B Reflection of a photon by an atom

I suppose the reason I am able to see myself in mirror and not in wood is the reflection off a mirror is specular whereas wood is diffuse.
In reflective material(e.g: silver) when the photon hits the frontline atom, the atom's electrons absorbs the energy and release it back with tiny energy loss.

My question is why the atom re-releasing the absorbed energy in same angle at it got hit? Why it didn't release the photon in a random direction or to the back side?
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vanhees71

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The electromagnetic field (and a photon is only a quite special state of the electromagnetic field) obeys the Maxwell equations, and the Maxwell equations tell you, how the field behaves given the matter (charge-current distribution) around it. Taking these equations and quite simple models for matter like a mirror shows that light is reflected in the way given in the picture. It's just the solution of Maxwell's equations subject to some quite simple boundary conditions given by the mirror.
 
The electromagnetic field (and a photon is only a quite special state of the electromagnetic field) obeys the Maxwell equations, and the Maxwell equations tell you, how the field behaves given the matter (charge-current distribution) around it. Taking these equations and quite simple models for matter like a mirror shows that light is reflected in the way given in the picture. It's just the solution of Maxwell's equations subject to some quite simple boundary conditions given by the mirror.
So Are you saying that the atom's re-emission path of photon can be predicted using maxwell's equation?
and, Why didn't the photon's emission spread in a wave like pattern?
 

ZapperZ

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A "mirror" is usually a thin film of a METAL. Let's say it is a thin film of Al, which is quite common.

Now, I will tell you that if you take an isolated Al atom, or simply a gas of Al atoms, you will NOT get the same mirror reflection. In fact, even if you take several Al atoms, say, 20, and string them along together, you will STILL not get your mirror reflection.

This is why the field of solid state physics is DIFFERENT than the field of atomic/molecular physics. As Phil Anderson says "More Is Different"! When atoms form a solid, many of the properties that we know of of the solid are NOT due to the properties of the atom, but rather to a COLLECTIVE property of the solid. Properties such as energy BANDS are not present in an atom. They are present in a solid!

An Al metal is different than an Al atom. The presence of the CONDUCTION BAND is one major example. The electrons that make up the conduction band do not belong to any individual atoms. They are like street gangs, moving and roaming about the entire solid without being tied to any one particular atom. So they do not have energy levels like an atom. They also can have very complicated properties, such as having this collective excitation called "plasmon", which often is attributed to the phenomenon of mirror reflection.

Moral of the story: you cannot describe a solid via the property of the atom that make up the solid. Otherwise, graphite will not look and feel so different than diamond.

Zz.
 

vanhees71

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Of course, for a single photon you can only describe the probability distribution for its deflection, which is given (to very good approximation) by Maxwell's equations.

Also a single atom has as good as nothing to do with a macroscopic object like a mirror though this mirror is built up by many such atoms. The point however is, that many is really many here, and you have to describe the mirror as a whole. For some details, see @ZapperZ 's posting #4, which just popped up when I started to write this :-))).
 
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My question is why the atom re-releasing the absorbed energy in same angle at it got hit?
As @ZapperZ indicated above, you cannot understand specular reflection in terms of an interaction between a single photon and an single atom. It takes roughly as many atoms as fits under one wavelength. The interaction can consider that collection of atoms to be a classical object which interacts with the photon essentially via the classical Maxwells equations
 

Vanadium 50

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when the photon hits the frontline atom, the atom's electrons absorbs the energy and release it back with tiny energy loss.
That's not true, so of course we can't explain "why" if it's not true. Where did you read this?
 

ZapperZ

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Can you point out what's wrong in this statement? I just read a "similar" fact from explainthatstuff.com
It is wrong based on what I just said! It has nothing to do with the atoms. "Reflection" of atoms due to an incoming photon has no direction. It can be reflected in any direction that the atom wants. It is also with specific frequency only because atoms has specific energy levels. Yet, we see a reflection over a continuous spectrum.

Don't you see a contradiction there already with that explanation?

Zz.
 
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Can you point out what's wrong in this statement? I just read a "similar" fact from explainthatstuff.com
Another thing that is wrong about it is the idea that a bit of the energy of the photon is lost. If that were the case then reflections would be redder rather than dimmer
 

Vanadium 50

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If reflections are properties of individual atoms, why does (polished) violet phosphorus have a mirror like surface but red or yellow phosphorus does not?
 

jbriggs444

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Another thing that is wrong about it is the idea that a bit of the energy of the photon is lost. If that were the case then reflections would be redder rather than dimmer
There is a way to recover a glimmer of truth from a claim that light loses energy when encountering a perfectly reflective mirror. Adopt the frame of reference in which the mirror is initially at rest. A pulse of light arrives and is reflected. The pulse carries momentum and as a consequence, the mirror rebounds very slightly. Doppler then calls for a red shift in the reflected light pulse.
 
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There is a way to recover a glimmer of truth from a claim that light loses energy when encountering a perfectly reflective mirror. Adopt the frame of reference in which the mirror is initially at rest. A pulse of light arrives and is reflected. The pulse carries momentum and as a consequence, the mirror rebounds very slightly. Doppler then calls for a red shift in the reflected light pulse.
That mechanism does exist, but it is insufficient to explain the amount of energy lost it the reflection from a standard mirror.
 

sophiecentaur

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Why didn't the photon's emission spread in a wave like pattern?
A single photon arriving at a 'flat surface' will interact with the whole surface and not just with one particular atom on the surface. The wave, of which the photon is a part, is the best way to describe what's happening. Many photons will turn up just where the wave calculation predicts - with a minuscule spread in angle which can also be calculated by a classical diffraction calculation. An Energy Change can be included if it is actually significant (see above: it isn't)
There is no point in asking "What really happens?". There is no 'really' involved.
 

Nugatory

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Can you point out what's wrong in this statement? I just read a "similar" fact from explainthatstuff.com
where we told that
Rays of light (which are really packets of light energy called photons, fired in a stream like bullets from a machine gun)...
That is pretty much entirely wrong - whatever photons are, they are nothing like that. There’s no reason to take this web page seriously.
 
My question is why the atom re-releasing the absorbed energy in same angle at it got hit? Why it didn't release the photon in a random direction or to the back side?
Because the (single) atom will not care of the angle in/out of the photon (roughly speaking). the atom(single) will diffuse the photon, as you imagine.
And to get the things worst, if you look at the mirror -very- close you will find it not flat at all at microscopic level, even the flattest mirror will be quite rought below 10E-8m scale. So why a photon will bounce like a ball in the wall?

More important of photons (particle) or rays (angles) is the (electromagnetic) field. The field will interact with more atoms in the mirror surface, and the reflection is the net result of this process, not a single atom.
Giving the wavelenght (very small) and the area of the surface (very large respect to wavelenght) the field will see an array of atoms, if enough regular, this will give the net constructive result we call a reflection.
Some part of the field can be absorbed (some photons lost) other part can diffuse depending on the substance (atoms) and flatness (geometry)

I remember a enlighteen explanation from Feynman about this, making mirrors reflect at the angle you want (quite, depending on wavelenght) stripping out parts of the coating: diffraction gratings.
 

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