# Relativistic Momentum Help With Equation Reduction

by jimbobian
Tags: equation, momentum, reduction, relativistic
 P: 50 Hi, so basically have been looking at this and working my way through the maths for myself. However I have hit this point and can't get past it: \begin{align} u = \frac{v - u}{1-\frac{uv}{c^2}} \end{align} Which should be able to become: \begin{align} u = \frac{c^2}{v(1-\sqrt{1-\frac{v^2}{c^2}})} \end{align} I have tried and tried but can't seem to get it to work. Can anyone help me out on this one please?! Thanks James
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 Quote by jimbobian Hi, so basically have been looking at this and working my way through the maths for myself. However I have hit this point and can't get past it: \begin{align} u = \frac{v - u}{1-\frac{uv}{c^2}} \end{align} Which should be able to become: \begin{align} u = \frac{c^2}{v(1-\sqrt{1-\frac{v^2}{c^2}})} \end{align} I have tried and tried but can't seem to get it to work. Can anyone help me out on this one please?! Thanks James
That wiki entry is not being clear with it's order of operations!

What you should be getting is:

$$u = \frac{c^2}{v}(1-\sqrt{1-\frac{v^2}{c^2}})$$
P: 50
 Quote by G01 That wiki entry is not being clear with it's order of operations! What you should be getting is: $$u = \frac{c^2}{v}(1-\sqrt{1-\frac{v^2}{c^2}})$$
Hi G01, thanks for your reply. Have tried aiming for that equation instead and still can't get there. I have tried all sorts of different approaches, such as getting it in the form of a quadratic or dividing the top and bottom of the original fraction by c^2, but I just end up nowhere. Could you perhaps give me a hand in the right direction, maybe the first step or two - but don't make it too easy for me!

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## Relativistic Momentum Help With Equation Reduction

Show us exactly what you've tried, and where you get stuck, and tell us why you're stuck.
 P: 50 Ok, have shut my computer down now and am replying on my phone. I will post my attempts tomorrow. Thanks, James