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Addition of velocities

  1. Jul 12, 2009 #1
    still don`t have an answer i can understand.
    If a moving emitter moves at .99 c, does the lead photon move away from the emitter such that it is c x time away from the emitter at any point in time? (not asking about the original position of the emitter when it emitted)
     
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  3. Jul 12, 2009 #2

    malawi_glenn

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    moving emitter with respect to what? Why just 0.99c ?

    and who's time? This is relativity you know.

    Light speed is always c to all frames, and the emitter will always be c times his proper time away from the photon.

    If there were 2 guys emitting the photon in a common rest frame in +x direction, and one of them starts to move in -x direction with a speed, say 0.1c, with respect his old rest frame. The distance between the photon and the moving guy as measured by the guy who has not moved from the position of emittance is equal to? Is this your question?
     
  4. Jul 12, 2009 #3

    Fredrik

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    If all of this is relative to a single observer, e.g. you, and the photon and the emitter are moving in the same direction, then the answer is no. The time derivative of the distance between the emitter and the photon (in your rest frame) is <c. It's =c in the emitter's rest frame.
     
  5. Jul 12, 2009 #4
    1 - moving emitter with respect to a parked train. .99 seemed like a good idea.
    2 - time with respect to the moving emitter.
    3 - i guess i don`t know what proper time is - is it time inside the emitter?
    4 - if proper time is time inside the emitter, then what will the parked train measure as the speed of emitter`s lead photon?
    (here`s my problem - if the time of the emitter and train guy are the same (big if ) then the photon would be in 2 places at "once".
     
  6. Jul 12, 2009 #5

    malawi_glenn

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    In Special Theory of relativity, time and space are relative, the moving emitters proper time and the proper time of the parked train are different. Do you want us to point you to introductory material/tutorials for Special relativity so you can work these things out for yourself and learn them? :-)
     
  7. Jul 12, 2009 #6
    sincere thanks but "been there, tried that". I was hoping for a sentence that would explain what happened (in a physical way, not mathematical) that resulted in the two clocks being different.
    It seems to me that the slowing of the moving clock is the result of the time it takes for the light to get to the train - the time gets longer than expected every second because every tick leaves a second later plus it takes a finite amount of time to travel to the train. This seems similar to an optical illusion to me.
     
  8. Jul 12, 2009 #7

    Fredrik

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    That's not how I interpreted your question. When you used the word "distance", was that also as measured by the moving emitter? Can you ask the question again, and this time make sure that every time you reference a time or a distance you make it perfectly clear what frame you're talking about?

    It's actually a mathematical property of a curve in spacetime. It's important because what a clock measures is the proper time of the curve in spacetime that represents its motion. For this reason, people in this forum usually answer the question of what proper time is by saying "it's what clocks measure". This is true, but the actual definition is more complicated.

    I don't understand what you're saying, but you're probably just mixing frames.
     
  9. Jul 12, 2009 #8

    malawi_glenn

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    But you said you knew what proper time is, and the original question can be interpreted in many ways since it is very ill-defined in terms of Special Relativity.

    What is a "physical way" and what is a "mathematical way"?

    The language and logic in physics is math, and since you don't know how time dilation is derived, then I sincerely invite you to take a study of Special Relativity
     
  10. Jul 12, 2009 #9
    glad to try again. Thanks for your patience.
    If a moving emitter moving at .99 c passes a stationary train (relative to the moving emitter) and emits a photon just at that point (simultaneously) and the train takes off in the opposite direction at, say .5 c relative to the train`s stationary starting position, what is the velocity of the photon relative to the moving emitter (is the photon c x time distance from the moving emitter at any point in time?) ( not asking about the distance or velocity from the original emission point).
    Also, since the train and the moving emitter are moving in opposite directions, do their measurements (c x time) of their distance from the photon yield different locations for the photon?
    Isn`t there a pretty definite location for the point of emission which is also the point where the train began it`s .5 c journey?
     
  11. Jul 13, 2009 #10

    Fredrik

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    This still isn't very clear.

    You say that the emitter is moving at 0.99c, but you don't say in what frame. Since you mention a train that's "stationary", I have to assume that the velocity of 0.99c is in the train's rest frame. Right after "stationary" you say "relative to the moving emitter" which contradicts the above since it means that the train is co-moving with the emitter. I'm pretty sure that's not what you meant, so I'll ignore that part.

    The emitter emits a photon at the event where it passes the train. There's no need to use the word "simultaneously", since the meeting between the emitter and the train is just one point in spacetime. (Simultaneity is only an issue when you consider at least two points). At this event, the train instantaneously boosts its velocity to -0.5c in its original rest frame. I would have preferred a thought experiment involving a train with a constant velocity, since we're not going to use the fact that the train had another velocity before the emission event, but it's OK for now.

    Now you're asking for "the velocity of the photon relative to the moving emitter", but since you used the phrase "relative to the moving emitter" very incorrectly earlier, I still don't know you mean. What you're actually asking for here (regardless of whether you meant it or not) is the velocity of the photon in the emitter's rest frame. It's c. Light always moves at c in the rest frame of any object. (Note that only massive objects have rest frames). So in that frame, the distance between the photon and the emitter is ct, where t is the time measured by a clock attached to the emitter.

    Maybe you meant to ask for the velocity of the photon in the train's original rest frame, or the velocity of the photon in the train's new rest frame. It's c in both of them. It's c in all inertial frames.

    What you're probably interested in is the distance in the train's frame between the emitter and the photon at a time t in the train's frame (i.e. measured by a clock attached to the train). This distance is 1.5ct, so it's increasing by 1.5*299792458 meters every second. Note however that even though this quantity has dimensions (i.e. units) of velocity, there's no object that moves with that velocity in any inertial frame.
     
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