The optimal atmospheric density

Andrew Usher wrote:
Yousuf Khan wrote:

Thanks for this, I have not had a chance to go into any more detail
about the details you've presented, but I'll look them over. Very
interesting reading. And yes, Venus puts a lie to all of our simplistic
assumptions of how a "terrestrial" planet must be.

Well, please get back to me then - As far as I know, the stuff about
optical effects has never been discussed before.

I'm not aware of any newsgroup discussion about it (and neither is the GoogleGroups search agent), but if book citations should count, then one could mention "The Inventions of Daedalus" (ISBN 3-87144-768-4) by David Jones, in which he devotes a few pages to the "optically flat Earth", although I think his formula for density and refrative index is flawed.

Jens.
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Jens Kleimann wrote:
Yes. Picture the situation in which the curvature of a tangential ray of light is still above that of the planet, but just barely so (i.e. C1, but almost =1). If the value of C is such that the total, integrated angle of refraction between the observer and infinity approaches 90 degrees, then at local midnight, when the Sun is actually at the nadir, you can get an Eastern/rising and a Western/setting image of the Sun at the same time!
Oops, I meant to say "curvature ... _below_ that of the planet" (i.e. radius of ray's curvature larger than R_p, else the tangential ray would immediately hit the ground when launched from the surface). Rest stands unchanged.

Jens.
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Jens Kleimann wrote:

Yes. Picture the situation in which the curvature of a tangential ray of light is still above that of the planet, but just barely so (i.e. C1, but almost =1). If the value of C is such that the total, integrated angle of refraction between the observer and infinity approaches 90 degrees, then at local midnight, when the Sun is actually at the nadir, you can get an Eastern/rising and a Western/setting image of the Sun at the same time!

As I mentioned in my original post, there is no way this could ever be
seen in visible light, due to excessive Rayleigh scattering. It is
possible in certain infrared wavelengths, but very marginal - I
calculated that it would be possible on a planet Earth's size, but
very little larger. Even in the absence of scattering, light does not
travel indefinitely around because the photon sphere is an _unstable_
equilibrium.

Both images, when both are visible, will be horizontally compressed to
a flat line. Its apparent width will be only due to seeing.

Bonus question: Ignoring the fact that the images will be very flat and very dim, can you get multiple images of the Sun above the, say, the Eastern horizon from light rays that travel around the globe multiple times in a tightly wound spiral before finally leaving for infinity?

For the same reason, they would be superposed.

Andrew Usher

Yousuf Khan wrote:

Okay, in the first paragraph you said that the C on Earth is 1/5, and in
the next paragraph you said it's 1/6. Which one is it?

No, I said G = 1/5 (sorry if it wasn't clear).

So the "true" horizon is below the visible horizon? Does this mean for
example being able to see the red light of the Sun even after it's gone
below the horizon?

No, the true horizon is always at or above the visible horizon. On
Earth, the true horizon is normally visible, so they are the same. The
visibility of the Sun has to do with the astronomical horizon's being
above the others.

Sunrise and sunset, following the usual definition, are simply the
times when the Sun's altitude below the astronomical horizon is the
sum of its angular radius and the refraction at the horizon; the Sun
at the horizon is vertically compressed by a factor of 1-C (strictly
that factor is correct at the horizontal; but for C1 the true horizon
is very near and the error is insignificant).

So on a world with a C that is greater than 1, there would be no
compression of the Sun as it rises or falls near the horizon?

No, there would be infinite compression - to a horizontal line. This
is mentioned a few paragraphs later.

The astronomical horizon, on the other hand, is always above the
horizontal, and we may well ask how much above. The refractive
invariant says that the ray, from a horizontal object, striking a
point on the surface, is that that would have passed over at a height
(n-1)R with no atmosphere (5,700 ft for Earth). The angle can be
estimated by supposing that the ray hits that height at a horizontal
distance corresponding to the scale height of the atmosphere, that is,
sqrt(2HR). The angle is thus (n-1) sqrt(R/2H) - this is 20' for
Earth , while the actual refraction is 28'. We can take as the formula
then (n-1) sqrt(R/H) for the upward deviation of this horizon; that
will be approximately true for small C.

So below the astronomical horizon, we can't see any stars?

No. The stars visible below that horizon are those that are below the
true horizontal, that is, a line halfway between the zenith and nadir
in the celestial sphere.

As C approaches 1, some of the above equations become singular. The
horizon distance becomes infinite - implying that the refraction at
the horizon must be also - and the flattening of the Sun at the
horizon becomes infinitely great. However, the Sun never actually sets
at C = 1, because of that infinite refraction! It will simply fade out
as it asymptotically approaches the true horizon (which is precisely
horizontal here) and approaches the state of a horizontal line.

Now the time of sunset become later as C increases (relative to the
'astronomical' sunset), while the time of the end of twilight should
not follow it, as due to extinction light becomes fainter the farther
it has to travel through the atmosphere. Therefore the darkness of the
sky (relative to noontime) will increase with C, and at some value of
C close to 1 the sky will be dark while the Sun is still visible,
although not of course as dark as at midnight.

Are you saying that the light of the Sun will still be visible up to and
beyond midnight, even if the Sun is on the other side of the planet?

No, the sky will always be dark at midnight. Again this was answered
below.

In this case the true horizon is above the horizontal, but owing to
its infinite distance will not be visible (the visible horizon is
never more than about 5' above horizontal to the human eye).

How is the true horizon infinite in this case? Shouldn't the horizon
curve down due to the curvature of the planet like in every other case?
Or are you saying due to refraction, even surface features below the
curve of the surface will get shown at the horizon?

It is infinitely far away because light rays originating from any
point on the surface can reach any other if C 1 - which means
exactly that light curves back to the surface.

As the distance
roughly reflects the visibility of stars below the astronomical
horizon, we can say that that visibility is greatest at C = 1 and
diminishes to nil as C becomes very large.

Because when C is higher than 1, the astronomical horizon gets washed
out by the lights on the surface of the world?

Light pollution is a separate issue. I am saying that the distance
between the two horizons is greatest at C = 1; both in fact climb
higher with increasing C but must coincide at C infinite.

Andrew Usher

On 21 sept, 14:58, Andrew Usher wrote:
Jens Kleimann wrote:
Yes. Picture the situation in which the curvature of a tangential ray of light is still above that of the planet, but just barely so (i.e. C1, but almost =1). If the value of C is such that the total, integrated angle of refraction between the observer and infinity approaches 90 degrees, then at local midnight, when the Sun is actually at the nadir, you can get an Eastern/rising and a Western/setting image of the Sun at the same time!

As I mentioned in my original post, there is no way this could ever be
seen in visible light, due to excessive Rayleigh scattering. It is
possible in certain infrared wavelengths, but very marginal - I
calculated that it would be possible on a planet Earth's size, but
very little larger. Even in the absence of scattering, light does not
travel indefinitely around because the photon sphere is an _unstable_
equilibrium.

Yes. But from a point of view at the photosphere, there will be a
precise direction where light will travel indefinitely around.

There will be a specific (very shallow) angle where light will slowly
climb, eventually getting higher and climbing faster (due to
decreasing curvature) and escaping to infinity after 90 degrees. There
will be another distinct (shallower) angle where the light also slowly
climbs and eventually escapes to space after 450 degrees. And yet
another angle where light takes 810 degrees to spiral out. Et cetera.

The angle where light takes 90,5 degrees to escape will be distinct
from the angle where light escapes in 90 degrees. The image of Sun
will have nonzero extension.

Every image of Sun will have nonzero extension. In a band of finite
angular width above the horizon, you will see infinitely many images
of the entire sky, each image being distinct and of nonzero angular
width. And in a band of finite angular width below the horizon, you
will see infinitely many images of the entire planetary surface,
including your head.

Both images, when both are visible, will be horizontally compressed to
a flat line. Its apparent width will be only due to seeing.

See above. Intrinsic nonzero width for each image.

Bonus question: Ignoring the fact that the images will be very flat and very dim, can you get multiple images of the Sun above the, say, the Eastern horizon from light rays that travel around the globe multiple times in a tightly wound spiral before finally leaving for infinity?

For the same reason, they would be superposed.

No, they won´t.

And some simple considerations.

G will depend not only on climate/weather but on atmospheric
composition.

If you double the radius of planet and leave surface atmosphere
unchanged along with planetary (uncompressed) density, then the
surface gravitational acceleration g scales with R, thus the gradient
of atmospheric density and refractive index scale with R, and C scales
with R2. If you leave atmospheric composition and temperature
unchanged and change atmospheric density then its gradient of
refractive index and C scale with rho.

Also, note that neon (and helium) have pretty weak refraction as well
as Rayleigh scattering compared to nitrogen and oxygen.

On Sep 21, 11:59*am, Crown-Horned Snorkack
wrote:

Yes. But from a point of view at the photosphere, there will be a
precise direction where light will travel indefinitely around.

You know that I don't disagree with you at all. I was only presenting
what you would actually see, not the theory that you are bringing up.

... The image of Sun
will have nonzero extension.

Every image of Sun will have nonzero extension.

But far smaller than the resolution of any telescope, let alone
atmospheric blurring!

In a band of finite
angular width above the horizon, you will see infinitely many images
of the entire sky, each image being distinct and of nonzero angular
width. And in a band of finite angular width below the horizon, you
will see infinitely many images of the entire planetary surface,
including your head.

Certainly, in theory. But you couldn't see these, not with any
equipment. E.g. on the Earth-size planet, the stip containing all the
images of your head would be less than a trillionth of a trillionth of
a degree, and likely not a single photon from it would be detected!

Both images, when both are visible, will be horizontally compressed to
a flat line. Its apparent width will be only due to seeing.

See above. Intrinsic nonzero width for each image.

What I said: its APPARENT WIDTH will be only due to seeing (a few
arcseconds, perhaps).

Bonus question: Ignoring the fact that the images will be very flat and very dim, can you get multiple images of the Sun above the, say, the Eastern horizon from light rays that travel around the globe multiple times in a tightly wound spiral before finally leaving for infinity?

For the same reason, they would be superposed.

No, they won´t.

For all practical purposes, they would be.

And some simple considerations.

G will depend not only on climate/weather but on atmospheric
composition.

I never said otherwise. However, since the exact value of G is not
very important, it was unnecessary to go into detail.

Also, note that neon (and helium) have pretty weak refraction as well
as Rayleigh scattering compared to nitrogen and oxygen.

No terrestrial planet will have neon or helium as a significant
portion of its atmosphere. This is because all terrestrial planets
lose almost all their primary atmosphere to hydrodynamic escape -
that's why they're terrestrial, and not gas giants. True, there must
be intermediate cases but I'd guess it's a pretty narrow range.

Andrew Usher

On 21/09/2010 7:58 AM, Andrew Usher wrote:
Jens Kleimann wrote:

Yes. Picture the situation in which the curvature of a tangential ray of light is still above that of the planet, but just barely so (i.e. C1, but almost =1). If the value of C is such that the total, integrated angle of refraction between the observer and infinity approaches 90 degrees, then at local midnight, when the Sun is actually at the nadir, you can get an Eastern/rising and a Western/setting image of the Sun at the same time!

As I mentioned in my original post, there is no way this could ever be
seen in visible light, due to excessive Rayleigh scattering. It is
possible in certain infrared wavelengths, but very marginal - I
calculated that it would be possible on a planet Earth's size, but
very little larger. Even in the absence of scattering, light does not
travel indefinitely around because the photon sphere is an _unstable_
equilibrium.

Both images, when both are visible, will be horizontally compressed to
a flat line. Its apparent width will be only due to seeing.

I don't expect we'll see the real disk of the Sun until well after the
Sun actually rises over the true horizon, but I think we will likely see
a flattened light kind of like what we see at sunset or sunrise just a
few minutes before the Sun rises here on Earth.

But will Rayleigh scattering in the visible spectrum not allow that upto
midnight? Why not?

Yousuf Khan

On 21/09/2010 12:59 PM, Crown-Horned Snorkack wrote:
Every image of Sun will have nonzero extension. In a band of finite
angular width above the horizon, you will see infinitely many images
of the entire sky, each image being distinct and of nonzero angular
width. And in a band of finite angular width below the horizon, you
will see infinitely many images of the entire planetary surface,
including your head.

With so much reflection within such an atmosphere, will light from space
even get in? Including from the Sun. If the photosphere reflects so much
light back down, it must also be redirecting some light from the Sun
back up.

Yousuf Khan

On 21/09/2010 1:24 PM, Andrew Usher wrote:
On Sep 21, 11:59 am, Crown-Horned
wrote:
Also, note that neon (and helium) have pretty weak refraction as well
as Rayleigh scattering compared to nitrogen and oxygen.

No terrestrial planet will have neon or helium as a significant
portion of its atmosphere. This is because all terrestrial planets
lose almost all their primary atmosphere to hydrodynamic escape -
that's why they're terrestrial, and not gas giants. True, there must
be intermediate cases but I'd guess it's a pretty narrow range.

Andrew Usher

Well, some Super-Earths will be approaching the mass of Uranus or
Neptune. Maybe not absolutely very close in mass, but within a few
multiples.

Yousuf Khan

Yousuf Khan wrote:

I don't expect we'll see the real disk of the Sun until well after the
Sun actually rises over the true horizon, but I think we will likely see
a flattened light kind of like what we see at sunset or sunrise just a
few minutes before the Sun rises here on Earth.

There is no moment of sunrise with C 1. I calculated that the Sun
would become visible to the human eye (Rayleigh depth = 30) to an
elevation of about -6 degrees (corresponding to civil twilight on
Earth). How much would it be flattened at this point? On the Earth-
size planet, about exp(30*6/120) = 5 times, on the large planet, about
exp(30*6/48) = 50 times. So indeed it would not quite be at the true
horizon, nor precisely a horizontal line. But if our eyes saw far
enough into the infrared, these would be true.

But will Rayleigh scattering in the visible spectrum not allow that up to
midnight? Why not?

Because the Rayleigh depth halfway around the planet would be about
30*180/6 = 900! Remember, an optical depth of 900 means light is
dimmed exp(900) (~ 10^400) times.

Andrew Usher