Relativistic Momentum Help With Equation Reductionby jimbobian Tags: equation, momentum, reduction, relativistic 

#1
Aug1711, 01:13 PM

P: 52

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 



#2
Aug1711, 02:38 PM

HW Helper
P: 2,688

What you should be getting is: [tex] u = \frac{c^2}{v}(1\sqrt{1\frac{v^2}{c^2}})[/tex] 



#3
Aug1711, 03:52 PM

P: 52





#4
Aug1711, 04:09 PM

Mentor
P: 11,255

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.




#5
Aug1711, 04:15 PM

P: 52

Ok, have shut my computer down now and am replying on my phone. I will post my attempts tomorrow.
Thanks, James 



#6
Aug1711, 04:59 PM

Sci Advisor
PF Gold
P: 4,863

Well, brute force use of the quadratic formula leads directly to the desired answer.




#7
Aug1911, 09:27 AM

P: 52

Haha, went back over my general quadratic attempt after PAllen's suggestion and realised that I had made a rather fundamental cross multiplication error! Fixed that and got to the desired equation. Lovely!
Last question, the equation that I get is: \begin{align} u = \frac{c^2}{v}(1\pm\sqrt{1\frac{v^2}{c^2}}) \end{align} I assume that we choose the negative version of the equation, because the positive version would yield a value for u which is greater than c? Cheers 


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