atomic watches!). In this case, observers moving with respect to us will disagree that we slapped ourselves at the same time. Some will say you slapped yourself before I did; others will say the opposite.
  14. Jun 19, 2004 #13
    Thanx for clearing that up. I make the semi-educated guess that the events at our A and B locations are a "distance" problem. However, what of equating the emission of the photons at A in the stationary frame at the same time the photons were emitted at A' in the moving frame, and simularly for the photons emitted at B and B', and M and M'. The photons are measured simultaneously in the moving frame at the thee positions indicated.

    The question here: Are the photons emitted from A' and B' simultaneously in the moving frame from symmety considerations.
  15. Jun 19, 2004 #14

    Doc Al

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    I'm not sure what you are talking about. You must describe the precise situation you have in mind. (Are you mixing this thread with another?)

    If A and B are the locations where photons are emitted simultaneously in the stationary frame (t = 0), and A' and B' are the observed locations of those emissions in the moving frame, then no, A' and B' are not simultaneous: the moving frame will measure those events (the photon emissions) to be at different times. (For details, see: https://www.physicsforums.com/showpost.php?p=231002&postcount=115)
  16. Jun 21, 2004 #15
    Doc Al, I have agreed that SR will predict whatever it predicts. The following does not change any of the parameters we have been using in this thread.

    All moving frame values are non-primed with the exception of M’, the consistent location of the observer O in the moving frame.

    At no time is there an inference that M’ was at the midpoint of the A and B photons emitted in the stationary frame.

    To demonstrate the following:

    Einstein’s moving train calculation indicating when the oncoming B photon is detected at t1 the A photon was located at a position consistent with –t1. Said in other words, as t1 is determined from t0 which locates M’ at t0, the A and B were equidistant to M’(t0) when t = t1.

    A moving observer located at M’ on a moving frame passes through the midpoint M of photon sources located at A and B in the stationary frame just as A and B emit photons. M’ is moving along a line connecting A and B, toward B.

    At this instant the moving source t = t0. Later the moving observer detects the photon from B at t1, and later the photon from A at t2. The observer has measured her velocity wrt the stationary frame as v. Determine the position of the A photon at tx in terms of t0, t1, t2, and v when the B photon was detected at t1.

    The photon from A must reach the position of M’ when t = t2. Therefore, the distance traveled by the A photon during Δt = t2 – t1, is Δtc. This is equal to the distance cΔt = vΔt + vt1 + vtx . Now we rearrange somewhat to arrive at, vtx = vΔt – cΔt + –vt1. Now as vΔt - cΔt is just -vtx - vt1

    vtx = -vtx - vt1 – vt1

    2tx = -2t1

    tx = -t1

    Therefore, in the moving frame the photon from A and the photon from B were equidistant from M’(t0) at t1.
  17. Jun 21, 2004 #16

    Doc Al

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    Sorry geistkiesel, but I think these discussions are pretty much a waste of time. I don't think you understand enough SR to have a meaningful discussion about it, never mind provide an intelligible critique.

    My offer remains: When you are ready to discuss Einstein's simple argument showing the relativity of simultaneity (no calculations to confuse you!) and you are prepared to point out the flaw in it, let me know.

    Right. As measured by the moving frame, the A photon travels a distance equal to (t2 - t1)c between photon detections by O'.
    No it's not. Where did you get this quaint relationship? Looks like someone's been swapping frames again!
  18. Jun 21, 2004 #17
    Doc Al, c(t2 -t1) is the distance A miust travel during t2 - t1 in the moving frame to arrive at M'(t2) when it does. v(t2 -t1) to get the distance the frame moves in t2 -t1. vt1 is the distance moved from the M'(t0) to M'(t1) and vTX is the distance A is from M'(t0), everything carefully restrained to the moving platform.

    Come on Doc Al, no one is swapping frames, that's my charge against you; I would appreciate it if you would get your own metaphors. If you thought someone was "swapping frames" , you wouldn't hesitate to point directly to the source, would you? Doc Al you are beginning to relapse into your old ways. Get a hold of yourself man. The world hasn't come to an end, yet. We may be close, but there is time left for plenty to do.
  19. Jun 22, 2004 #18

    Doc Al

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    That distance c(t2 - t1) is measured in the moving frame.
    But now you claim it's also the distance measured from the stationary frame?
    Gee... I hope you're not using times measured in the moving frame to do calculations in the stationary frame? :eek:

    Telegram to geistkiesel: Get a clue!

    Oh, and be sure to see my post in the other thread. I'll walk you through these calculations step by step. Don't worry... baby steps!
  20. Jun 22, 2004 #19
    Originally Posted by geistkiesel
    Doc Al, c(t2 -t1) is the distance A miust travel during t2 - t1 in the moving frame to arrive at M'(t2) when it does

    Right, that's what I said.

    v(t2 -t1) to get the distance the frame moves in t2 -t1.

    This is the distance calculated by the moving observer. And yes t0 has a stationary connection. She knows she is moving wrt stationary frame and this is what she calculates, using moving frame data. All she has to do is look out the window.

    Quote geistkiesel
    vt1 is the distance moved from the M'(t0) to M'(t1) and vTX is the distance A is from M'(t0), everything carefully restrained to the moving platform.

    Well Doc the A photon is in the stationary frame, anmd the moving frame as well. IS thee some jurisdictional problem we are javing her? All the calculaions two lines uop are all moving frame calculations. You got yourself confused when you tried to change the units and notation to confuse anyone following the thread, remember?

    What is this "I hope " stuff? You can see what I am doing. The calculations are in the moving frame. It just so happens the t0 is referenced to a stationary landmark. I can't help it if the gedunken was designed as it was. Talk to A if you have a complainl. If you want to claim that the moving values are going to be dilated then do it. Don't forget to squeeze time down to an appropriate value.

    Quote geistkiesel
    Come on Doc Al, no one is swapping frames,

    I told you all the parameters were in the moving frame and that is what I used. Are you angry at me for something? You seem rather piqued. Was it something I said?

    After that I'd be scared to read your post, . . . oh the horror, the horror . . .. Doc Al the 'slice and dice prof'', to baby step Geistkiesel through the throes of a dying theory. See the Final Chapters of, "Throw SR from the Frame".

    But you did that in the last post didn't you. You even stole the term "frame jumping " from me. Oh Doc Al how you do stress me out. I was going to coin the phrase about "frame jumping SR theorists" , but you stole my thunder Doc il what a dreadful day I am having.

    What clue? Tell me here and now. Embarrass me in front of my friends. Why threaten a dismal future. You might not have me to kick aorund anymore Doc Al, I may be turning professional; I caught a glimpse of thread where they were discussing something about, 'giving me the business'.
  21. Jun 22, 2004 #20

    Doc Al

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    v(t2 -t1) is the distance that the stationary frame is observed to move according to the moving frame. It's got to be a measurement made by the moving frame, since it uses times measured by the moving frame! It is not the distance that the moving frame moves with respect to the stationary frame.
    You are mixing measurements and frames left and right. Maybe this will help to clear your head:

    There are two events of interest in this problem. Event #1 is the detection of a B photon by moving observer O' (who is located at x' = 0). Event #2 is the detection of an A photon by moving observer O' (still located at x' = 0). You do realize that the two frames disagree as to when those events took place? You can't just mix and match measurements made in the different frames.
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