[aphextron]

This is a bit hand-wavey in its' description of light reflecting. One of the most mind blowing bits of physics I've ever learned is that photons do not actually "bounce" off of a surface like little balls. They are always absorbed and re-emitted. The actual photons that hit an object are not the same ones that eventually enter your eyes. The atoms in an object are stimulated by photons hitting them to then emit a new photon via the photo-electric effect, which we perceive as light "bouncing".

[whatshisface]

Having thought about it, I'm not sure if I agree. Light can reflect off of transparent crystals, which do not have the right electron energy levels for absorption (that's why they're transparent.) It's possible that the Feynman diagram of a reflection might involve the photon "going away and another one replacing it" (I don't know what the diagram looks like), but Feynman diagrams cannot be interpreted as sequences of events. Instead, they describe an instantaneous flow of amplitude between different quantum states. I would only go along with the absorption and re-emission interpretation if there was always a time delay between the two steps (which there cannot be if the mirror does not have any energy level deltas suitable for storing the energy).

The other side of the debate would be that if the mirror is moving towards or away from the light, the reflection will be Doppler-shifted to a higher or lower frequency. Does this mean that the reflected photons are not the same photons as the incident photons, or does it mean that the same photons have had their energy changed? I think there is no meaningful distinction because every two particles with the same name in quantum mechanics are identical anyway. There's no telling which are which. If I showed you a photon, then took it back and showed you another, you would never be able to tell whether I had opened the same box twice or if I had taken the old one out and captured a new one from my desk lamp.

[chemeng]

They are not the same photons. They are emitted by the electron changing energy states after excitation of the incoming photon. But your intuition that there is some quantum weirdness is correct. Quantum Electrodynamics is the theory you're looking for. It unifies quantum mechanics and special relatively in the context of light/matter interaction. Given you seem to be familiar with Feynman. You're in for a treat! His book on QED is meant to be an accessible explanation.[1]

[1]https://en.m.wikipedia.org/wiki/QED:_The_Strange_Theory_of_L...

[whatshisface]

Hydrogen atoms can't store photons that do not have the same wavelength as any of its spectral lines. Nonetheless, hydrogen gas has an index of refraction that is a continuous function of frequency, and as a result it can reflect light of any wavelength. Therefore, hydrogen does not store any energy for any length of time in the course of typical reflection. Even if the Feynman diagram for reflection looks like a combination of absorption and re-emission, if the absorption and emission process can't occur at separate times, then it is not right to say that one is followed by the other.

[boomboomsubban]

Isn't hydrogen gas almost exclusively H2, giving it plenty of possible electrons to interact with photons?

[whatshisface]

That's a good catch, but H2's spectral lines are still discrete. It's not that gasses can't absorb photons, it's that they can only do so at very specific frequencies.

To simplify the argument, you could use Helium instead. Here is a (discrete!) list of its spectral lines: https://physics.nist.gov/PhysRefData/Handbook/Tables/heliumt...

And here is a chart of its (continuous!) index of refraction: https://refractiveindex.info/?shelf=main&book=He&page=Mansfi...

[BeetleB]

You didn't address the concern. A given atom cannot absorb any photon. They can only absorb photons of certain energy levels/wavelengths. Electrons can occupy only certain energy levels. So they can only absorb photons that match the difference if two energy levels.

[empath75]

> They are not the same photons.

How do you define "the same" for two photons?

[tsbinz]

This is the question that should be higher up in the thread. This whole discussion makes no sense. Photons carry no hidden identity.

[zamadatix]

A photon doesn't need to carry a UUID for people to conceptualize a difference between a photon going through free space and being absorbed and re emitted. The parent comment asks the right question, what is being meant by "the same" is required to know before being able to say the discussion makes no sense.

Taking your first literal interpretation as the only way the discussion could have value is no more helpful than this exchange: "I'm going to the store, do you want the same hot dogs we have in the fridge?" "That doesn't make sense, the store can't have the same hot dogs we have in the fridge".

[izietto]

> "I'm going to the store, do you want the same hot dogs we have in the fridge?" "That doesn't make sense, the store can't have the same hot dogs we have in the fridge"

Yeah but hot dogs belong to classical mechanics, while photons to quantum mechanics, and in quantum mechanics you can actually have the same hot dogs in different places, until you observe them

[Enginerrrd]

>The parent comment asks the right question, what is being meant by "the same" is required to know before being able to say the discussion makes no sense.

This is true, but I think the comment you are replying to is correctly identifying a fundamental conceptual misunderstanding of the comments preceding it. In the sense that they likely mean, it's not really a valid question. Sort of like the "are you still beating your wife?" question. Asking it is making incorrect assumptions.

[zelphirkalt]

Who knows, maybe they actually do, but we do not know it yet. I mean, how do you know they do not?

[rosybox]

The absence of evidence doesn't make an idea more plausible, it makes it less plausible.

I can't disprove that a celestial teapot is circling the sun between Earth and Mars, so how do you know it isn't?

https://en.wikipedia.org/wiki/Russell%27s_teapot

[mhh__]

Modern physics is built on the idea of fundamentally indistinguishable particles.

We may not know but we can only know what we can see and what we can see is explained by what we know.

[ngvrnd]

it's all the same photon

[hanoz]

For those missing the reference: https://en.m.wikipedia.org/wiki/One-electron_universe

[mr_toad]

I wonder if there’s an antimatter universe going backwards in time from the Big Bang.

[viklove]

But that's just not true. The electron will only get excited if the incoming photon has a very specific wavelength, so most of the light that hit your house's walls, for example, is reflecting. The photoelectric effect is real, but it only applies in specific circumstances...

[chemeng]

There will be a transfer of energy to the system if there is an interaction, but will only absorb the incident photon at specific energies. In the case of reflection in this classical view, it simply re-emits the energy back out as there are no valid energy transitions. The reality is far more complex and this thread is mixing multiple theories together, wave and particle views and simultaneously comparing single atom mechanics with those of lattices/solids. This is all causing a lot of confusion that's hard to address in comments.

[ggggtez]

I think this is called, "knowing enough to be dangerous". You clearly have some understanding of light, but you are missing some key facts that would lead you to the correct understanding.

>Light can reflect off of transparent crystals

No, the point of the above post is that a photon is not "reflected", but captured and re-transmitted by photo-electric effect.

Light travels at maximum speed C, only when in a vacuum. Otherwise it travels through a medium. We can see that the different speeds of light cause dispersion such as when it enters and leaves a clear prism (changing mediums, thus spreading out the different speeds/wavelengths of light). Since we know that light is "traveling in a medium" when it travels in air or in the prism, then what do we mean by this? It is traveling slower, so it must have some information about the medium...

So how does the light "know" it's in a material, without interacting with the atoms of the material?

Of course, it couldn't know. It is interacting with the material. The behavior of the particle is fully explained by the photo-electric effect.

[whatshisface]

Absorption and emission are not the only two processes by which light can interact with matter, reflection is another. The argument is that if the material can't store the energy then you can't really call the event an absorption. Feynman diagrams aren't sequences of events so just because you see the photon coming to, disappearing, and then departing an electron doesn't mean it happens in that order.

[anonytrary]

> Light can reflect off of transparent crystals

If light is reflecting off of something, it's precisely because that thing isn't transparent[0].

> if the mirror is moving towards or away from the light, the reflection will be Doppler-shifted to a higher or lower frequency.

This does not refute OP's assertion. If the photon is absorbed and a "different"[1] one emitted, the emitted photon is still emitted from a reference frame that is moving with some velocity, therefore the emitted photon will still be Doppler-shifted... The photon is still emitted at the correct energy level. The photon appears to have higher energy in the stationary reference frame.

[0] Disregarding human-centric definitions that have to do with visible spectrum.

[1] Thinking of the photons as "different" doesn't actually mean much when it comes to quantum mechanics; they're indistinguishable bosons.

[dan-robertson]

It doesn’t really make sense to talk about “the same” photon because photons don’t really have identity. Even in a vacuum you can send a photon in one end and see one on the other and not really know that the one going in was “the same” as the one going out. It can turn into other particles on its way and then recombine into a photon and there is no way to know this based on the observation that a photon came out of the end of your vacuum.

[zelphirkalt]

That's one of the things, that annoys me with questions about physics, which me might not know well enough yet: When someone asks about whether something "is", people answer with what we can in general know at our current level of understanding and then make it seem as though that is the same, as answering the question, what really "is".

> Even in a vacuum you can send a photon in one end and see one on the other and not really know that the one going in was “the same” as the one going out.

I don't really care, if _I or current physicists_ cannot _distinguish_ them, when I ask, whether they _are_ the same. I would rather have an answer like as follows:

"We do not know this yet. We do not have the means of telling, whether it is really the same photon or not. However, even if they were not the same, it would make no difference (according to our current understanding of physics!), as their effect on the surroundings would still be the same, because ..."

[keithnz]

no, what dan said is better, it might annoy you, but physics is such that wording is important. Even then, nearly all wording is losing information and in some cases changes it from the reality. You are asking a question which you want an answer but the question itself has a troubling fit with reality as you try to relate to things in terms that seem familiar.

[StuffedParrot]

No, physics is such that semantics are important. What’s the observable difference between absorption/emission and reflection? If the latter doesn’t happen at all the distinction is meaningless.

[elcomet]

I don't think it's hand wavey. It's just a different (simpler) model. To describe this phenomena (wet things are darker), you don't need to go into details about photons, or quantum mechanics or whatever.

Geometrical optics is really sufficient here, it explains it perfectly, and is much more intuitive than other more complex models.

[aryankashyap]

Thanks!

[amluto]

> The actual photons that hit an object are not the same ones that eventually enter your eyes.

This part is philosophy, not physics. Photons are “indistinguishable particles”. You cannot tell two photons apart. This has a rigorous meaning: if you swap two photons, the state of the universe is unchanged. (This is in contrast to fermions — if you swap two fermions, the state of the universe has its phase shifted 180 degrees.) If photons we’re distinguishable, then swapping two photons would swap them, and the resulting state would be different. The matters from a statistical perspective. Look up Bose-Einstein statistics if you’re interested.

As for whether photons are absorbed and re-emitted or merely reflected, this surely depends on the surface on question.

[raldi]

> if you swap two fermions, the state of the universe has its phase shifted 180 degrees

Do you mean there are two kinds? Or if you swap any two?

[amluto]

Swap any two.

This is why two fermions can’t be in the same state: if you swapped them, you’d be back where you started, but that somehow also had to phase shift 180 degrees, and the only complex amplitude that is the same as its own negation is zero.

[mrfox321]

Electrons are indistinguishable as well.

[sjg007]

What surfaces reflect?

[amluto]

Metals and, generally, homogenous materials (dielectrics, conductors, etc). Light reflects off of a conductor or a dielectric due to a relatively straightforward solution to Maxwell’s equations.

In contrast, anything phosphorescent most definitely absorbs and re-emits, sometimes hours later. Imagine glow-in-the-dark pigment.

[ddingus]

It was fun to think about glow in the dark pigments that way.

[sjg007]

Thanks, yes makes sense. But do metals actually reflect or do they just absorb and just release spontaneously?

[amluto]

This probably depends on your perspective. If you think of a metal as a homogenous linear medium (which is a good approximation for long wavelengths but is not exact), then light is a wave, photos are an unnecessary detail, and the wave mostly reflects and is absorbed as heat. If you look the metal as a whole bunch of atoms and a sea of mobile, discrete electrons, then the incoming photons excite the sea of electrons and, when the interaction settles down, photons are leaving. If you insist on modeling the electrons with QED, congratulations, you’ve made your life unnecessarily complicated unless the photons have such high energies that electrons are created or end up with relativistic velocities. If you believe in string theory, maybe the photons were only an approximation in the first place.

Physics is all about choosing an appropriate model of the world that captures the detail you need without being more complicated than needed. Is a person on a playground swing a pendulum, or is it a bag of molecules sitting on a flexible piece of rubber suspended by a bunch of rusty metal links from a flexible steel bar that is, in turn, supported by many geological layers in the Earth’s crust?

[CrazyStat]

How does absorption and reemission preserve phase information?

What about total internal reflection? Why does the light travel through the entire body of a glass prism without issue and then suddenly get absorbed and reemitted at the surface? Glass (and water) are transparent specifically because they don't absorb photons in the visible spectrum, so how is it that they absorb and reemit?

[kurthr]

Because the electro-magnetic field polarization is preserved in many cases (specular). That doesn't mean there isn't a plasmon generated in between or that the sign of the electric field is maintained. What is interesting is that in the case of metalic (from low to high index of refraction) there is an inversion of the electric field. This does not happen at the reflection from a high index to low index (e.g. under water looking up to see total internal reflection the polarization remains the same). You can see this same effect bouncing water waves off of a concrete wall.

[maskedinvader]

IANAP but from my limited understanding (mostly reading Richard Feynman) light beam = stream of photons, they don't travel through entire body of glass prism without issue, they get absorbed and re emmitted along the way too (Which is what explains the different speed of light in different mediums)

[CrazyStat]

I'm not convinced but even if we accept that explanation, why do all the atoms along the way emit the light in the direction it was travelling and then the ones at the surface suddenly emit it in a different direction?

[georgeburdell]

I can recommend Hecht’s Optics textbook for this. I found it at my local library. I have had a number of electromagnetics courses over my life and none explained the unifying scattering principles of reflection, transmission, and refraction concept so clearly as that book

[dhimes]

Hecht is an amazing author IMHO. Even his intro physics book is remarkable for introducing Emmy Noether at the outset of dynamics.

[lonelappde]

They don't! That's why the thickness of th glass of affects it's reflectivity, and not the simple way you might guess. Read Feynman's QED book or watch the YouTube vidoes for a lay intro.

[hn3333]

Probably it's just how the math works out. The atoms emit in all directions, however the waves traveling in all the other non-correct (according to reflections and refraction laws) directions cancel each other out, and only the ones with the new correct direction are still there / visible.

[wizzwizz4]

WikiBooks really does have everything: https://en.wikibooks.org/wiki/A-level_Physics_%28Advancing_P...

This example is wrong, but useful, and it's (probably) less wrong than what you've been talking about so far. At the very least, it's more useful.

[hn3333]

Thanks for correcting me and thanks for the link. But what have I been talking about so far? I don't understand. Also I just watched this video I'd recommend that explains the idea: https://www.youtube.com/watch?v=NumSE2LvSmQ

[JoBrad]

This video addresses that, as part of a longer discussion.

https://youtu.be/CiHN0ZWE5bk

Also this one:

https://youtu.be/YW8KuMtVpug

[tigershark]

That doesn’t look right at all to me. Why the photon is then re-emitted in exactly the same direction that it would have if it was reflected rather than a random direction?

[tyre]

One constant in quantum electrodynamics is that it doesn't look right to me, but it seems to work!

Richard Feynman's "QED" is a great read on this.

[cameldrv]

You can also watch the original lecture. This is one of my very favorite explanations of anything:

http://www.vega.org.uk/video/programme/46

This explanation starts at about 30 minutes, but I highly recommend watching from the beginning.

[yesenadam]

Thanks! I hadn't seen these. There are 4 Feynman lectures on this page[0], the Douglas Robb Memorial Lectures, of which you linked to the second.

http://www.vega.org.uk/video/subseries/8

1. A gentle lead-in to the subject, Feynman starts by discussing photons and their properties. 2. What are reflection and transmission, and how do they work? 3. Feynman diagrams and the intricacies of particle interaction. 4. What does it mean, and where is it all leading?

The site has many other science videos too:

http://www.vega.org.uk/video/series/5

[yesenadam]

Update: have now watched 3 of the 4 videos. Can see why you loved it so much! I've watched a lot of Feynman talks and lectures, and read most of his books. ..but this has particularly good explanations of the material, distinctly clearer than a couple of talks of his I'd seen before. Plus a few priceless references to NZ and the NZ national character! (filmed Auckland 1979)

[improv32]

The photon has some amplitude for reflecting in all possible directions, it happens that for certain directions these amplitudes are very close to the same value and so reinforce and become more probable, and at wider angles tend to cancel out and become less probable. When this is not the case and wider angles instead reinforce you can observe this directly in for instance a diffraction grating.

[tigershark]

Yes, I agree. But what this has anything to do with the probability of emitting a photon in a specific direction? The parent was speaking about absorption and re-emission, not reflection.

[mac01021]

I think the two slit experiments indicate that a photon does exactly have a direction of travel?

[lonelappde]

Only after it arrives, so the situation's a lot more nuanced than that.

[whatshisface]

It has enough direction of travel for laser pointers to work.

[melon_madness]

Conservation of momentum, I think

[whatshisface]

If the emitted photon always had the same momentum as the absorbed photon, the light would go right through the mirror, passing on its way without changing direction.

[samatman]

Only if the momentum wasn't transferred to the mirror.

Which is, in fact, what happens. Solar sails exploit this fact:

https://en.wikipedia.org/wiki/Solar_sail

[whatshisface]

If momentum was transferred to the mirror then the emitted photon would not have the same momentum as the absorbed photon.

[melon_madness]

Conservation of momentum doesn’t mean that the momentum of the emitted photon has to be equal to that of the absorbed photon. It means the sum of the momentum of the emitted photon and the mirror must equal the momentum of the absorbed photon, assuming the mirror had no momentum to begin with.

[samatman]

> Why the photon is then re-emitted in exactly the same direction that it would have if it was reflected rather than a random direction?

That is, the equal and opposite direction, exactly as predicted by conservation of momentum.

[mr_toad]

Momentum is a vector. The sum of the vectors is preserved.

[8bitsrule]

>Everything in this essay comes from this source. https://fermatslibrary.com/s/why-some-things-are-darker-when...

It's less hand-wavey. Also, the reflectivity of water is very low at low angles of incidence, rising to 100% at normal angles. Less light reaches the wetted material. https://en.wikipedia.org/wiki/File:Water_reflectivity.jpg

[thelittleone]

If I recall a high school physics class correctly the colour of the photons that are emitted from the object (the colour we see) are the ones the object does not absorb. If so, is the object the colour we see or is it the colour it absorbs.

[ddingus]

Color may not actually be a basic physical property.

Purple is an artifact of our minds, for example. Does not exist as a distinct wavelength of light.

The object absorbs, does not absorb, emits, whatever wavelengths of light. To us, those have colors, where they are in the visible spectrum.

Take us out of the equation, and?

[thaumasiotes]

The little balls model perfectly explains why light reflects off a mirror at the same angle at which it arrived. How does the absorb-and-reemit model handle this?

[silvestrov]

"The Science Asylum" explains it: https://www.youtube.com/watch?v=cep6eECGtw4

[lonelappde]

That's almost exactly the same as Feynman's QED lecture from the 1980s(?), but with MSPaint animation instead of a chalkboard.

[melon_madness]

I don't think this is quite the photoelectric effect. The atoms absorb photons and re-emit photons--they aren't emitting electrons.

[mrlala]

Does that mean like, all the time?

For example.. if I'm looking at a star, and thinking it's the 'same' photon that came off of that star billions of years ago... that might be true in the vacuum of space, but since it's passed through the medium of the atmosphere, does that mean it's constantly been absorbed/new photons being emitted?

[empath75]

It’s not even true in the vacuum of space. It might have been created and annihilated billions of times on the way.

[marcosdumay]

> The atoms in an object are stimulated by photons hitting them to then emit a new photon via the photo-electric effect, which we perceive as light "bouncing".

That's how electrons and photons interact. It doesn't make sense to say that they are different photons or the same ones, since you can't observe the interaction.

[lonelappde]

> The actual photons that hit an object are not the same ones that eventually enter your eyes.

For anything opaque, you should expect this, because otherwise how could you see something's color if you only saw the same photons as the light source it reflected?

[anamexis]

Because the object only reflects photons of certain wavelengths.

[socceroos]

Dumb question: why do these ejected photons mirror the incoming photon's trajectory if they're a new particle?

[aryankashyap]

Could you please elaborate on how it's hand-wavey. Happy to take constructive criticism and improve :)

[ignoramous]

This might interest you: https://www.quora.com/What-are-the-physics-behind-colour/ans...

[sieabahlpark]

Except light is a wave and not a particle so reflection is explained when you consider that.

[reubenmorais]

Light is not a wave. Photons are particles. Light as a wave is a convenient abstraction that explains a lot of light's behaviors, but not all of them. I recommend Feynman's QED lectures (or the book), where he explains this.

[hutzlibu]

This debate is very old. Newton and Huygens fought long over it .. and eventually it was solved by: it is both or rather neither. It is a quant.

https://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality#...

[fennecfoxen]

A "quant" is someone who studies quantitative finance, typically found in investment fields. The word for this wave-packet with particle properties is "quantum". It is a singular noun; its plural is "quanta".

[hutzlibu]

Thx. Not a english native speaker. In german it would be just "quant".

[lonelappde]

That's half right.