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Homework Help: SR: Rapidity of two particles in uniform motion

  1. Jun 3, 2007 #1
    1. The problem statement, all variables and given/known data
    A particle leaves the spatial origin P of O at time [itex]t=0[/itex] and constant velocity. After a time t as measured by O, a second particle B leaves P at a different constant velocity and in pursuit of A. B catches A after proper-time t as measured by B. Show that the rapidity of B with respect to O is twice the rapidity of A with respect to O.

    2. Relevant equations
    I assign the coordinates [itex](t', x')[/itex] within this frame (slightly counter-intuitive I know, but the problem has already denoted some variables as unprimed, which I might otherwise want to use). I'm not adept at drawing diagrams in LaTeX, so I won't bother. However, as I envisaged the problem, the worldlines of particles A and B are as follows:

    [tex]t'_A(x')=x'/v_A[/tex]
    [tex]t'_B(x')=t + x'/v_B[/tex]

    I label the event where they meet as [itex](t'_1,x'_1)[/itex]. So we have the equality

    [tex]x'_1 \left(\frac{1}{v_A}-\frac{1}{v_B}\right)=t\,\,\,\,\,\,\,(1)[/tex]

    We have the fact that the proper-time as measured by the particle B also happens to be t. I could think of two ways to write that:

    [tex]t=\gamma_B (t'_1-t)\,\,\,\,\,\,\,(2)[/tex]

    and also as (which I think is correct)

    [tex]c^2(t'_1-t)^2-x'_1^2=c^2 t^2\,\,\,\,\,\,\,(3)[/tex]

    I list some handy rapidity based relations (where the rapidity, [itex]\phi,[/itex] is related to the velocity, v,):

    [tex]\beta = \frac{v}{c} = \tanh(\phi)[/tex]
    [tex]\gamma = (1-\beta^2)^{-1/2} = \cosh(\phi)[/tex]
    [tex]\beta \gamma = \sinh(\phi)[/tex]

    3. The attempt at a solution
    My method of attack was to use the equation (1) and either (2)/(3) and obtain some relation between the velocities of both particles, which I could then take the artanh of to get the rapidities.

    Processing equation (1) gave me

    [tex]t=x'_1\left(\frac{v_B - v_A}{v_A v_B}\right).[/tex]

    Substituting this into (2) gave me

    [tex]\left(\frac{v_B - v_A}{v_A v_B}\right) (1+\gamma_B)=\frac{t'_1}{x'_1}=\frac{1}{v_A}[/tex]

    Playing around with fractions ended up giving me

    [tex]v_A = v_B \left( \frac{\gamma_B}{1+\gamma_B} \right)[/tex]

    Using the definitions/relations regarding rapidity above, I still couldn't get to the answer I wanted, which is

    [tex]\phi_B = 2 \phi_A[/tex]

    Does anyone have any ideas? Thanks for reading this long question!
     
    Last edited: Jun 3, 2007
  2. jcsd
  3. Jun 4, 2007 #2

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    You're second to last equation is right. To derive the last one, note that since tanh is one to one, it's equivalent to show tanh(\phi_B)=tanh(2\phi_A), or:

    [tex] v_A = \tanh(\phi_A) = \tanh(\phi_B/2) [/tex]

    Try rewriting the last term using hyperbolic trig identities.
     
  4. Jun 4, 2007 #3
    StatusX:

    Thank you for your reply. Huh! The answer was right under my nose, all I needed to do was look up the trig. identities.

    This problem was the last part of a question in a previous undergrad. SR paper. How was I meant to know the identity? Is there an obvious route for its derivation? It's just that I'd never seen it before, and so didn't recognise it, even though I came across the expression

    [tex]\tanh(\phi_A) = v_A = \frac{\sinh(\phi_B)}{1+\cosh(\phi_B)}[/tex]

    I just didn't realise this was equivalent to [itex]\tanh(\phi_B/2)[/itex]

    Anyway, many thanks for pointing me in the right direction.
     
  5. Jun 4, 2007 #4

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    One easy way to work with hyperbolic trig functions is to use the change of variables y=e^x. Then you get, eg, sinh(x)=1/2(y-1/y), and:

    [tex] \tanh(x) = \frac{y^2-1}{y^2+1} [/tex]

    And since [itex]e^{x/2}=\sqrt{y}[/itex], we get:

    [tex] \tanh(x/2) = \frac{y-1}{y+1} = \frac{y^2-1}{(y+1)^2} [/tex]

    [tex]= \frac{y-1/y}{y+1/y+2} =\frac{\sinh(x)}{\cosh(x)+1}[/tex]
     
  6. Jun 4, 2007 #5
    So it does. Thanks again.
     
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