## 4D orbits again

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 orbits again

I'm still confused how orbits become impossible with steeper gravity.
If an object goes toward a star, it gains momentum and is saved from destruction.
dodecahedron
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### Re: 4D orbits again

It's not that orbits as such are imposible in 4d. Circular orbits are permitted in every dimension.

What happens is that elliptical orbits are stable only in 3d. In other dimensions, only circular orbits are stable, which means that if a planet is disturbed in its orbit, it has to assume a new circular orbit, or leave the system (plunge into the sun or ejected from the system). In three dimensions, it just assumes a different elliptical orbit, which means that it can remain in the system much longer.
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### Re: 4D orbits again

Not that I'd not be possible to do the required math of that physics, but so far I've just not lingered there. Whereas you obviously did. - So, could you give me some hints, what's wrong there?

--- rk
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### Re: 4D orbits again

most of the stuff i do is pre-calculus, and mostly pre-trignometric. i haven't done the mathematics here either, but i imagine that when one follows an elliptical orbit, the forces and inertias do not match what's needed to keep going in that orbit, which makes that orbit unstable. But the result is well known.

most of my hyperbolic geometry, is home grown from scratching around with polytope equations etc.
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### Re: 4D orbits again

Wendy, i'm saying that elliptical orbits are possible in 4D in the same way they are here.
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### Re: 4D orbits again

Elliptical orbits are only possible if you assume the inverse square law, that is, that the force of gravity decreases in proportion to 1/r^2 where r is the distance from the center of gravity.

Basically, to have an orbit, there are two factors at play: the inward pull of gravity, which draws the orbiting object closer, and the momentum of the object, which wants to fling it out into outer space. An orbit can only happen if these two opposing forces are balancing each other out.

Now, balancing these two forces isn't as simple as it sounds, because the strength of the forces change: the strength of gravity changes depending on how far away you are from the center of gravity, and the momentum changes as a result of the gravity. Think of a falling object: it starts from zero momentum because it's stationary, say when you're holding it out the window of a high tower; as gravity acts on it, it accelerates, so its momentum also increases.

In an ideal world, where there are no hidden perturbations, a (perfectly) circular orbit is always possible, no matter what. A circular orbit is where the momentum of the object exactly balances the force of gravity, so that the gravity is just enough to pull the object's path into a circle, and because the object's path is in a circle, its speed doesn't change, and so its momentum doesn't change, and because the circle is a constant distance away from the center, the force of gravity doesn't change either. So everything is in balance.

In the real world, however, this ideal scenario pretty much never happens. The object's momentum may almost balance the pull of gravity, but not quite. For example, the object may be moving a little too slow, so the force of gravity is stronger than its momentum, with the result that the object is pulled inwards towards the center of gravity. However, here an interesting thing happens: as it gets pulled in, it also speeds up, so its momentum increases. This means the momentum may be able to reach the point where it balances out gravity again. Of course, gravity also increases when the object is closer to the center, so here we have a crucial deciding question: does the momentum grow faster than the force of gravity, so that the object will acquire enough momentum to overcome the inward pull of gravity?

In some cases, the answer is no: if the object is moving too slowly to begin with, gravity pulls it inwards, and its momentum grows, but it's falling too fast, so the growing force of gravity completely overwhelms its momentum, and it crashes into the center (the central star, e.g.).

In other cases, the answer is yes: if the object is moving fast enough initially, even though it's not quite fast enough to completely overcome the gravity, as it gets pulled inwards the object's speed increases, and its momentum likewise, but it doesn't fall fast enough for gravity to overwhelm it. Instead, it moves faster and faster until its momentum overcomes the gravity, and it begins to move away from the center of gravity again. Here, there's another juncture: will the object, after its momentum overcomes gravity, be moving so fast that it will completely escape gravity? Again, remember that gravity is always pulling inwards; so once the object is moving away from the central star, the star's gravity will pull on it and it will slow down. But on the other hand, the object is also moving farther away from the star, which means the force of gravity is diminishing. If the object is moving fast enough, the decreasing force of gravity will never be able to pull it back, and the object flies out of orbit. However, if the object's initial momentum was not great enough, then after moving out some distance from the star, the force of gravity will manage to reduce its momentum to zero, and the object starts moving back towards the star.

Now, in 3D, gravity obeys the inverse square law, which means its strength decreases in proportion to 1/r^2 where r is the distance from the center. This exact proportion is very important, because if you do the math (which I'll skip here, as it's long, tedious, and boring -- you can look it up in any astronomy textbook on planetary motions), you'll discover that there are 4 possible cases:
- The perfect circular orbit: momentum exactly balances out gravity (never happens in real life)
- The elliptical orbit: the momentum is slightly off-balance, but not enough, so at one point, gravity is stronger and the object falls inward, but as it does so, its momentum increases past the balancing point, and it flies outward again. But it doesn't have enough momentum to completely escape; gravity will eventually reduce that outward momentum to zero and pull it back again. And so on. So the distance of the object from the central star oscillates -- you get an elliptical path.
- The crash course: the object doesn't have enough momentum to overcome gravity at all, so gravity overwhelms it and it crashes into the star.
- The hyperbolic escape path: the object is moving much too fast for gravity to capture it; gravity manages to pull its path into a curve (to be precise, a hyperbola, hence the name), but, as mentioned earlier, this only increases the object's momentum, past the point where gravity is no longer strong enough to pull it back, so the object flies off and escapes.

OK, so that's all in 3D. What if we go up one dimension?

Here, a very important assumption must be decided on: does gravity in 4D decrease in proportion to an inverse square law, or an inverse cube law? The question of why inverse cube or inverse square is a deeper issue -- but let's leave that for later. For now, let's suppose 4D gravity falls in accordance to the inverse cube law. What happens then?

I won't get into the mathematics of it -- Patrick Stein ('pat' on this forum) has already written a paper on this, which you can find if you search in this forum for it -- what I'll do is to give you an intuitive interpretation of why the results are the way they are.

First of all, let's all agree that circular orbits are both possible and completely impractical. Possible, because no matter how many dimensions of space you're in, a circle is always a circle -- it's a constant distance from the center, and so gravity, no matter how fast it falls with respect to radius, will always be constant, and therefore, there's an exact value of momentum along this circular path, that will exactly balance out the force of gravity, so that the object will remain in perpetual orbit. Impractical, because in the real world, nothing is perfect, and there will be perturbations, small or great, so you will in practice never be able to maintain a perfectly circular orbit. IOW, momentum and gravity will never match exactly, even if they come close; so circular orbits pretty much never happen (and if they did, a single speck of dust falling in from space will change that orbit instantly, so it will not stand the test of time).

Second, let's look at elliptical orbits a bit more closely. Why do elliptical orbits exist? It's because even though momentum and gravity never exactly balance each other out, they are in equillibrium -- sometimes gravity wins, but that only causes momentum to increase; sometimes momentum wins, but it gets slowed down by gravity eventually. So this tug-of-war is always at a stalemate: neither side completely wins over the other. But where does this balance come from? When momentum loses, for example, gravity will pull the object closer -- but when the object moves closer, the force of gravity also increases! If the force of gravity increases too fast, then the object's momentum won't have time to catch up. On the other hand, when momentum wins, the object will move away from the center. But that also means gravity becomes weaker -- if the force of gravity falls too fast, then it will never be able to cancel out the momentum and pull the object back. So the critical deciding factor of elliptical orbits is: how fast does gravity change when radius changes?

When gravity obeys an inverse square law, it so happens that the math just works out so that there's a range of momentums over which it will reach equillibrium in a simple harmonic motion -- that is, an elliptical orbit.

However, assuming 4D gravity obeys the inverse cube law, this balance never happens. Instead, what you get is one of the following:
1) The object is moving a little too fast: so it starts moving away from the central star. However, because gravity decreases too fast with increasing radius, it doesn't quite manage to counteract the momentum; the object's distance will gradually increase -- to be precise, for each revolution around its orbital path, its distance will increase by a constant amount. So its path is a spiral, spiralling outwards from the star. This is bad news, because a spiralling path means there is no orbit; it's moving away from the star at a constant rate.

2) The object is moving a little too slow: so it starts moving inwards towards the central star. However, because gravity increases too fast with decreasing radius, its increased momentum is unable to grow fast enough to match the growing pull of gravity: so its distance from the star will decrease a constant amount for every revolution around the star. This is bad news, since it means that in finite time, it will crash into the star.

3) The object is moving much too fast to begin with: it never gets captured by the star but just flies away. This is the same as in the 3D case of a hyperbolic path (no orbit).

4) The object is moving much too slow: so slow, in fact, that gravity completely overwhelms it from the get-go, so it just falls into the star within a short time. This is the same as the crash-course path in the 3D case.

The nagging question then, is why the inverse cube law has no possibility of balancing (except the perfect circle case, which is impractical)? The reason is that 1/r^3 grows too fast when you get closer to the star, and falls too fast when you get away from the star. In contrast, 1/r^2 also grows when you get closer to the star -- but not quite fast enough, so there's a small margin where momentum can eventually overcome gravity again; it also falls when you move away from the star -- but not quite fast enough so there's a small margin where it can eventually overcome momentum and pull the object back. So with 1/r^2, there's a margin of tolerance close to the ideal circular orbit, where momentum and gravity can reach equillibrium in an elliptical orbit even though they don't quite balance each other out. But with 1/r^3, this margin of tolerance is zero -- the slightest deviation from the ideal circle and either momentum is slowly winning the tug-of-war (the planet eventually flies off into space) or gravity is slowly winning (the planet eventually crashes into the star).

In fact, this conclusion is independent of dimension: whether you're in 4D or 3D or 5D or 100D doesn't matter; what matters is the fact that gravity decreases in proportion to 1/r^3. Similarly, elliptical orbits are possible in any dimension provided gravity decreases in proportion to 1/r^2.

So this leads to the question of why 1/r^2 or why 1/r^3, which I'll address in the next post.
Last edited by quickfur on Wed Feb 20, 2013 9:01 pm, edited 1 time in total.
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### Re: 4D orbits again

OK. So why does 3D gravity obey an inverse square law? Why is it 1/r^2, and not 1/r, or 1/r^3, or something else?

The truth of the matter is, we don't know for sure. But we have a very good guess: this has to do with flux. The idea is that any force of nature needs force carriers in order to exert its effects on distant objects. What the nature of these carriers are, is unimportant, just the fact that they are distinct entities, that, presumably, are "emitted" by the source of the force, and "responded to" by the objects that are affected by the force.

Unless there is a special situation, the source of a force, intuitively speaking, doesn't "know" where the objects affected by the force may be, so it simply sends out force carriers in all directions. So this means that if N force carriers are emitted per unit time, then at time 0, they are all concentrated at the source, and at time 1, they are evenly distributed on the surface of an n-dimensional sphere of radius 1 centered on the source, at time 2 they are evenly distributed on the surface of an n-dimensional sphere of radius 2, etc.. This means that the density of the force carriers at any given point must decrease with the surface area of the n-dimensional sphere -- since the number of force carriers N doesn't change, but the area they cover is increasing.

In 3D, the surface area of a sphere grows in proportion to r^2, so unsurprisingly, the strength of a force will also decrease in proportion to 1/r^2 -- you can think of it as the same number of force carriers that need to cover the surface of the sphere, so their density must decrease. Hence, in 3D we have the inverse square law.

In 4D, then, since the surface area of a 4D sphere grows in proportion to r^3, it would be reasonable to expect that the strength of a force should also decrease in proportion to 1/r^3. Hence, in 4D, we'd expect to have an inverse cube law, with all of its consequences -- such as the non-existence of elliptical orbits.
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### Re: 4D orbits again

OK, I found pat's post that has the link to his paper explaining why 4D planets can't have stable orbits (except the unrealistic perfect circle).
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### Re: 4D orbits again

Thanks for your interesting summary on that topic, quickfur. I found it very interesting.
You wrote:
The nagging question then, is why the inverse cube law has no possibility of balancing (except the perfect circle case, which is impractical)? The reason is that 1/r^3 grows too fast when you get closer to the star, and falls too fast when you get away from the star.

This is plain and obvious as long as we deal with the same sort of masses in 4D as in 3D.
But I was wondering:
If Newton's gravitational law F= G * ((m1*m2)/r^2) is changed to F= G * ((m1*m2)/r^3)
... then I wonder if it's OK to change something about the denominator part of the fraction (that is change r^2 => r^3)
but continue to calculate with the conventional 3D kind of masses of the stellar objects?
Because assumingly the 4D masses are of a different quality (in a way "heavier" than the 3D masses) and I thought this might be taken into account.
For example the 4D masses might be multiplied with a certain (unknown) constant or squared in a specific way to make up for their 4D extra "heaviness".
If we did this, the fast "growing &decreasing" of the gravitational force would be levelled out and orbitals would be possible in 4D, wouldn't they?
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### Re: 4D orbits again

ac2000 wrote:Thanks for your interesting summary on that topic, quickfur. I found it very interesting.
You wrote:
The nagging question then, is why the inverse cube law has no possibility of balancing (except the perfect circle case, which is impractical)? The reason is that 1/r^3 grows too fast when you get closer to the star, and falls too fast when you get away from the star.

This is plain and obvious as long as we deal with the same sort of masses in 4D as in 3D.
But I was wondering:
If Newton's gravitational law F= G * ((m1*m2)/r^2) is changed to F= G * ((m1*m2)/r^3)
... then I wonder if it's OK to change something about the denominator part of the fraction (that is change r^2 => r^3)
but continue to calculate with the conventional 3D kind of masses of the stellar objects?
Because assumingly the 4D masses are of a different quality (in a way "heavier" than the 3D masses) and I thought this might be taken into account.
For example the 4D masses might be multiplied with a certain (unknown) constant or squared in a specific way to make up for their 4D extra "heaviness".
If we did this, the fast "growing &decreasing" of the gravitational force would be levelled out and orbitals would be possible in 4D, wouldn't they?

I'm not sure what you mean by "3D kind of mass" vs. "4D kind of mass" here. Mass is directly proportional to the amount of matter in the object, regardless of its dimension, and regardless of its shape. It doesn't matter if the object is a 2D disk, 3D sphere, or 4D hypersphere, or even a tube or some other shape; the total amount of force that is "felt" by the object is directly proportional to how much "stuff" is in it -- that is, multiple of the number of atoms in it.

While it is true that the mass of a 4D object increases faster with increasing radius, that doesn't really matter in the end, because gravity doesn't depend on the radius of the object, only its mass, and how far away from the center of gravity it is. So while a 3D planet of, say, 1000km radius has less mass than a 4D planet of 1000km radius, that doesn't really affect the way gravity acts on it; the 4D planet will just behave like a bigger, heavier 3D planet. You could simply consider a smaller 4D planet that has the same mass as the 3D planet, and all the equations will work out the same way.

I bring up the matter of radius, because it's important not to confuse the radius of the planet (i.e., how big it is) vs. the radius of its orbit (its distance from the star), which is what the r in Newton's equation is referring to. The radius of the planet, besides determining how much mass it has, has no real bearing on the amount of gravitational force it feels. Gravity acts on the mass, not on the radius -- that's why in Newton's equation you only have m1 and m2, the masses of the objects in question, not their radius or any other size measurements.

Basically, in dealing with orbital mechanics, we're making the simplification of planets and stars of non-zero volume into zero-size point masses. I won't get into how this doesn't really change the overall behaviour of the system -- you can consult any introductory physics textbook for that -- the point is that at the kind of large distances that planets and stars are at, the volume of the bodies don't really matter; what makes the difference is their masses and relative distances. (Besides, trying to factor in volume and all that requires extremely complex math and physics modelling, which is kinda a waste of effort because their total effect on the results is miniscule at those distances, basically not more than just roundoff error.) Ultimately what matters in the large scale is the rate at which gravity diminishes with distance. And by rate, I'm not referring to just a constant factor; but the inherent shape of this diminishing -- i.e., the 1/r^2 or 1/r^3. Differences in constants don't really change much -- you just substitute the masses with different values (i.e., smaller/larger planets) to balance it out. They are just differences in quantity. But differences between the shapes of the 1/r^2 vs. the 1/r^3 curves -- that produces a qualitative change in the behaviour of the system. That's what causes 4D orbits to have a significantly different kind of behaviour than 3D orbits.
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### Re: 4D orbits again

The usual model for action at a distance, is the radiant model, in that the source radiates a flux of forcekins. These buffer anything that they chance to hit, which either pulls them or pushes them. A source radiates a shell of fluxions, which travel outwards at a constant speed. Since the total flux is proportional to the strength of the source, say GM, the fluxions are spread out ever-thinner as they go further, so eg, S = 2 pi² r³ in four dimensions.

The force is then GM/2pi2 r3, in four dimensions. The same relations hold for electricity and any other radiant force. With light and heat, the inverse cube still holds, but the action is not of vector relation (pull vs push), but of scalar relation (more). So the intensity of a light at 1 foot, is eight times that at two feet, since the candle has to light eight times the volume.
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### Re: 4D orbits again

wendy wrote:The usual model for action at a distance, is the radiant model, in that the source radiates a flux of forcekins. These buffer anything that they chance to hit, which either pulls them or pushes them. A source radiates a shell of fluxions, which travel outwards at a constant speed. Since the total flux is proportional to the strength of the source, say GM, the fluxions are spread out ever-thinner as they go further, so eg, S = 2 pi² r³ in four dimensions.

The force is then GM/2pi2 r3, in four dimensions. The same relations hold for electricity and any other radiant force. With light and heat, the inverse cube still holds, but the action is not of vector relation (pull vs push), but of scalar relation (more). So the intensity of a light at 1 foot, is eight times that at two feet, since the candle has to light eight times the volume.

Which leads to the conclusion that in very high dimensions, the universe would be very cold and dark, because light diminishes too quickly with distance, and so does heat. This applies both to stars and to artificial light/heat sources.
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### Re: 4D orbits again

It's even stranger with hyperbolic geometry. If you use the brightness of the sky as a guide to placing things, and a curvature of something like 4000 miles (so the earth fits inside a cell of x5o3o4o, the whole universe fits inside a sphere of the sun (ie less than 864000 miles diam). The sun is supposed to be 80,000 miles away, and the orbit of the earth will bring different stars into view in different seasons.

In hyperbolic space, there is little dependence on the actual number of dimensions: the thing approaches log(distance).
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### Re: 4D orbits again

I'm not sure what you mean by "3D kind of mass" vs. "4D kind of mass" here. Mass is directly proportional to the amount of matter in the object, regardless of its dimension, and regardless of its shape. It doesn't matter if the object is a 2D disk, 3D sphere, or 4D hypersphere, or even a tube or some other shape; the total amount of force that is "felt" by the object is directly proportional to how much "stuff" is in it -- that is, multiple of the number of atoms in it.

I meant by "3D kind of mass" vs. "4D kind of mass" that maybe there could be different kind of atoms in 4D matter. So that not only the whole volume extents in a 4th dimension but every single atom has an extentention in the 4th dimension and that this might somehow make a 4D mass heavier.
I don't know, maybe be it's nonsense.

So, if mass is proportional only to the amount of matter in the object regardless of the dimension, as you wrote, does this mean one can simply calculate the mass of a 4D object if we know what's material it is made of (or the density) and the size?
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### Re: 4D orbits again

ac2000 wrote:
I'm not sure what you mean by "3D kind of mass" vs. "4D kind of mass" here. Mass is directly proportional to the amount of matter in the object, regardless of its dimension, and regardless of its shape. It doesn't matter if the object is a 2D disk, 3D sphere, or 4D hypersphere, or even a tube or some other shape; the total amount of force that is "felt" by the object is directly proportional to how much "stuff" is in it -- that is, multiple of the number of atoms in it.

I meant by "3D kind of mass" vs. "4D kind of mass" that maybe there could be different kind of atoms in 4D matter. So that not only the whole volume extents in a 4th dimension but every single atom has an extentention in the 4th dimension and that this might somehow make a 4D mass heavier.
I don't know, maybe be it's nonsense.

Even if 4D atoms have heavier mass (and maybe they do), it doesn't change the fact each atom's mass will be some fixed scalar number, say m. Then given any 4D object, they will be made of some number of atoms, say N. So then the total mass of the object is just N*m, which is yet again another scalar number, say M. So, since the amount of gravitational force felt by the object is proportional to its mass, this means F = K*M, for some number K. Assuming the inverse cube law, then, K = G/r^3, for some constant G (the 4D gravitational constant), and r is the distance between the object and the star. Note that the only thing dependent upon the r^3 is K; the value of M is completely independent of r^3. So it doesn't matter what that value is; the qualitative behaviour of gravity is not affected by the value of M. It's the 1/r^3 part that changes its behaviour. Since M is a constant, there's nothing in it that can really make a qualitative difference to the equation.

So, if mass is proportional only to the amount of matter in the object regardless of the dimension, as you wrote, does this mean one can simply calculate the mass of a 4D object if we know what's material it is made of (or the density) and the size?

Yes. So if you have a 4D cuboid, for example, you'd have 4 length measurements along each of its dimensions, say a, b, c, d. Then, given a density of h, the mass of the cuboid is M = h*a*b*c*d.
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### Re: 4D orbits again

So it doesn't matter what that value is; the qualitative behaviour of gravity is not affected by the value of M. It's the 1/r^3 part that changes its behaviour.

I still don't understand this. Because on the wiki page about gravitation http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation it says in the first formula that m1*m2 is divided by r^2 (so r^3 in 4D).
And the bigger the masses are in this formula, the smaller would be the F force, when the multiplied masses are divided by r^2 or r^3, wouldn't they?
So I don't understand where the "number 1" in 1/r^3 comes from which you mentioned?
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### Re: 4D orbits again

ac2000 wrote:Thanks for your explanations, quickfur.

So it doesn't matter what that value is; the qualitative behaviour of gravity is not affected by the value of M. It's the 1/r^3 part that changes its behaviour.

I still don't understand this. Because on the wiki page about gravitation http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation it says in the first formula that m1*m2 is divided by r^2 (so r^3 in 4D).
And the bigger the masses are in this formula, the smaller would be the F force, when the multiplied masses are divided by r^2 or r^3, wouldn't they?

No, the bigger the masses, the larger the force.

So I don't understand where the "number 1" in 1/r^3 comes from which you mentioned?

1/r^3 is just a generic way of referring to the fact that the denominator is r^3. Obviously the full equation is F = G*m1*m2 / r^3, which you can also write as F = (G*m1*m2) * (1/r^3).

It may help to consider Newton's second law of motion: F = m*a. That is, force equals the mass of the object times the acceleration it's undergoing. Or, put another way, a = F/m, that is, the amount of acceleration an object undergoes is equal to the force exerted on it divided by its mass (the heavier the object, the more force it takes to change its velocity).

Putting this together with the above equation, we may write m1*a = G*m1*m2/r^3, where m1 = mass of planet and m2 = mass of star. Solving for a, we have: a = G*m2 / r^3. Note that the acceleration of the planet is independent of its mass; and since m2 is constant (the star's mass doesn't change -- at least not significantly), the acceleration is proportional to 1/r^3. In other words, the object's mass does not change how much it accelerates under the force of gravity (one recalls Galileo's experiment, where he dropped a light object and a heavy object simultaneously from a high tower, and both reach the ground at the same time -- contrary to popular expectation that the heavier object will hit the ground first).

Of course, this doesn't mean that the planet's mass plays no role in its orbital motion; the actual calculations to derive the orbital path are much more complicated, but this at least shows that mass does not play as big a role in orbital motion as one might be inclined to think at first glance. Rather, it is the r^3 denominator that pretty much dictates what kind of orbital motions will be possible.
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### Re: 4D orbits again

No, the bigger the masses, the larger the force.

Yes, of course, sorry.

In other words, the object's mass does not change how much it accelerates under the force of gravity (one recalls Galileo's experiment, where he dropped a light object and a heavy object simultaneously from a high tower, and both reach the ground at the same time -- contrary to popular expectation that the heavier object will hit the ground first).

Thanks for that example, that makes things clearer. I remember this experiment from the distant school days. Something with a feather and a lead ball being dropped in a vacuum. There was no difference in speed.

I don't know if I do fully understand those formulas. I guess it's too much math for my unmathematical brain. I probably should stay away from mathematical things because it's so frustrating. It's very interesting on the one hand but I always fail to really understand it. That's no fault of your explanations which are certainly very well written.
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### Re: 4D orbits again

ac2000 wrote:[...]
I don't know if I do fully understand those formulas. I guess it's too much math for my unmathematical brain. I probably should stay away from mathematical things because it's so frustrating. It's very interesting on the one hand but I always fail to really understand it. That's no fault of your explanations which are certainly very well written.

No, don't stay away from mathematical things. The fact you find them frustrating is not because they are inherently hard to understand, but simply that you haven't found the right frame of mind to correctly interpret them, and through no fault of your own: a lot of math is thrown about these days but often inadequately explained, filled with unstated assumptions or otherwise unhelpful over-simplifications. Your frustration is more a sign of inadequate explanation (on my part or others') than any inability to grasp mathematical things. I find that stubborn persistence in spite of all this eventually yields great insight.

On the other hand, though, things like planetary motion are not exactly the simplest subjects to tackle. It took many centuries of development before we could finally pin down what exactly governs their motion. It's a long road from the "wandering stars" of the ancients to today's understanding of elliptical orbits, and even then, I don't think we have fully captured all the intricacies of it yet, though we are pretty close. And while I've given a sort of intuitive, hand-waving sort of explanation previously, that doesn't really do justice to the actual math of it -- which involves complicated concepts such as differential equations and stuff like that. I can't say I fully understand it all either -- I had to rely on Pat's excellent derivation to prove the final results about 4D orbits. But that's where one starts: with the general road map of it all, and then work out the details as you go along.
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### Re: 4D orbits again

No, don't stay away from mathematical things. The fact you find them frustrating is not because they are inherently hard to understand, but simply that you haven't found the right frame of mind to correctly interpret them, and through no fault of your own: a lot of math is thrown about these days but often inadequately explained, filled with unstated assumptions or otherwise unhelpful over-simplifications. Your frustration is more a sign of inadequate explanation (on my part or others') than any inability to grasp mathematical things.

Thanks a lot for your encouragement, quickfur .
Although I think it really is an inability to grasp mathematical things on my part, because I can recognize if something is explained well or not. Our math teacher back at school could explain quite well, I thought, and others were of the same opinion, even those people who were not good at maths. But at some point things simply get to complicated for me. I think people do have different kinds of talents they're born with, and maths does not seem to be mine. It's probably because of that strange language of maths, all that formula stuff. However you're right of course. That should be no reason to stay away from it completely. There are many interesting aspect about maths that do not involve formulas, such as the history of maths. Or to some degree the 4th dimension (well, as long as it's not about orbits maybe.) .

I find that stubborn persistence in spite of all this eventually yields great insight.

Yes, I agree. Although it should remain somehow interesting or fun. It's strange, I'm easily frustrated if I don't understand a maths formula. But when I read something about 4th dimensions (such as the excellent texts on your website) it's not so frustrating, although I comprehend only part of it. Instead it's rather the feeling that the brain is trying hard to grow some new branches in order to be able to visualize some of these strange new things. So, it's kind of fun.

It's a long road from the "wandering stars" of the ancients to today's understanding of elliptical orbits, and even then, I don't think we have fully captured all the intricacies of it yet, though we are pretty close.

I don't know, maybe some ancients knew more then we think today. We can't know for sure because so many old texts are lost. And without any decent equipment, it must have been much harder to find out anything, so those ancients must have been very clever. For example Eratosthenes calculated the Earth's circumference as early as 250 BC (though it was not very accurate), but without all the instruments and knowledge people have today.
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### Re: 4D orbits again

I'd suggest that u're exaggerating with 4d by putting r^3 in r^2 place. Formulas for functions of time in 3d orbits (using r^2) can be reformulated to 2d (orbital plane) or 1d (distance line). Axises also get redefined though and it turns out that, in this way, 1d orbit appears dynamic in 2d as well as 2d in 3d. 3d orbit should also appear dynamic in 4d I guess.

4d orbits don't have too standardized axial positions yet so it remains open for future social astrophysicists, for today there is not much use of it (globally).
fallfromgrace
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### Re: 4D orbits again

The reason that r^3 is used in r^2, is that gravity is treated as a radiant vector flux. That is, one supposes that a source releases a measure of flux, which then radiates outwards. Since this happens in all directions (in any dimension), then its intensity reduces as the surface of a sphere for that radius. In four dimensions, the surface of a sphere of radius r is S = 2 pi^2 r^3, and it is this r^3 that is used in four dimensions.
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### Re: 4D orbits again

I can understand how u insist on gravitational force progression with dimension increase, but the way u put it to be defined by hypersphere-surface-only doesn't make sense (in reality we'd like to be divisible in any number of dimensions) as it doesn't yield long term orbit;

in 1d, 2d and 3d we can notice long term orbits (all reformable among each other and following inverse square law), so there should be the same in 4d;

that makes ur hypersphere-surface-only gravity dubious, not allowing 4d orbits too long in time
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### Re: 4D orbits again

The way field is defined is force = mass * field; and G.density = div(field). Because gravity is still behind in rationalisation, it's actually 4pi G.

When these are applied to a point charge in a given space, the equation comes to F = G M m / (surface of sphere). This is the gauss rule in that the flux across the surface is proportional to the mass inside the sphere, and that the field is proportional to the flux.
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### Re: 4D orbits again

quickfur wrote:Which leads to the conclusion that in very high dimensions, the universe would be very cold and dark, because light diminishes too quickly with distance, and so does heat. This applies both to stars and to artificial light/heat sources.

Our Universe is actually very cold and dark too! What we consider to be big differences in temperature here on Earth hardly register much in the 'chemistry' occuring at the temperatures on the surface of the Sun.
It is quite admirable the minor relative variance in temperature at which our chemistry occurs; and thank goodness it does! Only 100° temperature difference for water to go from being ice to steam! That is amazing.

quickfur wrote:Basically, to have an orbit, there are two factors at play: the inward pull of gravity, which draws the orbiting object closer, and the momentum of the object, which wants to fling it out into outer space. An orbit can only happen if these two opposing forces are balancing each other out.

Now, balancing these two forces isn't as simple as it sounds, because the strength of the forces change: the strength of gravity changes depending on how far away you are from the center of gravity, and the momentum changes as a result of the gravity. Think of a falling object: it starts from zero momentum because it's stationary, say when you're holding it out the window of a high tower; as gravity acts on it, it accelerates, so its momentum also increases.

It's been kind of vaguely reached at here but let's clear something up as it is confusing to use the term momentum when it comes to orbits.

I'll explain why. Momentum is a product of mass by velocity. Using the term momentum gives an impression that more massive objects will orbit differently to less massive objects.
As explained - here in this thread by other examples - a coin orbiting the Earth will do so in a similar fashion to how the moon does. Bringing density into it, a gold bar will orbit in a similar fashion to an equal sized stick of chalk.
I will clarify that a little bit more as it is of course a little more complex than that. But, using the term momentum does tend towards being misleading.
What actually primarily affects orbit is velocity and direction. These are the major facets. Yes, velocity is a part of momentum, but direction is not directly coded into it, and mass is mostly a side player.

Where mass does come into play is that two objects form a system. All orbits are actually two objects orbiting around each other.
If you take a gold bar and put it in orbit and take some chalk and put it in orbit you will see little difference. That is because the Earth comprises most of the mass when it comes to the joint orbit.
If you were to put a gold bar on one side of the Earth and a chalk bar on the other side of the Earth at equal distances the gold bar would meet the Earth first.
This is because the Earth would move more towards the greater mass - the gold bar - and cut down the distance the gold bar had to travel to reach it.

So mass does play its part but due to the the vast differences in size we can generally discount the mass of an object when it is circling the Earth.
We have to take its mass into account to change its course but we don't have to take the mass into account to work out the path it will follow around the Earth as anything we send up is a speck beside the Earth.

If we start to look at the major factors - speed and direction - we can more handily get a feel of the needs of an object to form an orbit.

quickfur wrote:The object is moving a little too fast: so it starts moving away from the central star. However, because gravity decreases too fast with increasing radius, it doesn't quite manage to counteract the momentum; the object's distance will gradually increase -- to be precise, for each revolution around its orbital path, its distance will increase by a constant amount. So its path is a spiral, spiralling outwards from the star. This is bad news, because a spiralling path means there is no orbit; it's moving away from the star at a constant rate.

In actuality, even our orbits tend to have a spiral nature to them. Our moon is moving away from our Earth at about 3.8cm per year. All orbits tend to be spirals.

quickfur wrote:OK, I found pat's post that has the link to his paper explaining why 4D planets can't have stable orbits (except the unrealistic perfect circle).

Let's look at what it takes to achieve a stable orbit.
What you need to do is send an object moving perpendicular to and at a distance from the planet at a certain speed such that its change of path will soon take it to a position where it is travelling perpendicular to the original path by the time it reaches the perpendicular point around the planet. At this point it will have zero velocity away from the planet and some commensurate sideways velocity.
This should then continue to return it to a point on the opposite side of the planet with its original relative velocity at the same surface distance but in the opposite direction of travel.
It should then continue on to the final perpendicular - opposite to the first - and finally all the way back to the original position with matching original speed, direction and distance.
Hey presto, stable orbit. There are things like frame dragging effects but I don't think we are worrying about them here.

You mentioned a circular orbit, and certainly at the right speed and altitude, such an orbit would for a time being be achieved.
Increase the initial speed slightly and you end up with an elliptical orbit with your origin point and its opposite point being at the closest point to the planet.
Decrease the initial speed slightly and you end up with an 90° elliptical orbit with your origin point and its opposite point being at the farthest point from the planet in the orbit.
The more you increase this speed the higher the orbit apex will be until you reach a point where the speed exceeds the gravity gradient rate and your object breaks out of a stable orbit and flies away.
Decrease the initial speed enough and the object will crash into the planet.

I think mathematically that we would tend to have to agree that a 4D world would also allow for increased initial perpendicular speeds that would not exceed this escape velocity.
Of course reaching this point would come a lot quicker in 4D - as the gravity gradient decreases at a much faster rate in 4D - but it doesn't cut out immediately beyond a circular orbit.
The factors I've provided here - speed, direction, rate of gravity decrease (per spatial dimensions) - should make it easier to show why this isn't so if that's not right.
Perhaps I've stated things correctly but if not could you use these factors to explain to me why it wouldn't be as I'm suggesting?
gonegahgah
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### Re: 4D orbits again

It dawns on me that all I've described are circular orbits.
For an orbit to be parallel to the surface at opposite sides of the planet at the same distance; it can only be so if the orbit is circular.
If you start at the same distance with higher velocity you will push into a higher orbit initially that becomes elliptical but not parallel at opposite sides of the planet.
Instead you probably require a tangential elliptical orbit to achieve a stable orbit or perhaps otherwise have one that precesses.

Still, surely slight tangential variations would be allowable in 4d instead of just a circular orbit?
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### Re: 4D orbits again

gonegahgah wrote:It dawns on me that all I've described are circular orbits.
For an orbit to be parallel to the surface at opposite sides of the planet at the same distance; it can only be so if the orbit is circular.
If you start at the same distance with higher velocity you will push into a higher orbit initially that becomes elliptical but not parallel at opposite sides of the planet.
Instead you probably require a tangential elliptical orbit to achieve a stable orbit or perhaps otherwise have one that precesses.

Still, surely slight tangential variations would be allowable in 4d instead of just a circular orbit?

The difference between 3D and 4D isn't a matter of quantity -- it's not because gravity is lower, or the planet moves too fast, etc.. It's a matter of quality: the way in which gravity falls with distance. That changes not just the quantity of the path (orbital distance, speed, etc.), but the quality of the path: you get a different kind of curve, not just a bigger or smaller orbital curve.

The peculiarity of 3D orbits under a 1/r^2 potential well is that there are so many possible stable paths. This is why 3D orbital systems are so stable: it's not because elliptical paths are "resistant" to orbital decay -- like you pointed out, variations happen all the time. But it's the fact that there are so many possible stable orbits that, excepting catastrophic events like a giant meteor crashing into the earth and sending it on a collision path with the sun, all the slight variations are just changing one stable orbit into another stable orbit. Technically, the earth isn't travelling along a single stable orbit; there's constant environmental perturbations going on, both on the small scale and the large scale. But all of these little disturbances are non-catastrophic, because it's just pushing the earth from one stable orbit to another stable orbit. So it's constantly switching to ever-so-slightly-different orbits, but since all of them are stable, there is no problem.

In 4D, however, most orbital paths are unstable. The margin of error in which a particular path changes from stable to unstable is razor-thin, so even the smallest perturbation will change an originally stable path into an unstable path. And because there are overwhelmingly more unstable paths than stable ones, the chances that an orbiting planet will be able to somehow recover from a slightly-unstable orbit back to a stable one is practically zero. In the vast sea of possibilities, the set of stable orbits is but a tiny island, and once you fall off, chances are you won't be able to find your way back. Whereas in 3D, the set of stable orbits is pretty big, so you have lots of room to get pushed around without falling into the waters of instability.

Another way to think about it is that the set of possible 3D orbits is like a mountain with a flat top, where the top of the mountain represents stable orbits. While it's possible to get knocked off the top and roll down the mountain (i.e. end up in an unstable path that either crashes into the sun or flies off into space), there's quite a lot of nearby stable orbits that, for the most part, something currently in a present stable orbit is quite likely to still be in some kind of stable orbit even after you knock it around a bit. In 4D, however, it's like a mountain with a sharp, steep, pointed tip, and you're standing tiptoe on top. All it takes is for a tiny nudge and you're tumbling down the mountain. You can hold on to the tip of stable orbits, but it's so much easier to fall off. There's simply no room for error. All it takes is a gust of wind and you're out, the planet is on a collision path with the star, or flying off into space.
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### Re: 4D orbits again

"It's more space
in four-space"

Well even when it is much sharper to fall out of a stable orbit into an unstable one,
the probability of collision also should decrease there: there are infinitely times more ways to miss a target!

--- rk
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### Re: 4D orbits again

Klitzing wrote:"It's more space
in four-space"

Well even when it is much sharper to fall out of a stable orbit into an unstable one,
the probability of collision also should decrease there: there are infinitely times more ways to miss a target!

--- rk

Yes, the probability of collision is lower. But the problem is that when the metastable state (i.e. stable orbit in the face of a gravity well) has a margin of error of zero, even a speck of dust will change a planet's stable orbit from a stable one to an unstable one. And I'm pretty sure that there will be much more than just one speck of dust colliding with the planet over astronomical time scales. And once you're knocked off the stable orbit, any further perturbations will, in all probability, only become more unstable. It's highly unlikely you'll ever get back into the (meta)stable state.

So even while the protoplanet is forming, it is already on its way to collide with the star or fly off into space. It would be impractically rare for a stable planet to even form in the first place, let alone survive long enough to host living beings for any meaningful amount of time.

So, unfortunately, dimensional analogy does not help us find what stable large-scale structures may exist in 4D, since they are in all likelihood fundamentally different from what exists in 3D, if they exist.

Of course, that doesn't stop us from speculating about 4D planets, because dimensional analogy does still give us much insight into how the extra dimension of space works. We can discover interesting structures like ring-poles and regions of perpetual twilight by postulating a stable planetary orbit that we artificially keep stable. It's just that things start to break down once we get to the astronomical scale. At which point, we either have to search for alternative solutions to the stable structure problem (very hard), or we can keep our small-scale dimensional analogy structures (like stable planets) and change the way 4D physics works so that it would allow such structures (easier -- but in the process we'd have to give up extrapolating 3D physics to 4D, since it will be fundamentally different).
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### Re: 4D orbits again

quickfur wrote:Another way to think about it is that the set of possible 3D orbits is like a mountain with a flat top, where the top of the mountain represents stable orbits. While it's possible to get knocked off the top and roll down the mountain (i.e. end up in an unstable path that either crashes into the sun or flies off into space), there's quite a lot of nearby stable orbits that, for the most part, something currently in a present stable orbit is quite likely to still be in some kind of stable orbit even after you knock it around a bit. In 4D, however, it's like a mountain with a sharp, steep, pointed tip, and you're standing tiptoe on top. All it takes is for a tiny nudge and you're tumbling down the mountain. You can hold on to the tip of stable orbits, but it's so much easier to fall off. There's simply no room for error. All it takes is a gust of wind and you're out, the planet is on a collision path with the star, or flying off into space.

This is a very visually easy explanation to follow; nicely done. Possibly a gravitational valley - like Einstein's - would help too with helping to visualise.

One of the things it might be interesting giving some thought to is the stable areas of those gravitational valleys and dimensional volumes?
The most important area for stable orbits is closer into the valley's centre; but not too close.
Bring planets much closer together and I imagine the number of stable orbits drops dramatically.
Move further away and it takes very little effort to achieve escape velocity.

I mention volume for a specific reason.
Objects in a 4D universe would tend to be shorter than in our universe.
Why? Because their is just so much more sideways space to work with.
Conversely a 2Der would have to be so much taller to try to achieve a similar complexity to us.

Take a brain. Our cells are arranged in a 3 dimensional space allowing for many connections from all directions.
A 4Ders brain cells have a whole extra realm of dimension to start spinning connections off into so you can fit many more cells within a much smaller across length of space with greater complexity.
The same goes for atoms and chemisty which would probably be very bizarre to what we have. The extra space for movement allows a whole different more compact way of existance.

Now, if we take this and scale it in to our question here we tend to notice that the gravitational curve created by 4D space can find a region that is more like our 3D gravitational curve closer in.
If we use the mountain or valley example again we can see that there is a nice region within to play for stable orbits.
For 4D objects, if they scale down proportionally, we can find a region closer in that is not so steep and at the smaller scale matches that of ours.

So if you have boulders rolling around a 3D valley and have marbles rolling around a 4D valley I feel you will find an equivalency zone?
Steepness or not, if you reduce your size you can orbit closer to the ideal 4D region which becomes equally large in comparison due to your relative smaller size.
Does that sound feasible at all?
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