[chris (OP)]

I am anticipating this question surfacing, based on what my children are learning at school at the moment, so I want to have a Boris Johnson "oven ready", Naked-Scientists-forum-approved, answer to this!

When considering a mirror surface, a light / electromagnetic ray reaches the mirror surface. It interacts with the cloud of electrons around the atoms on that surface and makes them vibrate. Those displacements of a charged particle - the electron - constitute a moving (accelerating and decelerating) charge, so electromagnetic waves are produced. This is the "reflected" ray. Am I right so far?

So what property of the atom / electrons means that the light ray that issues from the surface in this way have the same angle as the incoming ray?

[evan_au]

A simplified model could view it as a result of Lenz's Law.
- An electromagnetic wave approaching a metallic reflecting surface induces a current in the metal which opposes the external magnetic field. If it is a good reflector, the opposing field will be equal and opposite to the incoming wave.
- The incoming wave cancels the "mirror image" wave in the metal, so the incoming wave does not propagate into the metal surface.
- But the external wave and the wave in the metal surface does have constructive interference, producing an outgoing wave which is a mirror-image of the incoming wave, and the angle of incidence = the angle of reflection.

See: https://en.wikipedia.org/wiki/Lenz%27s_law

Of course, this is only part of the story, since non-conductive surfaces (eg glass) can also act as partial mirrors if they are in a medium with a very different speed of light.

[Bored chemist]

I want to have a Boris Johnson "oven ready", Naked-Scientists-forum-approved, answer to this!

Which one do you want?
Boris would say
"I  was at a hospital where there were a few coronavirus patients and I shook hands with everybody"
And a scientifically approved one would
 “advise against greetings such as shaking hands and hugging, given existing evidence about the importance of hand hygiene”.


I am not a good enough (or bad enough)  liar to guess what tosh Boris would spout, but the science version is that it's the only way to meet the requirements for conservation of energy and momentum.

[chris (OP)]

Thanks @evan_au but I'm struggling to follow; the bit about waves cancelling each other out has got me confused. Can you give me the gentler version first to get me started please?

[evan_au]

Can you give me the gentler version first to get me started please?

Particles like balls also obey angle of incidence = angle of reflection (if you ignore gravity, and the fact that balls tend to start spinning when they run into things). At least snooker gets around the gravity problem....
- The motion of the ball can be broken into two components of momentum:
- One parallel to the cushion, which is not affected by the collision
- One perpendicular to the cushion, which is reversed for an elastic (energy-conserving) collision
- This produces a scenario where angle of incidence = angle of reflection
- You could extend this to a photon striking a mirror...
- Or the molecules carrying sound, where the large-scale orderly motion of the sound wave (longitudinal wave) is superimposed on the general random movement of air molecules

Water waves (transverse waves) also obey the angle of incidence = angle of reflection if they run into a vertical wall.
- The wave front will move ahead at a certain velocity when in a uniform medium.
- When a water wave strikes a wall, the horizontal amplitude becomes zero perpendicular the wall, but the vertical amplitude increases (conserving energy)
- The energy stored in the increased amplitude is released by the water moving away from the wall, producing an outgoing reflected wave.
- The part of the wave moving parallel to the wall does not contribute to the reflected wave, so you end up with the traditional law of reflection.

After that, I'm out of analogies. Apply Maxwell's equations in a simulation package....

the bit about waves cancelling each other out has got me confused.

A moving wave front can be modeled as a new wave radiating from every point on the wave front. This myriad of point sources cancels out everywhere except for the solution where the wave moves ahead. This effect can be seen in the classic double-slit experiment, where both slits effectively become a point source of radiating waves.

When a wave hits a reflecting surface, you need to take into account the boundary conditions.
- For an electromagnetic wave, this includes the current induced in the metal by the electromagnetic wave
- Adding up the contributions from every point, the signal cancels everywhere except for an outgoing wave which obeys the law of reflection.
- If you modify the mirror so it is periodic (like a diffraction grating), there are multiple angles where the sources combine constructively, producing multiple outgoing waves. This does not obey the "traditional" law of reflection. (the exception proves the rule...)

See: https://en.wikipedia.org/wiki/Double-slit_experiment
https://en.wikipedia.org/wiki/Diffraction_grating

[chris (OP)]

Thanks @evan_au

[yor_on]

Sweet BC " I am not a good enough (or bad enough)  liar to guess what tosh Boris would spout, but the science version is that it's the only way to meet the requirements for conservation of energy and momentum. "

A perfect answer, and also one giving the kids very good questions to ask their teacher.

[alancalverd]

If I start approaching this from a point of view of physics, it becomes very difficult to answer.

The "wavelet"  model certainly gives the correct answer for a diffraction grating but only for very intense beams. If you reduce the intensity to a single photon it turns up somewhere as ordained by the model, but clearly cannot have spontaneously split into Evan's "myriad of point sources" in order to work out where to go.

Conservation of energy and momentum is an attractive starting point for a perfectly reflective mirror, but even the best-polished silver has a reflectance of around 95% at optical wavelengths, so conservation of energy doesn't really apply: 95% of visible photons are indeed reflected or retransmitted  with the same energy as they arrived but up to 5% is lost as heat, and if the initial interaction is between a photon and an electron which then re-emits a photon, the momentum transferred to the electron is surely going to be distributed among other electrons, so the emission angle is not determined solely by the angle of incidence.   

Depressingly, I am led to a reductio ad absurdam. If the angle of reflection is not equal to the angle of incidence, there is something asymmetric happening at the surface. But we have implicitly defined and explicitly made a reflecting surface as locally symmetric. It's a rubbish argument that stinks of philosophy, but it's the best I have at the moment!

[yor_on]

The difference between theory and practice :)
I lean more and more to the theoretical definitions myself

If we want it simple, and I do.
Piratically?

ahem, that should be practically.

(f* that spell checker)

[Bored chemist]

Conservation of energy and momentum is an attractive starting point for a perfectly reflective mirror, but even the best-polished silver has a reflectance of around 95% at optical wavelengths, so conservation of energy doesn't really apply: 95% of visible photons are indeed reflected

100% of the reflected photons are reflected, and you can apply the conservation of energy and momentum to them (even if you don't believe that angular momentum is a conserved quantity)

.

Or are you saying you can slow down the Earth's rotation without an external torque?

Yes.

if the initial interaction is between a photon and an electron which then re-emits a photon, the momentum transferred to the electron is surely going to be distributed among other electrons, so the emission angle is not determined solely by the angle of incidence.   

That's more or less what happens in fluorescence and, sure enough, that angle of fluorescence emission isn't the same as the angle of incidence.

[phyti39]

chris;

QED 'the strange theory of light and matter',
Richard Feynman, 1985
In his book, using the photon model of light, he explains the behavior of light involving the many paths concept.
In simple reflection from a flat surface, the photons take all possible paths EMD, from Emitter to Mirror to Detector, one path per photon, E and D in fixed positions.
The probability of detection is the sum of probabilities for each path. The analysis shows the greatest contribution comes from the middle section of the mirror.
I.e., the classical, 1st approximation, incidence angle equals reflection angle.

[alancalverd]

My problem with BC's explanation is that it is back to front or possibly circular. If we start with the observation that i = r we can deduce that the energy and momentum of the reflected photons are conserved. But that doesn't address the mechanism of reflection. We know that photons interact with electrons, but that interaction necessarily involves a transfer of energy and momentum to the electron, so it is not obvious why the reflected photon has the same energy and momentum as the incident one.

The billiard ball model is incomplete: obviously on collision with the cushion the ball transfers some energy and momentum to the table, otherwise it would continue to bounce around for ever. The analogous proposition is that any collision between a photon and an electron is perfectly elastic, but whilst all surfaces contain electrons, not all surfaces are equally reflective and some are strongly absorbent. So we can clearly distinguish between materials whose surface electrons are in a state that permits perfect elasticity and those that are fully absorbent, at least over a small spectral range.  But what about, say, platinum, with only 60% of the reflectivity of silver in the visible region? It's a pure metal with a fully delocalised conduction band, but not all the interactions are perfectly elastic (~ 40% of the photons are absorbed or scattered)  yet i = r.       

Phyti: "The probability of detection is the sum of probabilities for each path. The analysis shows the greatest contribution comes from the middle section of the mirror."  So you have calculated the probability of reflection at every  angle without reference to a specific electron-photon interaction? I think not: Feynman considers the probability of all paths from M to D, assuming that he already knows where D is!

[evan_au]

The "wavelet"  model certainly gives the correct answer for a diffraction grating but only for very intense beams. If you reduce the intensity to a single photon it turns up somewhere as ordained by the model, but clearly cannot have spontaneously split into Evan's "myriad of point sources" in order to work out where to go.

The fact that the dual-slit experiment works for single photons has been known for a long time. It's like a single photon goes through both slits at once, and then interferes with itself. The interference acts as if each slit is a point source.

Feynman's approach (as mentioned by phyti39) assumes that even a single photon takes every possible path (with various probabilities). The final probability distribution of where you will find the photon is the sum of all those probability distributions.
- Many of those probability distributions cancel each other out
- Leaving the most likely ones behind
- This works for diffraction gratings too...

PS: Slight overlap with Alan...

[alancalverd]

The "wavelet"  model certainly gives the correct answer for a diffraction grating but only for very intense beams. If you reduce the intensity to a single photon it turns up somewhere as ordained by the model, but clearly cannot have spontaneously split into Evan's "myriad of point sources" in order to work out where to go.

The fact that the dual-slit experiment works for single photons has been known for a long time. It's like a single photon goes through both slits at once, and then interferes with itself. The interference acts as if each slit is a point source.

My emphasis. Problem is that when we look at what comes through each slit, it's either a whole photon or nothing -  no evidence that the photon actually splits. So although the wavelet model  is adequately predictive, it isn't explanatory! But this is a digression

Feynman's approach (as mentioned by phyti39) assumes that even a single photon takes every possible path (with various probabilities). The final probability distribution of where you will find the photon is the sum of all those probability distributions.
- Many of those probability distributions cancel each other out
- Leaving the most likely ones behind
- This works for diffraction gratings too...

PS: Slight overlap with Alan...

Again my emphasis. What makes the reflected path more likely than any other? What makes all the others cancel? This time we have an a posteriori model that is neither predictive nor explanatory!

[Bored chemist]

. But that doesn't address the mechanism of reflection.

It also doesn't tell you how to make beer.
Or... lots of other things that weren't in the OP.

it's either a whole photon or nothing

Or it's a wave function.

What makes all the others cancel?

Path lengths- same as usual.

  But what about, say, platinum, with only 60% of the reflectivity of silver in the visible region? It's a pure metal with a fully delocalised conduction band, but not all the interactions are perfectly elastic (~ 40% of the photons are absorbed or scattered)  yet i = r. 

i=r is only defined for the 60%

Photon absorption is inherently statistical. You may not like that, but that's how reality works.

[evan_au]

What makes the reflected path more likely than any other? What makes all the others cancel?

This is getting beyond my pay grade, so I'll pose it as a query:

Is it true that traditional quantum wavefunctions can tell you the probability that you will find a particle at any given position, by taking the square of the modulus of the wavefunction?
- This probability is always positive, between 0 and 1.
- So if you add up many different probability distributions, it only takes one to have a non-zero value, and you will have a non-zero probability of finding the particle there; this appears to forbid cancellation.

Is it true that traditional quantum wavefunctions can also tell you the phase of the wavefunction at any given position?
- Unlike a probability, a phase can be positive or negative (relative to some reference)
- And that means the phase of the wavefunction can cancel at some points (and reinforce at other points)
- So you need to add all the wavefunctions (taking into account the phase) before you take the square of the modulus of the wavefunction to find the probability of the photon being there.
- This allows for both cancellation and enhancement in different places, producing the familiar diffraction patterns.

Where have I got this wrong?

See: https://en.wikipedia.org/wiki/Wave_function

[alancalverd]

. But that doesn't address the mechanism of reflection.

It also doesn't tell you how to make beer.
Or... lots of other things that weren't in the OP.

The OP explicitly asked

So what property of the atom / electrons means that the light ray that issues from the surface in this way have the same angle as the incoming ray?

and so far, nobody has answered it.

it's either a whole photon or nothing

Or it's a wave function.

No, what we detect at the slit exit is a single photon. One spatially located interaction with the photodetector we put next to the slit.  The wave function is a predictive model of what we find at some distance (> slit separation) from two slits.

What makes all the others cancel?

Path lengths- same as usual.

But they are presumed to radiate from the source point, so they diverge, do not overlap, and thus have no physical mechanism for cancellation. And if you put an absorber on the left, the paths on the right would not cancel, but this isn't what we observe.

 

  But what about, say, platinum, with only 60% of the reflectivity of silver in the visible region? It's a pure metal with a fully delocalised conduction band, but not all the interactions are perfectly elastic (~ 40% of the photons are absorbed or scattered)  yet i = r.

i=r is only defined for the 60%

Photon absorption is inherently statistical. You may not like that, but that's how reality works.

So how come this statistical process produces a precisely deterministic outcome? That was the original question!

[alancalverd]

This is getting beyond my pay grade, so I'll pose it as a query:

Frankly, it's a bit of a worry. The observation is Physics 101 Preparatory: for Imbeciles, now laid before chaps who presume to know an awful lot about science, and so far I haven't seen an explanation that holds water!

[Bored chemist]

and so far, nobody has answered it.

The properties concerned are energy and momentum.
If only someone had thought to mention this earlier...

No, what we detect at the slit exit is a single photon.

Yes.
So?
The question is  about what happens before we detect it.

So how come this statistical process produces a precisely deterministic outcome? That was the original question!

And you have both the classical and the QM/ QED versions of the answer.
Here's the answer in terms of Maxwell's eqns for the situation
http://pcwww.liv.ac.uk/~awolski/Teaching/Liverpool/PHYS370/AdvancedElectromagnetism-Part4.pdf
  you now have three answers.
Is that enough?

[alancalverd]

The properties concerned are energy and momentum.

Incoming photon transfers energy and momentum to an electron. Why does the electron then emit a photon with the same energy, same momentum vector parallel to the surface, but reversed momentum vector perpendicular to the surface? If the temperature is above absolute zero, the electrons are all jiggling about and quite capable of transferring energy and momentum to the mass of the mirror.

On the other hand http://pcwww.liv.ac.uk/~awolski/Teaching/Liverpool/PHYS370/AdvancedElectromagnetism-Part4.pdf  is clear and convincing, for which many thanks. But it relies on the bulk properties of the media, not individual photon-electron interactions, which makes it all very interesting: a quantum phenomenon (single photons are indeed reflected "properly") that is best described by classical continuum mathematics!

Since the OP arose from a reasonable guess at the questions Chris' kids will be asking, they are in for a very interesting evening - or possibly an after-dinner session at Christmas.