[Physics] How is the envelope of a wave derived from the wave equation

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I'm reading this book about electrical properties of materials where the electron is introduced as a wave. Using the equation of a wave, they bring about the "envelope" of a wave. So here is how the derivation goes:

  1. Consider the equation of a wave traveling in one dimension:
    $ u = ae^{-i(\omega t – kz)} $ where $\omega = 2 \pi f$ is the frequency, $k$ is the wave number, and $a$ is the amplitude. In addition to this information, the book refers to the phase velocity defined as: ${\partial z}/{\partial t} = {\omega}/{k}$

  2. Instead of 1 single wave, let there be multiple waves that are superimposed such that $$u = \sum a_n e^{-i(\omega_n t – k_n z)}.$$

  3. Consider the continuous case: $$ u(z) = \int_{-\infty}^\infty a(k)e^{-i(\omega t – kz)}dk'. $$

  4. Make two more assumptions. First assumption: the waves are zero everywhere except at a small interval $\Delta k $ and here the amplitude is unity such at $a(k)=1$. This brings the equation to $$ u(z) = \int_{k_0 – \Delta k /2}^{k_0 + \Delta k /2} e^{-i(\omega t – kz)}dk .$$

  5. Suppose that $t=0$, thus yielding the endpoint: $$ u(z) = \int_{k_0 – \Delta k /2}^{k_0 + \Delta k /2} e^{i kz}dk .$$
    When the integration is carried out, the result is $$ u(z) = \Delta k e^{ik_0 z} \frac {\sin\left(\tfrac12(\Delta kz)\right)}{\tfrac12(\Delta k z)}.$$

My question is how is the integration done here? I do notice that $u$ is a function of $z$ and that the integrand integrates over the wave number $k$. Is there some change of variables going on here? By the way I only posses knowledge of ODEs and only know how to do basic Separation of variables for PDEs.

Best Answer

You have to make a first-order limited development with the group velocity :

$\omega (k)\approx \omega ({{k}_{0}})+{{\left( \frac{d\omega }{dk} \right)}_{{{k}_{0}}}}(k-{{k}_{0}})={{\omega }_{0}}+{{v}_{g}}(k-{{k}_{0}})$ and $\omega (k)t-kz={{\omega }_{0}}t-{{k}_{0}}z+(k-{{k}_{0}})({{v}_{g}}t-z)$ so ${{e}^{j(\omega (k)t-kz)}}={{e}^{j({{\omega }_{0}}t-{{k}_{0}}z)}}{{e}^{j((k-{{k}_{0}})({{v}_{g}}t-z))}}$

Then, you just have to make the same change of variable $x=k-{{k}_{0}}$ that I have indicated but with $z-vgt$ and not z alone ! $\int\limits_{{{k}_{0}}-\Delta k/2}^{{{k}_{0}}+\Delta k/2}{{{e}^{j(\omega (k)t-kz)}}}={{e}^{j({{\omega }_{0}}t-{{k}_{0}}z)}}\int\limits_{-\Delta k/2}^{+\Delta k/2}{{{e}^{i}}^{x({{v}_{g}}t-z)}dx}$

$\int\limits_{-\Delta k/2}^{+\Delta k/2}{{{e}^{i}}^{x({{v}_{g}}t-z)}dx}=\frac{1}{iz}\left( {{e}^{i}}^{(\Delta k/2)({{v}_{g}}t-z)}-{{e}^{-i}}^{(\Delta k/2)({{v}_{g}}t-z)} \right)=\frac{2}{z}\sin \left( (\Delta k/2)({{v}_{g}}t-z) \right)=\Delta k\frac{\sin \left( (\Delta k/2)(z-{{v}_{g}}t) \right)}{(\Delta k/2)(z-{{v}_{g}}t)}$

Taking the real part :

$u(z,t)=\underbrace{{{u}_{0}}\frac{\sin \left( (\Delta k/2)(z-{{v}_{g}}t) \right)}{(\Delta k/2)(z-{{v}_{g}}t)}}_{\text{Envelope : propagate at group velocity}}\cos ({{\omega }_{0}}t-{{k}_{0}}z)$

The enveloppe is spatialy limited $\Delta z=\pm 2\pi /\Delta k$ and propagate at the group velocity.

Sorry for my english !

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