Modern Physics/Characteristics of Relativistic Waves

In classical physics, &omega; and k for light are related by


 * $$\omega = c k \,$$

In relativistic physics, we've seen that for waves with no special reference frame, such as light, &omega; and k are related by


 * $$\omega^2 = c^2 k^2 + \mu^2 \,$$

If &mu;=0 then the relativistic equation reduces to the classical, so we can assume that, for light, &mu; does equal zero.

This means that light does not have a mimimum frequency.

If &mu; is not zero then the wave being described are dispersive. The phase speed is
 * $$u_p = \frac{\omega}{k} = \sqrt{c^2 + \frac{\mu^2}{k^2}} $$

This phase speed always exceeds c, which at first may seem like an unphysical conclusion. However, the group velocity of the wave is
 * $$u_g = \frac{d\omega}{dk} = \frac{kc^2}{\sqrt{k^2c^2+\mu^2}}

= \frac{kc^2}{\omega} = \frac{c^2}{u_p}$$

which is always less than c. Since wave packets and hence signals propagate at the group velocity, waves of this type are physically reasonable even though the phase speed exceeds the speed of light.

Another interesting property of such waves is that the wave four-vector is parallel to the world line of a wave packet in spacetime. This is easily shown by the following argument.

The spacelike component of a wave four-vector is k, while the timelike component is &omega;/c. The slope of the four-vector on a spacetime diagram is therefore &omega;/kc. However, the slope of the world line of a wave packet moving with group velocity is c/ug, which is also &omega;/kc.

Note that when we have k is zero we have &omega;=&mu;. In this case the group velocity of the wave is zero. For this reason we sometimes call &mu; the rest frequency of the wave.