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(special relativity) derivation of gamma with approximation of v c

  1. Sep 20, 2007 #1
    1. The problem statement, all variables and given/known data
    "Use the binomial expansion to derive the following results for values of v << c.
    a) γ ~= 1 + 1/2 v2/c2
    b) γ ~= 1 - 1/2 v2/c2
    c) γ - 1 ~= 1 - 1/γ =1/2 v2/c2"
    (where ~= is approximately equal to)

    2. Relevant equations
    As far as I can tell, just
    γ = (1-v2/c2)-1/2


    3. The attempt at a solution
    Basically, I have no idea where to apply the v << c approximation, other than right after an expansion (keeping the first few terms and dropping the rest, if it were something like (c+v)n

    so all I have really is circular math starting with 1-v2/c2 and ending with c2-v2.. which doesn't help.

    for example..
    γ = (1-v2/c2)-1/2
    γ-2 = 1 - v2/c2
    at this point I would multiply by c2 or c, and attempt to continue, but I can't figure out what to do.

    Just a hint on where to go would be greatly appreciated.
     
  2. jcsd
  3. Sep 20, 2007 #2

    Doc Al

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    Replace [tex]v^2/c^2[/tex] with x, thus x << 1. Then do a binomial expansion of [tex](1 - x)^{-1/2}[/tex].
     
  4. Sep 20, 2007 #3
    I was under the impression that the binomial theorem worked only when n is a natural number.. so I rose the power of each side by 4, then 6 to see what would happen... (their negatives, actually)
    my result (if m is the number I'm raising the eqn to) is essentially:
    [tex]\gamma=(1-m/2*v^2/c^2)^{-m}[/tex]
     
  5. Sep 20, 2007 #4

    Doc Al

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    No. It works just fine for negative numbers and fractions.
     
  6. Sep 20, 2007 #5

    Dick

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    You can also think of it as the first two terms in the Taylor series for (1-x)^(-1/2).
     
  7. Sep 20, 2007 #6
    Working with the binomial expansion,
    if I want to evaluate [tex](1-x)^{-1/2}[/tex] I thought I would get something like...

    [tex](\stackrel{-1/2}{0})1-(\stackrel{-1/2}{1})x+(\stackrel{-1/2}{2})x^2...[/tex]

    I thought that was right, but [tex](\stackrel{-1/2}{1})[/tex] and the likes can't be evaluated, can they?

    Hope my attempt at binomial coefficients aren't too funky looking.
     
  8. Sep 20, 2007 #7

    Dick

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    Just do it as a Taylor series to avoid these complicated questions.
     
  9. Sep 21, 2007 #8

    Doc Al

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    It's simpler than you think. Read this: Binomial Expansion
     
  10. Sep 21, 2007 #9
    Doh... thank you. that solves all of them.
     
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