1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Special Relativity length contraction and velocity

  1. Apr 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Λ particle has a proper life-time τ = 2×10−10 s. After being born in the cloud chamber (a
    device to track energetic particles) of physics laboratory it left there a a 300cm long trail. Find
    the speed of this particle in the laboratory frame.
    2. Relevant equations

    Gamma(Lorentz Factor) = (1 - v^2/c^2)^-1/2
    v=l/t
    τ=tgamma


    3. The attempt at a solution
    Every time I try a solution, I seem to find I need more information, say the length travelled in the frame of the particle. Or get a speed above c! Urgently need help
     
  2. jcsd
  3. Apr 13, 2012 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Looks like you have all the right equations. Show what you did with them. You should end up with a single equation where the only unknown is the velocity.
     
  4. Apr 13, 2012 #3
    so I tried gamma = (1- v^2/c^2)^-1/2) and rearranged to get
    v = c(1- 1/gamma^2)^1/2
    and tried subbing in either
    gamma = τ/t

    so v = c(1 - t^2/τ^2)^1/2
    but I didn't really know where to go from there
     
  5. Apr 13, 2012 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Start here:
    Distance = v*time

    You're given the distance in the lab frame. What's the time in the lab frame in terms of the life-time τ? (Write an expression, don't try to give a numerical answer.)
     
  6. Apr 13, 2012 #5
    so v=l/t
    and t=gamma/τ
    so v = lτ/gamma

    and gamma=(1-v^2/c^2)^-1/2

    so v=lτ(1-v^2/c^2)^1/2

    v^2=(lτ)^2 (1-v^2/c^2)

    I rearranged it to get

    v^2(1 +(lτ)^2/c^2) = (lτ)^2

    So

    v^2= (lτ)^2 / (1 + (lτ)^2/c^2))

    Is that correct?
     
  7. Apr 13, 2012 #6

    Doc Al

    User Avatar

    Staff: Mentor

    OK.
    Redo this.
     
  8. Apr 13, 2012 #7
    oops that was a very silly mistake!
    So instead I should have got
    v = lgamma/τ

    so v=l/τ(1 - v^2/c^2)^1/2

    so v^2 = l^2/τ^2(1 - v^2/c^2)

    but now I'm a little stuck. How do I factorise out v from this? Or have I used a wrong substitution?
     
  9. Apr 13, 2012 #8

    Doc Al

    User Avatar

    Staff: Mentor

    Better. (At least time is in the denominator!) But your gamma is in the wrong position. Looking more carefully at your first post, I see that you have the time dilation formula backwards. (I should have caught that earlier.) Remember that the given half-life is the proper time.

    You're almost there. One more time.
     
  10. Apr 13, 2012 #9
    Yes that was a very silly mistake! Ah so I have! I always get them the wrong way round, something I need to get right for definite.

    So now I have

    v^2 = l^2/τ^2(1-(l/τc)^2)

    Is that more like it? When I tried to do it a few other ways I see that my problem was always factorising the v, although I found I needed more info (as I tried to do it with length contraction), it would have helped if I didn't have the equations the wrong way round!
     
  11. Apr 13, 2012 #10

    Doc Al

    User Avatar

    Staff: Mentor

    Almost. Do the algebra one more time. What's your starting equation before you isolated the v? (And be sure to use brackets so that it's clear what's on top or on bottom of any fraction.)
     
  12. Apr 13, 2012 #11
    okay so
    v = l/τgamma
    = (l/τ)(1 - v^2/c^2)^1/2
    so V^2 = (l/τ)^2 - (lv/τc)^2
    so v^2 + (lv/τc)^2 = (l/τ)^2
    so v^2(1 + (l/τc)^2) = (l/τ)^2
    so v^2 = l^2/(τ^2(1 + (l/τc)^2)

    Is that correct?
     
  13. Apr 13, 2012 #12
    also I've just seen some parts cancel so I'm left with

    v^2 = l^2/(τ^2 + l^2/c^2)
     
  14. Apr 13, 2012 #13

    Doc Al

    User Avatar

    Staff: Mentor

    Looks good.

    I'd add a bracket, just for added clarity:
    v^2 = l^2/[τ^2(1 + (l/τc)^2)]
     
  15. Apr 13, 2012 #14

    Doc Al

    User Avatar

    Staff: Mentor

    Good!
     
  16. Apr 13, 2012 #15
    that looks a lot more like the other formulas we've seen so assuming it is correct, thankyou very much!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook