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SR Simultaneous Lines Drawn in the Sand

  1. Jun 8, 2004 #1
    The Einstein Train Gedunken

    The observer O’ in the moving frame at M’ arrives at M, the midpoint of A and B in the stationary frame, at t0’ when A and B emit photons.

    Later a photon from B is detected when M’ is at t1’; the photon from A is detected when M’ is at t2’ in the moving frame. For convenience the velocity of the moving frame is v = 1 and dt = t2' – t1'.
    ||||||||||||---------------------------M’------------------------------------||||||||||||||| -> motion


    There are added sections of photo-sensitive strips ||||| such that A and B afre guaranteed to be within a section length when the photons are emitted at A and B. The midpoint of the photo-sensitive strips was determined using the same techniques used to determine M the midpoint of A and B. The strips are numbered starting from the inside positions and then consecutively to the ends of the sections. Each equally numbered pair of photo-sensitive strips have a common midpoint at M’. The resolution of the strips is in the sub-micron range.

    As photons are emitted at A and B the photo-sensitive strips located within one photon wave length of A and B, or less, are exposed.

    The postulates of special relativity theory state that the laws of physics and the measure of the constancy of the speed of light are invariant in all inertial frames. From special relativity theory observers in the moving frame conclude the events of the emitted photons were not simultaneous in the moving frame.

    Are the emitted photon events that are simultaneous in the stationary frame simultaneous in the moving frame?

    1.Comments on experimental arrangements or conditions are gratefully accepted.

    2.Comments on explicit or implicit stipulations and/or assumptions are gratefully accepted.

    3. Other comments..
  2. jcsd
  3. Jun 8, 2004 #2

    Doc Al

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    Einstein's argument

    I'll rephrase this a bit. In the stationary frame (O), M is the midpoint of two light sources A and B. Lights A and B are switched on at the exact moment that a moving observer O' passes M as observed in the stationary frame. All observations in the stationary frame confirm that the lights were turned on at the same time. For simplicity, let's call that time t = 0.

    Since the entire issue is whether or not observers in the moving frame will agree with the stationary frame's assessment that the lights were turned on simultaneously, we must take care to make no assumptions about simultaneity in the moving frame.

    Einstein's argument is simple. Rather than add things (like length measurements) which may only serve to confuse the issue, let's deal with Einstein's argument itself (as I recall it).

    Einstein reasons thusly:

    (1) If an observer is exactly between two lights when they emit pulses simultaneously (according to the observer), then
    (2) the pulses from both sources will be detected (received) by the observer simultaneously.

    Do you not agree then, that if statement (1) is true, that statement (2) must follow?

    Let us also stipulate, per Einstein, that since the speed of light is an invariant in any frame, this argument applies regardless of the speed of the observer. It equally applies to the stationary observer or to the moving observer.

    There is no question that for the stationary observer both statements are satisfied. But what can we conclude about the moving observer?

    Einstein argues that, from the viewpoint of the stationary observers, O' is moving towards B and away from A. Thus light from B must reach O' before light from A. Do you agree with this? Do you agree that every observer (in any frame) must agree that the light from A and B reached O' at different times?

    If you agree with this conclusion, which follows from the invariance of the speed of light, then you agree that statement (2) of Einstein's argument is denied.

    Since Einstein's argument is of the form:
    If A, then B.
    not B,
    thus not A.​
    We are forced to conclude that statement (1) is not true! Thus O' must conclude that the lights were switched on at different times (according to his observations).

    That's Einstein's gedanken experiment. Can you point out an error in that argument?
    Not sure what you are measuring with these photo-sensitive strips.
  4. Jun 8, 2004 #3

    before a photon even hits the moving observer he's already moved to t'1 (b photon). then he's already moved to t'2 before the second one hits him (a photon)

    all motion aside, put an observer at t'0 t'1 and t'2 and which observer does the photon from B hit first?

    the one from t'2 of course.

    it's quite rudimentary, i am not understanding the confusion behind all this
  5. Jun 8, 2004 #4
    are you sure that's what he's reasoning? because he would have to assume light travels instantaneously in order for that to be true.

    as another observation, if a person arrives AT the midpoint M at the exact time to be hit by photons from A and B, NO MATTER what direction speed acceleration, we can safely assume the photons were emitted from the sources simultaneously, even if they were not calibrated to do so.

    but this is simply an observation of speeds and distance midpoints known by any elementary school student
  6. Jun 8, 2004 #5

    Doc Al

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    Obviously the pulses are emitted and detected at different times. :smile:
    Not true. If you get hit by two bullets simultaneously, can you conclude that they were fired simultaneously? Of course not: it depends on where they were when they were fired.
    Time to move to a better school district.
  7. Jun 8, 2004 #6
    if the bullets ALWAYS travelled the same speed AND i was at the MIDPOINT distance between the shooters WHEN HIT.

    yes, yes i could
  8. Jun 8, 2004 #7

    Doc Al

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    Nope. It doesn't matter where the shooters are when you are hit, it only matters where they were when they pulled the trigger.

    Think about it. By the time the bullets reach you, the shooters have moved.
  9. Jun 8, 2004 #8
    the shooters don't move. they're stationary, the observer is the only thing moving in my example

    sorry if you were confused by that
  10. Jun 8, 2004 #9
    This is the same as the other two "SR questions of the century" that have been posted earlier on this forum. I'll admit I didn't read those all the way through cuz they were so long! But I'll take a shot at this one.

    Using M as the point of origin (so that AM = MB = AB/2, and A = -B), at the point when observer O reaches t1, the photon from B will also be at t1 (by definition) but the photon from A will be at -t1. Given the same duration of local time, both photons will have travelled an equal distance regardless of the frame of reference. At the point when observer O and photon A reach t2, photon B will be at -t2 (the photons have both passed M).

    #2 is true only if the observer is exactly between the lights when the photons are DETECTED. It does not matter where the observer is when they are emitted. In fact, the observer does not even need to exist when they are emitted. Look at the stars tonight. Consider how long it has been since the light you are seeing has been emitted. Have you really been alive that long? The events you are seeing are not occuring, they have already occured in the past.

    Time is local, so simultaneity is a matter of your frame of reference.
  11. Jun 8, 2004 #10
    exactly, which was the point i was trying to make, but i left out that the emitters do not move <woops>
  12. Jun 8, 2004 #11
    Have you ever looked at the so-called 'pole-vaulter paradox'? That's an interesting one as it shows you explicitely how according to SR you must lose simultaneity or face paradox.

  13. Jun 8, 2004 #12

    Doc Al

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    Incorrect. In fact, exactly the opposite is true. I don't care where the lights are when the photons are detected--they need not even exist by that time. :smile:
    It matters if you wish to make a deduction about whether the pulses are detected simultaneously. Which is the entire point.
    The thought experiment assumes that both observers have existed forever traveling at the same speed. The point is that statement (2) follows from statement (1). That's all.

    True, but I don't see the relevance to Einstein's argument.
    Now that I agree with! :smile:
  14. Jun 8, 2004 #13
    My apologies for not being clear. What I meant by the light, is A and B. Not the physical light, you're right, it could be removed once the photons have been emitted, it does not matter.

    My point is that in order for the photons to arrive at the observer at the same time, the observer must be equidistant from both originating locations at the point in time when both photons are observed. It does not matter where the observer is or is not when the photons are emitted. THAT was my point.

    Please don't waste your time by further tearing apart minor details of my responses, and focus solely on the point I'm trying to prove in response to the point you've tried to make. If we lose focus, it will turn into another 4-page discussion and there will be another identical thread created in a few days discussing the same principal.
  15. Jun 8, 2004 #14

    Doc Al

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    From the observer's point of view, the shooters are moving. Regardless, all that matters for the current argument (in analogy with Einstein's) is where the shooters were (with respect to the observer) at the moment the guns were fired. If the observer (victim?) was exactly between the shooters at the time they fired, then we can deduce that the bullets will arrive simultaneosly. (These are special photon bullets, of course, that always travel the same speed with respect to all observers. :smile: )
  16. Jun 8, 2004 #15

    Doc Al

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    I believe I understood your point the first time. You're still wrong. The position of the lights at the moment the photons are detected is irrelevant.

    Please focus on the argument at hand: If statement (1) is true, then statement (2) is true.

    I am completely focused. The point of yours that I am "tearing apart" is not a minor detail.
  17. Jun 8, 2004 #16

    Doc Al is right.

    If two lights flash simultaneously (in my frame) and the distance from me to each one is the same at the time of the flashes, then light from the two flashes reaches me at the same time. That's what it means for light speed to be a constant with respect to all observers.

    Your bullet analogy lets you down because the bullets are traveling at constant speed with respect to the shooters, but if you are moving wrt the shooters, the bullets are moving at different speeds wrt you.

    If you want to get to the point where constant light speed and all its consequences seem more intuitive, stop thinking about bullets!
  18. Jun 8, 2004 #17
    that makes NO sense.

    A and B flash. observer M is in the middle WHEN they flash. He immediately accelerates to light speed in the direction of B.

    Photon from A never reaches him.

    Photon B is intercepted halfway to B
  19. Jun 8, 2004 #18

    Sounds like you have a real solid understanding of this theory! :wink:
  20. Jun 8, 2004 #19
    what part of photons travelling the speed of light is hard to understand?

    if you travel AWAY from a clock AT LIGHT SPEED and look to see what time it says what do you see?

    NOTHING. no new photons are hitting your eyes from that direction :grumpy:
  21. Jun 8, 2004 #20

    Tom Mattson

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    I think what jdavel is hinting at is that you can't accelerate to light speed.
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