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(special relativity) derivation of gamma with approximation of v << c |
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| Sep20-07, 02:43 PM | #1 |
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(special relativity) derivation of gamma with approximation of v << c
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. |
| Sep20-07, 03:06 PM | #2 |
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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].
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| Sep20-07, 06:15 PM | #3 |
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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] |
| Sep20-07, 06:21 PM | #4 |
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(special relativity) derivation of gamma with approximation of v << c |
| Sep20-07, 06:31 PM | #5 |
Recognitions:
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You can also think of it as the first two terms in the Taylor series for (1-x)^(-1/2).
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| Sep20-07, 08:51 PM | #6 |
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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. |
| Sep20-07, 09:53 PM | #7 |
Recognitions:
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Just do it as a Taylor series to avoid these complicated questions.
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| Sep21-07, 05:08 AM | #8 |
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| Sep21-07, 09:59 AM | #9 |
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Doh... thank you. that solves all of them.
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