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Why doesn't my solution work?

  1. Aug 13, 2012 #1
    Hi all,

    I'm doing a practice problem and I thought I had a good solution, but it turns out I'm wrong and I'd like to know exactly why. The book seems to do it in different way, but I don't see why theirs is the right way.

    The problem is this:

    1vXRr.png

    LrD10.png

    So, here's the process I tried doing:

    -Find what the y component of the package velocity with respect to S1 (S1 is rocket 1's frame), using the velocity addition rule since we have the velocity of S with respect to S1 and the velocity of the package with respect to S

    -Find the velocity of rocket 2 with respect to S1, using the velocity addition rule since we have the velocity of S with respect to S1 and the velocity of rocket 2 with respect to S

    -Then, to make it so the package hits rocket 2 as it goes across the gap, we set these two velocities equal.

    This is where the solution and I seem to disagree, the solution says that the y component of the package velocity and the velocity of rocket 2 should match in frame S, not rocket 1's frame! I guess I don't actually see why one is preferable over the other, but theirs must be right. If I had to guess I'd say it's something involving proper time or proper velocity, but I don't see what exactly.

    Any help would be great, thanks!
     
  2. jcsd
  3. Aug 13, 2012 #2
    Working in the S frame:

    You have two trajectories for the package (P) and the second ship (2), which are

    [itex]s_p = \frac{4}{\sqrt{5}} \lambda \left(e_t + \frac{3}{4} [e_x \cos \alpha + e_y \sin \alpha ] \right) \\
    s_{(2)} = \frac{2}{\sqrt{3}} \tau \left(e_t + \frac{1}{2} e_y \right)\tau + d e_x[/itex]

    where [itex]\lambda, \alpha, \tau[/itex] are scalars. Perhaps by setting all 3 components equal you can solve this problem in the S frame and then take the package's four-velocity and boost it into ship (1)'s frame?

    It's easier to work in the S frame because you already know [itex]\beta = 3/4[/itex] for the package in the S frame.
     
  4. Aug 13, 2012 #3
    Thanks, but I see how they did it in the solution. I just don't know why theirs gives the right answer and mine doesn't.
     
  5. Aug 13, 2012 #4
    When you did this:

    I suspect this is the problem. Boosting four-velocities that have out-of-plane components can be tricky with the velocity addition formula. Could you show what you did for this part?
     
  6. Aug 13, 2012 #5

    pervect

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    Did you take into account that "simultaneous in frame S" is not the same as "simultaneous in the frame of rocket 1"?

    If you look at the problem in rocket 1's frame, the one that's launching the package, you need to account for both the relative velocity of rocket 2 changing (which you did), and the definition of simultaneity being different in frame S and frame R1 - which as far as I can tell, you did not,.
     
  7. Aug 13, 2012 #6

    PAllen

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    I see a big problem right here. The package is moving at an angle to x axis in S to get from rocket 1 to rocket 2. The normal velocity addition formula is only valid for colinear velocities. You can derive a more general formula, but it is easier to solve a problem like this all in one frame (e.g. S), then Lorentz transform to S'.

    I used different symbols than Muphrid, but the same general idea. You derive α in S, then Lorentz transform the package trajectory equation, and compute α' in the rocket(1) frame.

    [edit: I thought it would be good to give a simple example of why the simple velocity addition formula cannot be used on components of non-colinear velocities. Suppose frame F1 is moving at .99c in the x direction relative to F0. Suppose some object is moving at .99c at 45 degree angle to x in F1. You propose to apply velocity addition to x components (and I assume believe y component is unaffected). So you get that x component of object velocity in F0 is nearly c and y component is .99c/√2. This is absurd since the result is a speed way over c.]
     
    Last edited: Aug 13, 2012
  8. Aug 14, 2012 #7
    Damn, that makes sense. I forgot that even though the package is launched at the passing point in S, it's not the same time in R1... Thanks!
     
  9. Aug 14, 2012 #8
    Yeah, damn... I vaguely realized this afterwards. I think I just set it up in a way that makes it miserable to solve, and I didn't solve it right clearly. Thanks!
     
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