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Velocity addition formula for multiple velocities within each other

  1. Jan 24, 2012 #1
    I am having a problem coming up with an empirical formula for multiple objects moving with the same speed relative to each one up.

    I.e. there is a cart moving with speed u relative to me, and inside it is a cart moving speed u relative to the cart is inside it. and inside that cart is a cart moving speed u relative to the second cart. and so on...

    what i can't seem to come up with is an empirical formula that gives me the speed of the nth cart relative to me.

    so far i understand that if i used the tanh(λ) formula to derive an empirical formula because it just seems to get bigger for each additional cart.
    Last edited: Jan 24, 2012
  2. jcsd
  3. Jan 24, 2012 #2


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    You simply have to iterate the relativistic velocity-addition formula (setting c=1)

    [tex]v \oplus u = \frac{v+u}{1+vu}[/tex]

    Now assume that starting with u0 = 0 you have obtained un in the n.-th step; the next step is then

    [tex]u_{n+1} = u_n \oplus u = \frac{u_n+u}{1+u_n u}[/tex]

    where I used v = un

    You get

    [tex]u_0 = 0[/tex]

    [tex]u_1 = u[/tex]

    [tex]u_2 = \frac{2u}{1+u^2}[/tex]

  4. Jan 24, 2012 #3
    i got that much, but im trying to find a formula that will tell me the velocity of any nth cart, so i need something empirical instead of recursive.
  5. Jan 24, 2012 #4


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    "Empirical" means "experimental", so you're probably looking for some other word.

    I don't know if there's a simple way to write down such a formula, but if there is, the way to find it is to calculate the result for n=1,2,3,4,... as many as it takes for you to guess the result for an arbitrary n. And then you have to prove by induction that your guess is correct.
  6. Jan 24, 2012 #5


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    In order to do that you should express the Lorentz transformation in terms of the rapidity θ. Then you take the matrix L(θ) defining the Lorentz transformation and calculate the n-th power Ln(θ) of this matrix. Using the "double argument formulas" for hyperbolic sine and cosine you will find that L2(θ) = L(2θ) and therefore you can guess that the Lorentz transformation is additive in terms of the rapidity, i.e. Ln(θ) = L(nθ). Doing that allowes you to calculate θ(u) and n*θ(u) and invert this as un = u(n*θ(u1))

    Last edited: Jan 24, 2012
  7. Jan 29, 2012 #6


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    If we take the formula in post #2
    and substitute [itex]u = c\,\tanh\lambda;\ v=c\,\tanh\mu[/itex], you get[tex]

    (c\,\tanh\lambda)\ \oplus\ (c\,\tanh\mu)\ =\ c\,\tanh(\lambda + \mu)

    [/tex]which leads to the result[tex]

    u_n\ =\ c\,\tanh\left(n\,\tanh^{-1}\frac{u}{c}\right)

    [/tex]If u is very small compared with c, this can be approximated as[tex]

    u_n\ \approx\ c\,\tanh\frac{nu}{c}

    [/tex](All of the above is essentially what tom said in the last post, expressed in a different notation.)
  8. Jan 29, 2012 #7


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    Yes, this is what I tried to indicate in post #5
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