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## Homework Statement

In relativistic wave mechanics the dispersion relation for an electron of velocity [tex]v=\frac{\hbar k}{m}[/tex] is given by [tex]\frac{\omega ^2}{c^2}=k^2 + \frac{m^2 c^2}{\hbar ^2}[/tex] where c is the velocity of light, m is the electron mass (considered constant at a given velocity) [tex]\hbar=\frac{h}{2\pi}[/tex] and h is Planck's constant.

Show that the product of the group and particle velocities is [tex]c^2[/tex]

## Homework Equations

[tex]v_g=\frac{d\omega}{dk}[/tex]

## The Attempt at a Solution

From the dispersion relation I got

[tex]\frac{\omega ^2}{c^2}=k^2 + \frac{m^2 c^2}{\hbar ^2}[/tex]

[tex]\omega = c\sqrt{k^2 + \frac{m^2c^2}{\hbar ^2}}[/tex]

so that

[tex]v_g=\frac{d\omega}{dk}[/tex]

[tex]v_g=\frac{d}{dk}(c\sqrt{k^2 + \frac{m^2c^2}{\hbar ^2}})[/tex]

[tex]v_g=\frac{ck}{\sqrt{k^2 + \frac{m^2c^2}{\hbar ^2}}}[/tex]

But this answer, multiplied with the particle velocity will obviously not give c^2. What am I missing?

Thanks!

phyz