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

[msimmons]

Homework Statement

"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)

Homework Equations

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

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.

 

[Doc Al]

Replace $$v^2/c^2$$ with x, thus x << 1. Then do a binomial expansion of $$(1 - x)^{-1/2}$$.

 

[msimmons]

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:
$$\gamma=(1-m/2*v^2/c^2)^{-m}$$

 

[Doc Al]

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.

 

[Dick]

You can also think of it as the first two terms in the Taylor series for (1-x)^(-1/2).

 

[msimmons]

Working with the binomial expansion,
if I want to evaluate $$(1-x)^{-1/2}$$ I thought I would get something like...

$$(\stackrel{-1/2}{0})1-(\stackrel{-1/2}{1})x+(\stackrel{-1/2}{2})x^2...$$

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.

 

[Dick]

Just do it as a Taylor series to avoid these complicated questions.

 

[Doc Al]

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

 

[msimmons]

Doh... thank you. that solves all of them.