4D Planet: Angle of Sun

Ideas about how a world with more than three spatial dimensions would work - what laws of physics would be needed, how things would be built, how people would do things and so on.

4D Planet: Angle of Sun

Postby PatrickPowers » Fri Nov 16, 2018 9:55 am

Suppose I have a 4D planet, how does the height of the Sun over the horizon change during the day, particularly if the two rotational periods are not the same? I know how to do the calculation doing vectors but it is rather brute force and I don't have mathematical software.

I did this a year ago and got a simple answer but don't remember how, so maybe my simple answer was wrong. I keep thinking this is simple and obvious and I'm just not getting it, but maybe not.
PatrickPowers
Tetronian
 
Posts: 440
Joined: Wed Dec 02, 2015 1:36 am

Re: 4D Planet: Angle of Sun

Postby wendy » Sat Nov 17, 2018 9:21 am

Let's suppose that the sun moves around the earth, that is, on a great circle in the sky, once a year. It doesn't matter here because things are relative to the surface. We should note that this is the solar circle, and represent it as C.

The modes of rotation in 4d, is represented as a point in 6d space, where three of the 6d space forms the L axis (left-clifford) and the other three form the R axis. Any rotation, even at different speeds, can be formed by a simple sum of an L and R axis, where the two circular rotations are L+R and L-R resp.

When R and L are the same size, then either L+R or L-R is zero, specifically, the rotation is static on one of the axies. This is wheel or great-arrow rotation. The earth in 3d is a form of this.

Case of clifford-rotations

In a clifford rotation, we suppose R=0, and that C is not in L. At any given time, there is a great circle on the earth, represented by l,r which represents the path of the zodiac. Since l=r in size, we have a great circle that crosses the clifford parallels at a fixed angle. The sun moves around this zodiac circle once a year, and the zodiac circle is moved along the day-circle once a day. In other words, it's a diagonal on the torus sweeping across the surface once a day.

In terms of the what one might see from the ground. The sky is a hemi-globe, with an east and west pole. Everything rises on the east side, and sets on the west side. Because clifford-rotation is around the centre, things travel in 'straight lines' ie great circles, in the sky. Something that rises at x° E, (where the E/W divide is 0°), will rise to its highest point at x° above the horizon when at 0°, and set at x° W. In the course of being visible, it will travel through 180 degrees of the EW equator.

There is a thing called the 'lattitude sphere' or 'climata sphere'. This is the same for all locations, except in the way it stands on the ground. The mathematics comes from complex numbers, runs like this. The equation of a line is y = ax+b, where all of these are taken as complex numbers. We suppose we want the origin so b=0. We now introduce a term w = cis ( day * t), as a rotation of the argand diagram once a day. Then we see wy = awx, and thus the w's cancel out. Every point on the slope plane represents a great arrow on the clifford sphere.

We draw above the plane a second sphere, at (0,0,1), and draw lines from (Re, Im, 0) to (0,0,1) to intersect at this sphere. The plane is transformed into a sphere, where every pair of opposite points is a perpendicular rotation.

The zodiac or sun-track is represented by a circle, a line of lattitude, at twice the desired 3d tilt, eg 47° from the south.

This sphere also carries the tracks of the stars, which don't essentially move, and the sun, which moves on the track once a year. You place this sphere so that you are standing on the Nadir (where the sphere touches the ground), and the Zenith is directly overhead. This sphere is placed in the 3-space that divides the rising and setting parts of the sky. This sphere does not rotate, but is a kind of guide to the fixed stars. A star Sirius will always come to its cumulation at the point shown on this climata sphere.

A ray is drawn from the Nadir, through the lattitude-sphere to some point in the sky. This represents the cumulation of the star etc. The star rises 90 degrees before, and sets 90 degrees after this point, all in the same 2-plane through the nadir.

The lattitude sphere as 3d, can be seen as a globe. The 90 S represents 'equatorial', the 43 S is the tropics, the 43 N is the artic circle and the 90 N is the polar regions. This represents the climate different lattitudes have. If your nadir point is resting at 10 N, then that's 80 degrees from the N pole, which gives 40 degrees on the earth, or a lattitude of 50 Earthwise North, say London.

The sun travels around the line of the tropics (actually a torus in 4d), once a year. Any point on the 4-earth that falls on the tropics, the sun will be overhead once a year.

The cumulation of the sun will be found by drawing a line from the nadir, through the zodiac on the climata sphere, to the sky, with the usual rising and setting rules (90 before to 90 after). During the course of a year, the sun will track the image of the tropics on the lattitude-sphere on the sky. It reaches its highest mid summer and its lowest mid winter, the difference here is 94 degrees of the small sphere, or about 47 degrees of the sky,.
The dream you dream alone is only a dream
the dream we dream together is reality.

\ ( \(\LaTeX\ \) \ ) [no spaces] at https://greasyfork.org/en/users/188714-wendy-krieger
User avatar
wendy
Pentonian
 
Posts: 2014
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

Re: 4D Planet: Angle of Sun

Postby PatrickPowers » Sat Nov 17, 2018 1:21 pm

Good Lord! I feel like the Sorcerer's Apprentice. It's going to take a while to absorb this useful info.
PatrickPowers
Tetronian
 
Posts: 440
Joined: Wed Dec 02, 2015 1:36 am

Re: 4D Planet: Angle of Sun

Postby wendy » Mon Nov 19, 2018 10:21 am

Suppose you are lieing on your back, watching the night sky in fast motion.

In 3d, some stars will rise on the east, and make only a small arc before they set. If you are in the northern hemisphere, these will be around the south part of the horizon. Some stars in the north will never set. Their full path in the night sky remains visible at all times. Stars too far south are never seen in the north.

In 4d, all stars rise and set. You see exactly half the trace in the sky, of every star. Everything still rises on the east half of the sky, and sets in the west, but there is now a second motion. The sky turns around the east-west axis as well. This is not north/south, but two new directions, say 'forward' and 'reverse'. In the course of a star being visible, it will move 180 degrees forward, The sky is acting like a giant screw.

The stars cumulate, or come to their highest point, at 90 degrees forward of their rising. This always happens on the circle dividing East (or rising) and West (setting). If you imagine a globe, you could have it spinning once a day, and all stars rise in the north (at various lattitudes), and set in the south (at the same lattitude S). Because it takes 12 hours from rise to set, a star that rose at 150 deg E, will travel to 30 W (180 degrees), and set there. This is how the horizon looks in 4d. Stars that rise at the E pole, go straight overhead and set at the W pole. They still spiral, but it's very tiny. Likewise stars that 'rise' on the equator, always set on the equator, and cling to the horizon.

So stars always set on the diametric opposite point of the horizon.

The Gimbel

The curved bracket that holds a globe in 3d, is called the gimbel. In essence, the world spins and the points of the gimbel trace the lines of lattitude. The lines of longitude are broken into the parts of the day. In 3d, the gimbel runs from 90 S to 90 N. The 4d gimble is an entirely different thing, but still each point represents all of the places that have the same zenith-star.

Before the southern hemisphere was known, you could think of the year as a circle, and the north-pointer pointing to the season concerned. So December is in winter, and June is in summer. With the southern hemisphere, you have a second pointer on the opposite side, so December is summer and June is winter. If you replace this pointer with a full disk, you get what happens in 4D. You might have December at the beginning of spring, or late autumn. You would have season-zones like 3d time zones. (The time zones still exist, but we have rendered the whole circle to a point).

North and South still define the climate, but we move south to the equator, so on the circle above, S is the equator, N is the north pole, and the lattitude from 0 to 90 defines the climate. This becomes the climata direction. The more north you get, the colder it gets, sort of.

So our disk with the seasons, we can take one hemisphere, and shrink the equator to a point. This is the 'lattitude sphere' in 4d. The actual derivation is to suppose you are standing at a point (the nadir), and the lattitude sphere stretches from the nadir to the zenith. You draw a line from the nadir to some point in the sky, and this maps onto the lattitude sphere. Points opposite represent orthogonal great circles.

This sphere is the same for everyone. The main difference is which point is touching the ground. It's the same NS sphere as above, but your zenith point is a certian point on this sphere, and the forward/backwards match the cumulation of the stars, is the 2-plane between the zenith and nadir. This sphere stands so that the E-W axis is perpendicular.

Here comes the Sun

The sun's motion across the sky is the elliptic, or zodiac. In 3d and 4d, we take these as circles as the first approximation. It will cross some of the EW circles at some angle, as it is not parallel with the EW circles. There are two circles for which it is equidistant, one is the South-Polar, and one is the North-Polar. The trace of circles that allow the sun to be overhead, is the tropic torus. This is the margin on a duocylinder, or all points some specific distance from the south-polar. The path of the sun is on this torus, is to travel around both the long and short axis once a day, and slowly move across the torus. If you were to unfold the surface to a rectangle, the sun-path would run as a diagonal of the rectangle, and slowly edge to cover it.

The diagonals of all toruses of a sphere are in fact great circles, and as such are the same length.

In terms of our 4d gimbel, we have the N pole, as cold. Some distance down is the artic circle (where the sun might hover on the horizon for a day), the tropics (where the sun comes overhead), and the S pole (where the sun always gets to some same 'highest' point every day of the year).

To locate the sun, we should take this gimble, and set it up in the manner described (between the E and W hemiglobes), and note that the sun is a star that moves around the tropic once a year. It travels on a 94-degree circle on this, and sometimes it rises high in the sky, and other times it doesn't. The inclination to the ground changes through the year, rather as our sun does. It is this, rather than the shortening of the days etc, that make the summer and winter happen.
The dream you dream alone is only a dream
the dream we dream together is reality.

\ ( \(\LaTeX\ \) \ ) [no spaces] at https://greasyfork.org/en/users/188714-wendy-krieger
User avatar
wendy
Pentonian
 
Posts: 2014
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

Re: 4D Planet: Angle of Sun

Postby PatrickPowers » Sat Dec 08, 2018 5:27 am

"The modes of rotation in 4d, is represented as a point in 6d space, where three of the 6d space forms the L axis (left-clifford) and the other three form the R axis. Any rotation, even at different speeds, can be formed by a simple sum of an L and R axis, where the two circular rotations are L+R and L-R resp."

This I do not understand. It seems it could be quite useful. Care to offer a reference? I'm thinking there must be some buzzword that I can search on.
PatrickPowers
Tetronian
 
Posts: 440
Joined: Wed Dec 02, 2015 1:36 am

Re: 4D Planet: Angle of Sun

Postby wendy » Sat Dec 08, 2018 6:55 am

For the most part, I could not find any explicit reference to the rotation space of 4D or higher. The usual pattern is to discuss 'unitary space'. So these things one solves. That is, I did not read about this from a book, but sort of filtered the information about what I have read about quarterion multiplication, hopf fibulation, and a few emails with John Conway on the matter.

Having had a few discussions with Conway, and figuring out how he folds space, i come to the conclusion that the rotation-phase-space is a six-dimension thing. It comprises of two distinct 3d axies, which represent quarterions. The six-space then consists of l.I.r, where l and r are quarterions, and I is the initial. Since i don't have any advanced maths education (outside applied physics), i should put what i believe to be the closest term in brackets.

The radial dimension is still an overall intensity, but the relative ratios of l to r represent the ratios of speeds of the two axies, so a fast spinning l axis and a slow spinning r axis is closer to the L,L,L axis, and a small component on the R,R,R set. When L=R in intensity, the figure given is a bi-glomohedric prism (ie cartesian product of two 3d sphere-surfaces), each point on this represents a distinct 'wheel' rotation (+ direction; ie a great arrow rather than a great circle), the opposite orthogonal is not rotating.

In the notion of equipartion of energy, the rotation-point will tend to fall in the LLL or RRR hedrices (3-fabrics).

The salient features of Conway's model project onto this model, but we also keep things distinct, rather than to merge the time-opposites together.
The dream you dream alone is only a dream
the dream we dream together is reality.

\ ( \(\LaTeX\ \) \ ) [no spaces] at https://greasyfork.org/en/users/188714-wendy-krieger
User avatar
wendy
Pentonian
 
Posts: 2014
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

Re: 4D Planet: Angle of Sun

Postby mr_e_man » Sat Dec 08, 2018 11:52 am

PatrickPowers wrote: "The modes of rotation in 4d, is represented as a point in 6d space, where three of the 6d space forms the L axis (left-clifford) and the other three form the R axis. Any rotation, even at different speeds, can be formed by a simple sum of an L and R axis, where the two circular rotations are L+R and L-R resp."

This I do not understand. It seems it could be quite useful. Care to offer a reference? I'm thinking there must be some buzzword that I can search on.


There's always Wikipedia: https://en.wikipedia.org/wiki/Rotations ... _SO%284%29 . (I don't understand half of it, but there is some useful information there.)

You seem to be familiar with geometric algebra, so I will discuss 4D rotations in terms of bivectors.

Any rotation can be done by the geometric sandwich product with a rotor (a product of an even number of unit vectors), and any rotor is the exponential of a bivector. The set of rotations is a "Lie group", and the set of bivectors is the corresponding "Lie algebra". (But conventionally, these things are described with matrices instead of bivectors.)

The set of all bivectors in 4D is 6-dimensional, being spanned by {e1e2, e1e3, e1e4, e2e3, e2e4, e3e4}. A rotation by angle θ in the x1x2-plane corresponds to the bivector θe1e2. An isoclinic ("Clifford") rotation would be something like θ(e1e2+e3e4), and a general double rotation θe1e2+ϕe3e4.

Instead of writing everything as a sum of these simple bivectors, we could use isoclinic bivectors as a basis:

L1 = e1e2 + e3e4, L2 = e1e3 - e2e4, L3 = e1e4 + e2e3,
R1 = e1e2 - e3e4, R2 = e1e3 + e2e4, R3 = e1e4 - e2e3

For example, the simple rotation becomes θe1e2 = (θ/2)(L1+R1).

I haven't studied this in detail, but there is a non-trivial correspondence between operations in the Lie group and the Lie algebra. The Lie group operation is just composition of transformations (like applying several rotations consecutively). The Lie algebra operations are addition and the "commutator" cross product. If two bivectors commute (meaning A×B=0), then the corresponding rotors exp(A) and exp(B) also commute, and exp(A)exp(B)=exp(A+B).

The cross product of two bivectors can be calculated as follows, using the standard basis:

(e1e2)×(e1e2) = 0
(e1e2)×(e1e3) = - e2e3
(e1e2)×(e2e3) = e1e3
(e1e2)×(e3e4) = 0

etc., and using the isoclinic basis:

L2×L1 = 2L3, R1×R2 = 2R3,
L3×L2 = 2L1, R2×R3 = 2R1,
L1×L3 = 2L2, R3×R1 = 2R2,
L1×R1 = 0, L1×R2 = 0, L1×R3 = 0

Note that the cross product of any two R's is another R, so the right-isoclinic rotations form a group by themselves. (I believe this is the "swirlprism" symmetry group, and the "special unitary group".) And if each R is divided by 2, then these equations are the same as those for bivectors in 3D, or for pure-imaginary quaternions, or the ordinary vector cross product:

i×j = k,
j×k = i,
k×i = j

This isomorphism between bivectors in 4D and 3D is, of course, related to the Hopf fibration between the spheres in 4D and 3D.
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
Tetronian
 
Posts: 474
Joined: Tue Sep 18, 2018 4:10 am

Re: 4D Planet: Angle of Sun

Postby PatrickPowers » Sat Dec 08, 2018 1:45 pm

I never had much of a grip on geometric algebra. Even that weak grasp was over a year ago.

I guess I'll just rent out Mathematica and do it with linear algebra, which I know already. It should be pretty easy, just the usual dreariness of learning a big software package. To really do this properly I'll have to integrate the energy delivered to a region by the sun.

I did my approximations of the height of the sun by adding sine waves together. That's easy and works pretty well, but I didn't know what to do about the ecliptic and seasons. That's beyond such simple estimates. So Mathematica here I come. I need to get it anyway to illustrate the book. It won't be until February though, I'm busy.
PatrickPowers
Tetronian
 
Posts: 440
Joined: Wed Dec 02, 2015 1:36 am

Re: 4D Planet: Angle of Sun

Postby PatrickPowers » Sun Dec 09, 2018 4:39 am

My problem is solved! I was barking up the wrong tree.

My goal was to examine climate. This means, how much sunlight is a spot absorbing in one day. So what I really wanted was the dot product of two position vectors, the sun and the location on the surface of the planet, with the origin being the center of the planet.

I went to the graphing calculator at https://www.meta-calculator.com/. Not programmable, but it is good enough. I was able to graph the sun energy over the period of a "day" taking the obliquity of the ecliptic into account. Then I can integrate the thing by eye. (I don't need a number, a qualitative difference will do.) So far I've only done one graph. Next I have to figure out what exactly it is that I want to know. With a little luck, tomorrow I'll be able to draw all the graphs I need to answer my climate questions.

One question is this. I want the planet to be like Earth. On Earth the obliquity of the ecliptic is about 22 degrees. To be geometrically proportional I'd like to make it 11 degrees on 4D Earth. But it isn't clear to me that on 4D there is any such thing as obliquity of the ecliptic, or at least it is different. The difference is that on 4D Earth the equator is a torus. It doesn't lie in any sort of plane. It doesn't make sense to measure the angle between the orbital plane of the sun and the equatorial torus. All I can do is choose on place on the torus then locate the sun 11 degrees away from that. After a half a year the sun will be above the antipode. But that's colinear with the origin so it doesn't help me. I have to choose where the sun will be after a quarter of a year. Then I will have a 2D plane for the sun. I don't know how this choice affects the climate. I guess I'll have to try out different values and see whether it makes any difference. If so, then pick whichever alternative appeals.
PatrickPowers
Tetronian
 
Posts: 440
Joined: Wed Dec 02, 2015 1:36 am

Re: 4D Planet: Angle of Sun

Postby PatrickPowers » Sun Dec 09, 2018 4:17 pm

mr_e_man wrote:
This isomorphism between bivectors in 4D and 3D is, of course, related to the Hopf fibration between the spheres in 4D and 3D.


Aha, that seems to be what Wendy was talking about. Seems well worth learning when I get the time.
PatrickPowers
Tetronian
 
Posts: 440
Joined: Wed Dec 02, 2015 1:36 am

Re: 4D Planet: Angle of Sun

Postby mr_e_man » Tue Dec 11, 2018 8:45 am

I just found that it's simpler than I expected! The Hopf fibration seems to be more naturally defined with bivectors.

(Moved to a new topic: viewtopic.php?f=3&t=2351 )
Last edited by mr_e_man on Thu Dec 13, 2018 7:45 pm, edited 1 time in total.
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
Tetronian
 
Posts: 474
Joined: Tue Sep 18, 2018 4:10 am

Re: 4D Planet: Angle of Sun

Postby PatrickPowers » Tue Dec 11, 2018 11:36 pm

Aha! Another thing to learn when I get the time.
PatrickPowers
Tetronian
 
Posts: 440
Joined: Wed Dec 02, 2015 1:36 am


Return to Higher Spatial Dimensions

Who is online

Users browsing this forum: No registered users and 12 guests

cron