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(Sp. Relativity) Krel / Kcl in powers of (v/c)^2

  1. Oct 12, 2016 #1
    1. The problem statement, all variables and given/known data

    By expanding Krel / Kcl in powers of (v/c)^2, estimate the value of v/c for which Krel differs from Kcl by 10%.

    2. Relevant equations

    Kcl = classical Kinetic Energy = 1/2 m0 v^2
    Krel = relative Kinetic Energy = (y-1) (m0 c)^2

    3. The attempt at a solution

    I did a binomial expansion wherein x = (v/c)^2 and n = -1/2
    The result is...
    [1 - (v/c)^2]^(-1/2) = 1 + 1/2(v/c)^2
    so If I plug this value into the Lorentz factor of Krel,
    I can equate K rel to the Kcl equation.

    But at which step of this expansion can I apply the 10% difference?
    Do I need to set up an equation wherein Krel = 11/10 (Kcl)?

    Thanks!
     
  2. jcsd
  3. Oct 12, 2016 #2

    Simon Bridge

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    You use, in addition to what you have done, the definition of "percentage difference".
    What was your equation for Krel/Kcl ? (What were you exanding?)
    Does that satisfy the definition?
    Why not try it out and see what you get?
     
  4. Oct 13, 2016 #3
    Hi Simon
    Thanks for the help.

    I think I am stuck at the expansion of K rel / K cl = (y-1) (m0 c^2) / (1/2)(m0 c^2)

    Is there a website link or formula that can assist me with the expansion of the above equation in the power of (v/c)^2?

    *and would it be okay for me to assume Kcl is (1/2)( m0 (v/c)^2 ) rather than with just c^2?
     
  5. Oct 13, 2016 #4
    It's been solved. Thanks again.
     
  6. Oct 13, 2016 #5

    Simon Bridge

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    Well done.
    For the benifit of others stuck in the same way, how did you solve it?

    Presumably younfigured out that Kcl = (1/2)mv^2 = (1/2)mc^2 (v/c)^2
     
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