0
$\begingroup$

Assuming $c=1$, $v=\frac{\omega}{k}=\frac{\sqrt{k^2+m^2}}{k}>1$, for $m \neq 0$. Why is it not an issue that this $v$ is greater than the speed of light?

$\endgroup$
1

1 Answer 1

2
$\begingroup$

No. The physical speed is the group velocity $$ v_g= \frac{\partial \omega}{\partial k} = \frac{k}{\sqrt{k^2+m^2}} <1. $$ The phase velocity $$ v_\phi = \frac{\omega}{k} $$ is not relevant.

$\endgroup$
4
  • $\begingroup$ Why is $v_g$ interpreted as the physical velocity? It has the units of s/m $\endgroup$
    – Ryder Rude
    Jul 22, 2022 at 18:21
  • $\begingroup$ It has the same units as $\omega/k$ i,.e dimensionless. it is the speed at which energy and information travel. $\endgroup$
    – mike stone
    Jul 22, 2022 at 18:48
  • $\begingroup$ But if we didn't set $c=1$, this would have units s/m $\endgroup$
    – Ryder Rude
    Jul 22, 2022 at 18:58
  • $\begingroup$ I got it. The numerator would have a pc^2 if we don't set natural units. $\endgroup$
    – Ryder Rude
    Jul 22, 2022 at 19:04

Not the answer you're looking for? Browse other questions tagged or ask your own question.