I have some confusions regrading Group and Phase Velocity. Group Velocity exists for a group of waves, it's stated here in the video that Group Velocity the sum of the phase velocity of individual waves (Phase velocity is given by $\dfrac{ω}{k}$). If the Group velocity $\dfrac{dω}{dk}$is the sum of the phase Velocity of individual waves then why is it a differential quantity and not an integral quantity?
Question regarding Group and Phase Velocity
dispersionphase-velocitywaves
Related Solutions
Consider a wave
$$A = \int_{-\infty}^{\infty} a(k) e^{i(kx-\omega t)} \ dk,$$
where $a(k)$ is the amplitude of the kth wavenumber, and $\omega=\omega(k)$ is the frequency, related to $k$ via a dispersion relation. Note, if we wanted to track a wave, with wavenumber $k$, with constant phase, we would see that this occurs when $kx=\omega t$, i.e. $x/t = \omega/k = c$, with $c$ the $\textbf{phase}$ velocity.
We would like to know the speed at which the envelope $|A|$ is traveling.
For $\textbf{narrow banded}$ waves, the angular frequency $\omega$ can be approximated via the taylor expansion around a central wavenumber $k_o$, i.e.
$$\omega(k) = \omega(k_o) + \frac{\partial \omega}{\partial k} (k-k_o) + \mathcal{O}((k-k_o)^2),$$
where the scale of the bandwidth is quantified by the small parameter $(k-k_o)$. Therefore, we can rewrite $A$ as
$$A \approx e^{-i(\omega(k_o)t-k_o\frac{\partial \omega}{\partial k}t)} \int_{-\infty}^{\infty} a(k) e^{ik(x-\frac{\partial \omega}{\partial k} t)} \ dk.$$
Therefore
$$|A| = \left| \int_{-\infty}^{\infty} a(k) e^{ik(x-\frac{\partial \omega}{\partial k} t)} \ dk \right|,$$
which says that the envelope, $|A|$, travels at speed $\frac{\partial \omega}{\partial k}$, i.e. $$|A(x,t)| = |A(x-c_g t,0)|,$$ where we have defined $$c_g \equiv \frac{\partial \omega}{\partial k}.$$
The group velocity has dynamical significance, as it is the velocity at which the energy travels.
With a continuous wave you cannot transmit a signal. For a signal to be transmitted, you need a modulation of the wave, e.g. amplitude modulation. For example, to transmit acoustic frequencies (speech), you modulate the high frequency electromagnetic carrier wave (on the order of MHz for medium wave transmitters) with the acoustic frequencies(up to 20kHz). This modulation produces small variations called side-bands (plus and minus 20kHz) in the transmitted waves. The group velocity of a wave describes the velocity with which such modulation of the carrier amplitude, which transmits the signal, propagates. In free space, the group velocity of an EM wave is identical to the phase velocity $c$ because the dispersion is linear $\omega=c k$. Thus also a pulse shaped modulation propagates with unchanged form. On transmission lines, there can be significant nonlinear dispersion, i.e. the phase velocity $v_{ph}= \frac {\omega}{k}$ for different frequencies is not constant and, in general, different from the group velocity $v_{gr}=\frac {\partial \omega}{\partial k}$. This leads to a loss of shape of a pulse-like modulation of the carrier wave. However, the propagation speed of such a pulse modulation can still be obtained from the group velocity.
That the group velocity is opposite to the phase velocity happens only in systems with special nonlinear dispersion relations.
Best Answer
Your confusion is normal: phase and group velocity are the most poorly taught topics in physics.
Classical wave physics courses start with a plane wave:
$$ \psi (x,t)=U_{m}e^{i(k.x-\omega t)} $$
The plane wave extents from $x=- \infty $ to $x= \infty $ and oscillate in time from $t=- \infty $ to $t= \infty $.
In mechanics, the velocity of a point is precisely defined, but how do you define the velocity of a time changing curve whose domain is infinite?
The work around is usually done by introducing the phase of the plane wave:
$$ \phi (x,t)=k.x-\omega t $$
You could naively think the phase velocity is just the rate of change of the phase and calculate:
$$ \frac{d \phi (x,t)}{dt} = \frac{\partial \phi (x,t) }{\partial x}~ \frac{dx}{dt}+ \frac{\partial \phi (x,t)}{\partial t}=k\frac{dx}{dt}- \omega $$
But x and t are two independent variables. So $\frac{dx}{dt}=0$ and you get:
$$ \frac{d \phi (x,t)}{dt} = - \omega $$
Which is not the phase velocity people are talking about.
To get the good formula, you need to ask yourself what would be the velocity of a point M(x(t)) moving with the wave. The mental picture goes as follow. Imagine a surfer standing on the wave, its position at time t is x(t), and height of the wave is:
$$ \psi (x(t),t)=U_{m}~e^{i(k.x(t)-\omega t)}$$
As the surfer moves with the wave, he stays at the same height if the phase $k.x(t)-\omega t$ remains unchanged.
The phase velocity of a plane wave is thus the velocity of a point chosen on the envelope moving with the wave. Since $v_{\phi}=\frac{\omega}{k}$ does not depend on x(t), the phase velocity is independent of the chosen point and our definition makes sense.
Plane waves infinite extension in both space and time makes them unsuitable to model real waves. If we consider a more general signal $\psi(x,t)$ (red curve), we are back to the starting point concerning the definition of the signal velocity. How do you define the velocity of something whose shape changes both in time and space?
We can no longer chose a random point on the envelope and say that its velocity is the wave velocity. Such a choice would lead to a wave velocity that differs with the chosen point. Our choice of M must be guided by simplicity and universality. The simplest choice that summarizes the whole curve is the centroid (center of gravity) of $\psi(x,t)$. Its position $x_{c}(t)$ is given by:
$$x_{c}(t)= \frac{ \int_ {-\infty }^{ \infty} x ~\psi(x,t)~ dx }{\int_ {-\infty }^{ \infty} \psi(x,t) ~dx }$$
It can be shown that the velocity of the centroid is the group velocity:
$$ V_{g}= \big( \frac{d~ \omega (k)}{dk} \big)_{k_{0}} $$
Some Explanations of the paper
We shall need some results about Fourier transform. The Fourier transform of $f(x)$ is $\hat{f} (k)$ defined by:
$$\hat{f} (k)= \int_{- \infty }^{ \infty } f(x) e^{- i k.x}dx ~~\rightleftharpoons ~~f(x)= \frac{1}{2 \pi } \int_{- \infty }^{ \infty } \hat{f}(k)e^{i k.x}dk$$
Using this definition, we find: $$\hat{f}(0)= \int_{- \infty }^{ \infty } f(x) dx $$
and: $$ \frac{d\hat{f}(k)}{dk}= - i \int_{- \infty }^{ \infty }x ~f(x) e^{- i k.x}dx ~~\Longrightarrow ~~\big(\frac{d\hat{f}(k)}{dk}\big)(0)= - i \int_{- \infty }^{ \infty }x ~f(x) dx $$
Plugin these two results into the definition of the centroid, it comes:
$$x_{c}(t)= \frac{ \int_ {-\infty }^{ \infty} x ~\psi(x,t)~ dx }{\int_ {-\infty }^{ \infty} \psi(x,t) ~dx }= -i \frac{ \big(\frac{ \partial \hat{\psi}(k,t) }{\partial k}\big)(0,t) }{\hat{\psi}(0,t)} $$
The centroid is thus related to the first two terms of the Taylor expansion of the Fourier's transform.
If you do not have access to the paper, below are the screenshots.