On seasons and days on n-dimensional planetary systems

[quickfur]

The recent discussion about seasons and days on 4D planets made me realize something interesting about n-dimensional planets.

As we know, stable elliptical orbits exist only in 3D; in 4D and above, only perfectly circular orbits can exist (and if I'm not wrong, the sensitivity of the orbit increases sharply with dimension, so that in very high dimensions the slightest deviation spells almost instant doom for the planet). But for the sake of argument, let's assume that there's an n-dimensional planet in a perfectly circular orbit around a star. What kind of seasons and days would inhabitants of such a planet observe?

Supposing that there is some angular momentum (or whatever the n-dimensional equivalent is) left over from whatever formed the n-dimensional planet, it would have an initial rotation more-or-less parallel to its orbital plane. This initial rotation would redistribute itself over the planet to form clifford multiple rotations. But assuming that in the end, the planet is still approximately parallel to its orbital plane, we would have the following strange effect:

For clarity, let's say the orbital plane is specified by the first 2 coordinate axes. If the star is at the origin, then the position of the planet would be (R cos A, R sin A, 0, 0, 0, ...). The insight is that day/night cycles only happen in this plane (and in rotational planes close to this plane), corresponding to rotation of the first 2 coordinates. But when n is large, there are a lot of other coordinates in which multiple rotations may happen. Virtually all of them are almost orthogonal to the orbital plane.

Therefore, the region of the planet's surface that undergoes a day/night cycle is only a small region near the great circle that lies on the orbital plane. Most of the rest of the planet's surface is actually in an "arctic zone"; their rotational planes are orthogonal to the orbital plane so they remain perpetually at the boundary between the day and night parts of the planet.

In other words, as you go to higher dimensions, planets in an orbiting system would be mostly dark and cold except for a small band of the surface that lies in (or close to) the orbital plane. The higher the dimension, the colder the planet.

In summary, in 2D, circular planets have no arctic zone at all (the entire surface undergoes the same day/night cycle); in 3D, most of the surface undergoes a day/night cycle, with a small region around the poles that have arctic climate. In 4D, there is an entire toroidal ring around the planet that experiences arctic climate. In 5D, the arctic region grows into a 2-sphere while the region with day/night cycle remains a toroidal region around a single great circle. Now the percentage of the planet with arctic climate is larger than the percentage with a day/night cycle. The higher the dimension you go, the more of the planet's surface lies in that arctic zone between the day and night halves of the planet. Although they do get light from the star, it is very little compared to the "equatorial band" that gets a full day/night cycle. Thus, the majority of the planet's surface would be dark and cold.

[gonegahgah]

I wonder if there is any greater chance of a spiraling rotation through two axis - instead of one like in ours - in a 4D world. It is interesting that solar systems and galaxies tend to coalesce into 2 dimensional planes. Maybe the same thing would happen in a 4D world or maybe the galaxies and solar systems would form spheres instead. When you swing things around they tend to flatten so maybe it would be the same for both 3D and 4D. If you could get the world to have a spiral rotation through 2 axis it would certainly move the sun across more of the surface. There's also the question of the tilt of the axis. This would be a tilt in 3 axis instead of our 2. So the 'upper' two parts of the planet - in both directions perpendicular to the line of rotation would get the suns blast - being their summer.

Now that I think about it there is no reason why any of the two axis perpendicular to the line of rotation would be any further from the sun. The sunlight weakens at a cubed rate instead of a square rate but would still essentially hit both the perpendicular axis just as equally. We would have to come up with a new name. We would still have longitude which is along the line of rotation but we would have to call the others the lotitude and the lutitude. Neither could be considered to be our latitude as they would both equally have entitlement to the name.

But whether you travel across the planet across its lotitude or its lutitude both would be equally distant from the sun. So they would both still be able to enjoy the warmth of the sun. The planet may have to be a bit closer to the sun or the sun might have to be bigger to provide the equivalent amount of warmth that we get because it disperses at a cubed rate and weakens more than ours does with distance.

[quickfur]

The thing about orbits is that the shape of the orbit is an emergent phenomenon, not a fundamental phenomenon. At the fundamental level, you have two bodies, a large one (the star) and a small one (the planet), and the planet has a certain velocity/momentum w.r.t. the star. The star's gravity acts to pull the planet inwards to itself, but due to the planet's momentum, it manages to stay just ahead enough that it doesn't quite fall in. The path that it traces out as it does this happens to work out to be an ellipse (in 3D).

But the same kind of emergence doesn't happen for spiral orbits; it can only happen if somehow the star's gravity changes direction as the planet moves, or the planet itself has a changing momentum due to some unspecified external force. If left on its own, the planet would simply continue its present line of motion according to the law of inertia, and as long as the star's gravitational field is even, it will only result in orbits that lie within a 2D plane (the star and the planet plus the planet's inertial motion has only enough parameters to cover a 2D plane).

The only way I can see a spiral orbit happening is if the star has a significant velocity comparable to the planet, in a linearly-independent direction. But I'm not sure if it's possible to maintain this kind of configuration for very long. In any case, it's very unlikely that the resulting orbital path will work out to coincide with the mutually-orthogonal rings that we see in a duocylinder; that kinda thing is more the exception rather than the norm. (And in 4D, where even a simple orbital motion is unstable and sensitive to the slightest perturbation, I don't know if this kind of system will even yield a meaningful "orbit".)

[gonegahgah]

Oops. Sorry, not talking about the orbit; just the planet's rotation. Funnily enough if you were to travel in a space ship perpendicular to solar plane and were to chart the movement of the planets they would map out a spiral around a central line of space relative from yourself; but of course they remain in a circular orbit around the star.

I guess it would be difficult for our 3D planets to tumble end over end and spin at the same time. There are probably momentum issues at work against this. Simple rotational spin is probably the most stable as far as momentum is concerned. Perhaps the same would exist for a 4D planet. But still, the sun should shine on a 4D planet across the exposed hyperface in much the same way as it shines on one face of the Earth. The temperature variation between the hyperequator and the hyperpoles may be a bit more extreme but it shouldn't mean that some parts miss out on the sun.

This highlights one of the differences between our perspective and that of a 4Der. We would think of each pole as describing a square but a 4Der would still see it for what it is; a line still.

[wendy]

The model of bohr's atom solved the unstable orbit in the hydrogen atom, so i can't see why some quantum element could stablise the orbit of a planet. It has been suggested even for even three dimensions. Circular orbits or elliptic orbits do not make seasons: in our world, the shortest month is february, when the orbit theory says ought be summer. Well, yes, it _is_ summer here, but i am aware that some places it even snows in that month (gasp).

Regarding seasons etc.

In even dimensions, the length of the day is everywhere 24 hours, since the laws of equal energies will favour a clifford rotation in every case.

Seasons do come from the tilt of the ground to the sun. It's pretty cluey, but i will try to explain it in all its glorious detail. None the same, it should not be too hard, just not intuitive.

We start with the rising of the stars. Under clifford rotation, every point of a planet goes in a circle around the centre: there is no linear axis like in odd dimensions. For the most part, we shall assume that stars rise in the same point in the sky: that is the stars are so far away that they don't apparently move. The circle that a point goes around the planet-centre extends out to space, and may contain various stars, like stellus.

We look now at the horizon. This is the surface of a 3-sphere (like ours is a surface of a circle). There is a 'zenith-track', which goes from hard east to hard west. This is the track the zenith-star will follow here. You then have a line between hard east and hard west, which is the 'skirting track'. Stars on this track never rise or set, just crawl on the horizon. The rest of the circles rise and set, the height above ground is the same as the distance from the skirting track the rise-point is.

We now draw a 3-sphere, whose diameter is at the observer and at the the hard-east. It is drawn flat to the ground. A ray from a given star's rising point S to the observer O crosses this sphere at a point, say S'. One populates this sphere with different stars S', T', U', ...

This sphere is the same for all places, except the zenith and skirting points are different antipodes. Each of these points correspond to a 'day-circle', so there's a kind of projection that gives the zenith-point sphere × time-zones circle.

The sun, for not following one of the great circles of the earth (it is tilted, so to speak), runs anti-parallel to one that does. On the risings-sphere, the sun moves along a line of lattitude (eg 43s, where 90-23.5*2 = 43). This means that at any point, the sun describes a circle in the sky (of rising points, and of cumulations), that's 47 degrees in diameter.

If you take a sphere and draw in the 43s parallel, and mark it Jan-dec (like a clock), then you can put the sphere anywhere on the table. The circle drawn will have a top point (summer) and a bottom point (winter), which can fall in 'any' month. It is this change of angle, and a slight latency of inertia, that makes the seasons. If this were 43s, then you have the south pole as the 'solar circle', the n pole as the 'polar circle', and the various longitudes becoming the season-zones (the 3d case would only be a two-ended spinner on this, not the full circle).

We now take this sphere, and mark it from 0d (south pole) to 90 degrees (n), where the circle the sun rises on is the tropics (23.5 degrees). You then have pretty much like what you get at various lattitudes on earth. You have a 'equatorial' point, with no real seasons. It has less than even our equatorial regions, since here the sun does retreat and come to the equator, in 4d, it is always 23.5 degrees away. Seasons become apparent at the tropics, and most of the planet is in temperate zones (23.5d to 66.5d), with a polar type climate north of 66.5 degrees.

The areas between these are different. On the 3d earth, we have 0.398 tropics, .519 temperate and .083 polar. In 4d, the figures run .269 tropics, .462 temperate and .269 polar.

[Mrrl]

In summary, in 2D, circular planets have no arctic zone at all (the entire surface undergoes the same day/night cycle); in 3D, most of the surface undergoes a day/night cycle, with a small region around the poles that have arctic climate. In 4D, there is an entire toroidal ring around the planet that experiences arctic climate. In 5D, the arctic region grows into a 2-sphere while the region with day/night cycle remains a toroidal region around a single great circle. Now the percentage of the planet with arctic climate is larger than the percentage with a day/night cycle. The higher the dimension you go, the more of the planet's surface lies in that arctic zone between the day and night halves of the planet. Although they do get light from the star, it is very little compared to the "equatorial band" that gets a full day/night cycle. Thus, the majority of the planet's surface would be dark and cold.

If we consider straight-rotating planet (where one one rotation planes is an orbital plane and others are perpendicular to it), and look at area where the sun may rise higher than some angle alpha over the horizon, then this area covers cos(alpha)^N-2 part of the planet. In 2D case it's 1, in 3D - exactly cos(alpha) and so on. If we take alpha=30 deg (avegrage noon height of sun in St.Peterburg) as arctic zone limit, then in 4D we have 1/4 of planet in "arctic" zone, in 6D - 7/16, and only in 7D this area will be more than 50% of the planet (in 10D - about 2/3). So arctic area rises with more dimensions, but not so fast

[Mrrl]

The sun, for not following one of the great circles of the earth (it is tilted, so to speak), runs anti-parallel to one that does. On the risings-sphere, the sun moves along a line of lattitude (eg 43s, where 90-23.5*2 = 43). This means that at any point, the sun describes a circle in the sky (of rising points, and of cumulations), that's 47 degrees in diameter.

If you take a sphere and draw in the 43s parallel, and mark it Jan-dec (like a clock), then you can put the sphere anywhere on the table. The circle drawn will have a top point (summer) and a bottom point (winter), which can fall in 'any' month. It is this change of angle, and a slight latency of inertia, that makes the seasons. If this were 43s, then you have the south pole as the 'solar circle', the n pole as the 'polar circle', and the various longitudes becoming the season-zones (the 3d case would only be a two-ended spinner on this, not the full circle).

We now take this sphere, and mark it from 0d (south pole) to 90 degrees (n), where the circle the sun rises on is the tropics (23.5 degrees). You then have pretty much like what you get at various lattitudes on earth. You have a 'equatorial' point, with no real seasons. It has less than even our equatorial regions, since here the sun does retreat and come to the equator, in 4d, it is always 23.5 degrees away. Seasons become apparent at the tropics, and most of the planet is in temperate zones (23.5d to 66.5d), with a polar type climate north of 66.5 degrees.

The areas between these are different. On the 3d earth, we have 0.398 tropics, .519 temperate and .083 polar. In 4d, the figures run .269 tropics, .462 temperate and .269 polar.

I think that there may be different kinds of orbit inclination in 4D. Let the planet rotates in planes W-X and Y-Z. Orbit of the sun may be descired by two orthogonal vectors - directions to the sum in two moments with the same stellar time, but with 1/4 year interval. For one planet these vectors may be R*(1,0,0,0) in "spring" and R*(0,cos(a),0,sin(a)) in "summer", for another - R*(cos(a),0,sin(a),0) and R*(0,cos(a),0,sin(a)), with all possible configurations between. We can say that there are two inclination angles a>b, and sun goes through points R*(cos(a),0,sin(a),0) and R*(0,cos(b),0,sin(b)) (all coordinates are in (w,x,y,z) format). For such orbit zone where sun may be in zenith lays between "ring latitudes" b and a (i.e. for point (w,x,y,z) on the planet surface we have cos(a)<=sqrt(x^2+y^2)/r<=cos(b)). The problem is that there is only 2D area where sun may be in zenith (product of projection on sun orbit and rotational traces (cos(t+phi1),sin(t+phi1),cos(t+phi2),sin(t+phi2)). So only this area is "equatorial" on our planet, and when we go from it, the highest point of the sun will be less than 90 deg.

[quickfur]

[...] If we take alpha=30 deg (avegrage noon height of sun in St.Peterburg) as arctic zone limit, then in 4D we have 1/4 of planet in "arctic" zone, in 6D - 7/16, and only in 7D this area will be more than 50% of the planet (in 10D - about 2/3). So arctic area rises with more dimensions, but not so fast

OK, you're right, it doesn't rise that fast. But it does consistently rise. So if you get to high enough dimensions, most of the planet would be in the arctic zone. So the higher the dimension, the colder and darker it gets. At least on planets in orbital systems. But there might be other kinds of systems that we don't know about, where this is not true.

[quickfur]

The model of bohr's atom solved the unstable orbit in the hydrogen atom, so i can't see why some quantum element could stablise the orbit of a planet. It has been suggested even for even three dimensions. [...]

Yes Bohr solved the unstable orbit... in theory. But today we know that it's caused by the electron having wave/particle duality. For a planet to exhibit such effects, you'd have to magnify quantum effects to the macroscopic level. Which would cause really weird things to happen. Of course, this assumes that we're using a physics system similar (analogous) to our own.

If we consider more radical departures from "conventional" physics, though, there are many more possibilities. For example, i considered a radically different system of physics in this post where orbits are stabilized by a "symmetry force". There are, of course, many other possibilities. The problem then is, it becomes pretty much arbitrary, and there's no good reason why one system should be chosen over another, and exploring these kinds of systems may not necessarily yield deeper insight into 4D space itself.

[Mrrl]

We can say that there are two inclination angles a>b, and sun goes through points R*(cos(a),0,sin(a),0) and R*(0,cos(b),0,sin(b)) (all coordinates are in (w,x,y,z) format). For such orbit zone where sun may be in zenith lays between "ring latitudes" b and a (i.e. for point (w,x,y,z) on the planet surface we have cos(a)<=sqrt(x^2+y^2)/r<=cos(b)). The problem is that there is only 2D area where sun may be in zenith (product of projection on sun orbit and rotational traces (p*cos(t+phi1),p*sin(t+phi1),q*cos(t+phi2),q*sin(t+phi2)). So only this area is "equatorial" on our planet, and when we go from it, the highest point of the sun will be less than 90 deg.

Actually, a and b may have different signs. If a=b then sun always goes over the single rotational track (cos(a)*cos(t),cos(a)*sin(t),sin(a)*cos(t),sin(a)*sin(t)), and if a=-b then it goes across tracks with w^2+x^2=cos(a)^2, y^2+z^2=sin(a)^2. In the first case we have no seasons at all, and in the second there is slow drifting of "summer" by phase shift zones.

[wendy]

Any great arrow (ie a great circle with an arrow-head), that is not L-clifford-parallel with a given arrow, or the reverse of that arrow, is R-clifford-parallel with exactly one great-circle that is L-parallel to the first arrow.

Put simply, all great arrows correspond to a points on the surface of a bi-glomohedrix prism (the prism-product of two sphere-surfaces in 6D).

If X,Y represents a point on this surface, and -X the antipode of X, then X,-Y is the orthogonal rotating L-parallel, -X,Y is the orthogonal rotating right-parallel, and -X,-Y is the original rotating in reverse. Any random great circle W,Z, is left-parallel to X,Z and right-parallel to W,Y, these are right-parallel and left-parallel to X,Y.

One should note that where the sun-earth 2-flat, passes through W,Z, to a left-rotating world, the polar and solars are at X,Z and X,-Z, angle between X,Z and W,Z give the tropics. One should note that it is the polar and the progression through the tropics, that define the seasons and their zones.

In 4d, something like 29.1% of the world fall between the solar circle and the tropic torus, and 29.1% between the polar circle and the artic torus, giving equatorial and artic climates, while the remaining 42.8% fall in the temporate zone. There is just one of each zone, it is possible to follow the same time of day and season right around the world.

[Mrrl]

And what will this definition of tropic and arctic zones give for situation when the projection of sun is always in 23.5 degrees from one of rotational planes? Or when it migrates between 20 and 60 degrees?

[quickfur]

I considered the 4D case where the orbital plane is orthogonal to both clifford rotation planes. E.g., if the clifford rotation is in WX and YZ, then the orbital plane is in XY. In such a case, the climates will shift between tropical and arctic every 1/2 year. If you start in an artic zone, then after 1/4 year you're in the tropical zone (with full day/night cycle) then after another 1/4 year you're back in the arctic zone, etc..

This system gives the most even heat distribution across the planet, though I find it hard to justify how such a system came about, given the lack of corrspondence between the orbital plane and the rotational planes (or rather, the coincidental orthogonality of them).

[wendy]

I'm not sure if ye are following clifford rotations.

There are no special 'orthogonal' clifford parallels. You can't tell in a clifford rotation, anything other than the parities and intensity of rotation. If you set the line perpendicular to clifford-rotations, you get the same climate-zones as a sphere in 3d that is at an angle 45 deg tilt. The tropic torus and artic torus would coincide, but there would be a region where the sun is characteristicly higher and one where the sun is characteristicly lower. You still have the solar and polar circles.

Every point on a 4-planet rotates around the centre in 24 hours, so any point in the heavens is visible for 12 hours. This would suppose the daylight is 12 hours, regardless of the season.

The sun follows a great circle in the sky, but these do not have to rise high. Specificaly, if the sun rises x degrees from the east-point, then it will cumulate at 90-x degrees from the horizon. The number of degrees that the sun varies during a day depends on how the zodiac (the track of the sun in the heavens), projects from the rising-sphere (where it appears as a circle) onto the horizon-half-sphere. This is clearly not a fixed amount.

At the solar and polar circles, the zodiac is parallel to the horizon-middle, so it projects onto a line that is either 67degrees or 23 degrees from the horizon. This means that it will rise to 89 or 11 degrees every day of the year.

On the tropic torus, the zodiac-circle includes the zenith-point (ie the point opposite the zenith-point on the rising-sphere), so the sun will rise to variouly directly overhead down to something like 23 degrees off zenith.

Points in the temperate zone, the sun varies by 23 degrees, from L+11 to L-11, where L is the lattitude.

Points in the artic, reverse the situation of the polar, the sun rises between 0 to 23 degrees from the horizon.

It's the variation of this solination that gives rise to the seasons.

[Mrrl]

I see. So you always can find two orthogonal traces such that sun projection is always on the same distance from each of them ("a" degrees from one and 90-a from another?) Yes, it make the picture more understandable. So we select these traces as coordinate planes, sun is always over latitude "a", and if some point is at latitude "b", then maximal elevation of sun will vary between |a+b-90| and 90-|a-b|. And latitude 90-a is actually works as the polar torus, it is only area on the planet where sun may go along the horizon.
Something like that?

[wendy]

It is indeed correct.

The solar and polar represent great circles, that are 'anti-parallel' to the track of the sun.

Between these are toruses, or bi-cylinder limits, for which the solar and polar circles run up the centre. It's like candle, bent to a torus, with the solar as the wick.

The radius of the torus is the 'lattitude', or climate determining line. These are not spheres equidistant from a pole, like in 3d, but the thing is equidistant from a trace (like our lattitudes are equidistant from the trace of the equator).

Any torus, when unfolded, gives a rectangle, say r.cos(l) × r sin(l). The traces of the planet-rotation run from one diagonal to the other, and in 'broken diagonals', that is what you get when you slide a diagonal left on the torus. All traces are parallel.

On the tropic-torus, there is a different trace, that runs in the opposite diagonals. This is the zodiac, or trace of the zenith of the sun. You can mark this trace with the signs of the zodiac, when the sun is in Leo, then it rises above a point in some day-trace that crosses the year-trace in leo. These are the tropics of Leo. You get tropics for all of the signs, like cancer, virgo, etc. In atlases one sometimes sees a wavy line with signs of the zodiac on it, representing the lattitudes the sun is at in these seasons. The function is identical, the form is different.

Instead of artic circle, and antartic circle, you have the direct orthogonal of the zodiac-circle, where the sun can fall to the horizon, and crawl around this. You have, instead of artic and antartic, the various artics of Leo, cancer, virgo, etc. These are on orthogonals to the tropics of leo, cancer, virgo, the antipodes are on the same trace, but 12 hours away.

[Mrrl]

And what about timezones? At the every given moment we can select part of planet where the sun is in its highest point now. It has the shape of 2D hemisphere with the edge at the current polar circle (where sun slides along the horizon). These hemispheres divide planet to sectors, that have the same time of noon now. But after some time of year polar circle will shift (in zodiac direction?), so we'll have different timezone division. At the tropic torus it is not very significant, but at arctic torus this change of timezones may be polar nightmare...

[wendy]

The exact nature of the tropic torus is different to what is written. What i wrote is rather like the wavy line you see in atlases which show where the sun is overhead. In practice, there's a rather complex double-spiral that goes up and down the 3-sphere, where the lattitude is given by the wavy line, and the longitude by the time of day.

Well, in four dimensions, the points where the sun is zenith isn't a real straight line, but pretty much makes a fine grating over the tropic lattitude. But, just as in 3d, it serves just to show what lines it falls on, and use time-zones to place the second coordinate. On a torus, the time-zones are parallel to one set of edges, with (here), the same time running along the short edge.

To locate where the sun is zenith at any given moment, you need the GMT (which sets longitude), and the current tropic the sun is on (which is what the zodiac-line sets). The intersection of these gives the current point under the sun.

One should note the traces contain a full day, and that the time of day is set by the part of the trace nearest the sun (which would make the sun highest in the sky).