Quick rearrangement query

[abskebabs]

Hi everybody, I'm well into my relativity revision, and was just going through deriving the relativistic "velocity addition" forumula, but came across an algebraic stumbling block along the way. Basically, you could say I need to get the identity:

[math]c=a+b[/math]

From

[math]c=1+ab-{((1-a^2)(1-b^2))}^{1/2}[/math]

Any ideas? Help on this would be greatly appreciated.

[Klaynos]

I haven't worked it through (doing quantum statistical mechanics revision here) but...

 

(c-ab-1)²=(1-b²-q²+a²b²

 

Try working that through, hopefully everything will cancel nicely

[abskebabs]

I haven't worked it through (doing quantum statistical mechanics revision here) but...

 

(c-ab-1)²=(1-b²-q²+a²b²

 

Try working that through, hopefully everything will cancel nicely

 

Just from inspection I'm not sure if that'll work. lets have a go:

................................

Ok from trying that I get something like this:

[math]c=-1-ab+-\sqrt{1+a^2b^2-4a^2-4b^2}[/math]

I'm starting to doubt whether my initial equation was correct. In any case I'll just display the overall equation and what I need to get.

Ok, starting from:

[math]\gamma_{31}=(\frac{1}{c^2})\gamma_{32}\gamma_{21}(c^2+v_{32}v_{21})[/math]

Where: [math]\gamma=\frac{1}{\sqrt{1-\beta^2)}}[/math] and [math]\beta=\frac{v}{c}[/math]

I need to get:

[math]\beta_{31}=\frac{\beta_{32}+\beta_{21}}{1+\beta_{32}\beta_{21}}[/math]

Thanks for the help anyway:-)

[the tree]

I think Klay meant something like

[math]\begin{array}{rcl}

c & = & 1+ab-\sqrt{(1-a^2)(1-b^2)} \\

\sqrt{(1-a^2)(1-b^2)} & = & 1 + ab - c \\

(1-a^2)(1-b^2) & = & (1+ab-c)^2 \\

\end{array}[/math]

Which to be honest I can't see how it cancels down, I guess you might want to double check your original equation.

[Klaynos]

I Just worked it through from the original equation and I get:

 

c²-2c-2abc=-b²-a²

 

So I'd say rework out your original equation.