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Clocks when there is speed.

  1. Sep 30, 2013 #1
    Lets say that we have a spaship(with a man) that has speed,and a woman at ground that has not speed.
    Man throughs a light at ground and light comes again at spaceship. For man light follow a straight line,but for woman follow a V destination.
    Both see light speed to run with the same constant speed.But man see light to follow a smaller destination,so for woman man's clock run slower,but for man woman's clock run slower too.
    That is clear to me.
    That what I want to ask is:"Both will count the same time at their clocks. For example If man counts 2 second,what about woman?"

    Sorry for my bad english.
    Thanks a lot !!!
     
  2. jcsd
  3. Sep 30, 2013 #2
    How do you suppose they should compare their clocks?

    If the man jumps on another ship and comes back to earth than he counts fewer seconds. If the woman jumps on a faster ship and catches up with the man than she counts fewer seconds. Other possibilities might give different results
     
  4. Oct 1, 2013 #3

    HallsofIvy

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    What do you mean by "has speed" and "has not speed"? In what frame of reference? Certainly there exist a frame of reference (that of the woman) in which the space ship has speed v and the woman speed 0. But there also exist a frame of reference (that of the man) in which the space ship has speed 0 and the woman speed v.

    "throwing the light" stopped me for a moment!:tongue: You mean the man shone a light at the woman and it was reflected back to the man.

    Good. And if it had been the woman shining the light at the man, the situation would have been reversed.

    No, you just answered that! "So for woman man's clock run slower,but for man woman's clock run slower too." The woman, if she could see the man's clock, would see that it had counted a shorter time than her clock. The man, if he could see the woman's clock, would see that it had counted a shorter time than his. That's the whole point of "relativity". Every observation is relative to the observer. The woman sees the man's clock running slower, the man sees the woman's clock running slower.

    Don't worry about your English- it isn't perfect but it is understandable. My (put any language you choose here) is far worse!
    In "so for woman man's clock run slower,but for man woman's clock run slower too" the precise English would have been "so for the woman the man's clock runs slower, but for the man the woman's clock runs slower too".
    English is much stricter about "articles" than most languages. And the present tense, third person, of "run" is "runs". But you were 99% correct.
     
    Last edited: Oct 1, 2013
  5. Oct 1, 2013 #4
    Thank you a lot for everything. But I have one question too.
    Lets say that Man counts 2 seconds,so man will think that woman's clock counted less than 2 seconds.
    The same is for woman?? I mean,Woman counted 2 seconds,but will think that man's clock counted less than 2 seconds??

    I am correct?
     
  6. Oct 1, 2013 #5

    Nugatory

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    If they are moving at a constant speed relative to one another, then yes, you are right.

    The situation is symmetric; for both of them, they are at rest and the other one is moving relative to them, and the moving clock is running slow thanks to time dilation.

    Because of relativity of simultaneity, there is no paradox here. As far as woman is concerned, at the same time that she marked the end of the second second, man's clock was marking the end of the first second. But as far as man is concerned, the moment when his clock marked off the end of the first second is not the same time that woman's clock marked off the end of the her second second.
     
  7. Oct 1, 2013 #6
    I have one question more now.
    Lets say that man counts 10 years,so man if see at clock of woman will have pass 5 years for her.
    That is the same for woman. She will count 10 years,but will when see at man's clock will see that for him have pass 5 years only.

    I am correct?? If I am correct I will make my question.
     
  8. Oct 1, 2013 #7

    Nugatory

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    Is this not the same as your #4 above, except with years instead of seconds? If so, my answer in #5 still holds: "If they are moving at a constant speed relative to one another, then yes, you are right".
     
  9. Oct 1, 2013 #8
    Ok,perfect that what I mean.Lets say that man-woman are 20 years old. My question is:
    If man counts 10 year for him(he will be 30 years old),and if for him woman's clock counted 5 years,woman for the man will be(25 years old).

    That is the same from woman's point. She will see herself 30 years old,but man 25 years old.


    What happenes here??
     
  10. Oct 1, 2013 #9

    Nugatory

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    That's exactly what happens. Any time dependent process (the aging of the human body, sand falling through an hourglass, the decay of a sample of radioactive material, the motion of the earth around its axis and around the sun, ....) can be used as a clock, and they are all equally affected by time dilation.

    But there is still no paradox here; read what I said above about the relativity of simultaneity. Although in man's frame, his 30th birthday celebration is simultaneous with woman's 25th birthday celebration, but it does not follow that in woman's frame her 25th birthday celebration is simultaneous with his 30th birthday celebration.

    To get a paradox, you would have to bring them together at the same location, so that we could compare their ages at the same time in the same frame. You can't do that if they're both moving at a constant speed relative to one another; and if you drop that assumption to bring them together, that's a different situation, the classic and much-discussed twin paradox.
     
  11. Oct 1, 2013 #10
    Yea,I see but if man look at the mirror he will see hisself to be old,but If woman see him via "phone",or something else the man be younger for her??
    That is a bit strange :P
     
  12. Oct 1, 2013 #11

    Nugatory

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    As long as they are at different locations and traveling at high speeds relative to one another, the image that reaches woman's phone (traveling no faster than the speed of light) will have left man long before he looked in the mirror... so it would be even stranger if he didn't look younger in woman's phone.

    Now, if you had a way of transmitting an image instantaneously, then you could construct all sorts of horrible strange paradoxes. But there is no such thing.
     
  13. Oct 1, 2013 #12
    Thank you for your answers. Something final. Paradox of twins will happen only if spachip turns and come back again at earth. That gravitational force will make the man to think that woman's clock runs very fast.
    I am correct?? If yes,why?
     
  14. Oct 1, 2013 #13

    Nugatory

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    Neither gravity nor acceleration are the explanation (although you will find many careless explanations that get this wrong). This is the "Twin Paradox", worth studying in its own right. There are many good threads in this forum, and a decent online overview at http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html
     
    Last edited: Oct 1, 2013
  15. Oct 1, 2013 #14
    Thank you a lot my friend.
     
  16. Oct 2, 2013 #15

    BruceW

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    you can use 'gravity' to explain the twin paradox. i.e. in the travelling twin's 'reference frame' there will be a weird kind of 'gravity' which is due to the fact that this twin is undergoing accelerations. (note that I've used scare-quotes around the terms which are maybe not the proper terminology). But the problem is most easily explained by simply using an inertial reference frame.

    edit: This was such a bad explanation, I'll give it another go. In the 'reference mollusc' (as Einstein called it) of the travelling twin, the metric is going to have terms due to the travelling twin accelerating around in the universe. These terms are essentially fictitious forces (a.k.a. 'gravity'). You can then calculate the time that passes according to the travelling twin and stay-at-home twin. And you can loosely say that this is due to 'gravity'. But then you can do all this in the inertial reference frame instead, which is much simpler, and the metric is simple.

    edit again: I guess what I mean, is that if you want to see from both perspectives (the reference system of the stay-at-home twin and the reference system of the travelling twin) that everything works out the same and there is no paradox, then you can use 'gravity' to explain what happens in the perspective of the travelling twin.
     
    Last edited: Oct 2, 2013
  17. Oct 29, 2013 #16

    ghwellsjr

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    Let me illustrate with some spacetime diagrams. Since you described a scenario where the man and the woman are 20 years old when he takes off in a spaceship, I will first show a diagram depicting their mutual rest frame prior to launch and then the blue man takes off at 0.866c leaving the red woman behind. The dots represent 1-year intervals of time for both of them. We can call this the woman's rest frame:

    attachment.php?attachmentid=63416&stc=1&d=1383081317.png

    So like you said, when she is 30 years old, he is 25 years old. I have included some more comparisons: when she is 40, he's 30 and when she's 50, he's 35. He accumulates age at one-half the rate she does in her rest frame.

    Now let's look at a portion of the man's rest frame while he is traveling away from the woman:

    attachment.php?attachmentid=63417&stc=1&d=1383081825.png

    Again, just like you said, when he's 30 years old, she is 25. Their ages are symmetrical when compared from their own rest frames.

    I don't think it's strange if you see it on a spacetime diagram. Here is an example when the red woman sees the blue man via "phone" with a radio signal that traveled at the speed of light:

    attachment.php?attachmentid=63418&stc=1&d=1383082005.png

    When she is 35 and he is 27.5 (in her rest frame), she can't actually see him at that age, rather, she will see him as he was at age 24, whether it's via "phone" or whether she is just looking at him via a suitable telescope. This is a 4 to 15 ratio.

    Although we don't have to explain this with gravity, you are correct that the man will observe (not just think) that the woman's clock runs very fast after he turns around. Just as the woman saw the man's clock running at 4/15 of her clock's rate, he sees her clock running slower than his by the same 4/15 ratio during the outbound portion of his trip. But as soon as he turns around, he starts seeing her clock running faster than his own by the inverse of the ratio, namely 15/4.

    Here is a spacetime diagram of the entire trip in the rest frame of the woman indicating the signals going from the woman to the man showing that prior to his turn-around, he sees her clock running at 4/15 of his own and then after he turns around, he sees her clock running 15/4 of his own:

    attachment.php?attachmentid=63419&stc=1&d=1383082258.jpg

    Now I want to show you a non-inertial diagram for the rest frame of the man during the entire scenario based on the radar method of determining the distance to the woman as a function of the man's own clock:

    attachment.php?attachmentid=63420&stc=1&d=1383082460.png

    You can see that this diagram correctly depicts the signals traveling at c from the woman to the man and agreeing with all the observables in the inertial rest frame of the woman.
     

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  18. Oct 30, 2013 #17
    I thought I had understood the difference in clock rate for different observers, but I am unable to answer the following problem:-
    An observer O is stationary in an inertial frame of reference, with two stationary clocks which are some arbitrary distance apart.
    A rocket ship flies through the IRF at high speed, brushing past each clock as it goes. Each clock records the time of the "fly past", so the observer O measures a time interval 'To' for the rocket to pass between his two clocks.
    There is an astronaut A and a clock on board the rocket ship; this clock too records the time of each "fly past". So the astronaut A measures a time interval 'Ta' for the rocket to pass between the observer's two clocks..
    The observer O sees the clock on the space ship going at a slower rate than his own two clocks, but astronaut A sees O's clocks going more slowly than his own clock.
    How do the two time intervals 'To' and 'Ta' compare with each other?
    (Note that this is an extension of one of the standard thought experiments used to establish space contraction as seen by the speeding astronaut).
     
  19. Oct 30, 2013 #18

    BruceW

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    yep. there is no contradiction with what you are saying. According to observer O, the clock held by astronaut A ticks slower than his own. And according to astronaut A, the clock held by observer O ticks slower than his own. Both statements are true. The important thing is that it is different when we use different reference frames. So you can only ask how time intervals T0 and Ta compare with each other according to a specific reference frame. They will be different for different reference frames. So, for example, in a reference frame attached to the observer O, T0 > Ta. But according to a reference frame attached to the astronaut A, T0' < Ta' where the prime denotes that we are now talking about variables according to another reference system than before.
     
  20. Oct 30, 2013 #19

    ghwellsjr

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    And, we will assume that observer O has previously synchronized those two clocks according to Einstein's convention, correct?

    This is not correct. As the astronaut is approaching observer O and his clocks, they all see the other ones clock going at a faster rate than their own, and to the same degree. Then, as the clocks pass and retreat from each other, they will see them going at a slower rate than their own, again, to the same degree.

    'To' is always greater than 'Ta', assuming that the two clocks used to calculate 'To' have been synchronized in their mutual rest frame.

     
  21. Oct 30, 2013 #20

    ghwellsjr

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    I think you should read the scenario carefully and you will see that the question regarded the Proper Times displayed on inertial clocks, not the Coordinate Times of any reference frames. As such, it doesn't matter what reference frame is used, the measurements that are made are the same in all reference frames.

    Even if you are comparing the Proper Time interval between two events of one inertial clock to the Coordinate Time interval for those events, which can be different in different reference frames, the Proper Time interval will never be more than the Coordinate Time interval and they will be the same in the rest frame of the clock.
     
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