Hi everyone. So basically I am still struggling to find a description of the derivation of relativistic momentum (via relativistic mass) which explains itself properly (although that may be my fault for not understanding). So, I tried doing it with help from Feynman and can't work out what I'm doing wrong. It starts with a collision of two identical particles which, from what I've seen, seems to be the standard way to go:(adsbygoogle = window.adsbygoogle || []).push({});

[URL]http://www.cassiobury.net/images/momentum_diagram.jpg[/URL]

I'm also using the velocity addition formulae which were derived in the previous chapter:

\begin{align}

& v_x = \frac{v'_x+u}{1+uv'_x/c^2}\\

& v_y = \frac{v'_y\sqrt{1-u^2/c^2}}{1+uv'_x/c^2}\\

\end{align}

And the reverse:

\begin{align}

& v'_x = \frac{v_x-u}{1-uv_x/c^2}\\

& v'_y = \frac{v_y\sqrt{1-u^2/c^2}}{1-uv_x/c^2}\\

\end{align}

So, from the unprimed frame it is clear that:

\begin{align}

& v_{xA} = 0\\

\end{align}

Also let:

\begin{align}

& v_{yA} = w\\

\end{align}

Therefore:

\begin{align}

& v'_{xA} = \frac{0-u}{1-0/c^2}\\

& v'_{xA} = -u\\

& v'_{yA} = \frac{w\sqrt{1-u^2/c^2}}{1-0/c^2}\\

& v'_{yA} = w\sqrt{1-u^2/c^2}

\end{align}

As the velocity of B is simply the opposite of A:

\begin{align}

& v'_{xB} = u\\

& v'_{yB} = -w\sqrt{1-u^2/c^2}

\end{align}

Therefore, using the velocity addition formulae:

\begin{align}

& v_{xB} = \frac{2u}{1+u^2/c^2}\\

& v_{yB} = \frac{-w(1-u^2/c^2)}{1+u^2/c^2}\\

\end{align}

If we say that the mass of each particle is some function of its velocity and denote the velocity of B as v (as that of A is clearly w) then, by conservation of momentum:

\begin{align}

& 2m_ww = -\frac{-2m_vw(1-u^2/c^2)}{1+u^2/c^2}\\

& m_w = \frac{m_v(1-u^2/c^2)}{1+u^2/c^2}\\

\end{align}

Now according ot Mr Feynman what I should be getting is:

\begin{align}

& \frac{m_w}{m_v}\sqrt{1-u^2/c^2} = 1

\end{align}

Can somebody please either point out a mistake in my reasoning/understanding, or tell me why it hasn't worked.

Thanks,

James

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# Relativistic Momentum Derivation Help

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