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Einstein Velocity addition

  1. Feb 21, 2010 #1

    ok, so i'm looking at this, and i see that
    where c = speed of light
    if ship A is moving at .7c to the right ===> <=== and ship B moving .7c to the left

    and they are both on the same line and about to collide, at what velocity (in C) will they hit each other.

    the reference point is the collision point of impact not some distant observer.

    Theory i read appears that C + X = C but this seems to violate common sense.
    I can even imagine that you can't possible surpass the speed of light, but can someone make this a little more obvious.

    Thanks, and this is not a HW question, i'm in EE. i'm just asking for the fun of it.
  2. jcsd
  3. Feb 21, 2010 #2
    What is X exactly? By the way, common sense will lead you to dark alleys and beat you to death in physics.
    Last edited: Feb 21, 2010
  4. Feb 21, 2010 #3
    X is anything positive , so basically C+ 2C = C because i thought the speed of light is the max possible speed.
  5. Feb 21, 2010 #4


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    The problem is that "C+ V" is shorthand for adding the velocities according to Einstein's formula for addition of velocities. If two objects move directily toward one another, with speed u and v relative to some third frame of reference, then the speed of each, in the other's frame of reference ("U+ V") is
    [tex]\frac{u+ v}{1+ \frac{uv}{c^2}}[/tex].

    In particular, if one is a light beam, with speed u= c, then its speed ("C+ V"), relative to the other, is
    [tex]\frac{c+ v}{1+ \frac{cv}{c^2}}= \frac{c+v}{1+ \frac{v}{c}}= \frac{c(c+v)}{c+ v}= c[/tex]
  6. Feb 21, 2010 #5
    And we should say Lorentz velocities. He did the addition first. Physics is a team sport (sort of).
  7. Feb 21, 2010 #6
    In the reference frame where the collision point is stationary, clearly ship A and ship B are approaching each other with a relative velocity 1.4c. This doesn't contradict anything in special relativity though, since there is no single object being measured at a speed greater than c. (And of course if you are in one of the ships, the velocity of the other ship can be determined using Einstein's velocity addition formula, and it will be less than c, as expected.)
  8. Feb 21, 2010 #7
    You could (should for simplicity?) say one is moving -.7c and one .7c with respect to the collision point. You would then calculate their adjusted mass+momentum (and resulting change from collision) as the function F=2*f(0.7c), and not F=f(.7c+.7c) as you suggest
  9. Feb 21, 2010 #8
    Just use the relativistic addition of velocities formula and plug in the relative velocities of each reference frame...
  10. Feb 21, 2010 #9
    Halls of Ivy..glad you posted that formula...I was just thinking earlier today using that would make these repeated discussions on relative speeds easier...and if anybody searches they'll actually be able to do their own computation....
  11. Feb 22, 2010 #10
    Why isn't the formula and explanation posted in Frequently asked Questions??

    I can't find the input page for doing it, but if someone can identify it and it's not alread there, I'll draft a submission....
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