Light hitting a dielectric, increase of momentum

[Repetit]

When light propagates in a dielectric its wavevector is given by:

$$k = \frac{\omega n}{c}$$

where n is the refractive index. If light propagates in vacuum n=1 the momentum is $$\omega / c$$ but if light propagates in a dielectric of for example n=1.33 the momentum is increased beyond the value of the momentum in vacuum? Where does this extra momentum come from? It has to come from the dielectric somehow but how does this work?

 

[Claude Bile]

There is a momentum transfer to the medium. The momentum transfer is momentary since the light will recover that momentum (neglecting reflections) upon exiting the dielectric.

Do a google on radiation pressure and linear photon momentum conservation and you will probably get some papers that go into the sordid details.

Claude.

 

[olgranpappy]

When light propagates in a dielectric its wavevector is given by:

$$k = \frac{\omega n}{c}$$

where n is the refractive index. If light propagates in vacuum n=1 the momentum is $$\omega / c$$ but if light propagates in a dielectric of for example n=1.33 the momentum is increased beyond the value of the momentum in vacuum?

...but if the light propagates at a frequency that is much greater than the plasma frequency of the medium then 'n' is less than one... and if the light has a frequency that is exactly equal to the plasma frequency then n=0!

Anyways, the problem is that you can't really say that the light "has more momentum" in the medium since the momentum is not a local quantity for a field--we are not talking about photons here, we are talking about photons en masse as macroscopic electric fields--In this case, we have the poynting vector to tell us about the momentum of the field, but the pointing vector is not really a "local" quantity in the sense that it must be intergated over closed surfaces to obtain sensable results... You could integrate the poynting vector over the front and back faces of the dielectric slab to see that no energy is transferred to it.

This is to be contrasted with a "lossy" medium (with a complex dielectric function) where there will be energy transferred to the "slab" (which heats it up).

Cheers.