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It is well known that a prism can "split light" by separating different frequencies of light:

prism diagram

Many sources state that the reason this happens is that the index of refraction is different for different frequencies. This is known as dispersion.

My question is about why dispersion exists. Is frequency dependence for refraction a property fundamental to all waves? Is the effect the result of some sort of non-linearity in response by the refracting material to electromagnetic fields? Are there (theoretically) any materials that have an essentially constant, non-unity index of refraction (at least for the visible spectrum)?

  • This related question http://physics.stackexchange.com/q/19422/ provides dispersion as the answer but doesn't provide an answer to "why dispersion?". – Brandon Enright May 24 '13 at 22:07
  • Pardon my laziness to just Google it, but isn't Newton credited with having discovered and studied this aspect of light? If I'm remembering correctly, then Newton wouldn't have neglected to postulate a corpuscular argument, which I'd be very interested in reading. – David H May 24 '13 at 22:52
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    Relevant http://en.wikipedia.org/wiki/Dispersion_relation. – joshphysics May 24 '13 at 22:52
  • @joshphysics that article mentions "interaction of the waves with the transmitting medium". I'd like to understand if that's a special property of light or if all waves will have some sort of interaction. I assume no interaction implies an index of refraction of 1 (no refraction) so does that mean non-1 index of refractions always have dispersion regardless of the sort of waves? – Brandon Enright May 24 '13 at 23:03
  • See also http://physics.stackexchange.com/questions/65156/refraction-of-light-and-frequency-dependence/65205 – John Rennie May 25 '13 at 06:50
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    Ondřej and I have given brief answers, his with more depth on a certain aspect of the problem, mine with more breadth and less depth. For a full treatment, see the Feynman Lectures, I-31 and II-32. –  May 26 '13 at 17:28
  • @DavidH: Newton shoehorned a lot of observations into his corpuscular theory of light, e.g., he interpreted Newton's rings http://en.wikipedia.org/wiki/Newton%27s_rings as evidence against the wave theory. I don't think Newton knew $n=c/v$, but $n=c/v$ plus Snell's law equals trouble for a corpuscular theory, since a particle moving into a region of higher potential should slow down and be deflected away from the normal, which is contrary to what is observed. –  May 27 '13 at 21:26
  • For those who might be interested, this is actually what I was looking for when I came across this discussion. It’s a database of IoR for various materials. It provides raw data as well as visuals using graphs with adjustable parameters. Enjoy!! https://www.pvlighthouse.com.au/refractive-index-library – Digcoal Oct 26 '21 at 22:01

4 Answers4

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Lorentz came with a nice model for light matter interaction that describes dispersion quite effectively. If we assume that an electron oscillates around some equilibrium position and is driven by an external electric field $\mathbf{E}$ (i.e., light), its movement can be described by the equation $$ m\frac{\mathrm{d}^2\mathbf{x}}{\mathrm{d}t^2}+m\gamma\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}+k\mathbf{x} = e\mathbf{E}. $$ The first and third terms on the LHS describe a classical harmonic oscillator, the second term adds damping, and the RHS gives the driving force.

If we assume that the incoming light is monochromatic, $\mathbf{E} = \mathbf{E}_0e^{-i\omega t}$ and we assume a similar response $\xi$, we get $$ \xi = \frac{e}{m}\mathbf{E}_0\frac{e^{-i\omega t}}{\Omega^2-\omega^2-i\gamma\omega}, $$ where $\Omega^2 = k/m$. Now we can play with this a bit, using the fact that for dielectric polarization we have $\mathbf{P} = \epsilon_0\chi\mathbf{E} = Ne\xi$ and for index of refraction we have $n^2 = 1+\chi$ to find out that $$ n^2 = 1+\frac{Ne^2}{\epsilon_0 m}\frac{\Omega^2-\omega^2+i\gamma\omega}{(\Omega^2-\omega^2)^2+\gamma^2\omega^2}. $$ Clearly, the refractive index is frequency dependent. Moreover, this dependence comes from the friction in the electron movement; if we assumed that there is no damping of the electron movement, $\gamma = 0$, there would be no frequency dependence.

There is another possible approach to this, using impulse method, that assumes that the dielectric polarization is given by convolution $$ \mathbf{P}(t) = \epsilon_0\int_{-\infty}^t\chi(t-t')\mathbf{E}(t')\mathrm{d}t'. $$ Using Fourier transform, we have $\mathbf{P}(\omega) = \epsilon_0\chi(\omega)\mathbf{E}(\omega)$. If the susceptibility $\chi$ is given by a Dirac-$\delta$-function, its Fourier transform is constant and does not depend on frequency. In reality, however, the medium has a finite response time and the susceptibility has a finite width. Therefore, its Fourier transform is not a constant but depends on frequency.

domj33
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    That's a good summary! Basically the resonance is given by $k$, but your answer does not really say what is this $k$ from. Dipole oscillations around the atom? – fffred May 25 '13 at 08:31
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    @fffred Yes, $k$ comes from the dipole oscillations of the electron. I thought that the basic meaning was clear from the text and I don't want to go into much detail concerning finding specific values of the constants. I am certain one can find more details in the literature; this is just to give the main idea. – Ondřej Černotík May 25 '13 at 08:39
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    What about the atoms of the opaque materials?Oscillation is an assumption inconsistent to electronic cloud. – elias2010 Feb 28 '18 at 12:09
  • Hey I had a doubt. The reason of frequency dependence of $\epsilon_0$ is rooted in the dipole oscillations of electron. For non dispersive medium, where refractive index $n$ is constant, do we have $\gamma$ = 0? But this is not possible as there must be some damping. – Ruchi Feb 20 '21 at 09:11
  • This answer explains how light will affect an electron. Refraction is occurring within the transition from one medium containing multiple electrons into another medium also containing multiple electrons. Moreover, most of these electrons are bound within orbitals which, I assume, would limit their freedom to oscillate. So, this begs the questions: which electrons are oscillating, and in what direction(s) are they oscillating? My first guess is that the relationship between a materials valence and conduction bands plays an important role in this as well as reflection and absorption. – Digcoal Oct 26 '21 at 21:15
  • An addendum to this is that metals seem to be more reflective based on the amount of convergence between their valence and conduction bands. Since most transparent materials seem to be insulators, I would believe that the band gap would limit the oscillations to the more tightly bound valence electrons which would create refraction domains based on the crystal lattice structure of the material. Amorphous material would have more uniform IoF while regular lattices would develop multiple IoF causing instances of birefrigence. Pressure also affects IoF. – Digcoal Oct 26 '21 at 21:34
  • Oscillation of the free electrons (i.g. inside a TV antenna) is known. Oscillation of the orbital electrons inside a material is unknown (not aware of QM) – elias2010 Jan 04 '22 at 12:46
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    According to the equation above, by setting $\gamma = 0$ don't we still have a factor of $\Omega^2-\omega^2$ in the denominator? It would still seem to be frequency-dependent without the damping term. – Jono94 Oct 14 '23 at 10:07
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The simple explanation given in Hewitt's Conceptual Physics is that atoms in condensed matter have a high-frequency resonance, and the index of refraction for most substances is strongest at the blue end of the spectrum because that's the high-freqency end, which is closest to the resonance. The following is my attempt to flesh this out with a little more serious physics. It seems to capture some of the truth, but in some ways it's crude or wrong.

Index of refraction of silica glass as a function of wavelength (redrawn from Kitamura 2007).

Kitamura 2007 gives a summary of experimental data for silica glass over a wide range of wavelengths, along with a physical interpretation. The graph above is redrawn from Kitamura. What is observed is that the complex index of refraction has three prominent resonances with a shape that I think is referred to as a Lorentzian. At each resonance, the real part of $n$ swings low and then high, while the imaginary part has a peak, indicating absorption. They attribute each of these resonances to one or more qualitatively different physical phenomena. The visible spectrum lies between resonances at about 0.1 $\mu$m and 9 $\mu$m. The former is attributed to "interaction with electrons, absorption by impurities, and the presence of OH groups and point defects," the latter to "asymmetric stretching vibration of Si-O-Si bridges."

Although this is all pretty complicated, I think there's some fairly simple physics that can be extracted.

In the visible region, it looks like the decrease of the index of refraction with wavelength is due to a combination of two effects. This region of the graph picks up a negative slope from the 0.1 $\mu$m resonance on its left, and also a negative slope from the 9 $\mu$m on the right. This is a universal feature of any function formed by adding up a bunch of narrow Lorentzian resonances: far from resonances, it always has a negative slope. The bigger contribution to the slope seems to come from the resonance on the left, which is consistent with Hewitt's explanation.

Kitamura mentions several models that explain the resonances, of which the only one I'm familiar with is called the Lorentz model. In the Lorentz model, you take an electron to be a harmonic oscillator, like a little mass bound by a spring to a nucleus. The displacement of this driven harmonic oscillator (represented as a complex number to include its phase) is the Lorentzian $x=Af(\omega)$, where $f(\omega)= (\omega^2+i\gamma \omega-\omega_0^2)^{-1}$ and $A=(e/m)E$. As the electrons perform this oscillation in response to a plane wave, they generate their own coherenet plane wave. What is actually observed is the superposition of this wave with the incident wave. This superposition has two parts, a reflected wave and a transmitted one. In the limit of a low-density medium (such as a gas), the index of refraction is given by $n^2=1-\omega_p^2 f(\omega)$, where $\omega_p$, called the plasma frequency, is given by $\omega_p^2=Ne^2/m\epsilon_0$, where $N$ is the number density of electrons. The plasma frequency has an $e/m$ in it from the amplitude of the driven harmonic oscillator, and another factor of $e$ because the amplitude of the reemitted wave is proportional to the amount of charge oscillating. In the case of silica glass, I think the 0.1 $\mu$m resonance is probably what is described by the above mechanism, while the other resonances are similar mathematically but involve other effects than oscillation of bound electrons. E.g., the Si-O-Si bridges would resonate at a lower frequency due to the greater inertia of the nuclei compared to electrons.

An interesting feature of the graph is that there are broad plateaus, and as we go up in wavelength, each plateau is successively higher than the preceding one. This is explained by the Lorentz theory. In the limit the response of a driven harmonic oscillator approaches zero in the limit $\omega\gg\omega_0$, but approaches a constant (with reversed phase) for $\omega\ll\omega_0$. Adding the contributions from the various resonances produces an ascending staircase as observed.

Is frequency dependence for refraction a property fundamental to all waves?

The above does seem to suggest that there's some very universal behavior going on in the interaction of EM waves with matter.

Is the effect the result of some sort of non-linearity in response by the refracting material to electromagnetic fields?

No, I think it's basically the linear response of a driven harmonic oscillator.

Are there (theoretically) any materials that have an essentially constant, non-unity index of refraction (at least for the visible spectrum)?

I'm sure this would be a holy grail for people doing optics. AFAIK, the best way of getting rid of dispersion in real devices seems to be combining two materials so that the dispersion cancels out. Silica glass does seem to have a relatively constant $n$, and this would be because the visible spectrum is relatively far from the two nearby resonances. To get less dispersion in the visible spectrum, I guess you would want a substance in which the resonant frequency that glass has at 0.1 $\mu$m was displaced higher.

Kitamura, http://www.seas.ucla.edu/~pilon/Publications/AO2007-1.pdf

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    Does this imply that dispersion is somewhat temperature dependent? If we cooled a prism down near absolute zero would we changes in the dispersion? Edit: I just noticed the paper has "at room temperature" in the title. It seems it is temp dependent. – Brandon Enright May 24 '13 at 23:26
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    @BrandonEnright: To me, the Lorentz model seems to imply that $n$ would be roughly temperature-independent, except for maybe a slight change in $N$ due to thermal expansion...? I think the papers I came across on the web seemed to indicate that the temperature dependence was quite small, and hard to measure. –  May 24 '13 at 23:30
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    Also, it seems the motion of the electrons really matters. I thought refraction of light was based on the strength of the electric field but it looks like there is more to it than that. Hypothetically, if we replaced all electrons in a glass with muons instead then that material would have completely different refractive and dispersive properties? – Brandon Enright May 24 '13 at 23:33
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    The temperature dependence is straightforward as Ben surmised a matter of thermal expansion. Refractive index is more or less proportional to density (=electron density, but there are some exceptions, eg glasses containing titanium). If one looks at closely related substances only eg organics, the deviations of that proportionality contain some structural information, look for "Parachor". The temperature dependence is easily measured, as everybody knows who ever worked with a refractometer. – Georg Jul 16 '13 at 11:01
  • I don’t think this answers the question completely, but I think it’s the best one so far. I also appreciate the IoF refraction graph you posted. Do you know where I can find more? Those graphs were actually what I was searching for when I happened upon this question and answer. Thanks. – Digcoal Oct 26 '21 at 21:37
  • Reading through the article, I wonder if “air” affects the results any? Will light induce different reactions on the various constituents in the air? Will particular wavelengths cause rotations in O2, N2, or CO2 instead of linear oscillations, and will each be affected by different wavelengths? If some other process is occurring prior to reaching the interface, perhaps they will contribute to the various dips and spikes. I wonder if anybody ran experiments using a single gas with pure crystals. I also wonder if the graph has any correlation with a graph of solar radiance hitting Earth. – Digcoal Oct 26 '21 at 21:49
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I will hand wave here, looking at the problem a photon at a time.

We know from the double slit experiment that even individual photons impinging on the double slit geometry display an interference pattern, characteristic of the frequency/energy of the photon and the geometry of the slits.

One can think of a crystal as a very large number of three dimensional obstacles/slits ( 10^23 molecules in a mole give a huge number even for a one centimeter crystal in the path of your illustration).

crystal lattice

A photon impinging on the surface of the lattice, finds not two slits , but a depth of slits all the way through. The observed effect of the different angular distribution according to the impinging frequency of the photon must be the result of the quantum mechanical interference of the photon, which must be constructive in the angle of refraction given by its frequency and index of refraction and destructive everywhere else, otherwise we would be seeing interference fringes ( actually we do get a second rainbow, but that is a different story :) , though should be similar).

Then the problem is reduced to explaining the frequency dependence. I will hand wave again and say that the smaller the frequency the larger the distances in the interference pattern of the probability wave ; the photon will see the lattice gaps differently

enter image description here

according to its wavelength, as is true for the double slit experiment, so a fanning out is to be expected .

anna v
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    What you're describing is a mechanism for x-ray diffraction, not refraction. –  May 25 '13 at 15:13
  • @BenCrowell It certainly is an unusual explanation for dispersion but I guess it could still work. You'd just need to do the calculations in detail to check it. And there is one more thing - it explains dispersion only in crystals, not in, e.g., glass. – Ondřej Černotík May 25 '13 at 15:59
  • @OndřejČernotík as it HAS to work a photon at a time, i.e. single photons of the correct frequency should fall in the correct band of color, I do not see how it could be something else than a probability interference distribution pattern for photons one at a time. To get the color spectrum a calculation would include the distance of the unit cells of the crystal. In general the classical wave should merge with the photon ensemble frame except that calculating the latter would be too complex. – anna v May 25 '13 at 19:46
  • @BenCrowell X-rays have wavelengths much smaller than the distances of the unit of the crystal. They build up an interference pattern that can be fourier transformed to the crystal lattice. Here I am talking of larger wavelengths. My only argument is, as I said above, that it HAS to work a photon at a time, as the double slit experiment does too. – anna v May 25 '13 at 19:50
  • @BenCrowell I believe the diffraction Anna describes is the main one at work in opals. So "prismatic" (light splitting) behaviour is not only owing to refraction. – Selene Routley Jul 16 '13 at 11:38
  • This is why I have an issue with the general understanding of the double slit experiment: it assumes a single “packet of light” is a point-like entity when the light “acts like a particle.” That “packet” is interacting with the electrons of every piece of equipment from the source to the detector. How is it generated as a “packet,” and how does it maintain its “packet” state? I believe light is always a spherical wavefront which interacts with every ion it passes through. That wave front changes as it is absorbed and retransmitted by the ions it affects. The “packet” is always just the result. – Digcoal Oct 26 '21 at 21:27
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Provided that the electron & the atomic beams also exhibit refraction,it seems that this is a particle's property.Velocity and deflection angle depends on particle's mass/size for specific medium.Photon behaves as particle in this effect.Mass is given by de Broglie equation:m=hv/c^2 , v=frequency

elias2010
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    Dear Nikos Binis. It is usually frown upon to directly copy-paste identical answers. (The problem is if everybody start to copy-paste identical answers en mass.) In general in such situations, please consider one of the following options: (i) Delete one of your answers. (ii) Flag for duplicate posts and delete one of your answers. (iii) If you think the three posts are not duplicates, then personalize each answer to address the three different specific questions. – Qmechanic Dec 07 '14 at 16:21
  • You have right,forgive me.I ask for your compensation.I just want the science community to rethink about photon mass. – elias2010 Dec 11 '14 at 10:56
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    @elias2010 They are not going to rethink photon's mass because no experiment up to now indicates that the photon has mass. If you have any way to verify it, you've got a Nobel prize wating for you. Although that would be hard 1because the upper bound of the photon's mass is around $10^{-27} \mathrm{eV/c^2}$. – jinawee Dec 28 '14 at 18:00
  • Here is an experiment:Schmitt,Weitz,Vewinger,Klaers "Bose-Einstein condensation of photons in an optical microcavity”,Nature 468(7323):545-548 – elias2010 Dec 30 '14 at 08:41
  • There is no experiment for the opposite.Moreover,virtual photon has mass. – elias2010 Dec 30 '14 at 08:52