A New Look at Momentum — Question 1

Question 1

Momentum determines a bounded speed even when p exceeds mc

A massive particle has momentum magnitude $p=3mc$. Use $\mathbf p=\gamma m\mathbf v$, $\gamma=(1-v^2/c^2)^{-1/2}$, and $\mathbf F=d\mathbf p/dt$. Here m is invariant mass.

1. Invert the momentum relation to find its speed and Lorentz factor.

2. Compare its momentum with the classical value at that same speed.

Question 1 - Solution

Momentum determines a bounded speed even when p exceeds mc

(a) Method and calculation

$$\left(\frac p{mc}\right)^2=\frac{\beta^2}{1-\beta^2}\quad\Rightarrow\quad\boxed{\beta=\frac{p/(mc)}{\sqrt{1+[p/(mc)]^2}}}.$$$$\boxed{v=3c/\sqrt{10}},\qquad\boxed{\gamma=\sqrt{10}}.$$

(b) Interpretation and further calculation

$$p_{\mathrm{cl}}=mv=3mc/\sqrt{10}.$$$$\boxed{p/p_{\mathrm{cl}}=\sqrt{10}},\qquad v<c.$$

Physical check

Momentum has no finite upper bound, but a finite momentum always gives a subluminal speed for nonzero mass.