What would days and nights look like on this cylindrical world?

Question

This is yet another "literal worldbuilding" question, as in, building worlds. Inspired, in part, by this question.

What if this world is a very long (effectively infinite) hollow cylindrical cavity with the diameter roughly equal to that of the Earth's orbit. The day/night cycle is generated by the following arrangement: multiple suns falling through the center of the cylinder. The distance between the suns and their speed relative to the world surface are adjusted to generate approximately 24 hours cycle, just like on Earth.

While there are several issues with this idea, most importantly huge tidal waves generated by the gravity of the suns, and the fact that empty space inside will be quickly filled with matter both from the suns and the cylinder itself, let's ignore them for now.

This question is simple enough:

• What would the sky at some point on the surface look like during day and night, as well as dawn and twilight? By that I mean, how would the suns move, how would the lighting change, etc. How different would it be from our own experiences?

I would like the answer based on actual geometry and optics. I have poor spatial imagination which is why I need some help with that :)

The suns should be spaced enough so that the night is mostly dark, though of course we can't avoid some light, since there's no horizon. The size and the energy output of the suns can be modified as well, because the cylinder would collect all the energy, radiation and solar wind, which could be too much.

Optionally, I would also like to see what other problems arise with this arrangement, though I could ask a separate question for that.

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There's a similar "tube world" arrangement in this question, but it's a little too complicated and can't be used to answer my question.

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Just to clarify: I want this world to support flora and fauna as close to Earth-like (temperate climate) as possible.

Answer 1

Let me start with the additional problems that arise:

• The trajectory of the suns is not stable. If they are slightly off-center, gravity will pull them towards the side of the cylinder that they're closer to, analogous to the ringworld stability issue. You could work around this by using stellar engines of some sort to keep the stars centered, or making the cylinder a tiny bit flexible and using motors to change its shape dynamically....
• There is nowhere for the heat that is generated by fusion in the cores of the suns to escape, apart from conduction through the crust towards outside space. You could work around this by putting large holes in your cylinder through which outside space is visible, by making your crust very thin (on the order of meters), or by making it very conductive (by adding an active cooling system that pumps heat outside). This oversimplified illustration shows the relevant mechanisms that keep earth's surface at its equilibrium temperature, and how the inside of your cylinder would heat to over 2 million Kelvins without any countermeasures:

Now, to your actual question.

The only relevant parameter is the distance between the suns, in AU. The speed at which they move follows automatically from your requirement that one sun should pass every 24 hours. It will be rather high, though :)

You will, of course, always see an infinite number of suns, but most of them will be very dim and very close to the horizon. Here's what the sky will look like, with the apparent brightness of the suns (= the area they occupy in the sky) written next to the dots.

suns spaced at 1AU: suns spaced at 20AU:

To calculate the total illumination, some math is required. You need to calculate the infinite sum of the contributions of each sun. In this formula, d is the distance between the suns in AU, and o is the offset from mid-day, where o=0 means mid-day, and o=1 means mid-day tomorrow.

This gives the following equation for the momentary strength of illumination (assuming that the power output of one sun at 1AU distance is 1):

-(π sinh((2 π)/d))/(d (cos(2 o π) - cosh((2 π)/d)))

To find your preferred value of d, just plot this formula for various values.

Here's a quick python snippet that does exactly that, since I couldn't get nice plots out of Wolfram Alpha:

#!/usr/bin/env python3
from argparse import ArgumentParser
from math import sqrt, sinh, cos, cosh, pi
import numpy
from matplotlib import pyplot as plt

cli = ArgumentParser()
cli.add_argument('--distance', type=float, default=1)
cli.add_argument('--average-illumination', type=float, default=0.25)
args = cli.parse_args()
power = 0.31831 * args.average_illumination * args.distance

hours = numpy.arange(0, 24, 1/60)
illuminations = []
for hour in hours:
    offset = hour / 24 - 0.5
    illuminations.append(
        -power * pi * sinh((2 * pi)/args.distance) / 
        (args.distance * (cos(2 * offset * pi) - cosh((2 * pi)/args.distance)))
    )

fig = plt.figure()
ax = fig.add_subplot(1, 1, 1)
ax.set_xticks(range(25))
ax.set_xlim(0, 24)
ax.set_yscale('log')
ax.grid()
ax.plot(hours, illuminations)
ax.set_title(f'spacing: {args.distance} AU, '
             f'luminosity: {power} L0, '
             f'min: {min(illuminations):.5g}, '
             f'max: {max(illuminations):.5g}')
# from https://en.wikipedia.org/wiki/Lux#Illuminance
ax.annotate("moonless clear sky with airglow", (0.5, 0.002/100e3))
ax.annotate("full moonlight", (0.5, 0.3/100e3))
ax.annotate("dark limit of civil twilight", (0.5, 3.4/100e3))
ax.annotate("family living room lighting", (0.5, 50/100e3))
ax.annotate("very dark overcast day", (0.5, 100/100e3))
ax.annotate("sunrise or sunset on clear day", (0.5, 500/100e3))
ax.annotate("overcast day", (0.5, 1000/100e3))
ax.annotate("indirect daylight", (0.5, 10000/100e3))
ax.annotate("full daylight", (0.5, 1))
# from https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5718773/
ax.annotate("survivable for minutes in firefighter's clothing", (0.5, 2))
ax.annotate("survivable in aluminized clothing", (0.5, 4))

plt.show()

And plots for some distances:

Distances above 180AU are impossible because then the suns would be moving faster than the speed of light; decreasing the cylinder diameter would solve this.

In these cases, I try to maintain the same average heat flux that is experienced on earth, to allow meaningful photosynthesis. You can see that if you want proper darkness at night, there will be short hard bursts of heat which will only be survivable in underground bunkers.

If you're willing to reduce the average heat flux to say 1% of that experienced on earth, that is, around 3 W/m², you can achieve this:

With only 1% of the power flux, you will only have 1% of the photosynthesis, solar power, wind power, fossil fuel formation etc, so your land will generally only support 1% of earth's population density. Advanced civilizations may however harvest tidal power from the tidal accelerations of the passing stars, and "reverse geothermal" power from the heat flux through the crust. This heat flux will be far stronger than on earth.

Other interesting effects which I haven't considered:

• the light of very-far-away suns will travel a long path through the atmosphere; this means that their light will be scattered and they won't be actually visible properly. It's just like the sun gets distorted and reddish during sunset, only the effect will be literally infinitely stronger.
• there will be effects from special relativity: the light of approaching stars is blue-shifted, and their power output will appear different since time passes at a different rate in the star's cores.
• since the light of oncoming stars will be blue-shifted and the light of receding stars will be red-shifted, there will be a constant radiation pressure in the direction in which the stars are moving. this will accelerate the atmosphere, causing westward wind. I'm not sure how to calculate the strength, though. Solar wind particles will have a similar effect.

There's another great way in which you could achieve day and night, though: Your population could live in a narrow valley such that only suns that are above 30-or-so degrees over the horizon are actually visible. There will still be atmospheric scattering, but some tinkering with the atmospheric composition could fix that.

Answer 2

First pass: the geometry is simple

In the first pass, we disregard the cylindrical shape of the world, and we assume that the light sources are in free space, moving against a black backdrop.

Let's assume that:

1. Each of the light sources moving through the tube produces the same amount of light as our own Sun; and

2. At midnight we want to have the same illumination as that produced by a full Moon.

Good to know:

• The illumination produced by a full Moon (around 0.1 to 0.3 lux) is between 400,000 and 1,000,000 times weaker than the illumination produced by the Sun at noon (around 100,000 lux). (That's 19 to 20 exposure steps, in photographic terms.)

• The illumination produced by a light source is inversely proportional to the square of the distance between the light source and the illuminated object.

With these assumptions, it follows that:

• For the illumination produced by one of those moving light sources to decrease 800,000 to 2,000,000 times (the doubling is because we are illuminated by the next moving light source) it must move to a distance of 900 to 1400 astronomical units (= the radius of the orbit of the Earth, i.e., the radius of the cylinder assumed by the question).

• The distance between two consecutive light sources will then be 1,800 to 2,800 astronomical units.

                                   2000 a.u.
       <−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−>
                                                    1000 a.u.
                                      <−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−>
     \ | /  Light source                                           \ | /     
··· --(•)-- ····················································· --(•)-- ···
<<<  / | \  <<< Movement                   ^                       / | \
                                           |
                                           |
                                           | 1 a.u.
                                      ○    |
                            Observer /|\   |
Ground                               / \   v
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
////////////////////////////////////////////////////////////////////////////

What about the speed of those light sources?

Hmm, that's a bummer. Light travels a distance of one astronomical unit in 8 minutes 20 seconds, which means that in one hour light travels a distance of 7.2 astronomical units, and in 12 hours it travels 86.4 astronomical units. Since the moving light sources need to travel about 1,000 astronomical units in 12 hours, it follows that they must move about 11.6 times faster than light.

Clearly, Einsteinian relativity doesn't apply in this world.

What the observer sees

At noon, the observer sees the light source overhead bathing the landscape in a sea of light, very very similar to what we see at noon.

At midnight, the observer sees a dark sky, with two very luminous stars at opposing points near the horizon.

Unlike on Earth, where the difference between daytime and nighttime is clear as day and night, on this world illumination varies gradually from full day to full night, with no clear separation between them. Most of the time it will be quite dark:

• Illumination on a heavily clouded day is about 5 lux, or about 20,000 times lower than the illumination at noon on a clear day. Taking this as the threshold between day and twilight, the light source would have to be about 140 astronomical units distant, or one sixth of the 1,000 astronomical units which we considered midnight.

• Taking the threshold between twilight and night to be 1 lux, that corresponds to a distance of some 320 astronomical units between the observer and the light source, or about 1/3 of the 1,000 astronomical units which we considered midnight.

• All in all, in each 24 hours cycle, the observer will see about 4 hours daytime, about 16 hours night time, about 2 hours dawn and 2 hours twilight.

Second pass: but, but, but, but...

In the first pass we disregarded the cylindrical shape of the world, and we assumed that the light sources move against a black backdrop.

Now, that is perfectly fine as regards visible light. Assuming that the world has about the same albedo as Earth, the cylindrical shape of the world won't make a great difference. Yes, at noon there would be just a little more light than what the calculations in the first pass would suggest, etc. But the difference is utterly negligible, even for the keenest photographer.

The problem is not visible light, the problem is infrared light.

Earth likes very much to remain at constant temperature; see the great worldwide wailing at the prospect of increasing the average temperature by a measly one degree centigrade over a century.

Earth does this by radiating back into space all the energy it receives from the Sun. While the energy Earth receives from the Sun is mostly in the visible spectrum, the energy radiated by Earth is mostly in the infrared range.

And here comes the catch: those infinitely many sources of light will make the inner surface of the cylinder as hot as the Sun in a very short time. (Short time, geologically speaking, of course.)

Let's see what happens with a random square meter of ground in this cylindrical world:

1. During daytime, that square meter of ground is warmed up by the visible light falling on it.

2. At night, on our spherical Earth, that square meter of ground emits the heat in the form of infrared light. Most of the infrared energy is lost into outer space; some of it warms the air a little, and is then re-emitted by the air in the form of far infrared. Eventually, all the thermal energy dissipated by the square meter of ground as infrared radiation is lost into outer space.

3. But on this cylindrical world there is no outer space. All the energy ever received by that square meter of ground remains in the system forever. At night, the square meter of ground emits infrared light, but it does not help, because it absorbs the same amount of infrared light emitted by all those other square meters of land elsewhere on the inner surface of the cylinder.

4. Every 24 hours more and more energy is added to the square meter of ground, and it has nowhere to go except to warm up other square meters of ground. In a short time, every square meter of ground on the inner surface of the cylinder will be in thermal equilibrium with the sources of energy.

Answer 3

If the tube rotated to produce "gravity", that gravity would pull down to the outside of the tube. So people would look up at the suns passing by.

If the suns have the mass of Sol, the Sun, they will eventually swell into red giants and greatly increase their luminosity after about ten billion years. That will cook the inside of the cylinder and maybe evaporate it into gas escaping into space. Then the red giant stars would turn in to white dwarfs after shedding significant amounts of mass. The stellar mass loss would produced strong solar winds which might push apart the cylinder, destroying it and would certainly devastate the already devastated surface.

One way to avoid it would be to make the cylinder much narrower and make the suns correspondingly dimmer than the Sun to adjust for the closer distance to the surface of the cylinder. Those dimmer stars will have lower mass than the Sun and will have steady luminosity for a much longer length of time, hundreds of billions or maybe trillions of years depending on their mass.

Or the suns could be stars which were already white dwarfs and which would be very, very gradually dimming down to black dwarfs. That would also take a very long time, perhaps trillions of years.

Or maybe you could make the suns giant lamps moving down the cylinder. They would have giant fusion generators to generate the power for the giant lamps used to illuminate the inside surface of the cylinder.

Of course maybe you don't care abut whether your setting will last for one billion years, ten billion years, a hundred billion years, or a trillion years.

Have you thought about what material your world would be made out of? You might need some hyopthetical fictional super strong materials.

Have you read Larry Niven's article "Bigger than Worlds"?

https://en.wikipedia.org/wiki/Bigger_Than_Worlds1

http://www.isfdb.org/cgi-bin/title.cgi?1333022

Answer 4

I can talk about the math in the abstract but since we have no figures here, it will be rather abstract.

The two factors will be: 1. Absolute magnitude of the suns 2. Distance from the sun to the surface when the sun is directly overhead.

The suns' light falls off by the square of the distance. This will determine its relative magnitude, which will increase until it passes over ahead, and decrease as it goes away. This will have to factor in the overhead distance. If the sun passes 3 (units) overhead, when it reaches 4 (units) farther away from that point, it will be only 5 (units) away from the person on the ground. Hence, if we measure give the brilliance of the sun at directly overhead a measurement of 100, it will be 36 when 5 units away -- 5 divided by 3, result squared, used to divide 100.

If it were alone, the new sun will first appear in the sky when the relative magnitude rises to a high enough level that a human eye can see it against the ambient light. The human eye is capable of seeing quite dim objects, so the actual factor is more likely to be that the prior sun is still putting out enough light to drown it out. (The luminosity difference between the Sun and the dimmest stars visible to the naked eye on a clear night past twilight and with no ambient light sources, either the moon or artificial, is about 10 to the 14th power.)

There would be no "night" vs. "day" You would have the sun overhead in its full brilliance, and then it would move off, slowly dimming, until it was dim enough that the new sun could be seen, and then it would continue to dim as the new sun brightened. The peak darkness would be the point at which the two suns were equal in brilliance. Then one would brighten as the other faded.

Distance from the surface will be important because that will decrease relative magnitude in a way that is not entirely dependent on the motion. There could be some very dark periods but the variation in light and dark would be continuous.