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B Moving clock as seen by an observer at rest

  1. Mar 27, 2016 #1
    Please forgive me if this has been asked before. I understand that a clock that is moving relative to an observer at rest will slow down.
    If the moving clock is in a transparent rocket, would the stationary observer on the ground see the moving clock slow down physically?
     
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  3. Mar 27, 2016 #2

    Nugatory

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    That depends on whether the moving clock is moving towards him or away from him. If the clock is moving towards him, then with every second that passes the clock will be closer, so the light that leaves the clock when it reads 12:00:01 has farther to travel to reach the observer's eyes than the light that leaves the clock when it reads 12:00:02. Thus the time interval between the arrival of the light from 12:00:01 and the arrival of the light from 12:00:02 will be less than the it would be if the clock were not moving - the clock appears fast. If the clock is moving away, it works the other way and the clock appears slow. (This is the relativistic Doppler effect - google will find much more about it).

    However, to decide what rate the clock is ticking at you have to subtract out the light travel time. If the clock is three light-seconds away, then the observerknows that the light spent three seconds in flight to get to his eyes, so if he sees it reading 12:08:13 he knows that was what it read three seconds ago. When we make that correction, we find that no matter what we see the moving clock is running slow - this is time dilation.
     
  4. Mar 27, 2016 #3

    phinds

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    Just to be sure you are clear about this, let's say that A and B have a velocity of some significant percent of c relative to each other. We choose a frame of reference in which A is at rest and all of the motion is assigned to B. To A, B's clock is running slow but to B his clock is running just fine at one second per second but A's clock is running slow. That is, time dilation never happens to YOU, it is something you perceive to be happening to someone who is moving relative to you.

    EDIT: my point is that your statement that the clock "runs slow" is not really correct. What is correct is to say it "runs slow in the frame of reference in which he is at rest (but does not run slow in the frame of reference in which it is at rest)".
     
  5. Mar 27, 2016 #4
    @samirgaliz
    Perhaps I can help you.
    You are on earth, and there's an astronout (male, so I don't have to use his/her) travels and sending signal to you.
    If your clock is at 00:00 you receive his signal read 00:00
    Your clock is 00:06, his signal read 00:12
    Yours is 00:12, his 00:24
    Yours His
    00:18 00:36
    00:24 00:48, etc...
    So the rate is 2, let's say k = 2.
    His proper time runs faster than yours??
    So he must be traveling toward you.
    Can you work out how fast he travels (speed = v)?
    Remember to calculate his proper time is yours * gamma
    Gamma is Lorentz Factor
    ##\gamma = \sqrt{1-v^2}##
    His distance is receding over time.
    So when his distance is at VT he sends you his clock reading
    ##t-vt = kt\sqrt{1-v^2}##
    ##1-v = k\sqrt{1-v^2}##
    ##(1-v)(1-v) = k^2(1-v)(1+v)##
    ##k^2 = \frac{(1-v)(1-v)}{(1-v)(1+v)}##
    ##k = \sqrt{\frac{(1-v)}{(1+v)}}##

    Now if I see
    ##k = \sqrt{\frac{(1-v)}{(1+v)}}##
    this looks like doppler equation?

    Solution
    ##v = -0.6##
    Hi moves toward you
     
  6. Mar 28, 2016 #5
    I interpret "The moving clock in a transparent rocket" is the clock fixed in a moving transparent rocket.
    You are turning your head and arm to keep watching the clock to compare with your watch.
    Pushing away tedious changing consumed times for light to travel from the clock to his eyes, the stationary observer on the ground see the moving clock slow down physically. Furthermore the pilot see the observer's watch slow down physically.
     
  7. Mar 28, 2016 #6

    phinds

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    And you believe that this physical observation that both clocks are seen to slow down holds when the two are moving towards each other at a substantial fraction of c?
     
  8. Mar 28, 2016 #7
    I believe so, the direction of motion does not matter. I would be glad if you put me on a right understanding.
     
  9. Mar 28, 2016 #8
    No need to observe a clock on a transparent rocket with a telescope or something.
    You could set a clock on the rocket to send a radio beep once every second, yet it would appear to be a slower rate as received by the stationary observer.
     
  10. Mar 28, 2016 #9

    phinds

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    Really? Even if the ship is traveling towards you at .5c? You think the beeps would be MORE than 1 second apart?
     
  11. Mar 28, 2016 #10

    phinds

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    See post #9
     
  12. Mar 28, 2016 #11
    I think what would happen is that the 'raw data' would show a higher frequency, but that data would be blue shifted.
    After manipulating the signal so that it corresponds to the known/designed radio frequency of the beep, it would look slower.
     
  13. Mar 28, 2016 #12

    phinds

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    But we were talking about exactly the situation where you do not do any manipulations, you just go with what you see (or hear in the case of radio beeps) --- see post #5. You are positing a different process so naturally you are getting a different answer. Let's answer one question at a time.
     
  14. Mar 28, 2016 #13
    Changing the question, if the moving clock is in a transparent rocket, would THE STATIONARY OBSERVERS on the ground see the moving clock slow down physically?
    The answer is Yes. Tom reported the rocket clocks was adjusted when it passed by him, Dick reports the rocket clock is delayed now it is passing by. Direction of motion does not matter with slowness.

    Back to the original question, Tom, the only observer, should watch rocket at least twice when the distances between him and the rocket differ. Travel time of light, i.e. he see here now is what happened such a place in such a past time, and Doppler effect affects his sight perception.
    samirgaliz, would you go this way?
     
  15. Mar 28, 2016 #14

    Nugatory

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    The stationary observers would see the moving clock running fast when it was moving towards them and running slow when it was moving away. Only after they corrected for the light travel time would they calculate (not "see", but "calculate") that the clock was runnings slow all along.
     
  16. Mar 28, 2016 #15
    You are right. I should have written explicitly that the observers can observe rocket only when it passes by. They do not observe ( or at least do not report about ) the distant rocket.

    To test my understanding, is there a moment that the clock in the approaching rocket, not in just adjacent nor leaving position, SEEMS to have the same pace with the observer's watch ? I suppose "Yes".
     
    Last edited: Mar 28, 2016
  17. Mar 29, 2016 #16

    jbriggs444

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    So you are considering the relativistic Doppler effect as the passing rocket approaches, passes nearby and then recedes. The moving clock is first seen to be ticking rapidly as it approaches and later seen to be ticking slowly as it recedes. You ask whether there is a point (or an angle) where the moving clock can be seen to be ticking at its normal rate. The answer is, of course, "yes". The tick rate will be seen to vary continuously throughout the exercise. If it starts fast and ends slow, there must be a point where it is ticking normally. This is the intermediate value theorem for continuous functions.
     
  18. Mar 29, 2016 #17
    Thanks for your interest. The point is it happens before the rocket reaches the observer, before the rocket is adjacent to him or pass over. How about it?
     
  19. Mar 29, 2016 #18

    jbriggs444

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    Taking the point of view of the observer, it is clear that the Doppler contribution to the frequency will be zero at the point of closest approach. Near that point the transit time for light from rocket to observer will be nearly unchanging. So as the rocket is seen to reach its closest approach, the rocket clock rate that is seen will only be affected by relativistic time dilation. It will be seen to tick slowly.

    Accordingly, the point where it is seen to tick normally must be prior to that.

    Edit: By the time this is seen to happen, the rocket will "currently" be somewhat farther along in its journey. But I do not think you are asking about that.
     
    Last edited: Mar 29, 2016
  20. Mar 29, 2016 #19
    Thanks again for your investigation including Edit. The observer sees that the clock that ticks normally has not reached the closest point to him. That would be an alternative statement.
     
  21. Mar 29, 2016 #20
    Thanks to everyone who clarified this. My assumption is when the observer on the ground observes the moving clock in the rocket at the closest point, ignoring approaching or departing. That is ignoring Doppler effect.
    So this brings me to my next dilemma!
    If the observer on the rocket measures the half life of radioactive material to be "to" using his clock. The observer at rest on the ground will also see the moving clock with the recording time "to" and thus sees the radioactive material decays at half life "to". But we also know that the observer on the ground will measure the half life of the moving radioactive material as measured by his clock to be t >to.

    So it seems like there is a contradiction for the observer on the ground! He will think that his clock is wrong since physically he observes the radioactive material decays in a time "to"!!
    Maybe I am going into a circular argument. My apologies!!!
     
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