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

AI Thread Summary
The discussion focuses on deriving the gamma factor (γ) in special relativity using the binomial expansion for cases where velocity (v) is much less than the speed of light (c). Participants suggest substituting v²/c² with a small variable (x) to simplify the expansion of (1 - x)^(-1/2). The conversation emphasizes that the binomial theorem can be applied to negative and fractional powers, contrary to some initial beliefs. A Taylor series approach is recommended as a simpler alternative for evaluating the expansion. The thread concludes with a participant expressing gratitude for the clarification, indicating that the binomial expansion method resolves the derivation issues.
msimmons
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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.
 
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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}
 
msimmons said:
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.
 
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.
 
Just do it as a Taylor series to avoid these complicated questions.
 
msimmons said:
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.
 

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