abskebabs Posted June 3, 2007 Share Posted June 3, 2007 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. Link to comment Share on other sites More sharing options...
Klaynos Posted June 3, 2007 Share Posted June 3, 2007 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 Link to comment Share on other sites More sharing options...
abskebabs Posted June 4, 2007 Author Share Posted June 4, 2007 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:-) Link to comment Share on other sites More sharing options...
the tree Posted June 4, 2007 Share Posted June 4, 2007 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. Link to comment Share on other sites More sharing options...
Klaynos Posted June 4, 2007 Share Posted June 4, 2007 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. Link to comment Share on other sites More sharing options...
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