## (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.
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 Mentor Blog Entries: 1 Replace $$v^2/c^2$$ with x, thus x << 1. Then do a binomial expansion of $$(1 - x)^{-1/2}$$.
 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: $$\gamma=(1-m/2*v^2/c^2)^{-m}$$

Mentor
Blog Entries: 1

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

 Quote by msimmons I was under the impression that the binomial theorem worked only when n is a natural number..
No. It works just fine for negative numbers and fractions.
 Recognitions: Homework Help Science Advisor You can also think of it as the first two terms in the Taylor series for (1-x)^(-1/2).
 Working with the binomial expansion, if I want to evaluate $$(1-x)^{-1/2}$$ I thought I would get something like... $$(\stackrel{-1/2}{0})1-(\stackrel{-1/2}{1})x+(\stackrel{-1/2}{2})x^2...$$ I thought that was right, but $$(\stackrel{-1/2}{1})$$ and the likes can't be evaluated, can they? Hope my attempt at binomial coefficients aren't too funky looking.
 Recognitions: Homework Help Science Advisor Just do it as a Taylor series to avoid these complicated questions.

Mentor
Blog Entries: 1
 Quote by msimmons Working with the binomial expansion, if I want to evaluate $$(1-x)^{-1/2}$$ I thought I would get something like... $$(\stackrel{-1/2}{0})1-(\stackrel{-1/2}{1})x+(\stackrel{-1/2}{2})x^2...$$
It's simpler than you think. Read this: Binomial Expansion
 Doh... thank you. that solves all of them.