# Derivation of group velocity?

In the standard simplified derivation of group velocity (which can be found here) we use two waves $$y_1=A\sin(K_1x-\omega_1 t)$$ $$y_2=A\sin(K_2x-\omega_2 t)$$ In the proof we then get $$V_g=\frac{\Delta \omega}{\Delta k}$$ But I do not understand the step where this is then turned into $$V_g=\frac{d \omega}{d k}$$ why do we assume that $\Delta \omega$ and $\Delta k$ are small? The derivation is valid in the case where they are not small, which means that $$V_g= \frac{d \omega}{d k}$$ does not hold in this case and therefore does not hold in general.

Consider this example

Let $K_1=3$ and $k_2=1$ and let us say we have relationship $\omega=k^3$ using my first fromula we get $V_g=13$ but using the second (with $\bar k=2$) we get $V_g=12$, theses are diferent.

• The equation $V_g = \Delta\omega/\Delta k$ is a special case that arises because of the very simple model used. It isn't generally true. See this Wikipedia article and doubtless many more easily Googlable articles for more. – John Rennie Feb 13 '15 at 7:08
• @JohnRennie but why is it not true in the case of $\Delta \omega$ and $\Delta k$ been large, I can still get an equation of the form $y=2Acos(\Delta \omega t+ \Delta k x)sin( \bar \omega t+ \bar kx)$ which indicates a group veloicty of $\Delta \omega/ \Delta k$ which would not fit with the equation $d \omega / d k$ – Quantum spaghettification Feb 13 '15 at 7:20
• If $V_g = \Delta\omega/\Delta k$ then that means $V_g = d\omega/dk$ must also be true because you're just taking the limit of $\Delta\omega \rightarrow 0$ and $\Delta k \rightarrow 0$. However $V_g = d\omega/dk$ does not imply that $V_g = \Delta\omega/\Delta k$. – John Rennie Feb 13 '15 at 7:24
• @JohnRennie but $V_g=\Delta \omega/\Delta k$ does not mean $V_g = d\omega/dk$ if $\Delta \omega$ and $\Delta k$ are large! Which is what I am asking, why do we assume they are small? – Quantum spaghettification Feb 13 '15 at 7:26
• If $V_g=\Delta \omega/\Delta k$ is true for any size of $\Delta$ it must also be true for infinitesimally small $\Delta$. If the article you cite did the derivation for a specific value of $\Delta\omega$ and $\Delta k$ then you'd have a point, but since $\omega$ and $k$ can have any value you want you can make the deltas as small as you want. – John Rennie Feb 13 '15 at 7:30

The basic answer to your question is that the derivation is only valid for small $$\Delta k$$ (corresponding to $$\Delta \omega$$) so that the beats have a much smaller wavenumber and frequency than the two waves that you are adding. The group velocity only makes sense if you have well defined envelope. In this: https://www.desmos.com/calculator/ffaybnenut , you only have a clear envelope for about $$a \geq6$$, so this means $$\Delta k$$ must be a sixth as big as $$k$$ or smaller for the derivation to apply. Mathematically there's nothing invalid about it, but the point is to capture physics.