4D objects can rotate in two independent planes simultaneously, giving them 2 equators (great circles where the planes of rotation intersect the surface) and no poles.

Both equators are objectively identifiable, but nowhere near as perceptually salient as the equator and poles are on a 3D planet like Earth. And, well, I have no idea how to mentally visualize the motions of stars in a 3D sky while undergoing compound rotation...

So, is it possible to define some kind of celestial navigation system on a 4D world, such that observations of the sun and/or fixed stars can allow one to determine consistent reference directions?

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    $\begingroup$ This 'is' a four dimensional universe, do you mean a five dimensional universe? $\endgroup$
    – Pelinore
    Commented Apr 28, 2022 at 17:00
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    $\begingroup$ @Pelinore 4 spatial dimensions, 1 time dimension. $\endgroup$ Commented Apr 28, 2022 at 17:34
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    $\begingroup$ You're not likely to get stars or planets in 4D at all. $\endgroup$
    – Monty Wild
    Commented Apr 29, 2022 at 5:06
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    $\begingroup$ @MontyWild Not if you naively transfer our own physics into a higher dimensional space, but that's not specified as a precondition here, and how to manage it is a completely different question. Greg Egan sketches a solution for a 5+1D universe in Diaspora. $\endgroup$ Commented Apr 29, 2022 at 15:28
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    $\begingroup$ @N.Virgo That is a good point; I have been basing my intuition on Greg Egan's depictions of higher-dimensional space in Diaspora--but he doesn't directly address how things move in the sky, so perhaps I have that wrong. It would be great if I did! Fortunately, computers don't have my limitations, so I am writing a simulator to work out day lengths and solar elevations over time for various rotation states and points on the surface. $\endgroup$ Commented Apr 30, 2022 at 1:50

4 Answers 4


It is much worse than you think.

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Relativity by MC Escher

In 4-dimensions there is no guarantee we will get a fixed North Pole to our rotation. There might be no analogue of the equator. We're not talking about two equators as you say. We're talking about no equators.

Four-dimensional rotations can behave like 2-dimensional rotations. Everyone knows rotating the 2-dimesional sphere (circle) moves every point the same angle along the circle. There is no north or south pole. Likewise for your 4-dimensional planet.

The reason 3-dimensional rotations give rise to an equator is because of the fundamental theorem of algebra. It comes down to what are called the eigenvalues of the associated rotation matrix.

For each $n \times n$ matrix $M$ there is an $n$-degree polynomial called the characteristic polynomial. The solutions $\lambda_1,\ldots, \lambda_n$ are called the eigenvalues of the matrix. For each solution there is a vector $v_1,\ldots, v_n$ such that $M v_i = \lambda_i v_i$. These are called the eigenvectors.

Now if $M$ is a three dimensional matrix the fundamental theorem of algebra says the solutions of the polynomial look like $z , \overline z, w$ for some real $w$ and possibly complex $z$. Hence there is a real eigenvector $v$ such that $Mv$ points in the same direction as $v$. But since a rotation matrix cannot change the length of a vector we have $Mv = v$. Hence $v$ is the North pole of the rotation and the other plane behaves as a two-dimensional rotation.

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    $\begingroup$ That is, in fact, exactly as bad as I think it is! $\endgroup$ Commented Apr 28, 2022 at 17:45
  • $\begingroup$ Hmm. . . Maybe it would be good to think about navigation on a two dimensional planet first. Is there any reason to think that is impossible? $\endgroup$
    – Daron
    Commented Apr 28, 2022 at 19:42
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    $\begingroup$ Navigation on the 1D surface of a 2D planet is trivial, as long as you have some kind of clock that can track the orientation of the world over time. It's the same as the longitude problem on Earth. To reach any given destination, you only need to know whether you are currently east or west of it, which does not require a particularly accurate clock. If you have an accurate clock, then you can fix your exact current location by comparing the elevation of the sun to where the sun should be at your coordinate reference point at the current time. $\endgroup$ Commented Apr 28, 2022 at 23:43
  • $\begingroup$ Good answer, although as the op says I think they had noted this in their question. Couple minor nitpicks: 1) I think you accidentally made a typo in the definition of an eigenvector-- should be $Mv_i=\lambda_i v_i$. 2) Your argument doesn't exclude the possibility that $Mv=-v$. To do so you need the additional requirement that rotation matrices preserve handedness, ie have positive determinant $\endgroup$ Commented Apr 29, 2022 at 21:17
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    $\begingroup$ @Goodies Relativity by Maurits McEscher. $\endgroup$
    – Daron
    Commented Apr 30, 2022 at 13:11

Please note this answer has been heavily edited in light of a comment by quarague, who pointed out an error. It ended up being incomplete as a result of that, but maybe it's helpful as a starting point for someone else.

Your 4D planet has two equators. I'm assuming the sky is similar to ours in that it has a number of fixed stars, although of course the sky appears to an observer on the ground to be a three dimensional surface instead of a 2D one.

To specify a point on your planet's surface we need three coordinates. I originally thought these would consist of two latitudes and one longitude, but quarague pointed out that it's actually one latitude and two longitudes, as I'll explain.

We can define the latitude as the angular distance from one of the two equators, just as we do on Earth. The reason there is only one latitude instead of two is that they determine each other - if you're 10 degrees away from one equator then you're 90-10 = 80 degrees away from the other. (This means that each equator acts like the "pole" for the other, so if you're 90 degrees from one equator then you're on the other equator. This explains the lack of poles.)

Now consider one of the fixed stars. Think of its position as a vector, pointing out from your planet's centre to the star's position, so that it passes through the surface at a point where the star is directly overhead. As your planet rotates, the position of this point will change, since we fixed our coordinate system to the planet's surface.

However, I believe that the latitude of the point will not change. This is because rotations preserve angles, and the rotation of your planet also preserves the position of each equator. So the star's latitude stays fixed over time, just like it would on Earth.

The first step in finding your position on your world would be surprisingly similar to how it is on Earth. You simply take out your 4D equivalent of a sextant (whatever that is exactly) and measure the exact position of a star in the sky, or more likely several stars. From this you can determine your latitude.

I'm skipping over a lot of details here of course, mostly because I can't easily visualise them, but I suppose the process would be similar in principle to how it is on Earth: you measure the distance of the star from the horizon and calculate your latitude from that. Maybe you need a compass as well, which will make things tricky given the absence of poles (and also of electromagnetism as we understand it). But in principle it should be possible to get your orientation from star positions as well, if needed.

Just knowing your latitude is probably much less helpful on your world than it is on Earth, because there are still two dimensions in which you don't know your position. These are the two longitudes.

I do not know how those would be calculated, however. My initial guess was as follows: your world has two different day lengths, which correspond to its rate of rotation around each of its equators. (The two day lengths are necessarily different, otherwise there would be an infinite set of equators instead of two.) I thought that maybe the two longitudes would correspond to the time of day according to each of the two types of day. But I don't think that can be right, because in the end the data you get from a chronometer is only one number, whereas we need two independent numbers in addition to latitude in order to specify a position.

So currently this answer is incomplete - I will have to think more about what the two longitude-like coordinates are and how one might go about measuring their value.

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    $\begingroup$ Just to help out with the math. One easy way to get coordinates on a 3-sphere is to use a pair of complex numbers $z_1$, $z_2$ such that $|z_1|^2+|z_2|^2=1$. The two equators are the points where one of the two variables is zero and the other has norm 1. To specify a point you would use one latitude that is distance to the equator (this determines the other latitude) and two longitudes which both require an arbitrary zero line (ie Greenwich on earth). $\endgroup$
    – quarague
    Commented Apr 29, 2022 at 7:51
  • $\begingroup$ @quarague oh, the latitudes determine each other? I can imagine that being the case. That means this answer is mostly wrong, then! I guess it will end up being the opposite of what I thought: you get just one latitude from measuring the stars, then you need a chronometer to measure the two longitudes, which I now guess correspond to the time of day according to the two different lengths of day that the planet has. $\endgroup$
    – N. Virgo
    Commented Apr 29, 2022 at 8:35
  • $\begingroup$ @quarague I might edit my answer later in light of what you say, but you could also post your comment as an answer yourself - I think it basically is the answer to the OP's question. $\endgroup$
    – N. Virgo
    Commented Apr 29, 2022 at 8:38
  • $\begingroup$ I edited my answer, but I realised that for the moment I don't know how you'd measure the two longitudes. I think the idea in my comment above probably isn't quite right - see the answer for details. $\endgroup$
    – N. Virgo
    Commented Apr 29, 2022 at 8:56
  • $\begingroup$ My current guess is that navigation is just fundamentally harder in a 4D world than a 3D one. You can get your latitude from the stars and, I'm guessing, one dimension of longitude from a chronometer. But then there's a whole other dimension that you haven't pinned down, and it might be that there's simply no way to determine it other than using landmarks. $\endgroup$
    – N. Virgo
    Commented Apr 30, 2022 at 2:07

Just like in 2 or 3 dimensions you'd still be using trigonometry to triangulate your position, based on measured angles between objects of known position (celestial bodies). Do this enough times and you'll be able to determine your 4-space coordinates. Knowing the angle between two points in N-space narrows down your position to a N-surface, additional points gets you additional surfaces, narrowing down your location to the intersections between them. Take enough measurements and you will have an exact fix on your position. Then it's just a matter of converting from some arbitrary coordinate system to whatever surface coordinate system is used on your 4-spheroid.

There are some shortcuts you can take to reduce the number of measurements needed. You aren't located at an arbitrary point in 4-space you're located on the surface of the 4-sphere. If there is a magnetic field, knowing the direction of the local surfacewise component, (if it's fixed enough to be reliable) can give you a reference vector to get two measurements from the same object. Instead of just angle from horizon, you can also get a bearing from "north".

As long as the orientation of the planet can be derived at a moment in time the complications of rotating in 4 dimensions doesn't meaningfully affect the difficulty.

It should be mentioned that poles are not necessary to locate yourself on a surface. Your surface coordinate system will need a number of points to create a reference frame. These are points with known locations in physical space and defined locations in your coordinate space. In a N dimensions you need N + 1 points to fix a reference frame.

There is an established procedure for defining a set of reference coordinates for other celestial bodies, but that's a useful convention for astronomers, not any sort of requirement.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – L.Dutch
    Commented Apr 29, 2022 at 15:36

North line

My friend has this funny 3D toy diorama. Inside, there's a species called "humans", all blind chicken folded up, of course.. but they try to find out how we navigate. They are smart little beggars, who invented endlessly complex mathematics we don't know anything about.

Ok I'll try to explain in 3D. Here, we have north-south, east-west, up-down and outward-inward. In math babble: the sea is nor outward nor inward, it is where we are. We could write down our 3 "coordinates" longitude, latitude and uptitude, but in daily practice we only need the proper direction, that is the plane in which our ship is heading forward.

It's easy. To travel north in a ship, we need to go forward in the plane containing the North line. On our planet, we see a North line, not a North star. As long as our plane of travel contains that line, we go north. To navigate the ship, we need to turn it into the right direction and tilt it to travel upward as well.

Strange thing with celestial lines is, you can't ever reach them.

Humans told us there exists a realm named "expanding space time" containing point-shaped stars on certain distances. But our stars look like lines. These lines all move outward ? in one direction ? How can that be ? Humans have strange ideas about stars, if you ask me.

  • $\begingroup$ Why would the stars be lines? $\endgroup$
    – N. Virgo
    Commented Apr 30, 2022 at 4:01
  • $\begingroup$ They see more than we do. $\endgroup$
    – Goodies
    Commented Apr 30, 2022 at 7:29

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