electromagnetic-radiationphase-velocitywaves
Moving charges produce oscillating electric and magnetic fields -we have an electromagnetic wave.
In terms of moving charges or at the level of charges, what is phase velocity and group velocity of an electromagnetic wave?
What is the origin of these velocities?
I am to not able to relate the definitions of phase velocity(how fast the phase of a wave is moving) and group velocity(velocity of the envelope) to moving charges.
I am trying to get an intuitive picture of this. Please help me.
Wave velocity, when not otherwise specified, usually means the phase velocity: the rate at which the phase travels in space.
This site animates the differences between phase and group velocities. The group velocity is the speed at which energy is transported.
So if the context implies that the wave is carrying energy, the wave velocity is also the group velocity.
In a dispersionless media group and phase velocity are the same. Special treatment is required for strongly absorbing regions near a resonance.
A detailed treatment of phase, group, and signal velocity can be found here.
Phase velocity and group velocity are determined by the dispersion relation, which is typically the angular frequency as a function of wavenumber: $\omega(k)$. The dispersion relation can be measured and used to infer properties of the wave and its medium; this is a very common thing to do in Condensed Matter physics. A wonderfully complex example of this is explored in this question: What information can we extract from the electronic band structure?
Because they don't teach this in chemistry or basic physics courses usually, it's likely that you didn't even know that band structure IS a dispersion relation. It is the dispersion relation for electrons in a crystal of that material. Except the dispersion relation is then usually written equivalently as $E(p)$. ($E=\hbar\omega$, $p=\hbar k$ for planes wave solutions of the Schrodinger Equation).
Based on your question, it seems to me that you don't know where the group and phase velocity come from, so I will illustrate it here. Phase velocity and group velocity are defined as $v_{phase}=\frac{\omega}{k}$ and $v_{group}=\frac{\partial \omega}{\partial k}$. To calculate these, we need $\omega(k)$, the dispersion relation. We can either measure this through experiment as I already mentioned, or if we have a wave equation, we can calculate it by plugging in a plane wave $e^{i(kx-\omega t)}$. This is one way a band structure diagram, like the first link in my post, can be read. The slope of the curve tells you have fast wave packets move that have different mean wavenumbers (since wavepackets have a range of wavenumbers, and they aren't plane waves).
I'll do a couple examples to show you how this goes. First, "the" wave equation: $(\partial_{tt} - c^2 \partial_{xx})\psi = 0$. If we plug in the plane wave, we get $(\partial_{tt} - c^2 \partial_{xx})e^{i(kx-\omega t)} = (-\omega^2 + c^2k^2)e^{i(kx-\omega t)} = 0 \implies \omega^2 = c^2k^2$. This is our dispersion relation. Now we can get the phase and group velocities: $v_{phase} = \frac{\omega}{k} = \pm c$ and $v_{group} = \frac{\partial \omega}{\partial k} = \pm c$. So in this particular case, the phase and group velocity are the same, and they are $c$. Really, $c$ is written in the original equation the way it is because of this fact. In general, these velocities depend on the wavenumber.
With a waveguide, the dispersion relation given in Quantum Physics (Gasiorowicz, 3rd edition) Problem 2.2 is $\lambda = \frac{c}{\sqrt{\nu^2 - \nu_0^2}}$. Substituting $\lambda = \frac{2\pi}{k}, \nu = \frac{\omega}{2\pi}$, we get $v_p = \pm \sqrt{c^2+\frac{\omega_0^2}{k^2}}$. As you said, the phase velocity has a singularity. The first reason this doesn't disagree with special relativity is that special relativity puts a limit on how fast wave packets can travel (the group velocity), not the phase velocity, which is $v_g = \frac{c^2}{\sqrt{c^2+\frac{\omega_0^2}{k^2}}}$, and doesn't ever go above $c$.
The second reason why it can be ok for a wave to travel faster than light is that the equation describing it is not relativistic, so of course it can. The Schroedinger equation is one such wave equation, so I will work out the velocities for it as an example. I will do the free particle in 1D for simplicity's sake, but the result is still generally relevant: $i\hbar \partial_t e^{i(kx-\omega t)} = -\frac{\hbar^2}{2m}\partial_{xx}e^{i(kx-\omega t)} \implies \hbar\omega = \frac{\hbar^2}{2m}k^2 \implies v_p = \frac{\hbar k}{2m}, v_g = \frac{\hbar k}{m}$. Note that k can be anything, so the velocity is unlimited, which is NOT ok in SR! Also note that the plane wave (a momentum eigenstate) has a definite momentum $p=\hbar k$. So $v_g = \frac{p}{m}$ which is the velocity that a wave packet will move at, and the velocity we expect. This is what a free particle wave packet looks like: https://giphy.com/gifs/dimension-coil-packet-ddYwgktdZxTA4 Note that the phase velocity is half the group velocity.
Ok, so what if we do want to be relativistic? For particles with spin, this gets very complicated, but for spinless particles, we can use the Klein-Gordon equation (from $E^2 - (pc)^2 = (mc^2)^2$): $(-\hbar^2\partial_{tt} + \hbar^2c^2\partial_{xx} - (mc^2)^2)e^{i(kx-\omega t)} = 0 \implies \hbar^2\omega^2 - \hbar^2c^2k^2 - (mc^2)^2 = 0 \implies \omega^2 = k^2c^2 + \frac{m^2c^4}{\hbar^2} \implies v_p = \pm c\sqrt{1 + (\frac{mc}{\hbar k})^2}, v_g = \frac{kc^2}{\sqrt{k^2c^2 + \frac{m^2c^4}{\hbar^2}}} = \frac{c}{\sqrt{1+(\frac{mc}{\hbar k})^2}}$. Clearly, the phase velocity is bonkers. It is ALWAYS greater than $c$. Also, it's nothing at all like the phase velocity in the Schroedinger equation when our momentum is small. However, the group velocity is always less than c, and has the same behavior as the waveguide from earlier! The reason for this correspondence is that the waveguide has symmetry that makes the problem one dimensional, and $\omega_0 = \frac{mc^2}{\hbar}$ depends on the mass of the particle. We can also see that $\beta = \frac{v_g}{c} \implies \hbar k= \frac{v_g m }{\sqrt{1-\beta^2}} = \gamma v_g m$, which is the relativistic momentum, and thus exactly what we expect. This is what a wavepacket solution to this would look like (color is phase, brightness is amplitude): https://en.wikipedia.org/wiki/Dispersion_relation#/media/File:DeBroglie3.gif What is most intriguing to me here is that the phase velocity is higher when the momentum is lower.
Another point about phase velocity: in quantum mechanics, the waves really are complex valued. A plane wave has uniform amplitude everywhere, which means it is equally likely to find the particle anywhere. The phase doesn't matter. So we should reasonably expect that the phase shouldn't affect observables, but clearly in the Schrodinger equation this is not true, since $\hat{p}=-i\hbar \nabla$ obviously changes if the phase is changed locally. To fix this, we instead use the Pauli-Schrodinger equation: https://www.youtube.com/watch?v=V5kgruUjVBs&t=625s So when phase is treated correctly, it doesn't affect any observable. That's why SR doesn't care about phase velocity.
I hope this helps, and I'd be happy to make any edits to further clarify things or straighten out any mistakes I may have made. Cheers!
Edit in response to comment: Phase velocity tells you how fast a particular phase travels. In a plane wave the phase is $e^{i(kx - \omega t)}$. So if we set the exponent to $0$, which gives a phase of $1$, then $x = \frac{\omega}{k}t$. That is what defines phase velocity. Deriving group velocity is a bit more slippery though, because it's how fast the packet or group moves, not how fast each peak moves (the phase velocity). Honestly, beyond this the Wikipedia page is pretty good: https://en.wikipedia.org/wiki/Phase_velocity.
As far as the group velocity indicating "how fast energy travels", look at wave packets like these: https://gfycat.com/gifs/detail/happyleafybighornsheep and http://www.math.uwaterloo.ca/~kglamb/course_animations/sech_packet_small.gif It should be evident that "where" the wave is and how fast the packet is moving is determined by the group velocity. Say you're considering a particle bouncing off a wall, it doesn't matter if the phase velocity is much higher than the group velocity, it won't bounce off the wall (i.e. transfer momentum to the wall) until it gets there, and that's determined by the group velocity. In addition, for quantum wavefunctions we can only ever directly observe local phase shifts through the Aharanov-Bohm effect, but the whole wavefunction's phase could change by the same amount and you'd have no way to measure this. You can't measure phase in other words, the only way it matters is with interference effects, which only depend on relative phase anyway. Please let me know if this helps clear it up.
Best Answer
Well, in vacuum the group velocity and phase velocity are identical. The velocity of light in vacuum is not really determined by the properties of the moving charges. It is a fundamental constant of nature: if you like, it is determined by the properties of the vacuum. The important feature of the charges that generate the EM wave is the frequency at which they oscillate, which gives the frequency of the resulting wave.
The distinction between phase and group velocity becomes important when you are in a dispersive medium, meaning a medium in which you have charges that are stuck in one place somehow, e.g. electrons bound to atomic nuclei. (An example of a dispersive medium is a piece of glass.) Now the speed at which light propagates will depend on its frequency, this is what dispersion means.
Basically, dispersion arises because the bound charges have their own frequency that they like to oscillate at. However, the incoming light forces them to oscillate at a different frequency. The way they react to the incoming light (the amplitude of their oscillation) depends on how the light's frequency compares to their own favourite frequency. The bound charges' reaction is really important because the total EM wave that you observe is the sum of the incoming wave and the contribution from the oscillating charges, which re-radiate EM waves at an amplitude depending on their amplitude of oscillation.
Given that different frequencies of light have different speeds (i.e. phase velocities), we can now see why the group velocity is different, in general. The phase velocity is $$ v_p = \frac{\omega}{k}, $$ and the group velocity is $$ v_g = \frac{\partial \omega}{\partial k}, $$ which are only the same if $\omega = c k$, with $c$ a constant (the speed of light). This is why $v_p = v_g$ in vacuum. (Here $\omega$ is angular frequency and $k = 2\pi/\lambda$ is wavenumber.) However, if different frequencies have different velocities, then $v_p(\omega)$ is a non-trivial function of $\omega$, which is only true if the relationship between $\omega$ and $k$ (the dispersion relation) is more complicated than just the linear one. In other words, in a dispersive medium, $\omega \neq c k$. But the condition $v_p = v_g$ implies that $$\frac{\partial \omega}{\partial k} = \frac{\omega}{k} \; \Leftrightarrow \; \omega = c k, $$ which shows that if dispersion is present then $v_p \neq v_g$. To see why group velocity is a useful measure of how a wavepacket moves, which is the usual interpretation, see e.g. Wikipedia.
The microscopic origin of dispersion is non-trivial, so I'll leave it to greater luminaries than I to give a proper explanation. This was really just a comment that got a bit long. Good question though, hopefully someone really clever will go into all the gory details :)