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The Spaceship and The Wrench (1/2 + 1/2 = 4/5)

  1. May 5, 2012 #1

    This started out as a question, but answered itself in the process of being written up. I thought I'd post it anyway, in case it might help clarify something for someone else. It is an alternative derivation of the solution to a (similar) question asked in Six Not-So-Easy Pieces. Of course, Feynman's solution is far more elegant; this is merely how I came across the solution "in my own words."


    1. A spaceship is moving with velocity [itex]v_s[/itex] as measured by an observer "at rest".
    2. A wrench within the spaceship is moving with velocity [itex]v_w'[/itex] as measured by an observer within the spaceship.
    3. At what velocity is the wrench moving, as measured by the observer at rest?


    We might summarize the problem as: Solve for [itex]v_w[/itex] in terms of [itex]v_s[/itex] and [itex]v_w'[/itex].

    We begin with the equivalence,

    $$v_w = \frac{x_w}{t_w}$$

    where [itex]x_w[/itex] is the displacement of the wrench after time [itex]t_w[/itex].

    Note: I prefer to subscript all variables with their object counterparts, even if painfully obvious, in order to be perfectly clear as to whose perspective we speak of. In fact, my own attempts to solve this problem failed, initially, because I mistakenly swapped time, as experienced by the spaceship, with time, as experienced by the wrench, as if they were interchangeable.

    We recall the reverse Lorentz transformation (i.e. the displacement of the wrench in the rest frame, in terms of its displacement in the moving frame):

    $$x_w = \frac{x_w'+v_st_w'}{\sqrt{1-v_s^2/c^2}}$$

    And we note that

    $$x_w' = v_w't_w'$$

    Combining the above three equations and factoring out the [itex]t_w'[/itex], we reach

    $$v_w = \frac{t_w'}{t_w}\frac{v_w'+v_s}{\sqrt{1-v_s^2/c^2}}$$

    Finally, we seek to rid the time variables, and to do so, we recall the Lorentz transformations for time:

    $$t_w' = \frac{t_w-v_sx_w/c^2}{\sqrt{1-v_s^2/c^2}}\hspace{1in}t_w = \frac{t_w'+v_sx_w'/c^2}{\sqrt{1-v_s^2/c^2}}$$

    Which shall we substitute? [itex]t_w'[/itex] or [itex]t_w[/itex]? It turns out that substituting the latter is much more productive (as substituting the former simply undos our work thus far).

    v_w &= \frac{t_w'}{\frac{t_w'+v_sx_w'/c^2}{\sqrt{1-v_s^2/c^2}}}\frac{v_w'+v_s}{\sqrt{1-v_s^2/c^2}} \\
    v_w &= \frac{t_w'(v_w'+v_s)}{t_w'+v_sx_w'/c^2} \\
    v_w &= \frac{v_w'+v_s}{1+v_s\frac{x_w'}{t_w'}/c^2} \\
    v_w &= \frac{v_w'+v_s}{1+v_sv_w'/c^2}

    And there you have it.

    If a spaceship were moving at [itex]\frac{1}{2}c[/itex], and a wrench within it at [itex]\frac{1}{2}c[/itex], then the wrench, to an outside observer, moves at [itex]\frac{4}{5}c[/itex].
    Last edited: May 5, 2012
  2. jcsd
  3. May 5, 2012 #2


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    Looks like you know what you're doing. So I won't check the details. Since the final answer is correct, I assume the derivation is too. I just want to say that I like to do this by considering the composition of two Lorentz transformations in matrix form. I think this is is the easiest way to do it.

    In units such that c=1, the Lorentz transformation with velocity v is
    $$\Lambda(v)=\gamma(v)\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix},$$ where

    Let u be the velocity of the ship relative to the external observer. Let v be the velocity of the wrench relative to the ship. We just need to find the velocity of the Lorentz transformation ##\Lambda(u)\Lambda(v)##, so we do the matrix multiplication. Well, half of it anyway. We're not going to need the second row, so I'll just put asterisks there, instead of writing out irrelevant matrix elements.
    $$\Lambda(u)\Lambda(v) =\gamma(u)\gamma(v)\begin{pmatrix}1+uv & -u-v\\ * & *\end{pmatrix} =\gamma(u)\gamma(v)(1+uv) \begin{pmatrix}1 & -\frac{u+v}{1+uv}\\ * & *\end{pmatrix}.$$ Now we just need to compare this with the first formula I wrote down to see that the velocity is
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