Wavelength of photon in a medium

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A photon goes from vacuum into a medium. This causes its momentum to reduce. Hence by the de Broglie formula, its wavelength should increase.

But, since the refractive index of the medium is greater than 1, and
[tex] n_2/n_1 = \lambda_1 / \lambda_2 [/tex], and hence, its wavelength should decrease.

What am I missing here?
 
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hi,
we know that f = v[tex]\lambda[/tex]
[tex]\frac{n2}{n1}[/tex] = [tex]\frac{v2}{v1}[/tex] = [tex]\frac{\lambda1}{\lambda2}[/tex]
hope that will help
 
sorry the n2 and n1 should be reversed
 

sophiecentaur

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Does de Broglie apply to photons in such a straightforward way?
 
De broglie says that the rules for wavelengths for photons are valid for other matter as well.
 

sophiecentaur

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Yes but I think the apparent paradox is because there is an assumption that momentum has to reduce. That assumes that all that is changing is the speed of the photon. Surely, when an em wave travels through a medium, there is a constant interaction with the bulk of the medium and the photon is not just 'travelling through space with a different value for c'.
Start with the observed fact that the wavelength decreases and, rather than looking for a violation, think of how a photon could be making its way through the medium and it's not straightforward. It isn't as simple as saying it interacts with individual atoms - that would be an absorption / re-radiation process, like in a gas. So there doesn't have to be a conflict.
Also, there is the point that, on emerging from the other side of the medium, the photon will have the same energy and momentum as when it went in; it hasn't 'lost anything' as it would if the simplistic argument were applied.
If you think about it classically - a wave entering and emerging from a length of waveguide, you have the same sort of thing and you can explain the changes in wavelength in terms of distributed capacity and inductance (transmission line) and, probably also, forces on the waveguide at the entrance and exit, which would balance out.
 
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You're basically saying that there is no simple way to apply conservation of momentum for a photon entering a medium. But Snell's law can be derived by considering conservation of the photon's momentum in the plane of the interface. Is it just a coincidence that it works? By the same argument, we should not be able to do that.
 

sophiecentaur

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Refresh my memory about that derivation. After all, the OP implies that it doesn't work that way, I think.
 
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Well if the incoming light is at angle [tex]\theta[/tex][tex]_{1}[/tex] to the normal, its momentum is h*n[tex]_{1}[/tex] /[tex]\lambda[/tex], and the component in the plane of the interface is h*n[tex]_{1}[/tex]*sin([tex]\theta[/tex][tex]_{1}[/tex])/[tex]\lambda[/tex]. This has to be the same as the component on the other side, h*n[tex]_{2}[/tex]*sin([tex]\theta[/tex][tex]_{2}[/tex])/[tex]\lambda[/tex]. Then you have n[tex]_{1}[/tex]*sin([tex]\theta[/tex][tex]_{1}[/tex]) = n[tex]_{2}[/tex]*sin([tex]\theta[/tex][tex]_{2}[/tex])
 
Also, there is the point that, on emerging from the other side of the medium, the photon will have the same energy and momentum as when it went in; it hasn't 'lost anything' as it would if the simplistic argument were applied.
That is a good point. But I still remain confused: if nothing happens to the momentum of the photon inside the medium, how is it that the wavelength changes? Or, since the wavelength is changing, doesn't the momentum have to necessarily change?
 

sophiecentaur

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I guess the flaw in the photon momentum derivation of snell's law could be that you can't automatically assume that the photon's momentum in the plane of a reflector is unchanged. That is a totally 'ballbearing against a steel wall" argument. Wouldn't that be dodgy?
 
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Momentum is based on mass and velocity.
Mass and energy (for some purposes) can be interchanged, this is one of those purposes.

When a photon moves from a vacuum to a medium it's velocity reduces, and it's wavelength shortens. A shorter wavelength means a higher energy photon (higher mass photon). So mass increases and velocity decreases so momentum is conserved.
 

sophiecentaur

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Is that really valid? Where would the increased energy come from?
 
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Momentum is based on mass and velocity.
Mass and energy (for some purposes) can be interchanged, this is one of those purposes.

When a photon moves from a vacuum to a medium it's velocity reduces, and it's wavelength shortens. A shorter wavelength means a higher energy photon (higher mass photon). So mass increases and velocity decreases so momentum is conserved.
A shorter wavelength doesn't mean a higher energy photon. You're mixing up spatial and temporal frequency.
 
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Wouldn't that violate conservation of energy?
E = hc/[tex]\lambda[/tex]

Couldn't we just say that since E initial = E final and that c decreases in a medium that [tex]\lambda[/tex] increases?

I'm not exactly sure for conservation of momentum though since one could argue that E = pc so with decreasing c, momentum would have to increase to conserve energy but like mrspeedybob stated this would have to be the effective "mass" portion but this seems to violate conservation of momentum anyway...
 
I think we have the answer here:
http://rsta.royalsocietypublishing.org/content/368/1914/927.full#sec-7

The first momentum, calculated through the de Broglie relation, represents the canonical momentum. While the second one represents the kinetic momentum. I'm not entirely clear with the details yet; but I'll try posting a short summary(or if anyone else can) once I'm conversant with it.
 

sophiecentaur

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Lovely reference there, Muthu.
In one blow, you have managed both to simplify / explain and yet to complicate the whole matter :cry:.
The final answer seems to be "it depends on how you define Momentum".
Ah well, perhaps that could indicate that treating photons as particles is a bit more dodgy than we all used to think and that life is much easier if we talk in terms of waves except when the actual 'interactions' occur. :smile:
 

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