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How does optical reflection work?

by davidyanni10
Tags: metals, photon, quantum, reflection, reflectivity
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davidyanni10
#1
Mar29-13, 08:11 PM
P: 19
If you take, for example, a polished brass surface (or whatever really) and an incident photon, is the photon being absorbed and re-emitted by an atom at the surface of the reflective brass?

If so, how is it that the re-emitted photon is always the same wavelength as the incident photon?

The only plausible way I can think to picture what's going on is that the incoming EM radiation causes electrons to oscillate and therefore re-emit EM radiation of the same frequency. I don't understand how this vibration process could be 100% energy efficient though. Also, can electrons oscillate at any given frequency? Aren't these vibrations quantized?

I become even more confused when thinking about non-metals that totally don't have free electron gases. (For instance how does reflection in water work like when I see my face in the pond?)

Edit: I want to be clear about what I'm asking and why I can't find it (at least not easily) in the archives here or on the google.

I have come across many answers that essentially point to the Huygens Fresnel Principle

such as this.
http://www.physicsforums.com/showthr...ght=reflection
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zje
#2
Mar30-13, 01:42 AM
P: 16
Conservation of energy - remember that the energy of a photon is inversely proportional to its wavelength. The incoming photon transfers its energy to the electron and the energy (photon) that is released out must be the same, therefore the wavelength is the same. This may be oversimplified, but I think it answers at least part of your question.
Simon Bridge
#3
Mar30-13, 02:24 AM
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Quote Quote by davidyanni10 View Post
If you take, for example, a polished brass surface (or whatever really) and an incident photon, is the photon being absorbed and re-emitted by an atom at the surface of the reflective brass?
A metal, like brass, shares some electrons across the whole material ... so the "surface" is covered by a cloud of electrons. Reflection of visible light happens at different depths in this cloud.
If so, how is it that the re-emitted photon is always the same wavelength as the incident photon?
It isn't always - but it is mostly.
It's due to quantum mechanics - what you get is an electron absorbs a photon, which gives it more energy for a bit. After that bit, it gets rid of the energy. The quickest way is to just dump one photon.
The only plausible way I can think to picture what's going on is that the incoming EM radiation causes electrons to oscillate and therefore re-emit EM radiation of the same frequency. I don't understand how this vibration process could be 100% energy efficient though.
You'll notice that brass has a yellow tint to it's reflections? Not all wavelengths are reflected with equal probability. There are quite a lot of things that can happen besides reflection.
Also, can electrons oscillate at any given frequency? Aren't these vibrations quantized?
In a metal the energy levels of the conduction electrons are very close together.

I've been a more detailed discussion of reflection:
http://www.physicsforums.com/showthread.php?t=570003
I become even more confused when thinking about non-metals that totally don't have free electron gases. (For instance how does reflection in water work like when I see my face in the pond?)
For refractive substances, reflections are more easily understood in terms of the refractive index.
http://scienceline.ucsb.edu/getkey.php?key=2816

Richard Feynman has a good set of descriptions of reflection on the quantum scale.
Fundamentally, reflection is an emergent behavior - something that happens on average.

Darwin123
#4
Mar30-13, 03:17 AM
P: 741
How does optical reflection work?

[QUOTE=davidyanni10;4326507]If you take, for example, a polished brass surface (or whatever really) and an incident photon, is the photon being absorbed and re-emitted by an atom at the surface of the reflective brass?

If so, how is it that the re-emitted photon is always the same wavelength as the incident photon?

The only plausible way I can think to picture what's going on is that the incoming EM radiation causes electrons to oscillate and therefore re-emit EM radiation of the same frequency. I don't understand how this vibration process could be 100% energy efficient though. Also, can electrons oscillate at any given frequency? Aren't these vibrations quantized?

The charge-carriers are effectively unbound. An unbound charge carrier can oscillate at any frequency. The quantized steps in frequency are in important in bound-states. An unbound charge-carrier can make transitions in what is effectively a continuum of energy states.

The amplitude of vibration of an unbound carrier is effectively quantized even though it can oscillate at any frequency. If light of a particular frequency is incident on the charge carrier, then the charge carrier has to oscillate at the frequency of that light. However, the energy and momentum of the light beam has to change in discrete steps of energy because of the "particle" nature of light. Energy and momentum are conserved. When the light beam loses energy or momentum to the charge carrier, the carrier gains energy or momentum. The energy and momentum of the carrier vary with the amplitude of the vibration. When a photon is absorbed or scattered, the amplitude of the vibration in the charge carrier has to respond consistent with the conservation laws.Therefore, the particle nature of light results in a quantization of amplitude in the charge carrier.

The charge carriers are bound to a limited extent even in a block of metal. Adjacent energy states of the charge-carriers in the metal are very close together because they are not bound to individual atoms. The charge-carriers are bound by the boundaries of that metal because the nuclei attract them. Obviously, the conduction-electrons in a brass door knob have to stay in the door knob. However, the boundaries of the metal are very large to compared to the individual atoms. So the individual steps in frequency are so small that they are insignificant.

Slight modification of this model. In a metal, it is the amplitude of the plasmon which is quantized. Understanding metals involves understanding plasmons.

The efficiency of metals in reflection is caused by plasmons. The reflectivity in metals is large because the charge carriers move together in a type of wave called a plasmon. The Pauli exclusion principle forces the charge carriers to move in unison. When a photon of low energy hits a charge carrier in the metal, all the carriers have to vibrate at the same time. Therefore, the energy can't be randomized in separate carriers. Therefore, most of the energy is not lost as heat.

The argument that I earlier made about the energy of the unbound carriers is still valid. The amplitude of the plasmon is quantized by the particle nature of light. However, massive objects don't absorb energy all that well in collision. Massive objects efficiently absorb momentum, not kinetic energy.

Think of a ball bouncing off a brick wall. The carriers in a metal are like the bricks in the wall. A low energy photon hitting a huge block of electrons bounces off it without giving up kinetic energy. It may give up momentum, but not energy.

I become even more confused when thinking about non-metals that totally don't have free electron gases. (For instance how does reflection in water work like when I see my face in the pond?)

Insulators are similar. However, here you have to realize that the entire atom or the entire solid isn't bound. Therefore, the energy states associated with the center of mass aren't quantized.

I will talk a bit about elastic scattering rather than reflection. They are closely related.

Think of an atom drifting in outer space. It has internal degrees of freedom with quantized energy. Relative to the atoms center of mass, the orbital diameters of the atom are quantized. However, the center of mass of the atom isn't bound. So the center of mass is not bound. To make an electron move away from the nucleus, you need a minimum amount of energy that can be very large. There is no minimum in energy required to make the center of mass of the atom move.


Electrons are still vibrating, but they are bound to to atoms. The atom can move. It is not bound in the same sense. Or if it is bound to a molecule, then the entire molecule can move. It is the brick wall, again. All the electrons in the atom move together because of the Pauli exclusion principle. So even if the atom doesn't absorb the momentum in its internal states, the center of mass of the atom is unbound. Therefore, it can absorb any amount of energy.
davidyanni10
#5
Apr1-13, 07:09 PM
P: 19
Thanks for the answers and the links guys!

This clears things up a lot.
Claude Bile
#6
Apr12-13, 11:52 AM
Sci Advisor
P: 1,477
The key here is to understand that photons do not undergo absorption and re-emission upon reflection. Atoms at a surface can be thought of as driven oscillators - the incoming wave(function) polarizes the atoms, which produce a response field - the reflected wave(function). The response field must be the same frequency to preserve continuity of the electric field. This applies whether the atoms are part of a conductor or dielectric.

In fact, there is not real need to invoke QM, as reflection is readily explained via classical optics.

If you insist on a QM description, think of the atoms at a surface as being in a virtual energy state during photon reflection. This is why there is no quantization; the states are virtual. This is quite different to absorption and re-emission from a real energy state, where the emitted photon is random in time, phase, frequency and direction.

Claude.


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