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

 
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Sep20-07, 02:43 PM   #1
 

(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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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].
Sep20-07, 06:15 PM   #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]
Sep20-07, 06:21 PM   #4
 
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(special relativity) derivation of gamma with approximation of v << c


Quote by msimmons View Post
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.
Sep20-07, 06:31 PM   #5

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You can also think of it as the first two terms in the Taylor series for (1-x)^(-1/2).
Sep20-07, 08:51 PM   #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.
Sep20-07, 09:53 PM   #7

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Just do it as a Taylor series to avoid these complicated questions.
Sep21-07, 05:08 AM   #8
 
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Quote by msimmons View Post
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]
It's simpler than you think. Read this: Binomial Expansion
Sep21-07, 09:59 AM   #9
 
Doh... thank you. that solves all of them.
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