Consider the answer here by Chad Orzel which explains how a monochromatic light can slow down in a medium. He explains,

You can think each of the atoms (of the medium) as being like a little dipole, consisting of some positive and some negative charge that is driven back and forth by the off-resonant light field. Being an assemblage of charges that are accelerating due to the driving field, these dipoles will radiate, producing waves at the same frequency as the driving field, but slightly out of phase with it...

Then he says,

If you go through a little bit of math, you find that this gives you a beam in the same direction as the original beam-- the waves going out to the sides will mostly interfere destructively with each other-- with the same frequency but with a slight delay compared to the driving field.

So the upshot of this answer is that the phase speed$^1$ in the medium $v_p$ becomes less than $c$. Same explanation can be found here with the excuse that the underlying mathematics is a mess.

Can anyone give the supporting mathematics? How does a superposition f different waves with same frequency but slightly out of phase lead to a wave with a changed phase speed?

$^1$ Note that, when such a wave travels through a medium, there is only one frequency which doesn't change during the propagation in the medium. Hence, there is no question of dispersion or group speed.

  • 3
    $\begingroup$ It depends a bit. On the high frequency side of a resonance, the phase of the induced field leads. That is why phase velocities are generally larger than $c$ for x-rays. This gives rise to total external reflection for x-rays at shallow incident angles. $\endgroup$
    – user137289
    Commented Feb 12, 2019 at 15:15

4 Answers 4


The key point is that the sum of two waves with different amplitude and phase but the same frequency gives you a wave of that frequency. This is shown in the picture where the black curve is $\sin(x)$, the red curve is $-.2 \sin(x+5)$ and the blue curve is the sum of the two. Same frequency, still a sine wave, but different phase.

addition of sine waves

Suppose there is an electric field $E=Ae^{i( \omega t-kx)}$. It falls on a molecule at (say) $x=0$. This is polarised by the field and oscillates and produces a field $Ere^{i \delta}$, where $r$ (which is small) depends on the polarisability of the molecule and the phase shift $\delta$ varies according to the difference between the applied $\omega$ and the natural resonant frequency of the molecule. So the combined field is

$Ae^{i \omega t }(1+re^{i \delta})$

The bracket can be written $(1 + r \cos\delta + i r \sin \delta)=Re^{i \phi}$

where $\tan \phi={r \sin \delta \over 1 + r \cos \delta}$ and $R^2=1+r^2+2 r \cos \delta$

So the effect of the polarisable molecule is just to shift the phase by a small $\phi$. (There is also a change in the amplitude but that's not at issue.)

If you have a series of such molecules then each has the same effect. The effect on the wave is multiplicative so the phases add. If there is a distance $a$ between molecules then in travelling a distance $x$ there are $x/a$ molecules giving a phase change due to polarisation of $\phi x/a$. The wave becomes $E=Ae^{i(\omega t - k x +\phi x/a)}=Ae^{i(\omega t -k' x)}$ with $k'=k-\phi/a$, i.e. the wavenumber (and thus the wavelength) change. The frequency is the same, so change in wavelength implies change in velocity.

This is the 'little bit of math' Orzel refers to.

  • $\begingroup$ This is an interesting argument. However, it is a little strange that the polarized medium has to respond with a delay to the field it experiences, because this means the medium is absorbing energy. So, this kind of explanation only works for cases where the wave is being absorbed. But the explanation based on the wave equation in polarized medium seems to work also in the case of no absorption - the fact that the medium is polarizable is enough. $\endgroup$ Commented Feb 18, 2019 at 21:43
  • $\begingroup$ No. There is no absorption. This is simply the equation of a forced oscillator. The oscillator responds but with a phase shift - zero at low frequencies, increasing to $\pi/2$ when the forcing frequency matches the resonance frequency and then to $\pi$ at high frequencies $\endgroup$ Commented Feb 19, 2019 at 11:12
  • $\begingroup$ If the phase shift is non-zero or non-pi, time average of power of electric force is non-zero. $\endgroup$ Commented Feb 19, 2019 at 19:14

The main idea really is that the changed phase velocity is a collective phenomenon that only manifests in macroscopic electric field, but is due to mutual microscopic interactions of the medium elements, while those interactions take place at unchanged vacuum light speed $c$.

The phase shift happens in the sense the farther the medium element is from the vacuum-medium interface, the larger the phase difference between the primary wave and the secondary wave generated at that element. The secondary waves cannot be generated in phase with the primary wave, because then they would, after some short path into the medium, screen the primary field completely and no macroscopic wave would penetrate deeper - we would be looking at total reflection of the wave, something that happens rather in metals, not in glass (at least not if wave direction is normal to the medium interface.

But the phase shift isn't something that immediately implies changed phase velocity, and it is not easy to find a detailed account of how that works.

Luckily, there is another and more clear argument based on macroscopic EM theory. In vacuum, the Maxwell equations imply the following wave equation for electric field:

$$ \frac{1}{c^2}\frac{\partial^2 E_x}{\partial t^2} - \frac{\partial^2E_x}{\partial z^2} =0. $$ If we restrict our attention to solutions which represent harmonic plane waves, all obey the universal relation: $$ \Omega / k = c $$ where $c$ is vacuum speed of light. I mention this because general definition of phase speed is $\Omega/k$. So, all such waves have phase speed $c$.

Similar wave equation can be derived for electric field in medium, but there is a difference - a new term. However, to derive this equation, we need to make some simplifying assumptions. These are:

1) second derivative of polarization of an element is proportional to instantaneous value of electric field there;

2) net macroscopic electric field inside medium can be expressed as harmonic plane wave, similarly to the vacuum case:

$$ \mathbf E = E_0 \mathbf e_x \sin(\Omega t - kz). $$

In such case, Maxwell's equations imply the following equation for electric field as a function of position $z$ and time $t$:

$$ \frac{1}{c^2}\frac{\partial^2 E_x}{\partial t^2} - \frac{\partial^2E_x}{\partial z^2} =-C\frac{\partial^2 E_x}{\partial t^2} $$ where $C$ is some constant quantifying how much the medium polarized in given macroscopic electric field.

Note the new term on the right-hand side, which was not present in the vacuum case. This terms accounts for the effect of polarized medium on the total electric field - the fact that polarized medium can generate its own, secondary radiation.

Solutions for $k$ as a function of $\Omega$ can be found by inserting the assumed form of electric field and some manipulation. It turns out that in general, due to the new term the ratio $\frac{\Omega}{k}$ does not have the vacuum value of $c$, but can be lower or higher, and is a function of $C$. This is described in terms of quantity $n$:

$$ \frac{\Omega}{k} = \frac{c}{n}. $$

In glass, usually $n>1$ so the waves are more densely packed along the path of propagation.

So, we can see how the changed phase speed is related to polarized medium and known laws. However, at the same time, we had to assume that waves of this simple kind are actually possible. This is not true in all cases, such as nonlinear media.


Edit : I tried to complete the Feynman calculation. Thank you for being kind to English, which is not my native language ! And sorry for possible miscalculation !

The problem is part of the Oseen extinction theorem. You'll find usueful reference in https://en.wikipedia.org/wiki/Ewald%E2%80%93Oseen_extinction_theorem In particular, the irremplace Born and Wolf. But it's complicated !

This is a simpler calculation. It is a macroscopic one. It is not perfect but I think it is interesting.

The field in a point is the sum of all the fields radiated by all the slices. But these radiated fields depend of the fied himself and so we are led to an integral equation.

In order not to be too long, I consider as known the field radiated by a plane of uniform polarization which oscillates sinusoidally.

A thin plate placed at $z=0$ and of length $dz$ is polarized : a volume $\text{d}\tau $ is equivalent to the dipole moment $\overrightarrow{\text{d}p}=\overrightarrow{P}\text{d}\tau =\overrightarrow{P}\text{d}z\text{d}S$. The polarization is supposed to vary sinusoidally in time and directed according to Ox: $\overrightarrow{P}=\overrightarrow{{{P}_{0}}}{{\operatorname{e}}^{j\omega t}}$with $\overrightarrow{{{P}_{0}}}={{P}_{0}}\overrightarrow{{{e}_{x}}}$ We work in complex representation but the complex amplitudes are not underlined to lighten the notations. All the vectors are along Ox.

The field radiated by this plate is ($k=\omega /c$):

$\text{d}{{E}_{x}}=-\frac{1}{2{{\varepsilon }_{0}}}(jk){{P}_{0}}dz{{\operatorname{e}}^{j\left( \omega t-kz \right)}}$ if $z>0$ and $\text{d}{{E}_{x}}=-\frac{1}{2{{\varepsilon }_{0}}}(jk){{P}_{0}}dz{{\operatorname{e}}^{j\left( \omega t+kz \right)}}$ if $z<0$

On can find the result in the Feynman's lectures : http://www.feynmanlectures.caltech.edu/I_30.html

I have simply adapted this result to a polarized medium.

Now consider a medium such that the polarization ${{P}_{x}}$ is related to the complex amplitude of the electric field by the relation ${{P}_{0}}={{\varepsilon }_{0}}\chi {{E}_{x}}$with the a priori complex quantity called susceptibility and dependent (a priori) of the frequency. The complex indices is linked to the susceptibility by ${{n}^{2}}=1+\chi $ . (This definition is easily adapted to a conductor with conductivity $\gamma $ )

The medium extends for z varying from 0 to infinity. For, $z<0$ we have vacuum. The medium is "lit" by an incident plane wave polarised along Ox, ${{E}_{i}}={{E}_{0}}{{\operatorname{e}}^{j}}^{(\omega t-kz)}$ with $k=\omega /c$. This wave will polarize the medium as in the previous question.

Each slice of the medium will then radiate. The total field at a point z, is the sum of the incident field and the fields radiated by the various slices.

By changing the origin for each slice, we have :

${{E}_{z'>z}}(z)=-\frac{1}{2{{\varepsilon }_{0}}}jk\int\limits_{z}^{+\infty }{{{P}_{x}}(z')\text{d}z'{{\operatorname{e}}^{j\left( \omega t+k(z-z') \right)}}}$

${{E}_{0<z'<z}}(z)=-\frac{1}{2{{\varepsilon }_{0}}}jk\int\limits_{0}^{z }{{{P}_{x}}(z')\text{d}z'{{\operatorname{e}}^{j\left( \omega t-k(z-z') \right)}}}$

Since ${{P}_{x}}(x')={{\varepsilon }_{0}}\chi {{E}_{x}}(x')$ :

${{E}_{x\,}}_{z'>z}(z)=-\frac{1}{2}jk\chi \int\limits_{z}^{+\infty }{{{E}_{x}}(z')\text{d}z'{{\operatorname{e}}^{j\left( \omega t+k(z-z') \right)}}}$

${{E}_{x\,0<z'<z}}(z)=-\frac{1}{2}jk\int\limits_{0}^{z }{\chi {{E}_{x}}(z')\text{d}z'{{\operatorname{e}}^{j\left( \omega t-k(z-z') \right)}}}$

At $z>0$ , the amplitude of the total field satisfies the integral equation: ${{E}_{x}}(z,t)={{E}_{0}}{{\operatorname{e}}^{j(\omega t-kz)}}\underbrace{-\frac{1}{2}jk\chi \int\limits_{z}^{+\infty }{{{E}_{x}}(z')\text{d}z'{{\operatorname{e}}^{j\left( \omega t+k(z-z') \right)}}}}_{{{E}_{x\,}}_{z'>z}(z)}\underbrace{-\frac{1}{2}jk\chi \int\limits_{0}^{z}{{{E}_{x}}(z')\text{d}z'{{\operatorname{e}}^{j\left( \omega t-k(z-z') \right)}}}}_{{{E}_{x\,}}_{z'<z}(z)}$

Since the medium is linear, we have ${{E}_{x}}(z,t)={{E}_{x}}(z){{\operatorname{e}}^{j(\omega t)}}$ and the field obey the integral equation, valid if $z>0$ :

${{E}_{x}}(z)={{E}_{0}}{{\operatorname{e}}^{-jkz}}-\frac{1}{2}jk\chi {{\operatorname{e}}^{jkz}}\int\limits_{z}^{+\infty }{{{E}_{x}}(z')\text{d}z'{{\operatorname{e}}^{-jkz'}}}-\frac{1}{2}jk\chi {{\operatorname{e}}^{-jkz}}\int\limits_{0}^{z}{{{E}_{x}}(z')\text{d}z'{{\operatorname{e}}^{+jkz'}}}$

To solve this integral equation, we can try a solution of the form ${{E}_{x}}(z)=C{{\operatorname{e}}^{\left( -j\beta z \right)}}$with C and $\beta $ constants a priori complex, and with $\operatorname{Im}(\beta )>0$ to avoid divergence. (We could also differentiate the equation twice).

$\int\limits_{0}^{z}{\text{C}{{\operatorname{e}}^{-j\beta z'}}\text{d}z'{{\operatorname{e}}^{+jkz'}}}=C\frac{1}{j(k-\beta )}({{\operatorname{e}}^{+j(k-\beta )z}}-1)$ $\int\limits_{z}^{\infty }{\text{C}{{\operatorname{e}}^{-j\beta z'}}\text{d}z'{{\operatorname{e}}^{-jkz'}}}=-C\frac{1}{j(k+\beta )}({{\operatorname{e}}^{-j(k+\beta )\infty }}-{{\operatorname{e}}^{-j(k+\beta )z}})=C\frac{1}{j(k+\beta )}{{\operatorname{e}}^{-j(k+\beta )z}}$

If we replace in the original equation:

$C{{\operatorname{e}}^{-j\beta z}}={{E}_{0}}{{\operatorname{e}}^{-jkz}}-\frac{1}{2}jk\chi {{\operatorname{e}}^{jkz}}C\frac{1}{j(k+\beta )}{{\operatorname{e}}^{-j(k+\beta )z}}-\frac{1}{2}jk\chi {{\operatorname{e}}^{-jkz}}C\frac{1}{j(k-\beta )}({{\operatorname{e}}^{+j(k-\beta )z}}-1)$

$C{{\operatorname{e}}^{-j\beta z}}={{E}_{0}}{{\operatorname{e}}^{-jkz}}+\frac{1}{2}jk\chi {{\operatorname{e}}^{-jkz}}\frac{C}{j(k-\beta )}-\frac{1}{2}jk\chi \frac{C}{j(k+\beta )}{{\operatorname{e}}^{-j\beta z}}-\frac{1}{2}jk\chi \frac{C}{j(k-\beta )}{{\operatorname{e}}^{-j\beta z}}$

We identify the term in ${{\operatorname{e}}^{-j\beta z}}$

$1=-\frac{1}{2}jk\chi \frac{1}{j(k+\beta )}-\frac{1}{2}jk\chi \frac{1}{j(k-\beta )}=-\chi \frac{{{k}^{2}}}{{{k}^{2}}-{{\beta }^{2}}}\to {{\beta }^{2}}={{n}^{2}}{{k}^{2}}={{n}^{2}}\frac{{{\omega }^{2}}}{{{c}^{2}}}$

So, we have $\beta =n\frac{\omega }{c}$

Identifying with the term in ${{\operatorname{e}}^{-jkz}}$ we find a second relation

$0={{E}_{0}}+\frac{1}{2}k\chi \frac{C}{(k-\beta )}$ . It gives $C=\frac{2(n-1)}{{{n}^{2}}-1}{{E}_{0}}=\frac{2}{n+1}{{E}_{0}}$

The field in the region $z>0$ is ${{E}_{t}}=\underbrace{\frac{2}{n+1}}_{t}{{E}_{0}}{{\operatorname{e}}^{j}}^{(\omega t-nkz)}$ and so we have the coefficient of transmission and the change of phase velocity by adding radiated fields with variable amplitudes, all propagating at velocity c.

But we can also compute the reflected waves in the $z<0$ region by adding all the radiated waves in this direction :

${{E}_{r}}(z)=-\frac{1}{2}jk\chi {{\operatorname{e}}^{jkz}}\int\limits_{0}^{+\infty }{{{E}_{x}}(z')\text{d}z'{{\operatorname{e}}^{-jkz'}}}=-{{\operatorname{e}}^{jkz}}\frac{{{n}^{2}}-1}{(n+1)}\frac{1}{n+1}{{E}_{0}}=-{{\operatorname{e}}^{jkz}}\frac{n-1}{n+1}{{E}_{0}}$

The reflected wave is ${{E}_{t}}(z,t)=\underbrace{\frac{1-n}{1+n}}_{r}{{E}_{0}}{{\operatorname{e}}^{j(\omega t+kz)}}$ and we have the coefficient of reflection without using any condition at the interface.

We can follow Feynman, chapter 31 of volume 1. The summary is :

A sinusoidal progressive plane wave in the vacuum ${{E}_{0}}{{e}^{i\omega (t-z/c)}}$ passes through an infinite plane of small thickness $\delta z$ and index $n$.

In the usual formalism of refraction index, it is delayed by $\delta t=(n-1)\delta z/c$ since it advances at the speed $c/n$ in the plate instead of $c$: time of travel in the plate $\delta z(n/c)$ instead of $\delta z(1/c)$.

The wave after the plate is $E={{E}_{0}}{{e}^{i(\omega (t-\delta t)-\omega z/c)}}={{e}^{i\omega (\delta t)}}{{E}_{0}}{{e}^{i(\omega (t-\delta t)-\omega z/c)}}=\underbrace{{{e}^{i\omega ((n-1)\delta z/c)}}}_{1-i\omega (n-1)\delta z/c}{{E}_{0}}{{e}^{i(\omega t-\omega z/c)}}$

Finally $E'={{E}_{0}}{{e}^{i(\omega t-z/c)}}\underbrace{-i\omega (n-1)\delta z/c{{E}_{0}}{{e}^{i(\omega t-z/c)}}}_{\text{Field radiated by the plane}}$

You just have to reverse the procedure: the sum of the incident wave and the radiated wave is indeed a delayed wave.

I recommend reading Feynman! (Sorry for my english)

  • $\begingroup$ I agree with your recommendation of this reference, but the part you show in your answer is the warm-up only: he figures out what the resulting field will be if there is a delay. The important thing is that it is delayed (well, not always: see @Pieter's comment) and the delay is proportional to $(n-1) \delta\zeta/c$,but it takes the next section (31-2) to explain why $n > 1$. $\endgroup$
    – NickD
    Commented Feb 14, 2019 at 18:58
  • $\begingroup$ As you say, it was just a warm up. For the delay (if n > 1), I thought than you just have to reverse the reasonning : add the two waves.... But you also have to understand the computation of the field radiated by a thin plane, which is not always clear in Feynman's lecture ! What is more precisely the point which interest you ? A physical understanding or a mathematical one ? $\endgroup$ Commented Feb 14, 2019 at 19:28

Here are elements of an answer based on Sommerfeld's "About the propagation of light in dispersive media". This is chapter 2 of a book by Brillouin on "Wave propagation and group velocity" (which I highly recommend).

For the "Mathematics supporting the classical explanation of why the phase speed of light slows down in a medium": there is no such thing because phase velocity can be larger than $c$. In fact, even group velocity can exceed $c$ in cases of anomalous dispersion. Both phase and group velocities have specific interpretations valid in specific (i.e., limited) circumstances.

For "how a monochromatic light can slow down in a medium": a monochromatic light has infinite extension in time. Thus, it does not really "slow down"; the light has been there and will remain there for all times. Quoting Sommerfeld: "In order to be able to say something about the propagation, we must, instead, have a limited wave motion: nothing until a certain moment in time, then, for instance, a series of regular sine waves, which stop after a certain time or which continue indefinitely."

I think a major difficulty in answering this type of questions is in agreeing on what is meant by the speed of light, or of a wave more generally. For instance, Sommerfeld shows that the first light emerging from a medium, any dispersive medium, does so at exactly the speed of light in a vacuum $c$. He says "We will show here that the wave front velocity is always identical with the velocity of light in vacuum, $c$, irrespective of whether the material is normally or anomalously dispersive, whether it is transparent or opaque, [...]".

The proof given is a bit (only a bit) mathy and based on integrals in the complex plane. However, Sommerfeld (inspired by Voigt, it seems) also gives an intuitive mathless explanation of why it is so: "When the wave front of our signal makes its way through the optical medium, it finds the particles which are capable of oscillating originally at rest, [...]. Originally, therefore, the medium seems optically empty, only after the particles are set into motion, can they influence the phase and form of the light waves. The propagation of the wavefront, however, proceeds undisturbed with the velocity of light in vacuum [...]."

But the wavefront carries so little energy with it and it would seem more reasonable to attach the "speed" to the bulk of the signal rather than to its "forerunners". Brillouin (same book, chapter 3) then proves that the bulk of the signal arrives later often at (but not always) the group velocity using more involved integration techniques (namely, saddle-point method). Brillouin says (his emphasis) "Thus, it is seen that at that moment when the path of integration reaches the pole, the intensity of the oscillation increases very rapidly [...]. This moment, marking the arrival of the signal, permits one to define a signal velocity. This velocity will be shown to be equal to the group velocity, if [...]." So the wave was there already, but now it intensified.

If you are a fan of mechanical analogies (I am), here is a simple way of putting things (although, clearly, far from the rigor of Sommerfeld and Brillouin's methods): Consider a periodic laminate, that is a layering of space with two alternating phases of Young's modulus $E_i$, mass density $\rho_i$ and thickness $a_i$, $i=1,2$ (this could also be a dielectric in which case $E$ and $\rho$ become $\epsilon$ and $\mu$ (or is it the converse? Anyway)). An incident wavefront will propagate at velocity $v_i=\sqrt{E_i/\rho_i}$ in phase $i$ and, after transmission by $2n$ layers, will proceed at approximately the speed $$ c = \frac{x}{t} = \frac{n(a_1+a_2)}{n(a_1/v_1+a_2/v_2)} = \frac{1}{\langle 1/v_i\rangle} $$ where $\langle\rangle$ is an average wieghted by the $a_i$. But direct transmission through $2n$ layers leaves very little energy to that front. The bulk of the signal goes through many more reflections and transmissions (not unlike what happens in the greenhouse effect). To analyze these accurately, we have to make a complete and involved combinatorial account (excuses, I know). One way around that, is to suppose that the bulk of the signal has low frequencies so that the typical wavelength actually spreads over several layers. In that case, the laminate has an "effective" Young's modulus and an "effective" mass density given by $$ E = \frac{1}{\langle 1/E_i\rangle},\quad \rho = \langle\rho_i\rangle. $$ To get the first of these, imagine two springs connected in series. Therefore, such low-frequency signals go at the speed $$ v = \sqrt{E/\rho} $$ and indeed we have $v\leq c$ because of some well-known inequality (it's always Cauchy-Schwarz).

All in all, a wavefront goes at the one speed (the maximum speed) allowed in a given medium. All other things follow and therefore are somewhat slower. In that sense, there isn't much to prove.


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