In the Law of Reflection, the angle of incidence is equal to angle of reflection. Why is this true? This is clearly true experimentally, but how does one prove this true mathematically?
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It is mostly due to a perfect plane of the reflecting surface. If the surface is not plane, but wavy, you will have a "diffusion" rather than a reflection to some certain angle. – Vladimir Kalitvianski Jan 03 '19 at 13:29
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https://math.stackexchange.com/questions/1395215/proof-of-the-law-of-reflection-without-calculus – Daddy Kropotkin Jan 03 '19 at 13:30
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3The math behind the proofs linked in the above comments is somewhat straightforward; I'm not sure you're saying what you mean with 'prove this (mathematically)'. It's more likely that you're hoping for a derivation from first principles and fundamental characteristics of light and matter; wording which suggests the same would be helpful. – Jan 03 '19 at 13:30
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11This question is completely outside the realm of science, so I suggest to rephrase it. Science doesn't ever question "is XYZ true?" or "why is XYZ true?". Science doesn't ever prove an experimental result. Science uses experimental results and thinking as two inputs in order to produce its output: to predict future experimental results. That's it. If "math" stops working one day, science will probably trash it and look for other tools - but the light will still reflect the old way. – kubanczyk Jan 03 '19 at 16:22
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5Chapter 2 in Feynman’s QED book talks about the Law of Reflection. It’s interesting how time of travel plays a role. – Lambda Jan 03 '19 at 16:37
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3"how does one prove this true mathematically" - I just wanted to add; one does not - mathematics doesn't "prove" nature. Instead, you create a mathematical model, and check if it's predictions are in accordance with what can be observed in nature. – Filip Milovanović Jan 05 '19 at 01:20
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This is beautifully explained by Feynman using his path integrals.
I cannot hope to do it better, but just a quick non-mathematical overview. What is mind-blowing about the theory is that you assume that individual photon (on quantum electrodynamics level) is actually "reflected" in each possible direction by each atom of the mirror surface. If you calculate how all these "reflections" interfere with each other, you will see that it wouldn't result in chaos, because most of them tend to silence each other, except for one output angle. The silencing is because depending on timing of each possible path, the phases can be opposite at a place. According to the theory it means that the photon wouldn't probably appear there. What is great about it, is that "summing" (integrating) the phases of all these zillions paths doesn't require a supercomputer, but can be done in few minutes by drawing small pictures on a blackboard - see the video.
kubanczyk
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1You could add that this is the basis for why the Least Time Theorem seems to hold up - even for diffraction. – Stian Jan 04 '19 at 14:47
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6I feel the picture invoking photons and path integrals actually obscures the insight you offered in a comment to the question, namely that this is an experimental fact upon which we construct our theories of light, not something "explained" by science. The path integral depends on the action. If light always was off by 5° from equal angles when reflecting, we would have constructed a Lagrangian that, when fed into "the path integral" shows that off-by-5° path as the most likely. Note also that "the path integral" here is also just obscuring an application of the principle of least action. – ACuriousMind Jan 05 '19 at 11:45
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2@ACuriousMind Indeed. But a ray at -5° would not be reversible with time on a plane surface. It is the symmetry that underlies all the proposed mechanisms (of which this answer was the best). – Jan 06 '19 at 09:58
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The answer by harshit54 is very concise and clear and gives you answers in multiple layers of understanding. However, to quote Leonardo DiCaprio: we need to go deeper. Not because we must, but because we can! There's a TL;DR below.
A beam of light can be thought of as a stream of energy packets (photons, which are the quanta of light - lots of interesting words to look up in the dictionary already). Now let's zoom in and look at what happens when the photon hits any material. It runs into a wall of atoms - lots of nuclei surrounded by electrons (also energy packets - there's more to it but let's not write out all of quantum mechanics here). When a photon hits an electron, its energy gets absorbed and the electron goes into a higher energy state. This does not last long; the electron left an "empty" energy state below it, which is an energetically more favorable position for it. Thus, there is a chance that it spontaneously jumps back to a lower energy state. This chance increases over time, so it's pretty certain that it will jump back quite quickly. When it does, it needs to get rid of its extra energy. This energy is released as a photon!
If this is the only electron in the neighbourhood that releases a photon, it will go in any random direction. HOWEVER! There's a catch. Hint: this is where the wave nature of light comes into play. Let's assume that the beam of light hits the reflecting surface directly from above, so the angle of incidence is 0 degrees. Now you have many electrons that are being bombarded by even more photons, all emitting photons in many directions. The photons that are emitted at an angle however, will be out of phase with eachother (since there is a distance between the electrons, if two photons are emitted at any angle at the same time, there will be a slight delay between them). Photons that are out of phase will tend to cancel eachother out. Photons that are in phase (all the photons that are emitted upwards) will constructively interfere with eachother.
Now something interesting happens - something that also explains why lasers work. When an electron emits a photon, and there are many photons around it who all have the same phase and direction, the emitted photon will copy the phase and direction of the photons around it! So very quickly, all photons that are emitted in random directions die out and only photons that are emitted perfectly in phase with eachother remain.
Now tilt your light beam at an angle. No longer the photons that are emitted upwards are in phase with eachother, but only the photons that are emitted at the exact same angle as the incident photons are in phase. So they remain!
So why does this not happen at any surface? Well, the above only applies to surfaces with lots of electrons, found in materials where electrons are free - for example metals! Surfaces where all electrons are bound will not absorb the photons immediately - they'll penetrate the first few layers of atoms unhindered until by chance they are absorbed. When a new photon is emitted, it will run into other atoms (it's not at the surface anymore!) and keep the reaction going until at the surface, photons are reflected in random directions. Combine this with the fact that without free electrons it is VERY difficult to smooth a surface, it will give you no chance for a decent (specular) reflection.
TL;DR
Photon energy is absorbed by electron
Energy is emitted by electron in the form of a new photon
Photons that are out of phase with eachother die out
Only photons that are emitted at the same angle as the angle of incidence are in phase
Those remain. (Specular) reflection!
The above only works in materials with lots of free electrons, like metals
Mark Booth
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JadaLovelace
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2Is the converse true - that anything which provides clear reflections (like perhaps a highly-polished polyurethane-coated wood floor), has lots of free/loosely-held electrons? – HammerN'Songs Jan 03 '19 at 16:23
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1You said that for reflection we need a surface full of free electrons. However reflection also occurs on a surface of water which does not have any free electrons. How does this happen then? – Harshit Joshi Jan 03 '19 at 17:31
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3@harshit54 See here: https://physics.stackexchange.com/questions/210607/why-does-water-reflect-light the answer that enzolima gives is a specific case: the case of long wavelength light (relative to the atomic spacing) with metals. And in a crude approximation, a body of water can be treated as a weak conductor, depending on what you want to do with it – Daddy Kropotkin Jan 03 '19 at 20:41
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@Garima If this answers your query, please accept it by clicking the tick below it. – Harshit Joshi Jan 03 '19 at 20:56
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1@enzolima - I suggest you avoid the phrase 'photons die out', and explain how they get back in phase and take their energy in the right direction. – amI Jan 03 '19 at 22:04
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1This answer is not correct. What happens at reflection is that photons elastically scatter from collective electronic modes or optical phonons. – my2cts Jan 03 '19 at 23:13
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4All this misses the point. The details of the model (photons or waves, Huyghens or Fermat) do not matter. The basic reason is the symmetry of a flat surface and the reversibility of the rays. But when time symmetry does not apply (as in magnetic systems), the angles will not be equal. – Jan 04 '19 at 15:42
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Photons don't die out this way. And there's no need for a coherent source, even for a single photon the angle will be most likely identical to the angle of incidence. – M. Stern Jan 04 '19 at 19:12
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my2cts, Pieter and M. Stern are correct, there are some fallacies in my explanation. I've tried to compress some qm concepts into a single post that would be better off with more explanation - I may not have chosen the best words in my compression. I'll edit the post later when I have time.
In the meantime i'd like to respond shortly: my2cts, you are correct - that is the theoretical model used in solid state physics. It does not exclude the quantum mechanical model, both arrive at the same conclusions.
– JadaLovelace Jan 07 '19 at 15:35 -
@M.Stern true, a single photon will already have an identical angle. This is because its wavefunction will occupy multiple possible paths until it collapses - essentially allowing it to react to itself. I should probably not have omitted the momentum explanation in my model. I'll try to update it later. – JadaLovelace Jan 07 '19 at 15:38
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Well this can be proven in many ways. If you think of light as waves, then use Huygen's Principle.
A much easier proof can also be developed if you consider light as rays propagating in a line. For this we can use Fermat's Principle.
However if you think of light as particles then a much more intuitive proof can be created by considering a ball being hit on the ground. The part of its velocity parallel to the ground will not change (due to conservation of linear momentum) and the part perpendicular to the ground will flip(assuming an Elastic Collision).
Harshit Joshi
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8It might be worth mentioning that the reason the velocity doesn't change (for both the ball and the photon) is that this is the only solution that conserves both energy and momentum. – Harry Johnston Jan 03 '19 at 20:32
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Light as a classical (non QM) wave also have momentum so the argument is also valid for waves. – my2cts Jan 04 '19 at 14:10
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Some corpuscles get transmitted. According to Newton, refraction towards the normal implies that the speed of the corpuscles in that material higher. Very intuitive, but the theory is obsolete now. – Jan 04 '19 at 23:59
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@Pieter Albert Einstein also used the particle picture of light to explain the photoelectric effect. Did he explain the fact why bending towards the normal implied less velocity? – Harshit Joshi Jan 05 '19 at 02:44
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@harshit54 Newtonian corpuscles are intuitive, photons are not. Where are these entities supposed to "bounce"? What are they interacting with? Why would parallel momentum not change? How much momentum does a photon have compared to an electron or a vibrating atom? – Jan 05 '19 at 20:47
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And yet, Momentum Conservation is just another observed Law of Nature to which we have fitted mathematical formulas. So this is a good analogy but not really a "why", to which there is no answer. – Carl Witthoft Jan 05 '19 at 21:24
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@CarlWitthoft Momentum conservation in itself does not explain that the photon momentum does not change when interacting with electrons or atoms. Look at a grating, look at diffraction: photon momentum is not conserved. Law of Nature? – Jan 06 '19 at 09:43
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It is not necessarily true. A counterexample would be the thought experiment of internal reflection at the surface of a magnetic material, when incident and reflected waves experience different indices of refraction because of magnetic circular birefringence. I think MP Silverman wrote about it, but I cannot find a reference now.
In magneto-optics, one cannot assume Helmholtz reciprocity and reversibility of the rays. ("If I can see you, you can see me.")
The reason for this is that time reversal would also reverse the direction of electrical currents (in coils) and the direction of magnetic fields.
So the basic reason for the law of reflection is symmetry, the time reversibility of the light rays.
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This actually follows from the continuity relations of Maxwell’s equations at the interface of two media: the component of the field tangential to the surface must be the same by $\oint\vec E\cdot d\vec \ell=0$ while the normal component will have a discontinuity found by Gauss’ law and related to the ratios of permittivities at the interface.
ZeroTheHero
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As others have pointed out, you can look at this from ray optics (Fermat's principle), wave optics (consequence of phase matching from boundary conditions for the wave equation at an interface), or more complicated QM approaches.
There was a counterexample mentioned above that I wanted to add to - all you need is a common birefringent crystal (e.g. calcite). For a ray leaving the crystal through a face at some nonzero angle of incidence, there will be refraction out of the crystal as well as reflection at some angle. If your crystal axis is oriented appropriately, the reflected ray will see a different index than the incident ray, changing the reflected angle to be different than the incident angle. This is all just the consequence of phasematching at the interface.
danieldmulkey
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The extraordinary ray does not obey Snell's law at refraction, that is true. But angle of reflection is angle of incidence, also for calcite. – Jan 04 '19 at 22:49
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As a variation on harshit54 's answer, if you look at it classically, the surface exerts a force on the photon in the direction perpendicular to the surface. Thus, only the perpendicular component of the velocity vector changes. Since the magnitude doesn't change, it follows that the angles (as measured from the normal) are flipped.
Acccumulation
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What would a surface do to a photon? What is a surface anyway? What kind of force would it exert? What is your answer supposed to mean? – Jan 04 '19 at 22:51
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@Pieter From a classical perspective, it is intuitive that the surface would exert a force perpendicular to its surface. You don't need to get into deep philosophical questions about what a surface is to understand that. – Acccumulation Jan 04 '19 at 22:54
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Mixing classical mechanics with QED, I find that utterly fascinating. And surfaces, which is mostly a mathematical concept. Infinitely thin. – Jan 04 '19 at 22:57
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@Pieter I didn't do any mixing. I presented a wholly classical perspective. And the idea that there is something strange about including math in classical mechanics is bizarre. – Acccumulation Jan 04 '19 at 23:03
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Ah, you just did Newtonian optics, with his corpuscles. Why did not you say so? Venerable but obsolete theory. How large is a corpuscle supposed to be, by the way? – Jan 04 '19 at 23:06
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4@Pieter Your tone has repeatedly been mocking, sarcastic, and oblique rather than constructive. – Acccumulation Jan 04 '19 at 23:13
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I think this can be shored up to be reasonably accurate (as far as treating photons as particles and not field excitations can be, which is the level of the other answers). All you need to do is note the symmetry of the reflecting plane, and possibly consider that there might be some scatter that in a beam of many photons will average out. – jacob1729 Jan 04 '19 at 23:25
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A force on a photon would simply mean an impulse applied. On average there is no reason for this to be in any direction except the normal to the surface from symmetry considerations. (This might need slight care since the impulse isn't really applied, but photons are emitted and re-absorbed.) – jacob1729 Jan 04 '19 at 23:26
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@jacob1729 That "scatter", why do you expect that? Why would it "average out"? Should not that be noticeable as an increased angular spread? How do you envisage the interaction? Between the photon and what exactly? – Jan 05 '19 at 21:56
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@Pieter I envision it as there being a large collection of classical particles which have some interaction with photons. Each particle on its own would produce a spherical field, the collection of particles is infinite in the x-y directions and so the net field also possesses that symmetry. The law of reflection is a statement that this setup can't affect the incoming momentum of any particle; I was only considering a distribution of reflection angles since in fact there are small scale variations in the x-y plane that I would like to average over. – jacob1729 Jan 06 '19 at 16:01
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@jacob1729 So what you are saying is that the law of reflection is a statement that light cannot affect the momentum of the material? That is not true - there is this thing radiation pressure (and the idea of solar sails). So why could a beam of light not induce an electron current parallel to the surface, absorbing some momentum? – Jan 06 '19 at 22:01
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@Pieter what direction would the current flow and what would determine it? I still don't see how you break that symmetry? – jacob1729 Jan 06 '19 at 22:04
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@jacob1729 In a metallic mirror, electrons are mobile. At oblique incidence one could argue that the light would exert a force on (transfer momentum to) the electrons with a component parallel to the surface, giving rise to a current, to conserve total momentum. Why could that not happen to those bouncing photons? Mechanical newtonian corpuscles are trouble. – Jan 06 '19 at 23:23
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@Pieter your attachment to QED in this question is not really useful. The law of reflection works for lots of linear wave equations: be it Maxwell's, Schrödinger's, hyperbolic wave equation and all the others which can be reduced to eigenproblems for a Helmholtz equation in some environments (like vacuum with some boundaries). And the intuition about conservation of wave vector (its unimpeded component) is applicable here simply due to Floquet's theory. – Ruslan Jan 29 '19 at 07:10
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Snell's law expresses that for a perfectly plane reflecting interface the momentum parallel to the plane is conserved.
Note that one of the answers above is incorrect (Enzolima's).
my2cts
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I'd -2 if I could. Your answer is tautologic. The law of reflection basically is Snell's Law, you are basically saying that because of Snell's Law we have Snell's law. AND you are finding time and space to erroneously claim another (more detailed) answer is wrong, which it isn't. – Stian Jan 04 '19 at 14:56
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2@Stian Yttervik Please read my answer again. I say that Snell's law expresses momentum conservation parallel to the reflecting plane. – my2cts Jan 04 '19 at 15:10
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2@Stian Yttervik I reiterate that Enzolima's answer is most certainly wrong. Absorption and reemission destroys coherence. Reflection is coherent. – my2cts Jan 04 '19 at 19:32
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There's scant evidence that reflection does not occur because of coherent absorption and re-emission. – Carl Witthoft Jan 05 '19 at 21:26
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@ Carl Witthoft Where are the references for this idea that keeps floating up on this site? How does this theory account for the spectral dependance of the complex refractive index in materials ? – my2cts Jan 05 '19 at 21:30
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Why is everyone so angry on this question? I thought this was supposed to be a light discussion!! – corsiKa Jan 06 '19 at 09:20
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