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The dispersion relation can be expressed as:

$$w=\frac{\hbar k^2}{2m}$$

The energy of a free particle is:

$$E=\sqrt{p^2c^2+m^2c^4}$$

For a highly energetic electron we have $E\approx pc.$

The group velocity of the waves associated to the electron is:

$$v_g=\frac{\partial w}{\partial k}$$

and $$\beta=\frac{\partial^2w}{\partial k^2}$$

Why is it true that for a highly energetic electron, $\beta\approx0$? Is it because $v_g\approx c$ which is constant and so $\frac{\partial v_g}{\partial k}=0$?

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Basically yes. Firstly instead of your dispersion relation you must now use that $E = \hbar \omega$ and that $p = \hbar k$, then your new dispersion relation (valid for energetic electrons) is \begin{equation} \omega = k c \end{equation} and the result follows.

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