B Why Does This Algebraic Identity Work in Relativistic Doppler Calculations?

[Daniel Sellers]

I seem to remember this Algebra identity being covered in one of my classes years ago, but it has cropped back up in studying the relativistic doppler effect for light.

Can anyone please show me the intermediate steps to show that:

(1+x)/(sqrt(1-x^2) = sqrt((1+x)/(1-x))

or similarly

(sqrt(1-x^2)/(1+x) = sqrt((1-x)/(1+x))

I can solve problems well enough by factoring gamma out of these equations but it is bugging me that all the texts I can find keep taking this for granted and I can't see why.

 

[Charles Link]

$1-x^2 =(1-x)(1+x)$. Comes from $a^2-b^2=(a-b)(a+b)$. The rest is just things like $\frac{u^1}{u^{1/2}}=u^{1/2}$ etc. where $u^{1/2}=\sqrt{u}$.

 

[Daniel Sellers]

I knew it was something obnoxiously simple and obvious! Thanks very much!

 

[Ssnow]

Suggestion: to see quickly you can take the square from both side $\frac{(1+x)^2}{1-x^2}\,=\, \frac{1+x}{1-x}$, now it is quite obvious ...
Ssnow

 

[Mark44]

Suggestion: to see quickly you can take the square from both side $\frac{(1+x)^2}{1-x^2}\,=\, \frac{1+x}{1-x}$, now it is quite obvious ...
Ssnow

Easy -- just cancel the exponents!
$$ \frac{(1 + x)^2}{1 - x^2} = \frac{(1 + x)^{\rlap{/}2}}{1 - x^{\rlap{/}2}} = \frac{1 + x}{1 - x}$$