Relativistic Momentum Help With Equation Reduction


by jimbobian
Tags: equation, momentum, reduction, relativistic
jimbobian
jimbobian is offline
#1
Aug17-11, 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
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G01
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#2
Aug17-11, 02:38 PM
HW Helper
G01's Avatar
P: 2,688
Quote Quote by jimbobian View Post
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:

[tex] u = \frac{c^2}{v}(1-\sqrt{1-\frac{v^2}{c^2}})[/tex]
jimbobian
jimbobian is offline
#3
Aug17-11, 03:52 PM
P: 52
Quote Quote by G01 View Post
That wiki entry is not being clear with it's order of operations!

What you should be getting is:

[tex] u = \frac{c^2}{v}(1-\sqrt{1-\frac{v^2}{c^2}})[/tex]
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!

jtbell
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#4
Aug17-11, 04:09 PM
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P: 11,232

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.
jimbobian
jimbobian is offline
#5
Aug17-11, 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
PAllen
PAllen is offline
#6
Aug17-11, 04:59 PM
Sci Advisor
PF Gold
P: 4,862
Well, brute force use of the quadratic formula leads directly to the desired answer.
jimbobian
jimbobian is offline
#7
Aug19-11, 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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