time of flight between two scintillators I found this page on Wikipedia about finding distance between time of flight of two particles passing past two scintillators, https://en.wikipedia.org/wiki/Time_of_flight_detector, but I can't find any other proofs or anything other than simply showing this equation. Does anyone know where this is derived? Or possibly shown for relativistic speeds.
 A: The equation is
$$\delta t = L\left(\frac{1}{v_1}-\frac{1}{v_2}\right)$$
This is just saying that the difference in time recorded between the "start" and the "stop" depends on the velocity of the particle: the time it takes a particle with $v_1$ to travel a distance $L$ is $\frac{L}{v_1}$, obviously.
The next bit is trickier. We are supposed to prove that this reduces to
$$\delta t \approx \frac{Lc}{2p^2}\left(m_1^2 - m_2^2\right)$$
when the particles have the same momentum $p$, and we are told in the article that this is a relativistic approximation. 
Looking at the units, we have $L L T^{-1}M^{-2}L^{-2}T^2M^2 = T$ - which is encouraging.
Relativistically,
$$E^2 = m^2c^4 = p^2 c^2 + m_0^2 c^4$$
Assuming that the mass in the equation is the relativistic mass, and that this is much larger than the rest mass, we find 
$$p^2 = m^2 c^2 - m_0^2 c^2$$
Now the relationship between the mass $m$ and the rest mass $m_0$ is of course given by
$$m= \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$
Or
$$m^2~\left(1-\frac{v^2}{c^2}\right)=m_0^2$$
and the momentum reduces to
$$p^2 = m^2 v^2$$
which is the same as the classical equation.
Right now I'm blocking on how to take the last step here... I am marking this "community wiki" and leaving this for someone else to complete. Maybe I will be able to do it myself with fresh eyes in the morning.
A: What you need is $v$ in terms of $m$., $p$ and $c$.
Starting from $E^2 = p^2c^2+m^2c^4 = \gamma^2 m^2c^4$ or $p=\gamma mv$ where $\gamma^2 = \dfrac{1}{\left ( 1-\frac{v^2}{c^2}\right)}$ 
some algebra gives $\dfrac 1 v = \dfrac 1 c \cdot \sqrt{\left (1+\dfrac{m^2c^2}{p^2}\right) }$
Because $p\gg mc$ the square root can be expanded to the first term  
$\sqrt{\left (1+\dfrac{m^2c^2}{p^2}\right) } \approx 1+ \dfrac{m^2c^2}{2p^2} + . . .$ 
and this leads to the required relationship because the momentum $p$ is the same for both particles.
PS
I am afraid I do not understand what @Floris meant by "I am marking this "community wiki" and leaving this for someone else to complete."  An explanation would be appreciated.
