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Relativistic Addition of Electron Velocities

  1. May 7, 2009 #1
    1. The problem statement, all variables and given/known data
    An electron moves to the right in a laboratory accelerator with a speed of 0.822c. A second electron in a different accelerator moves to the left with a speed of 0.424c relative to the first electron. Calculate the speed of the second electron (in c) relative to the lab. Do not enter unit.


    2. Relevant equations
    [tex]
    V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2}
    [/tex]


    3. The attempt at a solution
    Relativity is difficult for me to get a handle on. The way I am setting this up, is i'm using the velocity .822c for Va/b and -.424c(since it this one goes left) for Vb/c, and Vac would be the velocity of the second electron relative to the lab, however i'm not getting the right answer. I'm guessing that I don't have the velocities substituted in the proper places in the formula, any help?
     
  2. jcsd
  3. May 8, 2009 #2

    tiny-tim

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    Hi Cheezay! :smile:

    You're using a for the 1st electron, and b for the lab, then b for the second electron and c for the 1st electron :frown:

    fiddle around with it, to get it consistent, and remember that Va/b = -Vb/a :smile:
     
  4. May 8, 2009 #3
    Ok. So considering that a = electron 1, b = electron 2, and c = lab.... I now have Va/c= .822c, and because Va/b=-Vb/a.. i use -.424c for the speed of electron 2 relative to the lab. Now i solve for Vb/c. I'm fairly confident i have my equation set up correctly now, which means i'm making an algebraic error now... any help?

    Using V as a variable, in place of Vb/c

    .822c=[-.424c + V]/[1-(.424c x V)/(c^2)] c's cancel...

    .822c=[-.424c + V]/[1-(.424 x V)/c] i move the whole term...

    .822c[1-(.424 x V)/c]=-.424c + V distribute .822c (c's cancel again)

    .822c - .348528 -.822V = -.424c +V

    -.348528 - .822V = -1.246c + V

    -.348528 - V = -1.51582c + V

    -2V = -1.16729c so V= .583c which isn't correct


    Any more help would be greatly appreciated! I have been on this problem for 3 days now!
     
  5. May 8, 2009 #4

    tiny-tim

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    Hi Cheezay! :smile:
    (you mean relative to electron 1 :wink:)
    However did you get an unknown on the RHS? :redface:

    Choose a b and c (it might be easier if you call them 1 2 and L) so that the RHS contains your two knowns, and the unknown is on the LHS! :smile:
     
  6. May 9, 2009 #5
    Ok.. i've figured it out. Thanks for the help!
     
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