- #1
jimbobian
- 52
- 0
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:
[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
[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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