Theory of relativity - proton entering a medium

[Saitama]

Start with
$\sqrt{c^2p^2+m_0^2c^4} - \sqrt{c^2p_1^2+m_0^2c^4} = c_m(p - p_1)$

and divide through by $m_0c^2$. You will find that the equation simplifies to

$\sqrt{[p/(m_0c)]^2+1} - \sqrt{[p_1/(m_0c)]^2+1} = (c_m/c)[(p/(m_0c) - p_1/(m_0c)]$

So, if you measure p in units of $m_0c$ and measure $c_m$ in units of $c$, you get

$\sqrt{p^2+1} - \sqrt{p_1^2+1} = c_m(p - p_1)$

Going back to the original equation and letting $E(p) = \sqrt{c^2p^2+m_0^2c^4}$, you have $E(p) - E(p_1) = c_m(p-p_1)$. Or,
$$\frac{E(p) - E(p_1)}{p-p_1} = c_m$$
The left-hand side is the slope of the line that connects two points on the $E(p)$ vs $p$ graph corresponding to initial and final momenta ($p$ and $p_1$) of the proton. So the above equation says that the slope of the line must be equal to $c_m$ if energy and momentum are to be conserved. Thus if you draw any straight line of slope $c_m$ such that it passes through two points of the graph of $E(p)$, the two points will correspond to possible initial and final momenta of the proton such that energy and momentum will be conserved in photon emission. If you draw the line of slope $c_m$ such that it is tangent to the graph of $E(p)$, then you will have the case where the initial and final momenta of the proton are equal and the emitted photon would have zero energy. The slope of the tangent line is the derivative $E\:'(p)$. But $E\:'(p)$ can easily be shown to equal the speed of the proton. So, the minimum speed of the proton for photon emission is $c_m$.

Great explanation. Thanks a lot TSny!