# I Conservation of energy in refraction

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1. Jan 12, 2017

### Jigglypuff

Hello,

This has been bugging me for some time now, so I would be interested to see what I have been missing so far.

Imagine a single ray of light (made up of many photons) hitting a perfectly non-absorbing (for this wavelength of light) spherical dielectric object, which has finite mass. The outside medium is a vacuum - i.e. where the photons are coming from - and there is no influence of gravity, friction etc. The object has some refractive index that is not 1, and the ray hits it at some angle that is not zero with respect to its axis of symmetry (i.e. it doesn't go through the centre of the sphere). The ray enters and exits the object (although a single interaction is enough to demonstrate my point).

The result would be a change in direction of the ray with respect to its original path, and therefore a change in momentum. A force is required to do this (one exerted by the object's reactionary electromagnetic field, caused by the oscillating electrons in the atoms it consists of). There must also be an opposite force exerted on the particle, which would then start to move in that direction.

As far as I am aware, the scattered field of the ray, whilst having had its direction changed, will have the exact same magnitude of momentum, and therefore the exact same energy as it did before the interaction with the object. However, the object has has work done on it, therefore it seems to me like energy has been created from nowhere, which cannot be the case.

If anyone has any insight as to what I have not thought of in this scenario, I would greatly appreciate the help! Thanks in advance for your time!

2. Jan 12, 2017

### Simon Bridge

Stop there. You either have a ray or you have photons. The two descriptions are part of different models. You can sometimes get away with mixing them up but do so at your peril. I have a feeling this is what is happening here - you may be thinking in terms of Newton's "corpuscle" model ... where light is an object that bounces off things etc.

... so the sphere is not made of matter or energy?
I'm just teasing, lets say the influence of gravity can be safely neglected.

The incoming ray we are considering does not have normal incidence - which, for a sphere, would be radial.
We are only worried about the interaction at the first surface encountered - OK.

For light this is problematic - it is not usually useful to model these processes in terms of force. However, in order for momentum to be conserved, the change must come from someplace... you are correct.
You can calculate the change in momentum given to the ball due to the reflection and refraction of the ray.

We don't normally worry about it because we clamp the ball to the bench, but there are experiments where light is used to push an object.

... your description allows for reflection and refraction. Are you conflating wave and particle models?

Energy and momentum are conserved in the interaction of light an matter.

What you have noticed is that the common descriptions of ray optics are incomplete and approximate.

Pick a model - try to stay inside it.
With rays: a ray is a single line along a wave ... the wave has an extent in space and delivers energy and momentum continuously.
In photons - what happens is that the incoming photon gets absorbed by part of the material, then re-emmitted in a different direction. The laws of refraction and reflection you are used to are what happens on average.

For a more complete lay-accessible description, see: http://vega.org.uk/video/subseries/8

3. Jan 12, 2017

### Staff: Mentor

This is not correct. The energy/momentum/frequency/wavelength can certainly change during scattering. A closely related phenomenon is responsible for any speeding tickets you may have received.

4. Jan 13, 2017

### Jigglypuff

Firstly, thank you for the replies!

Yes you are right, sorry for the confusion; however, this problem should be able to be explained using both models, right? If we take a single photon instead of a ray, it will be absorbed, and re-emitted (lets say in the average overall direction of refraction), shoving the atom in the opposite direction, therefore doing work on it. The photon should be the same, except for the direction of travel, or is that where the model becomes too much of an approximation? Is the momentum of the re-emitted photon lower (i.e. the difference being equivalent to the momentum given to the atom)? In that case, was there not at least partial absorption?

If we take the wave theory, then if the scattered field has a reduced amplitude, the difference in intensity should be that of the energy gained by the object. Would that be due to the reactionary field created by the object? Is that then not also absorption in a sense?

I wanted to not have the object stationary, in order to highlight the fact that work may be done on it as a result of radiation pressure. We can ignore the reflection when talking about the one photon example I mention above.

For the wave theory, perhaps it is better to change the example slightly. If we have an incident plane wave interact with a flat object (with a finite mass) at Brewster's angle, with the polarisation of the incident field being parallel to the plane of incidence, then the reflection will be zero (we can take the scattered field after the first interaction only, to avoid having to have a second Brewster angle afterwards), leaving this as a purely refractive process.

Thank you for the link by the way, anyone interested in science should listen to Richard Feynman!

I think that perhaps my problem may be in the approximation. But if light loses energy after a purely refractive event, then does that not make refraction a type of partial absorption? This is in response to the comment from Dale also.

I'm sorry if I sound a bit silly asking this, but already the discussion has helped me, so thanks in advance again!

5. Jan 13, 2017

### CWatters

6. Jan 13, 2017

### Staff: Mentor

Energy can be lost in refraction, scattering, reflection, or absorption. So energy loss does not imply absorption.

Last edited: Jan 13, 2017
7. Jan 13, 2017

### Simon Bridge

In the wave model, refraction, scattering, and reflection, do not involve absorption but can involve a transfer of energy.
In the photon model, all interactions involve absorption right? Otherwise we'd need to have a virtual particle to mediate the charge-photon interaction.
I suppose photon-photon scattering...

8. Jan 23, 2017

### ORF

Hello

Trying to answer the original question... Yes, light will loss a bit of its momentum because of the refraction. You can think in the refraction as an absorption+emission. Both processes change the momentum of the material: