The Wikipedia page on group velocity uses the superposition of two cosines (of slightly different wavenumbers and angular frequencies) to show that group velocity $v_g = \frac{d\omega}{dk}.$

The sum yields to the product of two cosines $ 2 cos(kx-\omega t)cos(\Delta k x - \Delta \omega t)).$

The right-hand cosine plays the role of modulator and it's phase velocity is $ \frac{\Delta \omega}{\Delta k}$

So that all makes sense and hence, we have $v_g = \frac{d\omega}{dk}.$

What I quite haven't understand is the continuous superposition.

For the continuous superposition the formula is $$ \int dk \Phi (k) e^{i(kx-\omega t)} .$$

$\Phi (k)$ being the amplitude (we make the assumption that it narrowly peaks around a certain value $k_0$), and the $exp$ preferably used because matter waves can't be made of $cos$ ...

The (continuous) summation is made between different $exp$ at slightly different wavenumbers (around $k_0$) BUT I can't see how the summation is made at slighly different angular frequencies since $\omega$ is now a function of k whose shape is absolutely unknown (if $k_1$ and $k_2$ are close we don't have the guanrantee that $\omega (k_1)$ will be close to $\omega (k_2)$).

So why wouldn't, similarly to the discrete case, be a summation over slightly different $\omega$'s?


1 Answer 1


The continuity of the function $\omega(k)$ guarantees that $\omega(k_1)$ and $\omega(k_2)$ will be close if $k_1$ and $k_2$ are sufficiently close. Formally, the continuity of $\omega(k)$ at $k_1$ means that you can make $|\omega(k_2) - \omega(k_1)|$ arbitrarily small by choosing $k_2$ such that $|k_2 - k_1|$ is sufficiently small.

If $\omega$ is a monotonic function of $k$ (at least in an interval where $\Phi(k)$ is non-zero), it is invertible, so you can express $k$ as a function of $\omega$ and use frequency as the integration variable via a change of variables. Namely, given the function $k(\omega)$ with $k_1 = k(\omega_1)$ and $k_2 = k(\omega_2)$,

$$ \int\limits_{k_1}^{k_2}dk\ \Phi(k)e^{i[kx-\omega(k)t]} = \int\limits_{\omega_1}^{\omega_2}d\omega\ \frac{dk}{d\omega}(\omega)\ \Phi\left(k(\omega)\right)e^{i[k(\omega)x-\omega t]} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.