What happens to a photon's energy when it is refracted?

[maxelcat]

TL;DR Summary

does a photon lose energy when it is refracted?

I believe that when a photon is refracted it slows. Why is it that the wavelength decreases but that its frequency stays constant?

Does this imply that the photon has not lost any energy in the process of slowing down given that E=hf?
Thanks

 

[Ibix]

Continuity of the electromagnetic field leads to the frequency being the same. If the wave in the medium is driven by the wave incident on the surface, when the incident wave just outside the surface is at a maximum then the wave just inside the medium must also be at a maximum. Likewise when the wave just outside the medium is at a minimum, so must the wave just inside be at a minimum. Thus the frequency has to be the same.

I don't believe the wave does lose energy (in an ideal situation anyway). Electromagnetic waves in a medium are rather more complex beasts than in vacuum, since they are an interaction between the electromagnetic fields of the electrons and protons of the medium with the travelling EM wave.

If you want an answer in terms of photons you'll need to ask in the quantum physics forum. I suspect the answer will be to hand you a textbook, though - photons are objects in relativistic quantum field theory and they aren't easy to understand.

 

[maxelcat]

thanks Dale. I figured this wouldn't be an easy answer

 

[tech99]

TL;DR Summary: does a photon lose energy when it is refracted?

I believe that when a photon is refracted it slows. Why is it that the wavelength decreases but that its frequency stays constant?

Does this imply that the photon has not lost any energy in the process of slowing down given that E=hf?
Thanks

Some of the energy incident on the boundary is reflected, so the refracted energy is less than the incident energy. In quantum terms I suppose we say that occasional photons are reflected at the boundary.

 

[Andy Resnick]

TL;DR Summary: does a photon lose energy when it is refracted?

I believe that when a photon is refracted it slows. Why is it that the wavelength decreases but that its frequency stays constant?

Does this imply that the photon has not lost any energy in the process of slowing down given that E=hf?
Thanks

Conservation of energy (constant frequency) still holds, as does conservation of momentum (wavelength/wavevector); during refraction/reflection, there is momentum exchange between light and medium. This is measurable:

https://journals.aps.org/pr/abstract/10.1103/PhysRev.50.115
https://royalsocietypublishing.org/doi/10.1098/rsta.2009.0207

 

[sophiecentaur]

Conservation of energy (constant frequency) still holds, as does conservation of momentum

One way to look at this is that the momentum of a single photon is small. The Momentum transferred to a macroscopic body (with significant mass) by collision with a photon is too little for a change in frequency to be measured (easily) There is a mechanical analogue with a small ball bearing bouncing off a steel wall (an elastic collision); the wall just won't absorb any significant momentum. (In this case, of course, both objects have mass but it gives some idea of the numbers involved.)

 

[Andy Resnick]

One way to look at this is that the momentum of a single photon is small. The Momentum transferred to a macroscopic body (with significant mass) by collision with a photon is too little for a change in frequency to be measured (easily) There is a mechanical analogue with a small ball bearing bouncing off a steel wall (an elastic collision); the wall just won't absorb any significant momentum. (In this case, of course, both objects have mass but it gives some idea of the numbers involved.)

Don't mix up energy and momentum- momentum is a vector, for example. In an elastic collision, changing the direction of propagation requires a change in momentum but no change in energy.

 

[Steve4Physics]

TL;DR Summary: does a photon lose energy when it is refracted?

No.

It's not ideal to treat a photon as both a wave and particle at the same time. Nevertheless, for the purpose of explanation, consider this.

A photon moves from a vacuum into some medium. With the usual symbols...

During refraction, $f$ doesn’t change, as already nicely explained in Post #2 by @Ibix.

A photon’s energy is given by $E=hf$. That’s a fundamental relationship. Since $f$ doesn’t change, neither does $E$.

$v_m$ is the speed of propagation in the medium and $\lambda_m$ is the wavelength in the medium.

$f = \frac {v_m}{\lambda_m}$. On entering the medium, both speed and wavelength decrease by the same factor. $E = hf = \frac {hv_m}{\lambda_m}$, unchanged.

The relationship $E = \frac {hc}{\lambda}$ is true for EM radiation in a vacuum (or approximately true when $v_m \approx c$).

[Minor edit.]

 

[tech99]

I feel that some of the energy of the wave when in the medium is being carried by the motion of electrons and their accompanying fields.

 

[Steve4Physics]

I feel that some of the energy of the wave when in the medium is being carried by the motion of electrons and their accompanying fields.

I agree. My Post #8 explanation is (overly?) simplistic - but may be adequate to answer the OP’s question.

In a crude model. photons pass through a refracting medium. In an improved model, ‘polaritons’ are transmitted through the medium. Polaritons - quasiparticles - incorporate the (oscillating) fields of the medium’s charges.

But my knowledge of QED is somewhat (very!) limited, so that may not be an adequate description.

 

[tech99]

If the two waves are to be a single quasi-particle, that seems to imply that they travel at the same speed.

 

[Steve4Physics]

If the two waves are to be a single quasi-particle, that seems to imply that they travel at the same speed.

Depends what is meant by a 'polariton' I guess. E.g. see this old thread: https://www.physicsforums.com/threads/are-polaritons-simply-photons-propagating-in-medium.653925/

 

[sophiecentaur]

Don't mix up energy and momentum- momentum is a vector, for example. In an elastic collision, changing the direction of propagation requires a change in momentum but no change in energy.

This is true but it's hard to predict what will happen in a collsion if you don't consider momentum. The change in momentum of the photon will affect the momentum of the material. Energy change won't (?) give you the change in direction.
Refraction is a wave phenomenon and 'collision' ideas don't easily help to describe what happens on exit. There is nearly always some energy loss in interaction with a substance. A notable exception is the 'elastic collision' when a phootn strikes a tightly bound metal nucleus in the Mossbauer effect when the frequency change is 'zero'. That just adds confusion to the student of classical optics.

 

[Andy Resnick]

This is true but it's hard to predict what will happen in a collsion if you don't consider momentum. The change in momentum of the photon will affect the momentum of the material. Energy change won't (?) give you the change in direction.
Refraction is a wave phenomenon and 'collision' ideas don't easily help to describe what happens on exit. There is nearly always some energy loss in interaction with a substance. A notable exception is the 'elastic collision' when a phootn strikes a tightly bound metal nucleus in the Mossbauer effect when the frequency change is 'zero'. That just adds confusion to the student of classical optics.

I think this is making the situation too complex: light reflecting off a mirror is an example of an elastic collision, for example.