The rest mass of photon $m_0=0$ and photon travels at the speed of light in vacuum. So the energy of photon in vacuum is given by $$E_{vacuum}^2=(m_0c^2)^2+(pc)^2=(pc)^2$$ $$E_{vacuum}=pc=\gamma m_0c^2$$ As photon travels at the speed of light $\gamma= \frac{1}{\sqrt{1-c^2/c^2}}=∞$ and $m_0=0$, so energy takes the form $E_{vacuum}=∞\times0$ which can have a finite value.

But if light travels in glass ($μ=1.5$) the speed of light becomes $\frac{2c}{3}$ so $\gamma= \frac{1}{\sqrt{1-\frac{4c^2}{9c^2}}}=1.34$ a finite value. In this case, energy of photon in glass becomes $E_{glass}=pc=\gamma m_0\frac{c}{μ}c=1.34\times0\times\frac{c}{μ}\times c=0 $

Why am I getting the energy of photon in glass as zero when I apply the equation $E^2=(m_0c^2)^2+(pc)^2$?


2 Answers 2


When we say, the speed of a light in a medium, we are actually referring to the average velocity of light, I'll explain myself.

Think about only one photon reaching the "dense medium", as we know from quantum electrodynamics, light interacts with charged particles, and this dense medium, is nothing but a lot of charged particles: protons, neutrons(even they interact with light because they are "made of" charged quarks), electrons...

So if this photon enters the medium, it will interact, or not, with the rest of the particles, resulting that the time it takes the photon to escape from the medium, may, or may not fulfill the relation c=L/t with L the lenght of the medium, I mean, it may interact with no particle, it may interact "a bit" or it may interact "a lot". So the velocity of light in a medium is more like the average velocity it takes a photon to escape from the medium.

Now think about neutrinos, they are massive particles, so they do not travel at speed of light, but their interaction with matter is so weak, that in some mediums, we see that they go faster than light. https://physics.stackexchange.com/questions/94138/do-neutrinos-travel-faster-than-light-in-air#:~:text=Neutrinos%20will%20travel%20faster%20than,%2Fs%20and%20n%3E1.

Also I want to remark that your calculation is wrong. For massless particles, the relation $E = \gamma m$ does not hold, instead you should use $E =\frac{p}{\beta}$ with $\beta = v$ (all this with units c=1, I'm sorry).

What you should finally understand is that you are using microscopic arguments, to directly solve a macroscopic problem. Understand macroscopic behaviour from microscopic behaviour is possible but complicate, and needs to use average of certain properties between a lot of complicated stuff, for example the famous $T\propto mv^2$ from the kinetic theory of gases that relates the temperature of a gas with the squared average velocity of its particles.

I hope I made my answer clear, if not whe can still comment on this!

  • $\begingroup$ Photons do not interact "a bit" with a medium, they either get absorbed or not. If a photon interacts with a medium then it cannot "escape it" because it's been absorbed. What escapes are re emitted photons. The only way a photon could interact without getting absorbed would be by changing frequency, but that's not the case here. $\endgroup$ Apr 29, 2021 at 2:30
  • $\begingroup$ @ManudeHanoi When I said "a bit" I was meaning that they interact with "a bit" of particles in the medium. Yes, they can be absorbed and reemitted but this process may change photon frequency, changing the photon, so I think this is a different problem.There are other proceses with may change photon energy, in fact when photons interacts with matter, they may lose energy transferring it tho the matter, but the process we are interested here is that in which the energy of the photons does not change, just its direction, and this is elastic scattering of photons with charged particles. $\endgroup$ Apr 29, 2021 at 7:01
  • $\begingroup$ Please correct me if im wrong but even elastic scattering is done though absobtion and emission. The direction of a photon doesnt change, it gets absorbed and another one is re emitted. From raylegh scat. wiki: "The particle[of matter], therefore, becomes a small radiating dipole whose radiation we see as scattered light" $\endgroup$ Apr 29, 2021 at 15:55
  • $\begingroup$ @ManudeHanoi I think we are talking about different phenomena, I'm talking about scattering between to particles, for example photon + electron --> photon + electron in which photon momentum only changes its direction, and maybe you are talking about the photon being absorbed by an atom and then reemitted? In the case I'm talking, I'm not sure if I'm correct 100% but yes, you could say that photon and electron are absorbed, but, by the vacuum and then emitted also by the vacuum. $\endgroup$ Apr 29, 2021 at 22:35
  • $\begingroup$ @ManudeHanoi but when emitted by vacuum is this photon actually different? Well, quantum mechanically not really(open to discussion), but classically, we just see light entering the medium, and getting out with a different direction. $\endgroup$ Apr 29, 2021 at 23:08

from refractive index wiki

In optics, the refractive index (also known as refraction index or index of refraction) of a material is a dimensionless number that describes how fast light travels through the material. It is defined as

n = c/v , 

where c is the speed of light in vacuum and v is the phase velocity of light in the medium

Matter is mostly made of vacuum and light always goes at the speed of light so the speed of light doesnt change in an "optically dense medium", it's the phase velocity that changes. How you could experimentally distinguish the speed of light from the phase velocity in an optically dense medium, that I wish to know.

  • $\begingroup$ I don't quite follow you. Does photon travel in a optically dense medium with reduced speed or does it still travel at the speed of light ? $\endgroup$
    – Asher2211
    Apr 28, 2021 at 16:41
  • $\begingroup$ physics.stackexchange.com/q/466 $\endgroup$ Apr 28, 2021 at 16:46
  • $\begingroup$ physics.stackexchange.com/questions/1307/… $\endgroup$ Apr 28, 2021 at 16:47
  • $\begingroup$ Group velocity, $v_{gr}=d\omega/d k$, also changes along with the phase velocity. Moreover, if your refractive index does not depend on light frequency, group velocity and phase velocities are the same $\endgroup$
    – Pavlo. B.
    Apr 28, 2021 at 21:02

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