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Newbie relativity question

  1. Apr 20, 2008 #1
    Hello everyone, great to meet you all.

    My name is Alejandro Di Mare, a computer science bachelor from Costa Rica currently studying music in Pasadena, California. For the past few months physics have captured my attention so greatly I'm thinking about persuing a masters degree in the subject next year (I hope I'll eventually be able to help people in this forum instead of asking simple SR questions :P).

    I've been reading a few posts on relativity trying to find if this has been addressed, but was unable to find it (I'm pretty sure someone must have asked this to you guys/girls, so I apologize if I force you to repeat yourselves).

    I am observing 2 spaceships, A and B. Relative to me, A moves to my left at a speed of 0.75c, B moves to my right at a speed of 0.75c. I'm wondering if one of you guys/girls could explain me how a passenger in spaceship A perceives spaceship B and what results he would obtain in an attempt to meassure it's speed (in a reference frame where A is stationary.).
  2. jcsd
  3. Apr 20, 2008 #2
    Hi Alejandro, welcome to PF :)

    Have a read of this FAQ about relativistic velocity addition and see if it helps. Sorry for being lazy, but if you are still stuck, feel free to ask for further explanations/ clarifications ;)

    Here is the link I left out :O --> http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html
    Last edited: Apr 20, 2008
  4. Apr 20, 2008 #3


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    Welcome to PF!

    Hi Alejandro! Welcome to PF! :smile:

    In your example, we apply the rule for combining velocities: (u + v)/(1 + uv/c²).

    With u = v = .75c, that gives 1.5c/(25/16), = 0.96c. :smile:

    For more details, see wikipedia (which is very good on physics) …

  5. Apr 20, 2008 #4


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    I'm also going to give you the short version of the answer hoping that you will ask additional questions if you aren't completely satisfied:
    from the point of view of A (in which A is stationary), B is moving past it at high velocity. To find that velocity, one should add both speeds, using the relativistic formula,
    [itex]w = \frac{u + v}{1 + u v / c^2}[/tex]
    which gives back the standard formula w = u + v when u and v are much smaller than c. You can just use this formula to calculate the velocity of B, according to A (and will get something around 0.96c, rather than 1.5c).

    [edit]tiny-tim beat me... by a lot even (did it take me so long to type that short answer?)[/edit]
  6. Apr 20, 2008 #5
    ah, beautiful, just what I was looking for, for some reason I'm able to picture the addition of velocities with frames that are approaching each other (or the typical example of a guy walking on a train, it's not difficult to invision the length contractions that make sense of the adddition of velocities) but I'm still having trouble seeing it with reference frames moving away from each other :P. Thanks for the help :cool:
    Last edited: Apr 20, 2008
  7. Apr 20, 2008 #6
    If it helps any, the velocity that A sees B aproaching at, is exactly the same as the velocity of B that he measures when they are going away from each other (except for the sign).
  8. Apr 20, 2008 #7
    Is it safe for me (the observer that sees each ship moving in opposite directions at 0.75c) to say that they're moving away from each other at a speed greater than c?
    I made the space-time diagram and see no problems with this affirmation, hope you can check it out and tell me if there's anything wrong with it :P

    The ships are moving away from each other with an angle of 60° (greater than 45°, which to my understanding would mean they're moving away from each other at c.). Of course I understand that if we rotate the diagram using Lorentz transformations to see how the same situation looks using spaceship A as the reference frame spaceship B's line would not be at an angle greater than 45° from spaceship A's line.

    Attached Files:

  9. Apr 20, 2008 #8
    nevermind that last post, I just read ht tp://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/FTL.html and confirmed what I thought :P
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