# Refraction of light on a flat earth

#### MyMatesBrainwashed

##### Senior Member
IF the density of the atmosphere decreases with height (surely not that difficult to prove) and IF light refracts through those different densities (harder to prove, I think) then doesn't that mean that, on a flat earth, light would not refract for anything on your (sorry, a, assuming the earth's flat) horizontal. The density of the atmosphere across a horizontal is the same and therefore would not refract.

The point being you should be able to sight anything on a horizontal at any distance and whatever it is you've sighted should be EXACTLY the same height (above mean sea level) as you were when you sighted it. Every single time.

I can't quite get my head around how things should look with that kind of refraction either. I initially think things would look weird(er than they do) but I doubt that's actually the case.

There's also the interesting part of how do you determine your height above sea level? Like, where I am now, I have no idea how high above sea level I am. I have no way of telling. I've no idea how I'd go about it other than looking it up and trusting that someone else (or elses) knew how to do it and did it well. It's all well and good standing on the beach knowing you're at sea level, but if you sighted a building however far away and then measured from the ground to where you sighted to, how would you know how high the building is above sea level to add on to your measurement without resorting to a source you have to trust?

I also can't get my head around whether, assuming a globe, there might be conditions where this would be possible? Or if you were to do this on a globe you would likely get different results every single time and never get the two heights to be the same?

Lastly... if my assumptions here are correct and a someone believing the earth to be flat were to perform this on a globe, how might they explain the results they didn't like away or try to fudge the test in their favour?

IF the density of the atmosphere decreases with height (surely not that difficult to prove) and IF light refracts through those different densities (harder to prove, I think) then doesn't that mean that, on a flat earth, light would not refract for anything on your (sorry, a, assuming the earth's flat) horizontal. The density of the atmosphere across a horizontal is the same and therefore would not refract.
That's not how it works.

Think of light going towards the horizon as a wave, not a line. Look at it from the side. The top of the wave is in thinner air than the bottom of the wave, so it goes faster, so the wave bends downward.

Which is why the sugar fishtank bends a laser even when the beam is parallel to the bottom.

That's not how it works.

Think of light going towards the horizon as a wave, not a line. Look at it from the side. The top of the wave is in thinner air than the bottom of the wave, so it goes faster, so the wave bends downward.

Which is why the sugar fishtank bends a laser even when the beam is parallel to the bottom.

If I understand his reasoning, he is talking about light passage between observer and an object at the same height over a flat earth (with the assumption that one could expect the same density throughout that path), rather than sunlight entering from above. It sounds reasonable, but trivial.

If I understand his reasoning, he is talking about light passage between observer and an object at the same height over a flat earth (with the assumption that one could expect the same density throughout that path), rather than sunlight entering from above. It sounds reasonable, but trivial.
You understand his reasoning, and it's a misconception I had myself many years ago. But light bends downwards even when there's a horizontal path of all the same density.

You understand his reasoning, and it's a misconception I had myself many years ago. But light bends downwards even when there's a horizontal path of all the same density.
Would that be the case on our hypothetical flat earth with our hypothetical constant air density if the light could be polarized so a wave travels parallel to the surface?

...how do you determine your height above sea level? Like, where I am now, I have no idea how high above sea level I am. I have no way of telling. I've no idea how I'd go about it other than looking it up and trusting that someone else (or elses) knew how to do it and did it well. It's all well and good standing on the beach knowing you're at sea level, but if you sighted a building however far away and then measured from the ground to where you sighted to, how would you know how high the building is above sea level to add on to your measurement without resorting to a source you have to trust?

I also can't get my head around whether, assuming a globe, there might be conditions where this would be possible? Or if you were to do this on a globe you would likely get different results every single time and never get the two heights to be the same?
You must stand on the shoulders of giants.

https://thonyc.wordpress.com/2013/07/22/getting-the-measure-of-the-earth/
The same basic method is used to determine elevation.

https://thonyc.wordpress.com/2009/12/25/

And finally a textbook on Geodesy from 1880
https://archive.org/details/geodesy01clargoog/page/n298/mode/2up?view=theater

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Would that be the case on our hypothetical flat earth with our hypothetical constant air density if the light could be polarized so a wave travels parallel to the surface?
if the air density did not diminish with height, then yes. But it demonstrably does, as pilots and mountaineers can attest.

if the air density did not diminish with height, then yes. But it demonstrably does, as pilots and mountaineers can attest.
No, I understand that, which is why I asked about a polarized light. Wouldn't it have a wave function in one horizonal direction which would all fall within the hypothetical layer of constant air density? What am I misunderstanding?

It's all well and good standing on the beach knowing you're at sea level, but if you sighted a building however far away and then measured from the ground to where you sighted to, how would you know how high the building is above sea level to add on to your measurement without resorting to a source you have to trust?
Basically, surveyors have tools to measure elevation very precisely (theodolites, total stations, and special GPS receivers).

If I had to do an optical measurement, I'd take two readings, from A to B and back from B to A, and since they err in opposite directions, I'd average them to cancel the error. There's some more on technique in geodetic survey textbooks; I have a fairly current one downloaded on my other tablet.

No, I understand that, which is why I asked about a polarized light. Wouldn't it have a wave function in one horizonal direction which would all fall within the hypothetical layer of constant air density? What am I misunderstanding?
I don't think it works that way. I had a look at Monte Carlo method for polarized radiative transfer in gradient-index media, and while I can't say I understand it, they don't talk about that.

No, I understand that, which is why I asked about a polarized light. Wouldn't it have a wave function in one horizonal direction which would all fall within the hypothetical layer of constant air density? What am I misunderstanding?
IF in that hypothetical layer of constant air density the refractive index is constant, it will not pose any problem pointing a laser. As soon as there is a gradient, the laser will bend.

Or am I misunderstanding your question?

IF in that hypothetical layer of constant air density the refractive index is constant, it will not pose any problem pointing a laser. As soon as there is a gradient, the laser will bend.

Or am I misunderstanding your question?
Nope, that's what I thought. I understand @Mick West and his nice diagram about the wave function passing through different layers with different densities. And, like @Mendel, I "sort of" grasp the paper referred to, but as it mentions both polarization rotation and scattering, I think that simple yes-no answers are inadequate, and more complex than I choose to investigate at this time. Thanks!

External Quote:
Light transfer in gradient-index media generally follows curved ray trajectories, which will cause light beam to converge or diverge during transfer and induce the rotation of polarization ellipse even when the medium is transparent. Furthermore, the combined process of scattering and transfer along curved ray path makes the problem more complex.
https://www.sciencedirect.com/science/article/abs/pii/S0022407314004464#:~:text=Examples of participating media with gradient refractive index,non-contact measurement of temperature distribution in flames,,, etc.

@Ann K

Thanks for the link, very interesting read! Indeed there is more to it.

Apologies if I just haven't grasped this from the above reading, but is there a good liquid to use as a gradient indexed medium? For example, will either plain water or sugar water produce results similar to the atmosphere (which I understand can be a gradient-indexed medium as well)?

@Ann K

Thanks for the link, very interesting read! Indeed there is more to it.
I really didn't word my question well, which should be "is there a way to constrain the light to a single layer", but upon reading about scattering and rotation of polarization, it sounds like the answer to that is "no, not very well".

You understand his reasoning, and it's a misconception I had myself many years ago. But light bends downwards even when there's a horizontal path of all the same density.

Snell's law describes how a light beam is refracted when passing through the boundary between two media with different optical densities. Below is my old sketch (in Finnish) with three layers and a light source in the middle. I actually calculated the angles according to the refractive indices. So the sketch is quite realistic.

When layers are extremely many and they are extremely thin, the situation corresponds to a single medium whose optical density increases continuously when going up.

This situation can be managed mathematically with the differential form of Snell's law, see https://phys.libretexts.org/Bookshelves/Optics/Geometric_Optics_(Tatum)/01%3A_Reflection_and_Refraction/1.08%3A_Differential_Form_of_Snell's_Law

Below is a (not so realistic) sketch of this situation. It could describe the light rays in the ideal atmosphere on the ideal flat Earth.

In the picture, the light rays are curves in the xy coordinate system, where the light source is at the origin.

All these curves are mathematically speaking continuous and differentiable (they have derivative in every point x > 0). The derivative f'(x) is positive at every x > 0 at every positive departure angle α in the quarter I. The derivative f'(x) is negative at every x > 0 at every negative departure angle α in the quarter II.

So every ray with negative departure angle is bending down. On the other hand every down bending ray has negative departure angle α (the angle α can always be chosen as small as necessary).

In the same way can be concluded, that all rays in the quarter I have positive departure angle. The ray with departure angle α = 0 is not among these bending rays. So the horizontal ray must travel straight.

This is not some mathematical proof based on the differential form of Snell's law. I haven't even gotten to know it yet. My reasoning is just a feeling guided by intuition. It can be mistaken.

Still, my feeling is pretty strong. As Mick said, "It's a misconception I had myself many years ago". Looks like I still have it. Could Snell's law be wrong in this case? Probably just my intuition is wrong? I'm confused.

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Apologies if I just haven't grasped this from the above reading, but is there a good liquid to use as a gradient indexed medium? For example, will either plain water or sugar water produce results similar to the atmosphere (which I understand can be a gradient-indexed medium as well)?
Apparently this is a standard physics lecture demonstration.
An even layer of granulated sugar, about a centimeter thick, is placed on the bottom of a small water tank. Water is slowly added by dribbling from a hose that is constantly moved over the surface, to cause minimum disturbance of the sugar. The tank is left undisturbed for 48 hours. The sugar slowly dissolves, forming a solution with a gradient of refractive index.

Note that no such demonstration exists for bending the other way—the one FE "science" video I know that had this claim showed a still of the above experiment, with the picture flipped upside down.

When layers are extremely many and they are extremely thin, the situation corresponds to a single medium whose optical density increases continuously when going up
No, it doesn't.

Imagine a sine wave that crosses several layers at once and figure out how it would bend.

You will see bending all the way down (or until the beam leaves the atmosphere) no matter what the departure angle is. This is also experimentally demonstrable, look for videos of the "sugar water tank laser" experiment.
Or see Mick's version (there's a whole thread on the topic):
The curve is slight, but there. More visible with some compression.

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Snell's law describes how a light beam is refracted when passing through the boundary between two media with different optical densities. Below is my old sketch (in Finnish) with three layers and a light source in the middle. I actually calculated the angles according to the refractive indices. So the sketch is quite realistic.

When layers are extremely many and they are extremely thin, the situation corresponds to a single medium whose optical density increases continuously when going up.

This situation can be managed mathematically with the differential form of Snell's law, see https://phys.libretexts.org/Bookshelves/Optics/Geometric_Optics_(Tatum)/01%3A_Reflection_and_Refraction/1.08%3A_Differential_Form_of_Snell's_Law

Below is a (not so realistic) sketch of this situation. It could describe the light rays in the ideal atmosphere on the ideal flat Earth.

View attachment 54327

In the picture, the light rays are curves in the xy coordinate system, where the light source is at the origin.

All these curves are mathematically speaking continuous and differentiable (they have derivative in every point x > 0). The derivative f'(x) is positive at every x > 0 at every positive departure angle α in the quarter I. The derivative f'(x) is negative at every x > 0 at every negative departure angle α in the quarter II.

So every ray with negative departure angle is bending down. On the other hand every down bending ray has negative departure angle α (the angle α can always be chosen as small as necessary).

In the same way can be concluded, that all rays in the quarter I have positive departure angle. The ray with departure angle α = 0 is not among these bending rays. So the horizontal ray must travel straight.

This is not some mathematical proof based on the differential form of Snell's law. I haven't even gotten to know it yet. My reasoning is just a feeling guided by intuition. It can be mistaken.

Still, my feeling is pretty strong. As Mick said, "It's a misconception I had myself many years ago". Looks like I still have it. Could Snell's law be wrong in this case? Probably just my intuition is wrong? I'm confused.

Snellius! My fellow townsman. Alway a plus.

Bold part:
Mathematically yes. But practically, no. The layer shall be infinitely thin to make the light go straight, which of course in practice cannot be achieved. So the beam of light will always be diffracted down.

@Mendel, @Ravi

My sketch with bending rays is just an illustration how I think the differential form of Snell's law describes the situation. There's a constant, continuous optical density gradient and a perfectly flat plane ("ideal atmosphere on ideal flat Earth"). Hope I used the terms correctly. The point is that we can forget the layers.

This is therefore a purely mathematical consideration, which may not have any connection with reality.

Let's first exclude the ray with departure angle α=0. So all rays are bending down in the sense that f'(x) is decreasing when x is increasing. In quarter I the derivative f'(x) is always positive and in quarter II it's always negative.

Let's look at this series of rays from the bottom up. At a certain point x0 the derivative is first negative but increasing (quarter II). Then the derivative becomes positive and increasing when going up (quarter I). Is there not any ray with f'(x0)=0 in this infinite set of rays?

No, in this set there is not such a ray, because there the derivative is always positive or negative. In mathematics "a continuous function cannot change its sign without passing through zero". It seems rational to think that f'(x0) is a continuous function of the departure angle α. So there must be a ray with f'(x0)=0. The only possibility is that this ray is the excluded one with α=0.

This reasoning can be repeated for every x. So the ray with α=0 is straight.

Looks like I'm repeating myself. However, my purpose is only to show how I understand Snell's law in this case. if I have interpreted it correctly, it only shows that this law is not enough to describe reality.

The thread "Simulating Atmospheric Refraction" has really important and new information to me, https://www.metabunk.org/threads/simulating-atmospheric-refraction.7881/. There is a reference "Gutierrez, 2006", but the link seems broken. I was left thinking the sentence "The end result of this differential transformation is a curved path that distorts the normal view of the real scene, but that path cannot be calculated with traditional ray tracing." So the real life physics is much more complicated than I thought.

Mick's experiments in the video "Refraction Explained with Lasers and Sugar" are really convincing. Duncan Moore's demonstrations in the video "Refractive Properties of Gradient Index Optics" are also very interesting. His demonstration tool is a plastic plate made of gradient index material. As fitting in your pocket, it is much more convenient than a sugar water tank. Where can I buy one?

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My sketch with bending rays is just an illustration how I think the differential form of Snell's law describes the situation. [...] This reasoning can be repeated for every x. So the ray with α=0 is straight
Yes. That's why your mathematical model does not describe physical reality.

That's not how it works.

Think of light going towards the horizon as a wave, not a line. Look at it from the side. The top of the wave is in thinner air than the bottom of the wave, so it goes faster, so the wave bends downward.
Shoulda known it would never have been that simple

While I understand the crux of your explanation I must admit I'm struggling with the scales involved, like the air being thinner/thicker nano meters apart (assuming I've got the correct scale). But I suppose it must be.

And I understand the thinner going faster but why doesn't the thicker make it go slower too and cancel itself out? Why does thin win? Genuinely interested in that.

And I understand the thinner going faster but why doesn't the thicker make it go slower too and cancel itself out? Why does thin win? Genuinely interested in that
Imagine a tank, or a tracked bulldozer. If both tracks run at the same speed, the tank moves in a straight line, but when one track is faster and the other is slower, that doesn't "cancel out", it makes the tank turn toward the slower side ("differential steering").

And you're right, the distances are very small, and the difference is slight in atmosphere, more of a statistical phenomenon than anything, but the bend is also very very slight (even less than the Earth surface bends).
Unless you consider the sugar water tank, where the atoms are much denser and the density gradient is much steeper—that bends much more.

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Imagine a tank, or a tracked bulldozer. If both tracks run at the same speed, the tank moves in a straight line, but when one track is faster and the other is slower, that doesn't "cancel out", it makes the tank turn toward the slower side ("differential steering").
Again, totally get that but I'm struggling to fit the analogy into a wave.

Am I wrong to be imagining something (a proton) travelling along the wave? Going into thicker, then going into thinner, then back to thicker and so on etc. Or is that correct and I'm missing something else?

Again, totally get that but I'm struggling to fit the analogy into a wave.

Am I wrong to be imagining something (a proton) travelling along the wave? Going into thicker, then going into thinner, then back to thicker and so on etc. Or is that correct and I'm missing something else?

A proton is a particle and not the same as a photon, can be accelerated though. Photon is a weightless boson.

Wave or photon through two optical materials (at an angle) with different refractive indices (like reflection and transmission), is happening of course essentially on quantum level. Snellius's laws work on a different scale, but fundamentally it gets quite weird. For instance, why would a photon "decide" to reflect or transmit at the surface?
The "marching band" analogy of Snellius works great in the macro world, but on small subatomic scales it gets freaky.

Basically, quantum mechanics is the best to describe photon-matter interaction.

Again, totally get that but I'm struggling to fit the analogy into a wave.

Am I wrong to be imagining something (a proton) travelling along the wave? Going into thicker, then going into thinner, then back to thicker and so on etc. Or is that correct and I'm missing something else?
photon propagation at the subatomic level is weird, I can't say I fully understood it when I looked at refraction some time back (thanks, Metabunk!).

The high-school level idea is that photons move along guided by the electromagnetic field, which follows the principles of wave propagation. Refraction is essentially a wave phenomenon, and you can't understand it at the photon level unless you're prepared to go into particle physics. Maybe @markus can point us to an explanation, but it'll probably be kinda complicated, and thinking about it from the wave perspective works.

I found the same question as myself on Physics Stack Exchange site, https://physics.stackexchange.com/q...grin-gradient-index-lens-at-the-optical-angle

" – – The refractive index is only a function of the distance from the optical axis and the rays of light are all parallel to that. Without a change in refractive index, the rays of light should not change direction (Snell's law)."

Quote from one answer: "It's just that Snell's law doesn't work at the boundary, and in the limit of smooth refractive index profile all points will be boundaries."

These answers are sufficient for me, although there is another question thread on the topic: https://physics.stackexchange.com/q...tion-index-continuously-inc?noredirect=1&lq=1

So far, the ultimate authority on these questions is quantum mechanics. The answers derived from its formalism are not easy to explain or illustrate to the layman (or anyone else). Richard Feynman tries this mission impossible in his excellent book "QED: The Strange Theory of Light and Matter". If he can't do it, no one can.

This is the situation in modern physics. There is no return to the perceptual (or mechanical) description of phenomena (actually hasn't been in a hundred years). We just have to live with that.

Below is a passage from American Journal of Physics. https://aapt.scitation.org/doi/abs/10.1119/1.17876?journalCode=ajp

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I found the same question as myself on Physics Stack Exchange site, https://physics.stackexchange.com/q...grin-gradient-index-lens-at-the-optical-angle

" – – The refractive index is only a function of the distance from the optical axis and the rays of light are all parallel to that. Without a change in refractive index, the rays of light should not change direction (Snell's law)."

Quote from one answer: "It's just that Snell's law doesn't work at the boundary, and in the limit of smooth refractive index profile all points will be boundaries."

These answers are sufficient for me, although there is another question thread on the topic: https://physics.stackexchange.com/q...tion-index-continuously-inc?noredirect=1&lq=1

So far, the ultimate authority on these questions is quantum mechanics. The answers derived from its formalism are not easy to explain or illustrate to the layman (or anyone else). Richard Feynman tries this mission impossible in his excellent book "QED: The Strange Theory of Light and Matter". If he can't do it, no one can.

This is the situation in modern physics. There is no return to the perceptual (or mechanical) description of phenomena (actually hasn't been in a hundred years). We just have to live with that.

Below is a passage from American Journal of Physics. https://aapt.scitation.org/doi/abs/10.1119/1.17876?journalCode=ajp

View attachment 54416

Several of the incorrect descriptions commonly used to explain refraction are dismissed in these two vids, as well as them presenting the shut-up-and-calculate answer:
"Why does light slow down in water?"

"Why does light bend when it enters glass?"

Below is a (not so realistic) sketch of this situation. It could describe the light rays in the ideal atmosphere on the ideal flat Earth.
A more realistic view can be had with the refraction simulator

Use the "Flat earth" setting. The side view is highly compressed. This is with standard refraction

The green line simulates a perfectly horizontal laser.

Here's a more dramatic vertical gradient, showing how everything bends down, with the more dramatic bends on the horizontal "rays" (there's no such things as a ray)

A proton is a particle and not the same as a photon, can be accelerated though. Photon is a weightless boson.

Wave or photon through two optical materials (at an angle) with different refractive indices (like reflection and transmission), is happening of course essentially on quantum level. Snellius's laws work on a different scale, but fundamentally it gets quite weird. For instance, why would a photon "decide" to reflect or transmit at the surface?
The "marching band" analogy of Snellius works great in the macro world, but on small subatomic scales it gets freaky.

Basically, quantum mechanics is the best to describe photon-matter interaction.
Oh man, classic! Wrote proton when I meant photon. Thank you for not being mean about that.

I'm happy to accept "it's quantum", leave it there and have my photons driving bulldozers.