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$E = h\nu$ and $P = h\nu/c$ in vacuum. If a photon enters water, it's frequency $\nu$ doesn't change. What are its energy and momentum : $h\nu$ ? and $h\nu/c$ ? Since part of it's energy and momentum have been transferred to water, it should be less.

If water's refractive index is $n$, Are the energy and momentum equal to $h\nu/n$ and $h\nu/c/n$ ?

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2 Answers

It's a non-trivial problem, which also involves how you define a photon in a medium - as a interacting particle and treating excitation of medium separately, or as a "dressed particle", including the interaction.

From Abraham–Minkowski controversy Wikipedia page:

The Abraham–Minkowski controversy is a physics debate concerning electromagnetic momentum within dielectric media.


  • The Minkowski version: $$p=\frac {n h \nu}{c}$$
  • The Abraham version: $$p=\frac {h \nu}{n c}$$


A 2010 study suggested that both equations are correct, with the Abraham version being the kinetic momentum and the Minkowski version being the canonical momentum, and claims to explain the contradicting experimental results using this interpretation

Look also (Google Scholar?) at "electromagnetic momentum in a medium" or "electromagnetic energy in a medium", as essentially its related to a classical problem.

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A photon doesn't transfer part of its energy to water. Either it is absorbed or it is not. The energy is always $E=h\nu$.

A photon doesn't transfer part of its momentum to the water either. If it is absorbed, it transfers all its momentum to an electron of course.

If not, then there are several explanations about what happens and none of them are particularly enlightening. One is the microscopic view, put forth by Mark in the comments below, that the photon is traveling in mostly empty space, punctuated every now and then by a charged particle, and so its momentum doesn't change at all.

This is technically the most correct, but in my opinion not much use if you are looking at macroscopic scales. In that case, we have the Abraham-Minkowski controversy about whether the photon's momentum is higher or lower in a medium. Steve Barnett purports to have solved this controversy in a 2010 paper, as mentioned in the Wikipedia article, and I find that paper easily readable and enlightening. According to Barnett, the Abraham momentum, $P=h\nu/cn$, corresponds to the kinetic momentum of the photon (which is the momentum one usually thinks of when considering a macroscopic body in motion); and the Minkowski momentum, $P=nh\nu/c$, is the canonical momentum (which is defined as Planck's constant divided by the de Broglie wavelength of the body).

The answer is really that "the momentum" of a photon in a medium is not a well-defined concept, so you need to specify what you are talking about.

Working mostly with plane waves myself, I prefer to say $\vec{p}=\hbar\vec{k}$ (which indeed grows by a factor of $n$ in a medium) since this allows me to intuitively explain several other phenomena in terms of conservation of momentum. I will freely admit that this is a gross oversimplification, and also not intuitive in the sense that there is no good explanation why the photon's momentum should be higher, only the unsatisfying mathematical explanation that since the speed of light is "effectively" lower in water, the momentum is "effectively" higher.

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Sorry but I don't think this can be right. What mechanism could give extra momentum to the photon propagating inside the water? The photon itself just sees vacuum interspersed with charged particles. So, it propagates with energy $h\nu$ and momentum $h\nu/c$ at all times. However, as it propagates through the water it will scatter off electrons many times. The dominant effect is elastic scattering, which basically just gives the photon a phase shift. If you average these phase shifts over many mean free paths you will see oscillations of the EM field with an effective wavelength $\lambda/n$. –  Mark Mitchison Nov 18 '12 at 23:24
Alternatively, one can actually quantise the field inside a dielectric using the formalism of Macroscopic QED. Then the "elementary excitations of the EM field" probably will have momentum higher than in free space by a factor $n$ (although you should check this, I'm no expert in Macroscopic QED). But these are not really photons in the usual sense, rather quasiparticles corresponding to oscillations of both the EM field and the bound electrons in the dielectric medium. –  Mark Mitchison Nov 18 '12 at 23:34
@MarkMitchison I'm aware that photons appearing to travel slower in water is purely due to averaging. However, the momentum of a single photon is h-bar times the wave vector, and the wave vector's magnitude is larger in a denser medium with a slower speed of light. This is how phase matching works in nonlinear optics - conservation of momentum. Also, surface plasmon resonance excitation mechanisms (such as the Kretschmann configuration) are explained by different photon momenta in different materials. –  ptomato Nov 19 '12 at 9:54
Indeed, the Abraham-Minkowski controversy is divided between whether the momentum of a single photon is larger or smaller in a dielectric, but it certainly doesn't stay the same! –  ptomato Nov 19 '12 at 9:55
I've thought and discussed about it some more and come to two insights: –  ptomato Nov 19 '12 at 10:47
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