Mixing
Why is angle of incidence equal to angle of reflection?
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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?
optics reflection
edited Jan 3 at 14:29
N. Steinle
1,503117
asked Jan 3 at 13:15
GarimaGarima
13516
$endgroup$
|
show 1 more comment
$begingroup$
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?
optics reflection
edited Jan 3 at 14:29
N. Steinle
1,503117
asked Jan 3 at 13:15
GarimaGarima
13516
$endgroup$
$begingroup$
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.
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– Vladimir Kalitvianski
Jan 3 at 13:29
3
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The 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.
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– Chair
Jan 3 at 13:30
9
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This 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.
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– kubanczyk
Jan 3 at 16:22
5
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Chapter 2 in Feynman’s QED book talks about the Law of Reflection. It’s interesting how time of travel plays a role.
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– Lambda
Jan 3 at 16:37
3
$begingroup$
"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.
$endgroup$
– Filip Milovanović
Jan 5 at 1:20
|
show 1 more comment
$begingroup$
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?
optics reflection
edited Jan 3 at 14:29
N. Steinle
1,503117
asked Jan 3 at 13:15
GarimaGarima
13516
$endgroup$
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?
optics reflection
optics reflection
edited Jan 3 at 14:29
N. Steinle
1,503117
asked Jan 3 at 13:15
GarimaGarima
13516
edited Jan 3 at 14:29
N. Steinle
1,503117
asked Jan 3 at 13:15
GarimaGarima
13516
edited Jan 3 at 14:29
N. Steinle
1,503117
edited Jan 3 at 14:29
N. Steinle
1,503117
edited Jan 3 at 14:29
N. Steinle
1,503117
1,503117
asked Jan 3 at 13:15
GarimaGarima
13516
asked Jan 3 at 13:15
GarimaGarima
13516
asked Jan 3 at 13:15
GarimaGarima
13516
13516
$begingroup$
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.
$endgroup$
– Vladimir Kalitvianski
Jan 3 at 13:29
3
$begingroup$
The 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.
$endgroup$
– Chair
Jan 3 at 13:30
9
$begingroup$
This 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.
$endgroup$
– kubanczyk
Jan 3 at 16:22
5
$begingroup$
Chapter 2 in Feynman’s QED book talks about the Law of Reflection. It’s interesting how time of travel plays a role.
$endgroup$
– Lambda
Jan 3 at 16:37
3
$begingroup$
"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.
$endgroup$
– Filip Milovanović
Jan 5 at 1:20
|
show 1 more comment
$begingroup$
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.
$endgroup$
– Vladimir Kalitvianski
Jan 3 at 13:29
3
$begingroup$
The 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.
$endgroup$
– Chair
Jan 3 at 13:30
9
$begingroup$
This 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.
$endgroup$
– kubanczyk
Jan 3 at 16:22
5
$begingroup$
Chapter 2 in Feynman’s QED book talks about the Law of Reflection. It’s interesting how time of travel plays a role.
$endgroup$
– Lambda
Jan 3 at 16:37
3
$begingroup$
"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.
$endgroup$
– Filip Milovanović
Jan 5 at 1:20
$begingroup$
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.
$endgroup$
– Vladimir Kalitvianski
Jan 3 at 13:29
$begingroup$
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.
$endgroup$
– Vladimir Kalitvianski
Jan 3 at 13:29
3
3
$begingroup$
The 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.
$endgroup$
– Chair
Jan 3 at 13:30
$begingroup$
The 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.
$endgroup$
– Chair
Jan 3 at 13:30
9
9
$begingroup$
This 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.
$endgroup$
– kubanczyk
Jan 3 at 16:22
$begingroup$
This 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.
$endgroup$
– kubanczyk
Jan 3 at 16:22
5
5
$begingroup$
Chapter 2 in Feynman’s QED book talks about the Law of Reflection. It’s interesting how time of travel plays a role.
$endgroup$
– Lambda
Jan 3 at 16:37
$begingroup$
Chapter 2 in Feynman’s QED book talks about the Law of Reflection. It’s interesting how time of travel plays a role.
$endgroup$
– Lambda
Jan 3 at 16:37
3
3
$begingroup$
"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.
$endgroup$
– Filip Milovanović
Jan 5 at 1:20
$begingroup$
"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.
$endgroup$
– Filip Milovanović
Jan 5 at 1:20
|
show 1 more comment
8 Answers
8
active
oldest
votes
$begingroup$
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.
answered Jan 3 at 17:06
kubanczykkubanczyk
727411
$endgroup$
1
$begingroup$
You could add that this is the basis for why the Least Time Theorem seems to hold up - even for diffraction.
$endgroup$
– Stian Yttervik
Jan 4 at 14:47
4
$begingroup$
I 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.
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– ACuriousMind♦
Jan 5 at 11:45
1
$begingroup$
@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).
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– Pieter
Jan 6 at 9:58
add a comment |
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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
edited Jan 4 at 16:19
Mark Booth
1074
answered Jan 3 at 14:20
enzolimaenzolima
8641710
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Is 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?
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– HammerN'Songs
Jan 3 at 16:23
1
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You 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?
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– harshit54
Jan 3 at 17:31
3
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@harshit54 See here: physics.stackexchange.com/questions/210607/… 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
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– N. Steinle
Jan 3 at 20:41
1
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@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.
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– amI
Jan 3 at 22:04
4
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All 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.
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– Pieter
Jan 4 at 15:42
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show 6 more comments
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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).
answered Jan 3 at 13:31
harshit54harshit54
82111
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It 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.
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– Harry Johnston
Jan 3 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.
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– my2cts
Jan 4 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.
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– Pieter
Jan 4 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?
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– harshit54
Jan 5 at 2:44
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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?
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– Pieter
Jan 5 at 20:47
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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.
answered Jan 3 at 23:03
PieterPieter
7,72631431
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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 $ointvec Ecdot dvec ell=0$ while the normal component will have a discontinuity found by Gauss’ law and related to the ratios of permittivities at the interface.
answered Jan 3 at 20:09
ZeroTheHeroZeroTheHero
19.1k52957
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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.
answered Jan 4 at 8:31
DanielDaniel
412
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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.
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– Pieter
Jan 4 at 22:49
add a comment |
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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.
answered Jan 4 at 18:29
AcccumulationAcccumulation
1,886210
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Those fascinating forces on photons.
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– Pieter
Jan 4 at 22:42
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@Pieter What is that supposed to mean?
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– Acccumulation
Jan 4 at 22:43
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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?
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– Pieter
Jan 4 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.
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– Acccumulation
Jan 4 at 22:54
2
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@Pieter Your tone has repeatedly been mocking, sarcastic, and oblique rather than constructive.
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– Acccumulation
Jan 4 at 23:13
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show 10 more comments
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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).
answered Jan 3 at 23:09
my2ctsmy2cts
4,8432618
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Do you care to motivate of the minus sign?
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– my2cts
Jan 4 at 14:07
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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.
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– Stian Yttervik
Jan 4 at 14:56
2
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@Stian Yttervik Please read my answer again. I say that Snell's law expresses momentum conservation parallel to the reflecting plane.
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– my2cts
Jan 4 at 15:10
2
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@Stian Yttervik I reiterate that Enzolima's answer is most certainly wrong. Absorption and reemission destroys coherence. Reflection is coherent.
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– my2cts
Jan 4 at 19:32
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Newton's optical theory of corpuscles, it sounds like.
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– Pieter
Jan 4 at 23:43
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show 4 more comments
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8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
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active
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$begingroup$
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.
answered Jan 3 at 17:06
kubanczykkubanczyk
727411
$endgroup$
1
$begingroup$
You could add that this is the basis for why the Least Time Theorem seems to hold up - even for diffraction.
$endgroup$
– Stian Yttervik
Jan 4 at 14:47
4
$begingroup$
I 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.
$endgroup$
– ACuriousMind♦
Jan 5 at 11:45
1
$begingroup$
@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).
$endgroup$
– Pieter
Jan 6 at 9:58
add a comment |
$begingroup$
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.
answered Jan 3 at 17:06
kubanczykkubanczyk
727411
$endgroup$
1
$begingroup$
You could add that this is the basis for why the Least Time Theorem seems to hold up - even for diffraction.
$endgroup$
– Stian Yttervik
Jan 4 at 14:47
4
$begingroup$
I 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.
$endgroup$
– ACuriousMind♦
Jan 5 at 11:45
1
$begingroup$
@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).
$endgroup$
– Pieter
Jan 6 at 9:58
add a comment |
$begingroup$
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.
answered Jan 3 at 17:06
kubanczykkubanczyk
727411
$endgroup$
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.
answered Jan 3 at 17:06
kubanczykkubanczyk
727411
answered Jan 3 at 17:06
kubanczykkubanczyk
727411
answered Jan 3 at 17:06
kubanczykkubanczyk
727411
answered Jan 3 at 17:06
kubanczykkubanczyk
727411
727411
1
$begingroup$
You could add that this is the basis for why the Least Time Theorem seems to hold up - even for diffraction.
$endgroup$
– Stian Yttervik
Jan 4 at 14:47
4
$begingroup$
I 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.
$endgroup$
– ACuriousMind♦
Jan 5 at 11:45
1
$begingroup$
@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).
$endgroup$
– Pieter
Jan 6 at 9:58
add a comment |
1
$begingroup$
You could add that this is the basis for why the Least Time Theorem seems to hold up - even for diffraction.
$endgroup$
– Stian Yttervik
Jan 4 at 14:47
4
$begingroup$
I 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.
$endgroup$
– ACuriousMind♦
Jan 5 at 11:45
1
$begingroup$
@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).
$endgroup$
– Pieter
Jan 6 at 9:58
1
1
$begingroup$
You could add that this is the basis for why the Least Time Theorem seems to hold up - even for diffraction.
$endgroup$
– Stian Yttervik
Jan 4 at 14:47
$begingroup$
You could add that this is the basis for why the Least Time Theorem seems to hold up - even for diffraction.
$endgroup$
– Stian Yttervik
Jan 4 at 14:47
4
4
$begingroup$
I 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.
$endgroup$
– ACuriousMind♦
Jan 5 at 11:45
$begingroup$
I 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.
$endgroup$
– ACuriousMind♦
Jan 5 at 11:45
1
1
$begingroup$
@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).
$endgroup$
– Pieter
Jan 6 at 9:58
$begingroup$
@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).
$endgroup$
– Pieter
Jan 6 at 9:58
add a comment |
$begingroup$
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
edited Jan 4 at 16:19
Mark Booth
1074
answered Jan 3 at 14:20
enzolimaenzolima
8641710
$endgroup$
2
$begingroup$
Is 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?
$endgroup$
– HammerN'Songs
Jan 3 at 16:23
1
$begingroup$
You 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?
$endgroup$
– harshit54
Jan 3 at 17:31
3
$begingroup$
@harshit54 See here: physics.stackexchange.com/questions/210607/… 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
$endgroup$
– N. Steinle
Jan 3 at 20:41
1
$begingroup$
@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.
$endgroup$
– amI
Jan 3 at 22:04
4
$begingroup$
All 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.
$endgroup$
– Pieter
Jan 4 at 15:42
|
show 6 more comments
$begingroup$
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
edited Jan 4 at 16:19
Mark Booth
1074
answered Jan 3 at 14:20
enzolimaenzolima
8641710
$endgroup$
2
$begingroup$
Is 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?
$endgroup$
– HammerN'Songs
Jan 3 at 16:23
1
$begingroup$
You 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?
$endgroup$
– harshit54
Jan 3 at 17:31
3
$begingroup$
@harshit54 See here: physics.stackexchange.com/questions/210607/… 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
$endgroup$
– N. Steinle
Jan 3 at 20:41
1
$begingroup$
@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.
$endgroup$
– amI
Jan 3 at 22:04
4
$begingroup$
All 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.
$endgroup$
– Pieter
Jan 4 at 15:42
|
show 6 more comments
$begingroup$
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
edited Jan 4 at 16:19
Mark Booth
1074
answered Jan 3 at 14:20
enzolimaenzolima
8641710
$endgroup$
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
edited Jan 4 at 16:19
Mark Booth
1074
answered Jan 3 at 14:20
enzolimaenzolima
8641710
edited Jan 4 at 16:19
Mark Booth
1074
edited Jan 4 at 16:19
Mark Booth
1074
edited Jan 4 at 16:19
Mark Booth
1074
1074
answered Jan 3 at 14:20
enzolimaenzolima
8641710
answered Jan 3 at 14:20
enzolimaenzolima
8641710
answered Jan 3 at 14:20
enzolimaenzolima
8641710
8641710
2
$begingroup$
Is 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?
$endgroup$
– HammerN'Songs
Jan 3 at 16:23
1
$begingroup$
You 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?
$endgroup$
– harshit54
Jan 3 at 17:31
3
$begingroup$
@harshit54 See here: physics.stackexchange.com/questions/210607/… 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
$endgroup$
– N. Steinle
Jan 3 at 20:41
1
$begingroup$
@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.
$endgroup$
– amI
Jan 3 at 22:04
4
$begingroup$
All 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.
$endgroup$
– Pieter
Jan 4 at 15:42
|
show 6 more comments
2
$begingroup$
Is 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?
$endgroup$
– HammerN'Songs
Jan 3 at 16:23
1
$begingroup$
You 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?
$endgroup$
– harshit54
Jan 3 at 17:31
3
$begingroup$
@harshit54 See here: physics.stackexchange.com/questions/210607/… 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
$endgroup$
– N. Steinle
Jan 3 at 20:41
1
$begingroup$
@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.
$endgroup$
– amI
Jan 3 at 22:04
4
$begingroup$
All 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.
$endgroup$
– Pieter
Jan 4 at 15:42
2
2
$begingroup$
Is 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?
$endgroup$
– HammerN'Songs
Jan 3 at 16:23
$begingroup$
Is 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?
$endgroup$
– HammerN'Songs
Jan 3 at 16:23
1
1
$begingroup$
You 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?
$endgroup$
– harshit54
Jan 3 at 17:31
$begingroup$
You 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?
$endgroup$
– harshit54
Jan 3 at 17:31
3
3
$begingroup$
@harshit54 See here: physics.stackexchange.com/questions/210607/… 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
$endgroup$
– N. Steinle
Jan 3 at 20:41
$begingroup$
@harshit54 See here: physics.stackexchange.com/questions/210607/… 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
$endgroup$
– N. Steinle
Jan 3 at 20:41
1
1
$begingroup$
@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.
$endgroup$
– amI
Jan 3 at 22:04
$begingroup$
@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.
$endgroup$
– amI
Jan 3 at 22:04
4
4
$begingroup$
All 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.
$endgroup$
– Pieter
Jan 4 at 15:42
$begingroup$
All 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.
$endgroup$
– Pieter
Jan 4 at 15:42
|
show 6 more comments
$begingroup$
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).
answered Jan 3 at 13:31
harshit54harshit54
82111
$endgroup$
8
$begingroup$
It 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.
$endgroup$
– Harry Johnston
Jan 3 at 20:32
$begingroup$
Light as a classical (non QM) wave also have momentum so the argument is also valid for waves.
$endgroup$
– my2cts
Jan 4 at 14:10
$begingroup$
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.
$endgroup$
– Pieter
Jan 4 at 23:59
$begingroup$
@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?
$endgroup$
– harshit54
Jan 5 at 2:44
$begingroup$
@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?
$endgroup$
– Pieter
Jan 5 at 20:47
|
show 3 more comments
$begingroup$
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).
answered Jan 3 at 13:31
harshit54harshit54
82111
$endgroup$
8
$begingroup$
It 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.
$endgroup$
– Harry Johnston
Jan 3 at 20:32
$begingroup$
Light as a classical (non QM) wave also have momentum so the argument is also valid for waves.
$endgroup$
– my2cts
Jan 4 at 14:10
$begingroup$
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.
$endgroup$
– Pieter
Jan 4 at 23:59
$begingroup$
@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?
$endgroup$
– harshit54
Jan 5 at 2:44
$begingroup$
@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?
$endgroup$
– Pieter
Jan 5 at 20:47
|
show 3 more comments
$begingroup$
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).
answered Jan 3 at 13:31
harshit54harshit54
82111
$endgroup$
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).
answered Jan 3 at 13:31
harshit54harshit54
82111
edited Jan 3 at 20:48
answered Jan 3 at 13:31
harshit54harshit54
82111
answered Jan 3 at 13:31
harshit54harshit54
82111
answered Jan 3 at 13:31
harshit54harshit54
82111
82111
8
$begingroup$
It 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.
$endgroup$
– Harry Johnston
Jan 3 at 20:32
$begingroup$
Light as a classical (non QM) wave also have momentum so the argument is also valid for waves.
$endgroup$
– my2cts
Jan 4 at 14:10
$begingroup$
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.
$endgroup$
– Pieter
Jan 4 at 23:59
$begingroup$
@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?
$endgroup$
– harshit54
Jan 5 at 2:44
$begingroup$
@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?
$endgroup$
– Pieter
Jan 5 at 20:47
|
show 3 more comments
8
$begingroup$
It 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.
$endgroup$
– Harry Johnston
Jan 3 at 20:32
$begingroup$
Light as a classical (non QM) wave also have momentum so the argument is also valid for waves.
$endgroup$
– my2cts
Jan 4 at 14:10
$begingroup$
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.
$endgroup$
– Pieter
Jan 4 at 23:59
$begingroup$
@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?
$endgroup$
– harshit54
Jan 5 at 2:44
$begingroup$
@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?
$endgroup$
– Pieter
Jan 5 at 20:47
8
8
$begingroup$
It 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.
$endgroup$
– Harry Johnston
Jan 3 at 20:32
$begingroup$
It 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.
$endgroup$
– Harry Johnston
Jan 3 at 20:32
$begingroup$
Light as a classical (non QM) wave also have momentum so the argument is also valid for waves.
$endgroup$
– my2cts
Jan 4 at 14:10
$begingroup$
Light as a classical (non QM) wave also have momentum so the argument is also valid for waves.
$endgroup$
– my2cts
Jan 4 at 14:10
$begingroup$
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.
$endgroup$
– Pieter
Jan 4 at 23:59
$begingroup$
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.
$endgroup$
– Pieter
Jan 4 at 23:59
$begingroup$
@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?
$endgroup$
– harshit54
Jan 5 at 2:44
$begingroup$
@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?
$endgroup$
– harshit54
Jan 5 at 2:44
$begingroup$
@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?
$endgroup$
– Pieter
Jan 5 at 20:47
$begingroup$
@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?
$endgroup$
– Pieter
Jan 5 at 20:47
|
show 3 more comments
$begingroup$
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.
answered Jan 3 at 23:03
PieterPieter
7,72631431
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 3 at 23:03
PieterPieter
7,72631431
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 3 at 23:03
PieterPieter
7,72631431
$endgroup$
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.
answered Jan 3 at 23:03
PieterPieter
7,72631431
edited Jan 4 at 0:41
answered Jan 3 at 23:03
PieterPieter
7,72631431
answered Jan 3 at 23:03
PieterPieter
7,72631431
answered Jan 3 at 23:03
PieterPieter
7,72631431
7,72631431
add a comment |
add a comment |
$begingroup$
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 $ointvec Ecdot dvec ell=0$ while the normal component will have a discontinuity found by Gauss’ law and related to the ratios of permittivities at the interface.
answered Jan 3 at 20:09
ZeroTheHeroZeroTheHero
19.1k52957
$endgroup$
add a comment |
$begingroup$
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 $ointvec Ecdot dvec ell=0$ while the normal component will have a discontinuity found by Gauss’ law and related to the ratios of permittivities at the interface.
answered Jan 3 at 20:09
ZeroTheHeroZeroTheHero
19.1k52957
$endgroup$
add a comment |
$begingroup$
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 $ointvec Ecdot dvec ell=0$ while the normal component will have a discontinuity found by Gauss’ law and related to the ratios of permittivities at the interface.
answered Jan 3 at 20:09
ZeroTheHeroZeroTheHero
19.1k52957
$endgroup$
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 $ointvec Ecdot dvec ell=0$ while the normal component will have a discontinuity found by Gauss’ law and related to the ratios of permittivities at the interface.
answered Jan 3 at 20:09
ZeroTheHeroZeroTheHero
19.1k52957
edited Jan 5 at 4:39
answered Jan 3 at 20:09
ZeroTheHeroZeroTheHero
19.1k52957
answered Jan 3 at 20:09
ZeroTheHeroZeroTheHero
19.1k52957
answered Jan 3 at 20:09
ZeroTheHeroZeroTheHero
19.1k52957
19.1k52957
add a comment |
add a comment |
$begingroup$
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.
answered Jan 4 at 8:31
DanielDaniel
412
$endgroup$
$begingroup$
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.
$endgroup$
– Pieter
Jan 4 at 22:49
add a comment |
$begingroup$
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.
answered Jan 4 at 8:31
DanielDaniel
412
$endgroup$
$begingroup$
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.
$endgroup$
– Pieter
Jan 4 at 22:49
add a comment |
$begingroup$
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.
answered Jan 4 at 8:31
DanielDaniel
412
$endgroup$
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.
answered Jan 4 at 8:31
DanielDaniel
412
answered Jan 4 at 8:31
DanielDaniel
412
answered Jan 4 at 8:31
DanielDaniel
412
answered Jan 4 at 8:31
DanielDaniel
412
412
$begingroup$
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.
$endgroup$
– Pieter
Jan 4 at 22:49
add a comment |
$begingroup$
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.
$endgroup$
– Pieter
Jan 4 at 22:49
$begingroup$
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.
$endgroup$
– Pieter
Jan 4 at 22:49
$begingroup$
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.
$endgroup$
– Pieter
Jan 4 at 22:49
add a comment |
$begingroup$
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.
answered Jan 4 at 18:29
AcccumulationAcccumulation
1,886210
$endgroup$
$begingroup$
Those fascinating forces on photons.
$endgroup$
– Pieter
Jan 4 at 22:42
$begingroup$
@Pieter What is that supposed to mean?
$endgroup$
– Acccumulation
Jan 4 at 22:43
$begingroup$
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?
$endgroup$
– Pieter
Jan 4 at 22:51
$begingroup$
@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.
$endgroup$
– Acccumulation
Jan 4 at 22:54
2
$begingroup$
@Pieter Your tone has repeatedly been mocking, sarcastic, and oblique rather than constructive.
$endgroup$
– Acccumulation
Jan 4 at 23:13
|
show 10 more comments
$begingroup$
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.
answered Jan 4 at 18:29
AcccumulationAcccumulation
1,886210
$endgroup$
$begingroup$
Those fascinating forces on photons.
$endgroup$
– Pieter
Jan 4 at 22:42
$begingroup$
@Pieter What is that supposed to mean?
$endgroup$
– Acccumulation
Jan 4 at 22:43
$begingroup$
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?
$endgroup$
– Pieter
Jan 4 at 22:51
$begingroup$
@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.
$endgroup$
– Acccumulation
Jan 4 at 22:54
2
$begingroup$
@Pieter Your tone has repeatedly been mocking, sarcastic, and oblique rather than constructive.
$endgroup$
– Acccumulation
Jan 4 at 23:13
|
show 10 more comments
$begingroup$
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.
answered Jan 4 at 18:29
AcccumulationAcccumulation
1,886210
$endgroup$
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.
answered Jan 4 at 18:29
AcccumulationAcccumulation
1,886210
answered Jan 4 at 18:29
AcccumulationAcccumulation
1,886210
answered Jan 4 at 18:29
AcccumulationAcccumulation
1,886210
answered Jan 4 at 18:29
AcccumulationAcccumulation
1,886210
1,886210
$begingroup$
Those fascinating forces on photons.
$endgroup$
– Pieter
Jan 4 at 22:42
$begingroup$
@Pieter What is that supposed to mean?
$endgroup$
– Acccumulation
Jan 4 at 22:43
$begingroup$
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?
$endgroup$
– Pieter
Jan 4 at 22:51
$begingroup$
@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.
$endgroup$
– Acccumulation
Jan 4 at 22:54
2
$begingroup$
@Pieter Your tone has repeatedly been mocking, sarcastic, and oblique rather than constructive.
$endgroup$
– Acccumulation
Jan 4 at 23:13
|
show 10 more comments
$begingroup$
Those fascinating forces on photons.
$endgroup$
– Pieter
Jan 4 at 22:42
$begingroup$
@Pieter What is that supposed to mean?
$endgroup$
– Acccumulation
Jan 4 at 22:43
$begingroup$
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?
$endgroup$
– Pieter
Jan 4 at 22:51
$begingroup$
@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.
$endgroup$
– Acccumulation
Jan 4 at 22:54
2
$begingroup$
@Pieter Your tone has repeatedly been mocking, sarcastic, and oblique rather than constructive.
$endgroup$
– Acccumulation
Jan 4 at 23:13
$begingroup$
Those fascinating forces on photons.
$endgroup$
– Pieter
Jan 4 at 22:42
$begingroup$
Those fascinating forces on photons.
$endgroup$
– Pieter
Jan 4 at 22:42
$begingroup$
@Pieter What is that supposed to mean?
$endgroup$
– Acccumulation
Jan 4 at 22:43
$begingroup$
@Pieter What is that supposed to mean?
$endgroup$
– Acccumulation
Jan 4 at 22:43
$begingroup$
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?
$endgroup$
– Pieter
Jan 4 at 22:51
$begingroup$
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?
$endgroup$
– Pieter
Jan 4 at 22:51
$begingroup$
@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.
$endgroup$
– Acccumulation
Jan 4 at 22:54
$begingroup$
@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.
$endgroup$
– Acccumulation
Jan 4 at 22:54
2
2
$begingroup$
@Pieter Your tone has repeatedly been mocking, sarcastic, and oblique rather than constructive.
$endgroup$
– Acccumulation
Jan 4 at 23:13
$begingroup$
@Pieter Your tone has repeatedly been mocking, sarcastic, and oblique rather than constructive.
$endgroup$
– Acccumulation
Jan 4 at 23:13
|
show 10 more comments
$begingroup$
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).
answered Jan 3 at 23:09
my2ctsmy2cts
4,8432618
$endgroup$
1
$begingroup$
Do you care to motivate of the minus sign?
$endgroup$
– my2cts
Jan 4 at 14:07
$begingroup$
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.
$endgroup$
– Stian Yttervik
Jan 4 at 14:56
2
$begingroup$
@Stian Yttervik Please read my answer again. I say that Snell's law expresses momentum conservation parallel to the reflecting plane.
$endgroup$
– my2cts
Jan 4 at 15:10
2
$begingroup$
@Stian Yttervik I reiterate that Enzolima's answer is most certainly wrong. Absorption and reemission destroys coherence. Reflection is coherent.
$endgroup$
– my2cts
Jan 4 at 19:32
$begingroup$
Newton's optical theory of corpuscles, it sounds like.
$endgroup$
– Pieter
Jan 4 at 23:43
|
show 4 more comments
$begingroup$
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).
answered Jan 3 at 23:09
my2ctsmy2cts
4,8432618
$endgroup$
1
$begingroup$
Do you care to motivate of the minus sign?
$endgroup$
– my2cts
Jan 4 at 14:07
$begingroup$
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.
$endgroup$
– Stian Yttervik
Jan 4 at 14:56
2
$begingroup$
@Stian Yttervik Please read my answer again. I say that Snell's law expresses momentum conservation parallel to the reflecting plane.
$endgroup$
– my2cts
Jan 4 at 15:10
2
$begingroup$
@Stian Yttervik I reiterate that Enzolima's answer is most certainly wrong. Absorption and reemission destroys coherence. Reflection is coherent.
$endgroup$
– my2cts
Jan 4 at 19:32
$begingroup$
Newton's optical theory of corpuscles, it sounds like.
$endgroup$
– Pieter
Jan 4 at 23:43
|
show 4 more comments
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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).
answered Jan 3 at 23:09
my2ctsmy2cts
4,8432618
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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).
answered Jan 3 at 23:09
my2ctsmy2cts
4,8432618
edited Jan 4 at 14:06
answered Jan 3 at 23:09
my2ctsmy2cts
4,8432618
answered Jan 3 at 23:09
my2ctsmy2cts
4,8432618
answered Jan 3 at 23:09
my2ctsmy2cts
4,8432618
4,8432618
1
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Do you care to motivate of the minus sign?
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– my2cts
Jan 4 at 14:07
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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.
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– Stian Yttervik
Jan 4 at 14:56
2
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@Stian Yttervik Please read my answer again. I say that Snell's law expresses momentum conservation parallel to the reflecting plane.
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– my2cts
Jan 4 at 15:10
2
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@Stian Yttervik I reiterate that Enzolima's answer is most certainly wrong. Absorption and reemission destroys coherence. Reflection is coherent.
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– my2cts
Jan 4 at 19:32
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Newton's optical theory of corpuscles, it sounds like.
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– Pieter
Jan 4 at 23:43
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show 4 more comments
1
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Do you care to motivate of the minus sign?
$endgroup$
– my2cts
Jan 4 at 14:07
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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.
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– Stian Yttervik
Jan 4 at 14:56
2
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@Stian Yttervik Please read my answer again. I say that Snell's law expresses momentum conservation parallel to the reflecting plane.
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– my2cts
Jan 4 at 15:10
2
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@Stian Yttervik I reiterate that Enzolima's answer is most certainly wrong. Absorption and reemission destroys coherence. Reflection is coherent.
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– my2cts
Jan 4 at 19:32
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Newton's optical theory of corpuscles, it sounds like.
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– Pieter
Jan 4 at 23:43
1
1
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Do you care to motivate of the minus sign?
$endgroup$
– my2cts
Jan 4 at 14:07
$begingroup$
Do you care to motivate of the minus sign?
$endgroup$
– my2cts
Jan 4 at 14:07
$begingroup$
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.
$endgroup$
– Stian Yttervik
Jan 4 at 14:56
$begingroup$
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.
$endgroup$
– Stian Yttervik
Jan 4 at 14:56
2
2
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@Stian Yttervik Please read my answer again. I say that Snell's law expresses momentum conservation parallel to the reflecting plane.
$endgroup$
– my2cts
Jan 4 at 15:10
$begingroup$
@Stian Yttervik Please read my answer again. I say that Snell's law expresses momentum conservation parallel to the reflecting plane.
$endgroup$
– my2cts
Jan 4 at 15:10
2
2
$begingroup$
@Stian Yttervik I reiterate that Enzolima's answer is most certainly wrong. Absorption and reemission destroys coherence. Reflection is coherent.
$endgroup$
– my2cts
Jan 4 at 19:32
$begingroup$
@Stian Yttervik I reiterate that Enzolima's answer is most certainly wrong. Absorption and reemission destroys coherence. Reflection is coherent.
$endgroup$
– my2cts
Jan 4 at 19:32
$begingroup$
Newton's optical theory of corpuscles, it sounds like.
$endgroup$
– Pieter
Jan 4 at 23:43
$begingroup$
Newton's optical theory of corpuscles, it sounds like.
$endgroup$
– Pieter
Jan 4 at 23:43
|
show 4 more comments
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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.
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– Vladimir Kalitvianski
Jan 3 at 13:29
3
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The 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.
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– Chair
Jan 3 at 13:30
9
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This 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.
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– kubanczyk
Jan 3 at 16:22
5
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Chapter 2 in Feynman’s QED book talks about the Law of Reflection. It’s interesting how time of travel plays a role.
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– Lambda
Jan 3 at 16:37
3
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"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.
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– Filip Milovanović
Jan 5 at 1:20