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Both frames of reference predict the other is slower -- proof

  1. Feb 26, 2015 #1
    While trying to understand how each frame of reference predict that the other is slower, I discovered what seems to be something unsymmetric. It looks as if it is possible to determine that a frame of reference is not one truly standing still. But that's impossible right? I'm really interested in where my mistake is.

    So here it goes. Imagine that a rocket is traveling away from Earth at speed v=282647 km/s which is the speed required for the time in the rocket to be slowed by one third. When 6 seconds passes in the rocket, 18 seconds passes at Earth.

    A person at Earth wants to measure that the time is slower in the rocket. How does he do that? When the rocket is next to him, he reads that the clock in the rocket is 2:00 sharp. 6 seconds later, he wants to do the next reading such that he will see that the time is really one third slower in the rocket. After 6 seconds, the person at Earth knows that the rocket has traveled a distance of 6*v kilometers. The light showing 2 (6/3=2) seconds on the clock in the spaceship needs to travel this distance at the speed of light before they will hit the eyes (or measurement instrument) on Earth. This time that light needs to travel is 6s*v/c=5,65s. After 6+5,65 seconds the person at Earth uses his knowledgeto predict that the light from the clock at the spaceship will reach him.

    So far so good. Now the question is, how does the person within the rocket predict that the time on Earth is going slower? He should follow the same procedure by assuming that he is standing still and that Earth is moving.

    He predicts that Earth is moving at the same speed. After 6 seconds (which is really 18 on Earth), the rocket has traveled three times longer than he is aware of. After this, he will wait 5,65s in his frame, which is 5,65*3=16,95s in Earths frame of reference. If everything is correct, the lightrays showing 2:02 from the clock on Earth, should reach the rocket after 18 seconds+16,95. This equals to 6 seconds + 5.65 seconds in the rocket. If that was true, it would be symmetrical and indistinguishable who the time is really going slower for.

    However I did the calculations and it turns out that the lightrays from the clock showing 2:02 on Earth need a lot more time to catch up to the rocket, such that the rocket predicts that time is slower than it is on Earth.

    Then I am wondering what is wrong? Is it possible to use this method to figure out differences between frames of references?

    Here are my calculations:

    If the light from the clock showing 2:02 on Earth should reach the rocket after 18+16,95=34,95s the following equation should be true:

    c*28,95=v*34,95

    Light showing 2:02 on Earth needs to catch up with the rocket while the rocket has a headstart of 6 seconds, hence the 28,95=34,95-6.

    It turns out that the equation doesn't add up.

    v*34,95=9878512
    c*28,95=8678978

    Why doesn't it add up? If the person at Earth did the measurement he would observe the correct lightrays that time is one third slower in the rocket. If the person at the rocket does the same procedure he gets wrong result?

    A big thanks to anyone who can help me out here.
     
  2. jcsd
  3. Feb 26, 2015 #2

    Orodruin

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    This is wrong, in the rocket frame, 18 seconds have not passed on Earth when he sends the signal. You need to look up and understand relativity of simultaneity. Events that are simultaneous in one system are not necessarily simultaneous in another.
     
  4. Feb 26, 2015 #3

    Nugatory

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    That's your mistake right there. If the relative speed between rocket and earth is ##v##, after 6 seconds of rocket time the earth has moved a distance of ##6v## according to rocket-guy - there's no "really 18" going on here. It is also true that after 18 seconds of rocket time the earth has moved a distance of ##18v## according to rocket-guy, and that after 6 seconds of earth time the rocket has moved a distance of ##6v## according to earth-guy, and that after 18 seconds of earth time the rocket has moved a distance of ##18v## according to earth-guy.

    It is essential to remember the relativity of simultaneity here. According to earth-guy, his clock reads 18 seconds from launch at the same time that rocket-guy's clock reads 6 seconds - the two events "rocket clock reads 6" and "earth clock reads 18" are simultaneous after allowing for the light travel time. But because of the relativity of simultaneity, these two events are not simultaneous according to rocket-guy - according to him, the "rocket clock reads 6" event happens at the same time as the "earth clock reads 2" event, not the "earth clock reads 18" event.
     
  5. Feb 26, 2015 #4
    I think that taking account for the light travel time is what I did, and it didnt add up. That's what I'm trying to understand.

    I concluded that Earth guy can assume that after 6 seconds, he needs to wait for light travel time and then he will observe that his clock is faster.

    If rocket guy does the same, it doesn't add up in Earth's frame. I know that the Earth guy believes that he is in another position when 18s has passed on Earth than what rocket guy believes. That is why I said that rocket guy has traveled further than he is aware of according to how the Earth guy predicts that the rocket guy predicts. I'm trying to understand how Earth frame understands that rocket frame predicts his results in his frame.

    Let's look at it this way. Earth guy knows about relativity, and that both of them measure the other as slower. However yet he can assume that he is the one standing still, since any frame of reference believe that they are standing still. Then I am trying to figure out how does Earth guy predict that rocket guy predicts that Earth is slower?

    Now when the Earth guy takes into account light time travel he still cannot predict that rocket guy will predict that time on Earth is 1/3 as slow as in the rocket.
     
  6. Feb 26, 2015 #5

    Orodruin

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    Relativity always assumes that light travel times are accounted for. It is a question not only about relativity of positions but also, as we have already pointed out, relativity of simultaneity. If the Earth and rocket observers agree beforehand to send light signals after 6 seconds have elapsed on their clocks, they both will think the other sent the light signal after 18 seconds in their own frame (after accounting for travel times).
     
  7. Feb 26, 2015 #6

    Nugatory

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    He knows about relativity, so he uses his knowledge of relativity to answer the question "using a reference frame in which the rocket is at rest, which events are simultaneous with the event 'rocket clock reads six seconds'?".

    One of those events is "earth clock reads two seconds, distance between earth and rocket is ##6v## (and by the way the light from this event will reach rocket-guy's eyes when the rocket-clock reads ##6+6v/c##)". The fact that earth-guy is on earth while he does this calculation is irrelevant; anyone anywhere in the universe moving at any speed can do it and conclude that using a frame in which the rocket is at rest the earth clock is running slow by a factor of three.

    Likewise, anyone anywhere in the universe and moving at any speed can work on the problem "using a frame in which earth is at rest, what events are simultaneous with the event "earth clock reads 18 and the rocket is a distance ##18v## from earth?". One such event is "rocket clock reads 6" and the light from that event will reach the earth when the earth clock reads ##18+18v/c##. This allows us to conclude that the rocket clock is running slow by a factor of three.
     
  8. Feb 26, 2015 #7
    I tried to calculate exactly that but I failed to get the correct results. After light travel time I calculated that the person in rocket frame calculates time as slower than one third on Earth. If anyone knows how I failed I'd love to know why.

    It is still good to know that it is true that both will predict one third of the other frame after light travel time, I was starting to question how it really is and doubt if they really did. I appreciate your help thanks.
     
  9. Feb 26, 2015 #8

    Nugatory

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    Are you familiar with the Lorentz transforms? If not, google for them.
     
  10. Feb 26, 2015 #9
    Thanks for your reply. I am trying to figure out how does Earth guy predict that rocket guy predicts that Earth guy is slower.

    I used the ##18+18v/c## equation in this line:

    The Earth guy predicts that when 18 seconds is his clock, it is simultaneous that rocket guy has 6 seconds on his clock and after light travel time will predict 2 seconds on Earth clock. Somehow I calculated that it doesnt add up after light travel time.
     
  11. Feb 26, 2015 #10

    Orodruin

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    Again, this is your problem, the assumption that simultaneous means simultaneous in all frames, which is not true in SR.
     
  12. Feb 26, 2015 #11
    Then how does the the person on Earth know that the person in the rocket observes Earth time as slower, if it is impossible for Earth guy to predict how?

    I'm not saying that it is simultaneous in all frames. I am just talking about the Earths frame, and explaining why the person in the rocket predicts Earth to be one third slower. As such it is not wrong to say that in Earths frame of reference, it is simultaneous that Earth clock show 18 seconds when rocket clock show 6. I am well aware of that in the rockets frame the opposite case is simultaneous for him.
     
  13. Feb 26, 2015 #12
    I got it!!! I did the mistake of thinking that when Earths clock shows 6 seconds the light reflecting information about 2 seconds on Earth is distance 6*v behind. But ofcorse in Earths frame, when the Earths clock shows 2 seconds, it is obviously 2*v distance behind the rocket. Then the equations add up. This means that Earth guy can predict how rocket guy predicts that time is slower on Earth all purely based on sticking to Earths frame of reference.

    The correct equation for this scenario is then c*32,95=v*34,95
     
    Last edited: Feb 26, 2015
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