# Group velocity of relativistic wave packet

1. Jul 24, 2006

Hi, I'm trying to figure out what the group and phase velocities of a wave packet describing a particle are in the relativistic case.

I started with the relationship between energy and impulse : E squared = p squared X c squared + rest mass squared X c to the fourth. In this, I input the Planck-Einstein relationships : E = h bar X omega and p = h bar X k.

I divide each side by h bar squared, take the square root to get omega, then divide by k to get the phase velocity or take the derivative relative to k to get the group velocity.

Here is what I get : phase velocity = c X square root of (one + (rest mass X c/p)squared)

group velocity = c / square root of (one + (rest mass X c/p)squared)

So that v phase X v group = c squared

I could not find this result anywhere so I have a doubt : can anyone tell me if I'm doing this correctly?

If it's correct, it's pretty interesting : we have a group velocity that is always less than c (as it should) and a phase velocity that is always greater than c, as if the particle was governed by waves that can't travel at less than the speed of light, just like tachyons...

Anyway, I'd be happy just knowing if I have the correct result for this, thanks for helping.

2. Jul 24, 2006

### Staff: Mentor

Yes, that result is correct. Many introductory modern physics textbooks derive

$$v_{phase} = \frac {c^2} {v_{particle}}$$

$$v_{group} = v_{particle}$$

which are equivalent to your result.

By the way, you can make it easier for others to read your equations by studying this thread: