Invariant spacetime and time dilation

Discussion in 'Physics & Math' started by Confused2, Jun 11, 2020.

  1. Confused2 Registered Senior Member

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    607
    I think this may be causing some confusion.
    Below I have attempted to draw the same clock seen either in the same frame as the clock or moving past it at velocity -v. I don't find it easy to draw diagrams - it isn't perfect so please forgive the imperfections.

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    In the A frame photon Alice has reached the detector. In the B frame (after the same length of time ct) Alice has reached the point marked Colin .. so Colin is the Alice photon. If v=-0.5 c then I think we are all agreed that Alice is only 0.866 of the way to the detector - hence in the B (moving) frame the same clock ticks at 0.866 times the rate at which the clock ticks in frame A.
     
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  3. Neddy Bate Valued Senior Member

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    If that diagram shows one single clock, then the event marked Alice and the event marked Derek are the same event, both of which occur when the one single clock has elapsed 1.000 tick. I don't understand why you are assigning human names to locations and photons, but okay. There is little point in calling the photon Alice when you could just call it the photon, but okay.

    On your left diagram, you need to mark a point which is 86.6% of the way to the detector, and call it Colin. There was an earlier time in that frame, when the photo was located there, and we could safely say that the clock had only elapsed 0.866 ticks at that time. Well, that is the same event that you show in your right diagram, called Colin, where that same clock has only elapsed 0.866 ticks at that time.
     
    Last edited: Jul 14, 2020
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  5. Neddy Bate Valued Senior Member

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    Sorry but, ct' does not become ct. The variable t is assigned to the time of some event in the UNPRIMED frame, and the variable t' is assigned to the time of a certain event in the PRIMED frame. Your diagram B is the frame in which the clock is moving (UNPRIMED FRAME), and your diagram C is the frame in which that same clock is stationary (PRIMED FRAME).
     
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  7. Confused2 Registered Senior Member

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    Alice? Not all of my ideas are good - probably not even the best ones. What I think we should be looking for is the same event in both frames - when Alice the photon reaches the detector we have a tick. The point of interest about the clock observed by a moving observer is "Does does a clock viewed by an observer moving relative to it tick at the same rate as the clock in its own frame?" We, well I, am looking for a tick which is the time taken for the photon 'Alice' to reach the detector in each frame.

    I disagree. After time t (having travelled distance ct) in the clock frame Alice has reached the detector. The moving observer sees that after time ct the photon Alice has only got as far as Colin and has not reached the detector. The moving observer has a particular interest in how long it takes the clock he is observing to 'tick' which is when the photon Alice reaches the detector. We could mark 0.866 tick seen by the moving observer as 0.75 tick in the clock frame but without (I think) any particular physical significance. As things stand we have an interim report by Colin that clock viewed by an observer moving past at 0.5c is 0.866 of the way to ticking and unless anything goes wrong it will be tick after ct/0.866 or (obviously) slower than the same clock viewed in its own frame.
     
    Last edited: Jul 14, 2020
  8. Neddy Bate Valued Senior Member

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    Okay, so when the photon (Alice) reaches the detector, that clock registers a tick. Let there be a digital display (Derek) on top of the clock which is connected to the detector at the top, and which displays the number of ticks detected. When the photon is emitted, the display at the top of the clock displays a clock time of 0.000. When the photon is detected, the display at the top displays a clock time of 1.000.

    Hopefully you will agree that we could also put another detector (Colin) inside the clock at 86.6% of the way to the first detector, and have that detector connected to another digital display inside the clock which would display a clock time of 0.866 when the photon (Alice) passes by the inside detector (Colin).

    Why are you using the same variable, t, in both frames? Let t' be the time displayed on the clock's digital display. You only allowed one clock, remember? So when the photon (Alice) has reached the Colin the clock displays the clock time t'=0.866. And when the photon (Alice) has reached Derek, the clock displays the clock time t'=1.000.

    Except you only allowed one clock, so a time of 1.000 represents the time when Alice reaches Derek.

    I guess you are saying that it is obvious to you that the time in the "moving frame" (unfortunate name for a frame) would be t=1.000 at the time when your one and only clock displays t'=0.866 and Alice is next to Colin. If so, that is fine, but it would have been nice if you had a second clock at rest in the "moving frame" which t could represent.
     
    Last edited: Jul 15, 2020
  9. Confused2 Registered Senior Member

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    607
    I'm starting to like that idea.
    You copied Phyti's diagram - I copied your diagram. No t'. I think we have to assume the observer has his own clock (a sophisticated wristwatch) so he can time t (a tick) in the clock frame and without difficulty also see what happens after the same time t as the same observer moves past the same clock at velocity -0.5c. I think it has been shown to be unhelpful to compare a stationary light clock with a moving light clock. I agree some clarification of 'wristwatch time' is in order. I have tried to be very clear about the observer moving past the clock (no 'moving frame') - which was intended to stress that moving past a clock won't affect the clock in any way.

    Once you go back to the idea of having a clock in a 'moving frame' you are in the realms of finding out what effect movement has on the clock rather than the effect of motion relative to the clock.

    phyti - are we agreed that motion has no effect on a light clock? What we do see is that motion relative to the clock frame causes the observed effects?
     
  10. phyti Registered Senior Member

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    492
    NeddyBate;

    Yes they do.
    Sue is the static twin, Max is the moving twin.We can make it more realistic with a smooth transition for the reversal, as shown in the static twin frame on the left. But this replaces one problem with another.
    The axis of simultaneity (aos) is established with the clock synch method of SR. All measurements are made with round trip light signals. When Max starts the transition at t=3, his inertial motion ends, and the aos does not apply. The curvature alters the time of return for the signals.
    Sue's aos is constant, parallel to the x axis, and the frequency of Max's clock is 1/2 of hers, except from t=6 to t=10, which corresponds to the curve in Max's path.
    As Max decelerates to 0 in Sue's frame, he gains .5t.
    In the moving frame on the right, Max's aos is horizontal parallel to his x axis. The frequency of Sue's clock is 1/2 of his from 0 to 1.6 then increases to 4.5, 1/2 way in the trip. The 2nd half of the trip is a mirror image of the 1st half.
    The acceleration for Max from t=3 to t=6, produces an equivalent g-field, and distorts the perceived path of Sue.

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    Last edited: Jul 16, 2020
  11. phyti Registered Senior Member

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    NeddyBate;

    The left diagram shows simultaneous measurements without length contraction are not reciprocal.
    Length contraction is a real/physical requirement for reciprocity.
    The length being measured has to be inside the length used as a standard when they pass each other.
     
  12. phyti Registered Senior Member

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    492
    NeddyBate;

    For gamma =2, v=.866
    Each clock sends a light signal (blue) for each tick.
    Home receives 4 to her 15, and 4 to her 1. Total 8 in 16.
    Traveler receives 1 to his 4, and 15 to his 4. Total 16 in 8.
    They observed different clock rates while apart, but age depends on accumulated time.
    Inspecting both clocks at reunion both agree, the traveler clock and traveler accumulated less time, and are younger than the home clock, and home.
    The clocks are identical and run independently of each other. The clocks could have been sent in probes without observers.
    There is no need of observation, counting, etc. Just record the times at the reunion.
     
  13. Neddy Bate Valued Senior Member

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    2,121
    phyti;

    What problem is replacing what problem? I don't see any problems.

    What does any of this have to do with your bizarre claim that my office is length contracted in my own reference frame as I sit in it, but that I am just unable to detect it because my tape measure is also length contracted?
     
  14. phyti Registered Senior Member

    Messages:
    492
    NeddyBate;

    For gamma =2, v=.866
    Each clock sends a light signal (blue) for each tick.
    Home receives 4 to her 15, and 4 to her 1. Total 8 in 16.
    Traveler receives 1 to his 4, and 15 to his 4. Total 16 in 8.
    They observed different clock rates while apart, but age depends on accumulated time.
    Inspecting both clocks at reunion both agree, the traveler clock and traveler accumulated less time, and are younger than the home clock, and home.
    The clocks are identical and run independently of each other. The clocks could have been sent in probes without observers.
    There is no need of observation, counting, etc. Just record the times at the reunion.

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  15. Neddy Bate Valued Senior Member

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    phyti;

    What does any of this have to do with your bizarre claim that my office is length contracted in my own reference frame as I sit in it, but that I am just unable to detect it because my tape measure is also length contracted?

    Considering that length contraction is the result of relative motion, and considering that there is no relative motion between myself and my office, don't you agree that there should be no length contraction in my own reference frame?
     
  16. phyti Registered Senior Member

    Messages:
    492
    NeddyBate;

    For gamma =2, v=.866
    Each clock sends a light signal (blue) for each tick.
    Home receives 4 to her 15, and 4 to her 1. Total 8 in 16.
    Traveler receives 1 to his 4, and 15 to his 4. Total 16 in 8.
    They observed different clock rates while apart, but age depends on accumulated time.
    Inspecting both clocks at reunion both agree, the traveler clock and traveler accumulated less time, and are younger than the home clock, and home.
    The clocks are identical and run independently of each other. The clocks could have been sent in probes without observers.
    There is no need of observation, counting, etc. Just record the times at the reunion.

    View attachment 3546
     
  17. Confused2 Registered Senior Member

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    607
    I do think the basic idea in Phyti's diagrams is brilliant - but maybe needs tweaking to make it more comprehensible.

    Edit - as timeout from the current discussion..
    If the light ray is vertical (on the y axis) in the clock frame - does an observer travelling past the clock (motion on the x axis) also see the ray travelling vertically?
     
    Last edited: Jul 16, 2020
  18. Neddy Bate Valued Senior Member

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    2,121
    No. The observer traveling past the clock sees the clock itself moving relative to him, so the light ray travels the same diagonal path as if it were the clock moving. In fact, the clock is moving in that observer's reference frame. All motion at constant velocity is relative. Put all your thought experiments in deep space, far away from any planets etc, and you will start to see how it makes no sense to call the reference frames "stationary frame" and "moving frame." The frames are supposed to be called something like K and K' where frame K uses the coordinates (x, y, z, t) and frame K' uses the coordinates (x', y', z', t'). Then each frame can consider itself stationary and the other moving. It is completely reciprocal.

    Edit: In the usual twin scenario, it's okay to call the twins "stay home" and "traveling" because the traveling twin actually does an acceleration at the turnaround point, and therefore the entire situation is not completely reciprocal. However, during each inertial leg of the journey everything is reciprocal, as far as clock tick rates are concerned.
     
    Last edited: Jul 16, 2020
  19. Confused2 Registered Senior Member

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    607
    Hi Neddy,

    Many thanks for that explanation.

    Hopefully it has become clear that my ultimate aim is to understand 'everything' about the elementary consequences of SR - your help (or anyone's) is greatly appreciated.

    I may be overthinking this.

    Let us imagine we are travelling at velocity v along a railway track with sleepers (track separators, whatever) arranged like so:

    IIIIIIIIIIIIII

    For coordinate convenience y is upwards and the track separation is Y.
    We (the observer) are on the lower track. The upper track is a distance Y away. The upper track lags in time (relative to us) by T=c/Y (?). When the train started (an impulse) the upper track started T=c/Y later from our viewpoint and is 'instaneously' x=vT behind us. If we fire a pulse (event A) vertically we hit the other track after t=c/Y which is just when the other track arrives at the same x position as our track when we fired the pulse. In our frame the 'light clock' works perfectly.

    Now for a diagram of the path of the pulse in the track frame.
    As far as I can make out the path of the pulse (the angle) is the result of drawing from an expedient location half way between detector and emitter. If we drew the same diagram from the location of the detector, the emitter now lags - and I'm not sure what the diagram would look like.

    Any thoughts welcome.
     
  20. phyti Registered Senior Member

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    492
    Anyone interested;

    Ann and Bob have identical clocks.
    There are 4 instances of Ann observing Bob's clock, for 1 tick on her clock, while beside her. A tick is an arbitrarily defined unit of time. The vertical progress of the signal defines the time within the light clock.

    1. Bob's v=0, blue signal moves vertically 1 tick.
    2. Bob's v=.3, blue signal moves vertically .95 tick.
    3. Bob's v=.6, blue signal moves vertically .80 tick.
    4. Bob's v=.9, blue signal moves vertically .44 tick.

    Ann observes Bob's clock running slower than hers as a function of v.

    t'=t[sqrt(1-v^2)]

    where t is time on Ann's clock,
    t' is time on Bob's clock,
    v is speed as a fraction of c,
    and c=1 as a standard.

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  21. Neddy Bate Valued Senior Member

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    There is no need to worry about lag times. You are setting up the thought experiment, so you can establish where event A is located, (I assume it would be at the bottom of one of your "I" shapes. And you can likewise establish where event B is located (I assume that would be the event where the light reflects off the top of the next "I" shape), and you can likewise establish where event C is located (I assume that would be the event where the light arrives back at the laser that fired the pulse at the bottom of the next "I" shape.

    So the drawing in the track frame is easy to draw, it goes diagonally up and down like this:
    I/I\I/I\I/I\I/I\I/I\I/I\I/I\I/I\I/I\I/I\I/I\I/I\

    So once again one frame has the vertical up-and-down light path, where the clock ticks at the standard rate, and once again the other frame has the diagonal up-and-down zig-zag light path, where the clock ticks at the slower, time-dilated rate. All of these thought experiments are giving us the same result, it is time (no pun intended) to just accept that the constant speed of light in both inertial frames results in time dilation (for moving clocks, not stationary ones).

    Yes, I think you are.
     
  22. Confused2 Registered Senior Member

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    607
    I'm interested. Let us imagine that the v=0 light clock is based on a single short pulse fired by a narrow beam laser. The time for the pulse to reach the detector is (we define) ct=1. There is no expanding sphere of light - only that single pulse from a narrow beam laser aimed precisely at the detector. With relative velocity > 0 you/we/I draw the path of that narrow pulse (fired vertically by the source) as no longer vertical (call it angle alpha) - can you explain the origin of that angle alpha? If we were to bounce the pulse off a mirror (where angle of reflection = angle of incidence) in the moving frame the detector would be long gone and not detect the pulse and the light clock would fail for a moving observer. Any thoughts (anyone?).
     
  23. Confused2 Registered Senior Member

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    607
    No, the same I shape. Pretty much the same question as to Phyti in the preceding post.
     

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