PX155 - I2 - relativistic momentum

derivation

$p$ is conserved for $v << c$ , and we need a linear transformation
try $m = f (v) m_{0}$ , where $m_{0} =$ "rest mass", $f (v) \to 1$ as $v \to 0$
now, $p = m v = f (v) m_{0} u = 0$
conserve momentum and mass in the centre of mass frame, $S^{'} :$
$f (u) m_{0} u - f (- u) m_{0} u = 0 . . . [i]$ $⟹ f (u) = f (- u)$ $f (u) m_{0} + f (- u) m_{0} = M_{0} . . . [i i]$
do the same in $S$ :
$f (v) m_{0} v = f (u) M_{0} u . . . [i i i]$ $f (v) m_{0} + m_{0} = f (u) M_{0} . . . [i v]$
$[i]$ in $[i i]$ :

2 f (u) m_{0} = M_{0}

f (v) m_{0} v = 2 f^{2} (u) u m_{0} ⟹ f (v) v = 2 f^{2} (u) u . . . [v]

o r, f (v) + 1 = 2 f^{2} (u) . . . [v i]

divide $[v]$ by $v$ and subtract from $[v i]$ : $$f(v)+1-f(v)= 2f^{2}(u) - 2f^{2}(u) \frac{u}{v}$$

o r, f^{2} (u) (2 - 2 \frac{u}{v}) = 1

o r 2 \frac{u}{v} = 1 + \frac{u^{2}}{c^{2}}

o r, f (u) = \frac{1}{\sqrt{\frac{1 - u^{2}}{c^{2}}}} = γ (u)

p = γ (v) m_{0} v

- this gives us relativistic mass and momentum

F = \frac{d p}{d t} ⟹ Δ p = \int_{t_{0}}^{t_{f}} F . d t^{'}

γ (v) m_{0} v = γ (w) M_{0} w

\frac{5}{3} m_{0} \frac{4}{3} c = \frac{8}{3} m_{0} w

∴ w = \frac{1}{2} c

γ (w = \frac{c}{2}) = 1.1547

- from $[i]$ :

M_{0} = \frac{8}{3 \times 1.1547} = 2.309 m_{0}