I recently stumbled across a small peculiarity I don't understand:

According to de Broglie, the frequency of a matterwave can be written as $f=\frac{E}{h}$, and its wavelength as $\lambda = \frac{h}{p}$.

However, relativistically for a particle with matter the following formulae are valid: $E = \gamma m_{0} c^{2}$, and $p = \gamma m_{0}v$, where $m_{0}$ is the particle's rest mass.

So, with $v=\lambda f$ (with the particle's velocity v, unequal the speed of light!), using de Broglie we would end up with $E=v p$, which is clearly not valid when inserting the relativistic energy and momentum equations:

$v*p = \gamma m_{0} v^{2} \neq E = \gamma m_{0} c^{2}$.

Where am I mistaken?

  • 3
    $\begingroup$ One must distinguish between group velocity and phase velocity. Possible duplicates: physics.stackexchange.com/q/34214/2451 and links therein. $\endgroup$
    – Qmechanic
    Mar 2 '14 at 13:45
  • $\begingroup$ I don't understand your problem sorry: how can you put v=λf and then arrive to E=vp? I don't know if it could help you but remember that the dispersion relations are different in light and matter, this is the kernel of the discussion. $\endgroup$
    – LC7
    Mar 2 '14 at 13:50
  • 1
    $\begingroup$ $v=\lambda f$ is valid for the wave, and if I insert f and $\lambda$ according to de Broglie, I end up with $E=vp$. But I already found my problem (thanks Qmechanic): I confused phase and group velocity. $E/p = v_{phase} = \frac{c^{2}}{v_{group}}$, which also results when inserting the de Broglie formulae. $\endgroup$ Mar 2 '14 at 14:06

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