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.


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 !

| cite | improve this answer | |
  • $\begingroup$ The envelope of the signal should depend on time, and not on the coordinate. It is necessary to count in dependence on time. $\endgroup$ – Evgeniy Yakubovskiy Jan 26 '19 at 8:40
  • $\begingroup$ I was at $t = 0$. I have changed adding the group velocity ! $\endgroup$ – Vincent Fraticelli Jan 26 '19 at 11:40
  • $\begingroup$ As you imagine the envelope measurement. It is necessary that the envelope meter move along with the wave and take readings for different z. How real it is. $\endgroup$ – Evgeniy Yakubovskiy Jan 26 '19 at 11:56
  • $\begingroup$ I have completed with the complete signal ! (sorry for my english) $\endgroup$ – Vincent Fraticelli Jan 26 '19 at 12:24
  • 1
    $\begingroup$ What are the limits of applicability of your formula and what it describes and what does not. $\endgroup$ – Evgeniy Yakubovskiy Jan 27 '19 at 11:32
  1. 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?

You're overthinking this. The only thing that's happening here is the integral of an exponential function, $$ \int e^{i kz}dk = \left. \frac{1}{iz} e^{ikz} \right. , $$ taken over the limits as specified, \begin{align} \int_{k_0 - \Delta k /2}^{k_0 + \Delta k /2} e^{i kz}dk & = \left. \frac{1}{iz} e^{ikz} \right|_{k_0 - \Delta k /2}^{k_0 + \Delta k /2} \\ & = \frac{ e^{i(k_0 + \Delta k /2)z} - e^{i(k_0 - \Delta k /2)z} }{iz} , \end{align} and then simplified by (i) taking out the common factor of $e^{ik_0 z}$ and (ii) multiplying by $\frac12 \Delta k$ on the numerator and denominator to clean things up: \begin{align} u(z) = \int_{k_0 - \Delta k /2}^{k_0 + \Delta k /2} e^{i kz}dk & = \frac{ e^{i(k_0 + \Delta k /2)z} - e^{i(k_0 - \Delta k /2)z} }{iz} \\ & = e^{ik_0 z} \frac{\Delta k}{\frac12 \Delta k \, z}\frac{ e^{i\frac12 \Delta k \, z} - e^{- i\frac12 \Delta k \, z} }{2i} \\ & = \Delta k \, e^{ik_0 z} \frac{\sin\left(\frac12 \Delta k \, z\right)}{\frac12 \Delta k \, z} . \end{align}

On the other hand, it's important to make a few more observations about the argument as you've reported it.

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 $$ u(z) = \int_{k_0 - \Delta k /2}^{k_0 + \Delta k /2} e^{-i(\omega t - kz)}dk .$$

The assumption that the wave amplitude $a(k)$ is only nonzero over a small interval is perfectly reasonable, but it's important to note that the two subsequent assumptions,

  • that it looks like a 'boxcar' function, with $a(k) = 1$ over that interval and then immediate, sharp cuts at the edges, and
  • that the phase is flat,

need not represent reality. In particular, the sharp discontinuities at $k_0 \pm \Delta k/d$ mean that $a(k)$ is not continuous, and that has a strong effect on the time-domain shape of the pulse: it introduces a lot of ringing, i.e. the lobes at $|z|\geq 2\pi/\Delta k$, with a weight that goes down as $\sim 1/z$. That looks reasonable, and indeed it's square-integrable, but it's not even fast enough for a standard deviation to be well-defined, let alone higher-order moments.

To see what I mean, consider swapping out your envelope for something that's completely smooth, like e.g. $a(k) = N\exp(-(k-k_0)^2/2\Delta k^2)$ (with some suitable normalization $N$), and then calculate and compare what $\int_{-\infty}^\infty |u(z)|^2 z^2 \mathrm dz$ looks like for the two cases.

The boxcar-spectrum wavepacket is an important example, and it works just fine for a number of applications, but it is also important to keep in mind that it is not the only viable option and that it can only be "derived" by imposing some additional hypotheses that are inappropriate in a wide array of real-world (as well as purely theory-driven) circumstances.

| cite | improve this answer | |
  • $\begingroup$ It is necessary to consider not the integral $\int_{-\infty}^{\infty}|u(z)|z^2dz$, but the integral $\int_{-\infty}^{\infty}|u(z)|z^2 exp(-z^2/2\sigma^2)dz$ for calculating the variance and other moments of the signal. Perhaps you had in mind the feature at zero coordinates z, but it is not there either. $\endgroup$ – Evgeniy Yakubovskiy Jan 28 '19 at 2:23
  • $\begingroup$ @EvgeniyYakubovskiy I'm not sure how else to put this, but that's utter rubbish. The first integral in your comment has very little physical meaning (I guess it could be classed as the second moment in $L^1$ norm, but we're working in $L^2$ norm, so the $L^1$ moment is meaningless) and the second integral doesn't have any at all. $\endgroup$ – Emilio Pisanty Jan 28 '19 at 2:38
  • $\begingroup$ The first integral is taken from your message, and the second integral is my correction of the first integral. The second integral is based on determining the signal dispersion. Your message is interesting if it were not wrong. Moments that, when using your first integral, do not exist, converge perfectly if the first integral is correctly written. $\endgroup$ – Evgeniy Yakubovskiy Jan 28 '19 at 3:39
  • $\begingroup$ @Evgeniy Your first integral is nowhere in this answer. $\endgroup$ – Emilio Pisanty Jan 28 '19 at 4:03
  • $\begingroup$ You were right I did overthink it. Also I understand that this is not very applicable at all in many cases. However I would just like to ask, how applicable is this when speaking of electrical properties in materials. I am reading a book about that subject and this is merely the beginning of the book and this is derivation for electron as a wave. The book claims to be a basic and simple overview but I am hoping that this is at least good enough for understanding materials. $\endgroup$ – Jonathan Aguilera Jan 29 '19 at 5:33

Non-Russian scientists are ahead of Russian in many branches of science, because since 1990 year Russian science slowed down. But hurt in some areas remained. In particular, in signal processing. That is why I managed to get a general formula for the signal envelope. I looked at the research on this site concerning the signal envelope and did not find a general analytical formula for the signal envelope. Calculated using the mathlab envelope of the high-frequency signal with sinusoidal modulation, but there is no general formula.

It must be said that in the theory of signals, the envelope is searched for using a conjugate signal $\sigma(t)$ if there is a high-frequency carrier. This conjugate signal is practically defined as a magnitude $\sigma(t)=\frac{ds(t)}{d\Omega t};\sigma(t)=s(\Omega t-\pi/2)$. It turns out the approximate formula for the envelope. With the exact hard to count formula of Hilbert transform $\sigma(t)=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{s(\tau)}{t-\tau}d\tau$. The result is an envelope $s^2(t)+\sigma^2(t)$.It turns out that the formula is valid for a high-frequency carrier and a smooth modulation function. The envelope of the low frequency signal has no physical meaning. I use a complex conjugate signal to find the envelope. $$\sigma(t)=s^*(t);|s(t)|^2=a(t)exp(i\Omega t)a^*(t)exp(-i\Omega t)=|a(t)|^2$$. The envelope is a smooth positive function, but should smoothly touch zero. But in the case of the envelope for solving the wave equation has its own characteristics. The line of the stationary phase of the signal is determined by the formula $$\frac{\partial z}{\partial t}=c(\omega)=(\Omega_0+\omega)/k(\omega)$$ Then the received signal is determined by the formula, where the high frequency of the signal for which we are looking for the envelope $$s(t,z)=[\int_{0}^{\infty}a[t-z/c(\omega)]exp[i(\Omega_0+\omega)(t-z/c(\omega))]d\omega+\int_{-\infty}^{0}a[t-z/c(\omega)]exp[i(-\Omega_0+\omega)(t-z/c(\omega))]d\omega]$$ Multiply by the complex conjugate signal and integrate over time, we get $$|s(t,z)|^2=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} a^*[t-z/c(\omega_1)] a[t-z/c(\omega)]exp[i(\pm\Omega_0+\omega_1)z/c(\omega_1)-i(\pm \Omega_0+\omega)z/c(\omega))]\delta(\pm\Omega_0+\omega_1-\omega\mp\Omega_0) d\omega d\omega_1/2\pi; (1)$$ Where we choose the sign of the frequency $\Omega_0$ equal to the sign of the frequency $\omega$. Where used the formula $$\delta(\omega_1-\omega)=2\pi\int_{-\infty}^{\infty} exp[i(\omega_1-\omega)t]dt$$ Formula (1) is converted to $$|s(t,z)|^2=\int_{-\infty}^{\infty}a^*[t-z/c(\omega)] a[t-z/c(\omega)]d\omega /2\pi$$ We take into account the smooth dependence of time $a(t)$, we get an envelope that depends on time, but the formula becomes approximate, since the delta function is determined approximately. The envelope calculation error equals $|\pm \frac{da}{d\Omega_0 t}+\frac{d^2 a}{d(\Omega_0 t)^2}|=|\frac{d^2 a}{d (\Omega_0 t)^2}|<<1$ This error is negligible for smooth envelopes and accurately takes into account the derivative of the envelope.

| cite | improve this answer | |
  • 3
    $\begingroup$ Please do not make many small edits to your posts. Instead address as many issues as possible with each edit. Edits bump posts in the active queue and many edits in sequence effectively spam the active queue, a behavior which is discouraged. $\endgroup$ – dmckee --- ex-moderator kitten Jan 26 '19 at 22:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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