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Relative Motion Leads To Impossibilities (or so it seems to me can you help?)

  1. Dec 25, 2011 #1
    I have a couple questions about relative motion, since I've been reading everywhere that motion is relative...

    (Feel free to point me to other threads that answer these questions if there are any - I searched and couldn't easily find any)


    So if I'm standing still, and someone travels away from me at almost the speed of light, noticeably less time will pass for him than for me, right? But his motion and speed are being measured relative to me, right? And that means that I am also traveling away from him at almost the speed of light. So time should slow down for me also. But it doesn't. How can that be explained?


    Let's say that me and 2 friend are hanging out in space. Friend #1 gets on his rocketship and zooms off to my right at 75% of the speed of light. Friend #2 gets in his rocketship and zooms off to my left at 75% of the speed of light. Relative to me, both friends are traveling at 75% of the speed of light. But relative to each other they are traveling at 150% of the speed of light. But nothing can travel faster than light. So something impossible is happening. How can that be explained?

    Thanks in advance for any answers.
  2. jcsd
  3. Dec 25, 2011 #2
    Question one:
    Yes. You observe your friend's time ticking by more slowly and he observes your time ticking by more slowly. It is odd. The thing that most people forget is that in order for you to get back together and compare your clocks, one of you will accelerate. So lets say that he suddenly stops and comes back to you. His clock will seem to have ticked by more slowly than yours because he has moved into your frame of reference.
    Question two.
    There is a velocity transformation just as there is a position and time transformation.


    That will probably do a better job of explaining than I will.
  4. Dec 25, 2011 #3
    In his rest frame time does run slower for you. Why do you consider this a problem?

    Velocities don't add that way in relativity. You are incorrectly applying a consequence of the Galilean (non-relativistic) Transformation to to a relativistic situation.

    In Newtonian physics velocities are added very simply. If you see person A moving to the right at velocity u and person B moving to the left at velocity v, then the velocity of person A with respect to person B (which we'll call w) is given by:


    In Special Relativity velocities are "added" in a slightly more complex way. The relativistically correct equation is:

    [tex]w= \frac{u+v}{1+uv/c^2}[/tex]

    When you use this equation you will always end up with an answer such that w<c. For example, in the problem you give:

    [tex]w= \frac{u+v}{1+uv/c^2} = \frac{0.75c+0.75c}{1+(0.75c)(0.75c)/c^2} = 0.96c[/tex]

    So friend 1 is only traveling at 96% the speed of light with respect to friend 2.
  5. Dec 25, 2011 #4
    re: QUESTION #1:
    It's a problem because I've heard that if someone travels away from earth very fast and then returns, less time will have passed on their clock than on mine here on earth. For example, to start both of our clocks say 5pm, but after he leaves and comes back, his clock says 6pm whereas mine says 7pm. Less time passed on his clock because he went fast. But relative to him I went fast too... so shouldn't less time have also passed on my clock?... so that our clocks will actually read the same time when the guy returns, no matter how fast he went? Which is correct?

    re: QUESTION #2: That is so cool. Elfmotat your explanation could not have been more clear, thank you.
  6. Dec 25, 2011 #5
    If your friend returns, he must accelerate. He is no longer in an inertial reference frame. Since he has accelerated, he has moved into your reference frame and the view of time that you had been seeing would be the correct one.

    If you had instead accelerated in order to catch him (in his view, stopping and turning around) then your time would have been behind.

    In your example above, indeed his clock would read 6 and yours 7. The absolute difference between your clock and his is not a contradiction of special relativity, because he has not been in a single inertial reference frame the entire time, but you have.
  7. Dec 26, 2011 #6
    Just to understand it better time=motion=energy. The limit to energy is "C". Conversation of energy is why the traveling twin does not age as fast. When you are at a resting state all the zero point energy is being used as time. When you are in motion energy is being shared between motion and time. If mass could go the speed of light the electron would not have energy left to cycle the proton. All the energy would be used up in the direction of motion leaving none for cycling.

    No one thing can travel faster than the speed of light. No two or more things can have a departing speed of greater than 2 "C".
  8. Dec 26, 2011 #7


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    This is absolute nonsense. You're going to get yourself banned. Please take this as another friendly warning to quit posting and spend your time reading so that you can learn or if you must post, ask questions instead of answering other people's questions.

    Time≠motion≠energy. Where do you get these weird ideas from?

    And it doesn't matter how much energy was used to put you in motion, electrons will cycle the proton exactly the same as they did when you were at the resting state.

    And this departing speed is simply the addition of two speeds in opposite directions and has nothing to do with any actual speed of any object with respect to any other object or with respect to any frame.
  9. Dec 26, 2011 #8

    At least read what others have posted before you give wrong replies.
  10. Dec 29, 2011 #9
    I looked up "inertial reference frame" on Wikipedia, which says "An inertial frame of reference is one in which the motion of a particle not subject to forces is in a straight line at constant speed." I'll assume that that is what you mean when you say "inertial reference frame"...

    So it seems to me that you're saying that your SPEED relative to mine has no effect on your time relative to mine; rather it is only your ACCELERATION relative to mine that will slow your time down. Is that right?

    If that's true, then the following two scenarios would also be true, I think:

    A) If I am floating in space and I saw a clock flying past me at a constant speed of 290,000 km/s relative to me, that clock would not seem to me to be running any slower than my own clock. (since both my clock and the other clock are traveling in a straight line at a constant speed, they are both in the same inertial reference frame, therefore time does not slow down for either one).

    B) Say I am floating in space and I see a clock flying towards me at a constant speed of 290,000 km/s, and then when the clock gets near to me it has a decelerating force applied to it that slows it down so that it is at rest beside me. If I look at that clock, I will see its time running super fast, right? (if the clock had accelerated relative to me I'd see its time slow down; so I assume that if it decelerates relative to me, I'll see its time speed up).

    Is my understanding of all of this correct? If not, can you correct any of my misunderstandings? Thank you!
  11. Dec 29, 2011 #10


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    No, your understanding is not correct. In any inertial reference frame, a stationary clock ticks normally and a moving clock ticks more slowly, the faster it moves, the slower it ticks.
  12. Dec 29, 2011 #11
    #1, from your standing still perspective you assume your velocity is 0. The same is true when the traveller calculates your velocity, they assume there velocity is 0. Everything you observe of the other is based on the assumption your velocity is 0. The concept of observation must be considered to understand relativity. For me the symmetry of it all came soon after.

    For this scenario I think you can safely ignore the FoR, instead considering that neither observer is accelerating.

    "So it seems to me that you're saying that your SPEED relative to mine has no effect on your time relative to mine;"
    That statement can be said as "What I observe doesn't effect/change/impact what you observe."
    Last edited: Dec 29, 2011
  13. Dec 29, 2011 #12
    Let me try to be more precise when I explain this. First, yes, an inertial reference frame is a frame (could be the inside of a space ship, a box, or a simple x and y axis) that has no forces acting on it.

    So lets set the stage...

    Alice and Bob are siblings living in the middle of space with nothing around them. They could either be moving or perfectly still and they would never know as long as they are not accelerating (this includes changing direction). On Earth it maybe difficult to convince yourself of this since even if you move in a car in a straight line, gravity pulls you and the car shudders a little, but you will definitely feel the difference between almost inertial movement and a sharp turn which is certainly accelerated.

    Now consider a slightly different situation. Alice is in frame A and Bob is in frame B. The two people (whose frames are defined by their motion) are moving past each other in a straight line and have equal claim to being stationary. As they go past each other Alice appears to see Bob moving near the speed of light from her right to her left side. Bob sees Alice move near the speed of light from his right to his left.

    Alice will see Bob's clock (which I forgot to mention is displayed in such a way that Alice can see it even though they are traveling near the speed of light) ticking more slowly than hers. Bob's Frame of Reference is also an inertial FoR and so we have to assume that he too would have a right to claim that Alice's clock is ticking more slowly compared to his own.

    I think you understood this from your first post. Informally we can think of the relativity of time like the fact that both Alice and Bob can claim that the other one is moving from right to left (not the same, but it might help?).

    Now the question:
    What happens when they come back together to compare clocks?

    If Alice and Bob are both moving inertially, they are both right. However, in order for them to get back together and compare time, one of them must accelerate! So lets say Bob decides to turn around and go see Alice. When they get back together Bob's clock will have measured less elapsed time than Alice's and they will both agree on that, but their histories will be different. Bob had to change direction which meant that for some period of time he was NOT in and inertial FoR and his claim to equal treatment under Special Relativity is lost.

    To sum up. When Alice and Bob are in an inertial FoR (like nitsuj says, they can consider themselves to be motionless), the rules of Special Relativity apply and both observe the other's time to be moving slowly. When one of them turns around to come back and meet the other, that person is in an accelerated FoR for some period of time and the transformations of Special Relativity do not apply. (We are assuming that they don't both turn around and accelerate back toward each other and meet up. They could possibly do this and make both clocks the same, but that would be even more unlikely than two sibling floating around in outer space)

    Does that make it clear? The key is understanding when the principles of Special Relativity apply and when they don't.
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