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.

Re: 4D orbits again

Postby quickfur » Wed Sep 04, 2013 3:02 pm

gonegahgah wrote:[...]
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?

Again, the difference between 3D and 4D is not a matter of scale (quantity), but an inherent difference in behaviour (quality). The reason I chose a flat-top mountain vs. a pointed tip was to try to emphasize this difference. But, like all analogies, it's bound to fail if you stretch it too far. :) The important point here is that in 3D, the "mountain of stability", so to speak, has a certain kind of shape, and that in 4D, it's a fundamentally different kind of shape. It's like the difference between a circle and a triangle. They are fundamentally different kinds of shape, and changing the relative scales of things doesn't change the fact that one is circular and the other is triangular. You can make it a bigger or smaller triangle, but it will never become a circle, and vice versa. The difference between 3D orbits and 4D orbits is a fundamental difference that doesn't change with scale. No matter how you try to play with scaling things up or down, a 4D orbit is fundamentally unstable ("triangular") and can never be made to behave the same way as a 3D orbit ("circular").

That is, unless you postulate a different kind of physics that allows a 1/r^2 decay of the force of gravity over distance. :) Then you will have the same kind of orbits that 3D has, except with an extra dimension of space. But then, it would mean that physics in this "modified" 4D space must behave in a fundamentally different way, and it won't be analogous to 3D physics at all.

Or, alternatively, you can postulate a momentum that is proportional to r^3, which could possibly balance out the 1/r^3 decay of gravity, but then it would imply an even bigger change to the physics of how things move, probably making the result completely unrecognizable to us in the process. :)

Unfortunately, this is the way the math works. A 1/r^2 gravitational decay is fundamentally different from a 1/r^3 decay. It may look like a small difference in writing, but actually, mathematically speaking, they are as different as a triangle is to a circle. Their fundamental properties are mutually incompatible.
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Re: 4D orbits again

Postby fallfromgrace » Thu Sep 05, 2013 10:45 am

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


also elasticity effect increases there; any shock wave gets absorbed faster
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Re: 4D orbits again

Postby gonegahgah » Sun Sep 08, 2013 12:43 pm

I won't seek to discount the possibility (maybe even the absolutivity) of the impossibility of 4D orbits as I don't know enough yet.
I thought I might explore some ideas along the way to see if these can help me to understand better.

First I just want to look at the area of 'momentum' as it applies to different dimensional universes.
As I've mentioned I don't think that momentum itself applies in our considerations of orbits but it still may be a useful idea to explore in itself and its applicability to different numbered spatial dimensions.

Momentum is a product of mass x velocity.
Mass in our 3D universe obviously has a spatial component to it but also has a density factor. So Mass is a product of volume x density.
This gives us L^3 x D (or L^3 x m / V; which works out to m anyway).

What we are measuring when we measure momentum is how badly a volume wants to maintain its speed through a point in space (note: not area) in its current direction of travel.
Momentum primarily comes into play when we are using a directed force to move an object or when we try to resist its movement - catcher style (ie. if it impacts with something).
If instead of presenting a point of resistance we present a surface of resistance we can improve the rate at which we slow down an object.

A rigid pole will have to resist a lot more something that is heading towards its point end it than will a rigid block will have to resist something heading towards its face.
Resistance has a surface component and, the more surface that can be interfaced between the mover and the catcher face and its base, the less deep the mover will impact into the catcher.
ie if you send a pole at a block of clay it will go deeper than if you send an equal mass pancake at the block of clay.
There is more contact surface area to catch the projectile in the later.

If we look at a 2D world where we are sending a 4x4 square, edge face on, toward a 4 line face then this will sink a certain distance.
If we look at a 3D world where we are sending a 4x4x4 cube, square face on, toward a 4x4 square face this sink a certain distance.
Actually, both of these will sink exactly the same distance.
The reason is probably easy to see, but it is because each back 4 length of the projectile will meet an equivalent resistance point before it.

Rockets are an example of the reverse type of momentum (or inertia).
As long as you have an equivalent ratio of surface push for each dimensioned rocket across the dimensions you will end up with an equal acceleration.
So if the rocket burns with a surface push in 2D is 1/3 and in 3D is 1/3 and in 4D is 1/3 you will achieve equal acceleration of the rocket.

So momentum really doesn't differ that much between dimensions.
I should qualify that with 'only when the surface push/resist ratios are equivalent' but due to the spatial nature of the different dimensioned worlds this is likely to be the case anyway.

So I would tend to think that momentum - or its effects at least - would tend to be fairly equivalent between our different dimensioned universes.
As I say, I don't think momentum has anything to do with orbits anyway, so I'll study that aspect next post...
Last edited by gonegahgah on Mon Sep 09, 2013 11:03 am, edited 2 times in total.
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Re: 4D orbits again

Postby gonegahgah » Sun Sep 08, 2013 12:55 pm

fallfromgrace wrote:also elasticity effect increases there; any shock wave gets absorbed faster

Hmmm, yes that would be true due to the shock wave radiating out in all surface directions rather than just straight down.
Cool. Interesting to consider. Again, probably not relevant to orbits but interesting to think about.
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Re: 4D orbits again

Postby fallfromgrace » Mon Sep 09, 2013 8:42 am

not relevant to orbital path isolated, but it counters the idea of orbital disruption with a speck of dust

here r a few details:
- if there were a multi layered 3-planet with it's 4d atmosphere it should have a greater capacity for any threats than one 2-planet have
- impact power of a smaller object towards greater object in 4d is lesser than in 3d, because difference in between 4d volumes is greater than difference in 3d case
- (already mentioned shock absorption)

along with Klitzing's argument, of lessened chances for collision, it makes 4d orbits even more difficult to disturb than in 3d
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Re: 4D orbits again

Postby fallfromgrace » Mon Sep 09, 2013 9:04 am

aside with mystical part of gravitational expression, there are thingies that allow fantastical features in 3-star system and it's followers

regressing from any whatsoever 4d 3-star system must always result in our 3d system, in my thoughts particularly our solar system; but revealing 4th spatial axis for sure can lead to multiple 3-star systems with very different shapes: beside 3-spheres, 4d cylinders with spherical base and 4d rings that all cross in central 3-star are also allowed along with many other shapes, as long as it yields 3d slice of 2-spheres known to us

this thought makes me feel like taking a break
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Re: 4D orbits again

Postby quickfur » Mon Sep 09, 2013 11:11 pm

gonegahgah wrote:[...]
First I just want to look at the area of 'momentum' as it applies to different dimensional universes.
As I've mentioned I don't think that momentum itself applies in our considerations of orbits but it still may be a useful idea to explore in itself and its applicability to different numbered spatial dimensions.

Momentum is a product of mass x velocity.
Mass in our 3D universe obviously has a spatial component to it but also has a density factor. So Mass is a product of volume x density.
This gives us L^3 x D (or L^3 x m / V; which works out to m anyway).

When dealing with physics equations, it's important not to confuse measurement (size) with mass, and it's important not to confuse volume with shape. And one should keep in mind that scalar values and vector values are fundamentally different.

Mass is basically "how much stuff" there is, a count of atoms, if you will (assuming that all the atoms have the same mass of course, just for simplicity). It doesn't really matter how these atoms are put together; if you start with 5 million atoms, say, then you can mold them into a ball, or a hollow sphere (which would be larger), or a long thin stick, or whatever other shape you fancy, but they are still 5 million atoms, no more, no less. Momentum is independent of shape; it only cares about how much mass there is -- how much "stuff" there is. It doesn't matter if this stuff is low-density sponge with lots of empty space in between, or if this stuff is a compact, dense blob of solid material. It doesn't matter if this stuff is a long line of single atoms (i.e., a 1D construct), or a flat 1-atom-thick sheet (i.e., a 2D construct), or any other shape of any other dimension. As long as there are 5 million atoms, that's how much mass the object has. That is to say, mass is independent of dimension, shape, and density.

Therefore, momentum is also independent of dimension, shape, and density.

You could think of it as every atom in the object having the same fixed amount of resistance to change in motion. (Well, this isn't strictly true, but for simplicity's sake, let's say all the atoms in the object have the same mass. If not, we can always divide them into subatomic particles of the same mass, at least in theory.) So if there are 100 atoms in the object, and the object is travelling at 5 km/h, that means each of those 100 atoms are travelling at 5 km/h, so in order to stop the object, you have to stop each of its 100 atoms. So that's 100*5 = 500 units of momentum. If there are 500 atoms in another object that's also travelling at 5 km/h, then to stop that object, you need to stop an additional 400 atoms. Since there are more atoms to "stop", it requires more effort; the momentum of the second object is 500*5 = 2500 units.

Notice that in the above scenarios, the momentum is independent of the shape of the objects. Two objects that have the same mass (e.g., both consist of 500 atoms) will have the same momentum, even if one is shaped like a long tube, and the other is a ball.

Now, in the real world, of course, atoms don't all have the same mass; but the same principles apply. Suppose object A consists of atoms of type X, and object B consists of atoms of type Y. Suppose atoms of type X have 5 units of mass, and atoms of type Y have 10 units of mass. Then if object A consists of 2000 atoms, and object B consists of 1000 atoms, then the mass of object A is 2000*5 = 10000, and the mass of object B is 1000*10 = 10000. That is, they have exactly the same mass. Again, it doesn't matter what their shape or density is; what matters is only "how much stuff" they contain. And since their mass is the same, that means if both of them are moving at, say, 20 km/h, then both of them have exactly the same momentum, even though they may have completely different shapes, and they are made of entirely different kinds of atoms.

Why is this important? It's because the same thing also applies across dimensions. Suppose object A actually exists in 2D space, and atoms of type X are 2D atoms roughly in the shape of circles. Suppose object B actually exists in 3D space, and atoms of type Y are 3D atoms occupying a roughly spherical volume. Does this make any difference to their behaviour? Not at all! Again, I emphasize that mass depends only on how much "stuff" there is in the object. It's completely oblivious to shape, density, and dimension. Under the same force, object A and object B will behave exactly the same way, because they have exactly the same mass, and if they are travelling with the same velocity, then they also have exactly the same momentum.

So, momentum is independent of dimension.

Moreover, any difference in density or volume between two objects can be made up simply by adding more mass to one of the objects. For example, let's say object C is a 2D disk of 2m radius, and object D is a 3D ball of 2m radius. Even though their radii are equal, they do not have the same mass, because there's a different amount of "stuff" in them. For simplicity's sake, let's say the atoms they are made of have the same mass, and that they both have density of 1. So then, the mass of object C is 2*pi*(2m)^2 = 8*pi (approx. 25.13), and the mass of object D is 4/3*pi*(2m)^3 = 32/3*pi (approx. 33.51). So even though in terms of measurement C and D are comparable, they actually have different masses.

But does this mean 3D objects have "inherently" more mass? Not at all! It just means the scale of things are different. For example, if we add more stuff to object C so that its radius increases to 2.309m (that is, sqrt(16/3)), then its mass would be 2*pi*(16/3) = 32/3*pi (approx. 33.51). That is, it now has exactly the same mass as D, and so it will behave the same way as D under the same forces.

I use this example to show that actually, any difference in atomic mass or density between 2D and 3D (or for that matter, 3D and 4D) is a difference merely in quantity; in terms of quality, there is no difference. Yes, a 3D object of radius 1 will have less mass than a 4D object of radius 1, but all you have to do is to increase the radius of the 3D object, and it will have the same mass as the 4D object. And as long as the amount of "stuff" in the object is the same, then it will have the same momentum when travelling under the same velocity. So there will be no difference in how momentum behaves between 3D and 4D.

[...]
So I would tend to think that momentum - or its effects at least - would tend to be fairly equivalent between our different dimensioned universes.
As I say, I don't think momentum has anything to do with orbits anyway, so I'll study that aspect next post...

Actually, momentum has everything to do with orbits. Kepler's equations of orbital motion are based directly on the momentum of the orbiting body. :)

In order to give more insight into this problem, let's, for now, ignore the whole issue of dimensionality in the first place. Let's forget about 3D vs. 4D, and consider how an object X, travelling with velocity V, behaves when it's subjected to some force F.

Let's say F is some kind of attractive force (not necessarily gravity) that pulls X towards some point S, and that furthermore, F changes depending on where X is. So F can be modelled as a "force field", or, specifically, a "scalar field", that is, given any point in space, we can assign a fixed number to it that represents how strong the force F is at that point. Note that we're not necessarily talking about gravity here, just some arbitrary force which has different strengths depending on where you are, but which doesn't change over time.

Now, as X travels through space (of unspecified dimension), it passes through points of different strengths of F. At one point, it may experience a stronger pull from F, and at the next point, it may experience a weaker pull from F. At each point along its path, X's velocity (and therefore momentum) changes according to how strong F is at that point. If F is very strong, then X's direction will change significantly, and it may slow down or speed up by quite a large amount. Buf if F is rather weak at that point, then X will, for the most part, continue travelling in almost the same direction and almost the same speed as before, except for just a slight change.

So the question is, what will be the path of X as it travels through this region of space under the force F? The first thing to note is that the path of X does not depend on the dimensionality of space! Here's the proof:

Suppose X is initially travelling in direction v1, for some vector v1. Suppose that, relative to X, the position of S is in the direction of v2. Since the force F always pulls X towards S, and never in any other direction, the result of F acting on X is that X will acquire a new direction that is some combination of v1 and v2. Let's call this new direction v_new. Then v_new = k1*v1 + k2*v2, for some scalar values of k1 and k2. If X is travelling very fast, then k1 has a large value -- that is, it has a stronger tendency to continue in the same direction it was going before. If X is travelling slowly, then k1 has a small value -- it's not as persistent in going in the same direction as before. Similarly, if F is very strong, then k2 has a large value -- F will succeed in pulling X towards S significantly, thus changing its direction of motion significantly. But if F is rather weak, then k2 has a small value -- F will still pull X towards S, but the overall effect is only slight. The relative values of k1 and k2 doesn't really matter here, though -- the important thing to note is that v_new is a linear combination of two vectors. This means that no matter what k1 and k2 are,, the resulting path of X can only be some combination of v1 and v2. It can only lie in the 2D plane defined by v1 and v2! This makes sense, because there is no other force which may move X out of this plane.

So, what this means is that the resulting path of X is always confined to the 2D plane defined by v1 and v2. It doesn't matter whether the ambient space is 2D, 3D, 4D, or even 5D. X's path always lies on this 2D plane.

So what defines what shape this path will take? Not the dimension of space, but how F acts on X. This is the key point. It's not the dimension of space that dictates how X will move; rather, it's how F acts on it. Or, more pertinently, the "shape" of the scalar field of F. That is, how the strength of F, as experienced by X, changes along its path. Note carefully here the distinction between the absolute strength of F (i.e., its strength is fixed given a fixed point in space) vs. the strength of F as experienced by X (depending on where X is, it will experience a different strength of F). If, along X's path, F remains consistently strong, then X's path will be very greatly changed from its original direction. If F remains consistently weak, then X's path will only deviate slightly from its original course.

Now let's consider various scenarios.

First, let's say X is travelling past S some distance away, not going directly away from or at S, but just passing by. So there are a few possibilities. Suppose the strength of F remains quite strong, and only diminishes slowly with distance. Then as X approaches S, there will be a slight increase in the strength of F. This increase causes X's path to veer in the direction of S. But let's say this veering isn't enough to put X on a collision path with S. So eventually, after passing the point of closest approach, X starts to move away from S again. As it does so, the strength of F begins to diminish, but only slightly. So F still exerts quite a strong effect on X, and its path continues to veer in the direction of S. But not strongly enough to stop X from moving away from S, so eventually, once X gets far enough, F is no longer strong enough to pull it back.

But suppose F changes very quickly with distance. Near S, F is extremely strong, but as you move away from S, the strength of F quickly falls to negligible levels. So if X is travelling towards S (but not directly), it will initially not feel very much pull from F, but as it gets closer, this pull increases very quickly, and soon, X finds its path being diverted drastically. So it's far more likely to get overwhelmed by the sudden increase in F's strength and end up on a collision path with S. But suppose its initial velocity is strong enough that this doesn't happen. So once it passes the point of closest approach, where it experiences a very drastic change in direction due to the fast increase in F's strength, it begins to move away from S. Since F's strength drops very quickly with distance, after a short time X finds itself only feeling a slight tug from F, and quickly, it gets far away enough again, that it no longer feels the force of F significantly.

Now let's come back to gravity.

What does it take for an orbit to happen? One possibility is that the initial velocity of X is such that when F acts on it, it only changes direction, but not speed. This implies that it must be travelling in a direction perpendicular to the direction of S. This in turn implies that the resulting path must be a perfect circle. Why? Because if it were not a circle, then X will be either farther away or closer to S after F acts on it. So at the next instant in time, X will experience either a weaker or stronger force, and since its velocity from the previous point in time hasn't changed, this means that now F acting upon it will cause a change in speed, since X's velocity and F's strength are no longer perfectly balanced. So if F acting on X always maintains its speed and only changes its direction, then X must be orbiting on a circular path. But, as we've pointed out on many occasions, even the slightest deviation from this perfect circle will cause F to either overwhelm X eventually (causing X to crash into the star), or to weaken so much that it's unable to keep X in orbit (X flies off into space). So in practice, this perfect balance pretty much never happens.

Another possibility is that X isn't travelling on a circular path, but initially, F is strong enough to pull X inwards towards S. So X begins to fall inwards towards S, but as it does so, its momentum increases. Since X is now closer to S, the strength of F increases also. If the strength of F increases too fast, like the second scenario above, then the increase in momentum is too weak, and X will be unable to escape the pull of F. So it will crash into the star. If the strength of F increases too slowly, then X's momentum will quickly increase and overcome the pull of F, and X will fly away from S again. Furthermore, F's strength hasn't grown enough to counterbalance X's momentum away from S, so it's unable to slow X down enough once it starts moving away from S, so X just flies off into outer space and never comes back. Only if the strength of F increase at the right rate, the increase in X's momentum will initially become strong enough to make X fly away from S, but since F is still quite strong, as X is moving away from S, F will succeed in slowing it down enough that eventually X doesn't have enough momentum to completely fly off; it will begin to fall back towards S again. And then this cycle repeats, and X begins to oscillate around S. This is what happens with elliptical orbits. It requires just the right balance of forces that this tug-and-pull oscillation happens, never so weakly that X just flies off into space, and never so strongly that X collides with S.

Notice that what makes the difference between a successful elliptical orbit and an unsuccessful one is the rate that the strength of F changes with distance. If F's strength changes too fast with distance, then there's not enough room for X's momentum to play the tug-and-pull "game" with F; it will either be quickly overwhelmed, or it will pull itself free easily and fly off into space. If F's strength changes too slowly with distance, then X's momentum never gets enough chance to increase fast enough when falling towards S to be able to escape collision, or, if it's flying away from S, never decreases fast enough for F to be able to pull it back eventually. Only when the rate change of F's strength is just right, can this tug-and-pull oscillation happen.

So, what decides whether orbiting motion is possible, is the rate of change of F over distance. The shape and density of the orbiting body ultimately is irrelevant. The dimension of space also doesn't have a direct bearing on this, because, as we've shown, orbital motion always happens in a 2D plane.

But now we come to the question, what decides the rate of change of F? If the flux theory of force propagation is correct, then the answer is the dimension of space. Under the flux theory, force is carried by "force carriers" that emanate equally in all directions from the source of the force. Since the source of the force doesn't "know" which direction the "targets" of the force will be, it sends out force carriers in all directions equally. What directions are available, then, relates directly to the dimensionality of space. So a 2D source of force will send force carriers in all directions along the 360° circle of possibilities, and a 3D source of force will send out force carriers in all directions through the surface of a sphere. Since the number of force carriers sent out doesn't change with distance, this means their density decreases in proportion with the surface area of the n-sphere. In 2D, this is just the circumference of the circle, which is 2*pi*r, so one would expect that forces in a 2D universe would obey a 1/r law. In 3D, this is the surface area of a sphere, which is 4*pi*r^2. So in 3D, forces obey a 1/r^2 law, which is what we observe in our world. Therefore, one expects that in a 4D universe, where the surface area of the 3-sphere is 2*pi^2*r^3, forces would obey a 1/r^3 law.

So this is where the dimensionality of space comes into play. Even though orbital in itself doesn't depend on the dimensionality of space, it does depend on the rate by which gravity diminishes over distance. And that, in turn, does depend on the dimensionality of space. Since 4D gravity is expected to obey a 1/r^3 law, this means that in 4D, gravity increase very quickly with decreasing distance, and likewise also decreases rapidly with increasing distance. The result, if you work out the math behind it, is that this rate of change is simply too drastic for the momentum of an orbiting object to be able to balance out in a the tug-and-pull oscillation. Either the force of gravity increases so fast that the object is overwhelmed and spirals inwards to collide with the star, or the force of gravity decreases so fast that, should the object manage to overcome the pull of gravity, gravity will rapidly lose its hold upon the object and it will just fly off into space.

In the end, it's all about the rate of change. The actual scalar quantities don't really matter that much -- if the planet is too small, you can just make it bigger. If it's too big, just make it smaller. But no matter what you do, you can't change the rate at which the force of gravity increases/decreases. Not unless you posit that the flux theory is wrong, and that forces obey an inverse square law regardless of dimension. In that case, stable orbits will be possible in all dimensions. What decides the game is the rate of change. All the other factors are merely parameters.
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Re: 4D orbits again

Postby gonegahgah » Tue Sep 10, 2013 4:18 am

quickfur wrote:When dealing with physics equations, it's important not to confuse measurement (size) with mass...

No worries, I don't do that. The equations are correct though. Eureka!

quickfur wrote:Therefore, momentum is also independent of dimension, shape, and density.

But the equation remains momentum = mass x velocity = (volume x density) x velocity.

quickfur wrote:Notice that in the above scenarios, the momentum is independent of the shape of the objects. Two objects that have the same mass (e.g., both consist of 500 atoms) will have the same momentum, even if one is shaped like a long tube, and the other is a ball.

This is why I expressed it as if the whole object were trying to maintain its speed through a point in space. I used volume only to mean that the whole volume effectively acts through that point to give our measure of momentum of the object.
Momentum independently is not spread over an area (though different subareas can have different momentum); it is a property of the whole object at once; hence its value is equal to the effect of passing the object through a point all at once.

quickfur wrote:I use this example to show that actually, any difference in atomic mass or density between 2D and 3D (or for that matter, 3D and 4D) is a difference merely in quantity; in terms of quality, there is no difference. Yes, a 3D object of radius 1 will have less mass than a 4D object of radius 1, but all you have to do is to increase the radius of the 3D object, and it will have the same mass as the 4D object. And as long as the amount of "stuff" in the object is the same, then it will have the same momentum when travelling under the same velocity. So there will be no difference in how momentum behaves between 3D and 4D.

Because of the missing dimensions in lower dimensions it's impossible to have the same amount of stuff. A circle in 2D with infinite radius has zero stuff in 3D so couldn't even beat a 1cm sphere in 3D.
That's always been one of the fun things about dimensions...
It's only when we look at equivalency of effect between the dimensions that we can find correspondence or divergence.

qickfur wrote:Actually, momentum has everything to do with orbits. Kepler's equations of orbital motion are based directly on the momentum of the orbiting body. :)

You'll have to point out the equations. I had a quick look and there are some that include a p but this stands for the "semi-latus rectum" not for momentum.

quickfur wrote:But now we come to the question, what decides the rate of change of F? If the flux theory of force propagation is correct, then the answer is the dimension of space. Under the flux theory, force is carried by "force carriers" that emanate equally in all directions from the source of the force. Since the source of the force doesn't "know" which direction the "targets" of the force will be, it sends out force carriers in all directions equally. What directions are available, then, relates directly to the dimensionality of space. So a 2D source of force will send force carriers in all directions along the 360° circle of possibilities, and a 3D source of force will send out force carriers in all directions through the surface of a sphere. Since the number of force carriers sent out doesn't change with distance, this means their density decreases in proportion with the surface area of the n-sphere. In 2D, this is just the circumference of the circle, which is 2*pi*r, so one would expect that forces in a 2D universe would obey a 1/r law. In 3D, this is the surface area of a sphere, which is 4*pi*r^2. So in 3D, forces obey a 1/r^2 law, which is what we observe in our world. Therefore, one expects that in a 4D universe, where the surface area of the 3-sphere is 2*pi^2*r^3, forces would obey a 1/r^3 law.

So this is where the dimensionality of space comes into play. Even though orbital in itself doesn't depend on the dimensionality of space, it does depend on the rate by which gravity diminishes over distance. And that, in turn, does depend on the dimensionality of space. Since 4D gravity is expected to obey a 1/r^3 law, this means that in 4D, gravity increase very quickly with decreasing distance, and likewise also decreases rapidly with increasing distance. The result, if you work out the math behind it, is that this rate of change is simply too drastic for the momentum of an orbiting object to be able to balance out in a the tug-and-pull oscillation. Either the force of gravity increases so fast that the object is overwhelmed and spirals inwards to collide with the star, or the force of gravity decreases so fast that, should the object manage to overcome the pull of gravity, gravity will rapidly lose its hold upon the object and it will just fly off into space.

I've read all your post quickfur but I apologise for bypassing much of it at the moment.
It is this last aspect you mention that may not be possible to discount. I don't know enough yet to know and want to tackle some other stuff first leading hopefully to that.
But before we work towards that I wanted to make sure we are on the same page when it comes to gravity itself.

I'll explain a scenario and see if we agree on what the result would be, if I may?
Let's say we were able to create a space ride and we had some magical gravity generators that worked exactly like real gravity.
We put the rider in the capsule. Unfortunately for the rider the capsule doesn't have windows and space is black anyway.
We set up the grav generators around the circuit and start turning them off and on in sequence and the capsule shoots around space changing in all sorts of directions passed each grav unit.
Finally it pulls to a stop back into the start with just the right amount of final gravity zing.
Knowing how fun some of the new theme park rides are that just drop you; what sort of ride would our riders have had?
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Re: 4D orbits again

Postby quickfur » Tue Sep 10, 2013 3:47 pm

gonegahgah wrote:[...]
quickfur wrote:Notice that in the above scenarios, the momentum is independent of the shape of the objects. Two objects that have the same mass (e.g., both consist of 500 atoms) will have the same momentum, even if one is shaped like a long tube, and the other is a ball.

This is why I expressed it as if the whole object were trying to maintain its speed through a point in space. I used volume only to mean that the whole volume effectively acts through that point to give our measure of momentum of the object.
Momentum independently is not spread over an area (though different subareas can have different momentum); it is a property of the whole object at once; hence its value is equal to the effect of passing the object through a point all at once.

Yes, that's correct.

quickfur wrote:I use this example to show that actually, any difference in atomic mass or density between 2D and 3D (or for that matter, 3D and 4D) is a difference merely in quantity; in terms of quality, there is no difference. Yes, a 3D object of radius 1 will have less mass than a 4D object of radius 1, but all you have to do is to increase the radius of the 3D object, and it will have the same mass as the 4D object. And as long as the amount of "stuff" in the object is the same, then it will have the same momentum when travelling under the same velocity. So there will be no difference in how momentum behaves between 3D and 4D.

Because of the missing dimensions in lower dimensions it's impossible to have the same amount of stuff. A circle in 2D with infinite radius has zero stuff in 3D so couldn't even beat a 1cm sphere in 3D.
That's always been one of the fun things about dimensions...
It's only when we look at equivalency of effect between the dimensions that we can find correspondence or divergence.

But this does not negate the point that mass is a scalar quantity, and that any discrepancy between relative scalar quantities can be adjusted by simply making the relevant objects bigger or smaller. I suppose comparing objects of different dimensions is moot, because a "true" 2D object (i.e., an infinitely thin one) can't exist in 3D anyway, so the comparison is invalid. But what I was trying to get at, was that the number of dimensions of the object doesn't matter when it comes to calculating the momentum, because the final value is just a scalar value. So in 4D you have an additional measurement (an additional dimension), but all that does is to add another scalar factor to your mass calculation. So the momentum that comes out is larger by some constant factor W. But that doesn't change the fact that it's a constant scalar value. Which means that if the resulting mass is "too big" for the purposes of orbital motion, we can simply shrink the object to make it the right size. If you assume some kind of equivalence in strength between 3D gravity and 4D gravity, then this just means that you make 4D planets have a slightly smaller radius so that their masses are comparable to 3D planets.

And like I pointed out, even the dimension of space itself is unimportant, because, like you said, momentum can be treated as though it is concentrated in a single point. And as I explained in my post, this means the resulting system has a single point S (the star that the planet is orbiting around), a single point X representing the orbiting body, and a force F that acts radially from S to F. The only two vectors at play at the initial velocity of X and the direction of F. So, boiled down to the essentials, this is just two points on a 2D plane. The orbiting body never moves outside this plane, since there is no external force that may cause it to do so. Which means the behaviour of this system depends entirely on how the initial velocity interacts with the force F.

Now the initial velocity can be freely adjusted, since we're looking for the optimal conditions in which we can obtain an orbiting system. So that already automatically excludes any initial velocity that would lead to a collision or X flying off into space. We don't have to consider those cases since we're interested in stable orbits. This leaves F as the only remaining variable. As I explained in my post, the rate of change of F with distance is what determines what kind of orbital paths will be possible. If it changes too quickly or too slowly with distance, then the momentum of the orbiting body will not be able to balance it out in a way that gives rise to harmonic motion (i.e., a stable orbit). So it's the rate of change of F with distance that decides whether or not stable orbits are possible, and what kind of orbits they will be.

qickfur wrote:Actually, momentum has everything to do with orbits. Kepler's equations of orbital motion are based directly on the momentum of the orbiting body. :)

You'll have to point out the equations. I had a quick look and there are some that include a p but this stands for the "semi-latus rectum" not for momentum.

Sorry, I was not accurate in what I said. Kepler's equations themselves describe what kinds of orbital paths exist, they don't directly relate to momentum. However, the paths described by Kepler's equations are possible only because in 3D, momentum balances out gravity in a nice way. I recommend reading this article on Wikipedia for the details:

http://en.wikipedia.org/wiki/Kepler_problem

One of our forum members, pat (who's unfortunately been inactive for a long time now), wrote up a derivation of 4D orbital mechanics based on the same principles, except with 1/r^3 gravity instead of 1/r^2 (so instead of F=(k/r^2)r, he uses F=(k/r^3)r). When he got to the end, he found that this extra factor of 1/r caused a drastic change in the final equation, that basically implied that the only stable orbit was a perfect circle. The analogues of the elliptical orbits in 3D turned out to be non-periodic in 4D: the extra 1/r factor caused them to turn into spiralling paths, where each period around the star would add/subtract a constant amount from their distance to the star. They are somewhat more stable than the really unstable orbits (which correspond with the unstable orbits in 3D), but unlike 3D, they are non-periodic. Each orbit brings them closer to the star or farther out from the star. So after a sufficient number of orbits, the planet would either crash into the star, or be so far away that it couldn't possibly be life-sustaining (there would be almost no heat received from the star, and it would be so far away observers on the planet wouldn't even notice it -- not to mention that it continues drifting away from the star).

quickfur wrote:But now we come to the question, what decides the rate of change of F? If the flux theory of force propagation is correct, then the answer is the dimension of space. Under the flux theory, force is carried by "force carriers" that emanate equally in all directions from the source of the force. Since the source of the force doesn't "know" which direction the "targets" of the force will be, it sends out force carriers in all directions equally. What directions are available, then, relates directly to the dimensionality of space. So a 2D source of force will send force carriers in all directions along the 360° circle of possibilities, and a 3D source of force will send out force carriers in all directions through the surface of a sphere. Since the number of force carriers sent out doesn't change with distance, this means their density decreases in proportion with the surface area of the n-sphere. In 2D, this is just the circumference of the circle, which is 2*pi*r, so one would expect that forces in a 2D universe would obey a 1/r law. In 3D, this is the surface area of a sphere, which is 4*pi*r^2. So in 3D, forces obey a 1/r^2 law, which is what we observe in our world. Therefore, one expects that in a 4D universe, where the surface area of the 3-sphere is 2*pi^2*r^3, forces would obey a 1/r^3 law.

So this is where the dimensionality of space comes into play. Even though orbital in itself doesn't depend on the dimensionality of space, it does depend on the rate by which gravity diminishes over distance. And that, in turn, does depend on the dimensionality of space. Since 4D gravity is expected to obey a 1/r^3 law, this means that in 4D, gravity increase very quickly with decreasing distance, and likewise also decreases rapidly with increasing distance. The result, if you work out the math behind it, is that this rate of change is simply too drastic for the momentum of an orbiting object to be able to balance out in a the tug-and-pull oscillation. Either the force of gravity increases so fast that the object is overwhelmed and spirals inwards to collide with the star, or the force of gravity decreases so fast that, should the object manage to overcome the pull of gravity, gravity will rapidly lose its hold upon the object and it will just fly off into space.

I've read all your post quickfur but I apologise for bypassing much of it at the moment.
It is this last aspect you mention that may not be possible to discount. I don't know enough yet to know and want to tackle some other stuff first leading hopefully to that.

Well, if you want to derive accurate results, you'll unfortunately have to grapple with the equations behind it. While it's fun to speculate about stuff, sometimes we can unconscious end up with conclusions that have no basis in reality, if we're not adequately grounded in how the math actually works out.

I'll be the first to admit that I was extremely disappointed to learn about the instability of 4D orbits. I downloaded pat's paper (I forgot the link now, I'll try to look it up again), and tried to find flaws in it, because the idea of 4D planets was just so appealing I wasn't ready to give it up. I dug through every line of his derivation to see if there was a mistake somewhere, or perhaps a missed possibility, but it turned out that it was actually a relatively straightforward derivation (though it does require some level of facility with calculus to follow). After going at it for many hours, I realized (and in the paper pat himself pointed this out) that the source of the problem was that extra 1/r factor introduced by having a 1/r^3 gravity. This is basically what I was getting at in my post -- it's the rate of change of gravity that dictates whether stable orbits exist, and what shape(s) they will take. A 1/r^3 gravity basically changes in a way that's fundamentally incompatible with how momentum works. This is actually very clear in pat's article; you can see that whereas in the 3D case (the wikipedia page linked above), the 1/r^2 factor is nicely balanced out by the angular momentum L=mr^2*w in such a way, that the solution to the differential equation takes the form of a conic section, which includes two periodic paths: the circular and elliptical orbits, in the 4D case the 1/r^3 factor is not balanced out by the angular momentum, with the result that the solution has an additional factor that, even under the most favorable circumstances, will add or subtract a constant amount from the planet's orbital radius every orbital cycle. The only time this factor can be 0 is when the orbit is a perfect circle. Even the smallest deviation causes the planet to drift inwards or outwards by a constant amount every single orbit, and a meteor hitting the planet (a very frequent occurrence on the astronomincal timescale) would cause large changes in its orbital radius -- in almost all cases knocking it into an unstable path.

I tried -- believe me, I tried -- to find ways of getting rid of this irritating extra factor, but eventually I had to accept that it was an inevitable consequence of gravity obeying a 1/r^3 law in 4D. That was basically the only thing different about pat's derivation than the derivation in the wikipedia page (note that the dimensionality of space isn't even part of the equations -- like I said, it's basically motion in a 2D plane, and depends only on the way gravity interacts with the planet's initial momentum). It's the fact that it's 1/r^3 instead of 1/r^2 that causes stable orbits to be impossible (except for the perfect circle, which is unlikely in practice because it's too sensitive to small perturbations).

But before we work towards that I wanted to make sure we are on the same page when it comes to gravity itself.

I'll explain a scenario and see if we agree on what the result would be, if I may?
Let's say we were able to create a space ride and we had some magical gravity generators that worked exactly like real gravity.
We put the rider in the capsule. Unfortunately for the rider the capsule doesn't have windows and space is black anyway.
We set up the grav generators around the circuit and start turning them off and on in sequence and the capsule shoots around space changing in all sorts of directions passed each grav unit.
Finally it pulls to a stop back into the start with just the right amount of final gravity zing.
Knowing how fun some of the new theme park rides are that just drop you; what sort of ride would our riders have had?

You won't feel a thing, actually. :) When you're in orbit (which is basically what's happening here if the gravity generators really worked exactly like real gravity), you're in "free fall", and there is no net acceleration, so you don't feel anything at all.

In fact, theme park rides, in theory, can be made such that the riders don't feel a thing -- engineers deliberately design it to be slightly off-balance (but within the limits of safety) so that you will feel a changing amount of acceleration, because it's more fun that way. Nobody would be interested in a boring ride where they don't feel anything! :)

But on another note, I like this metaphor. Suppose we have clever engineers who can design such a ride in 4D, such that the gravity generators create a force that obeys a 1/r^2 law instead of a 1/r^3 law. Then guess what would happen? Our space ride would start to behave like 3D orbits -- circular and elliptical orbits -- in 4D space. If we make the cars in this ride big enough -- assuming the gravity generators can cope with the additional load -- then they can essentially become "planets" in a stable orbit. So if we can somehow have these magical gravity generators in 4D space on an astronomical scale, then we will also have stable 4D planets too! :)

The only downside is that such a system wouldn't exist naturally, since there's no good reason why any naturally-occurring 4D system would obey a 1/r^2 law instead of a 1/r^3 law. Unless we posit that 4D physics works in a drastically different way from 3D physics.
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Re: 4D orbits again

Postby gonegahgah » Fri Sep 13, 2013 10:36 pm

I found a variable gravity simulator at: http://www.testtubegames.com/gravity.html
Looks like orbits are possible in 2D...
Getting anything in 4D.......... different question...
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Re: 4D orbits again

Postby quickfur » Sat Sep 14, 2013 10:58 pm

Interesting. Playing around with 1/r^3 gravity, it looks the vast majority of paths are cubic (i.e., y=x^3) paths; unfortunately, non-quadratic cubic (i.e. those that can't be reduced to y=x^2 type equations) paths are all aperiodic. So the chances of a permanent orbit seems to be really really slim. You'd be lucky to even get a spiralling path, much less an actual circular orbit!
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Re: 4D orbits again

Postby gonegahgah » Sun Sep 15, 2013 6:24 am

I actually managed to get the planet to go around almost twice before it shots off into space with:
Gravity Fun at TestTubeGames.com: [ForceG: -3,Qual: 1,Zoom: 1,xSet: 85,ySet: 85], [x0: -129,y0: -16,vx: 0,vy: 0,t0: 0,who: 1,m: 1000], [x0: -41,y0: -128,vx: 1.25,vy: 2.05,t0: 40,who: 2,m: 100]

I get the feel that there is some harmonic nature to our 1/x^2 gravity that gives us our nice elliptical orbits.
Anything else seems to produce flowers or pure crashes or pass-bys.
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Re: 4D orbits again

Postby gonegahgah » Sun Sep 15, 2013 1:07 pm

This one goes around 4 times before it shoots off into space...
Gravity Fun at TestTubeGames.com: [ForceG: -3,Qual: 1,Zoom: 1,xSet: 85,ySet: 85], [x0: -129,y0: -16,vx: 0,vy: 0,t0: 0,who: 1,m: 1000], [x0: -41,y0: -128,vx: 1.25,vy: 2.02,t0: 40,who: 2,m: 100]
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Re: 4D orbits again

Postby fallfromgrace » Sun Sep 15, 2013 4:38 pm

I can't follow differential calculation through http://www.csh.rit.edu/~pat/lj/orbit4.pdf, especially its ending, so I'll take a break for sure. I guess that this calculation is the back up some of u take. There was an attempt to try to hop in 4d orbit though.
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Re: 4D orbits again

Postby quickfur » Mon Sep 16, 2013 5:21 am

gonegahgah wrote:I actually managed to get the planet to go around almost twice before it shots off into space with:
Gravity Fun at TestTubeGames.com: [ForceG: -3,Qual: 1,Zoom: 1,xSet: 85,ySet: 85], [x0: -129,y0: -16,vx: 0,vy: 0,t0: 0,who: 1,m: 1000], [x0: -41,y0: -128,vx: 1.25,vy: 2.05,t0: 40,who: 2,m: 100]

I get the feel that there is some harmonic nature to our 1/x^2 gravity that gives us our nice elliptical orbits.
Anything else seems to produce flowers or pure crashes or pass-bys.

Yep, now you're starting to understand why 4D orbits are so unlikely. :)

Not to mention, there is a fundamental connection between 1/x^2 gravity and elliptical orbits, in the sense that the ^2 exponent connects it with conic sections (which are curves of the form Ax^2 + By^2 + Cy = D). An ellipse, for example, has an equation of the form (Ax)^2 + (By)^2 = C. The quadratic nature of these equations link them to circles and ellipses (also parabolas and hyperbolas, but those are the non-orbiting cases).

Once you throw in a 1/x^3 factor, though, it's totally a different ballgame, 'cos now you're dealing with cubic (i.e., x^3) curves, which are of a fundamentally different sort: almost all cubic curves are aperiodic.
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Re: 4D orbits again

Postby gonegahgah » Mon Sep 16, 2013 11:35 am

quickfur wrote:Yep, now you're starting to understand why 4D orbits are so unlikely. :)

Not to mention, there is a fundamental connection between 1/x^2 gravity and elliptical orbits, in the sense that the ^2 exponent connects it with conic sections (which are curves of the form Ax^2 + By^2 + Cy = D). An ellipse, for example, has an equation of the form (Ax)^2 + (By)^2 = C. The quadratic nature of these equations link them to circles and ellipses (also parabolas and hyperbolas, but those are the non-orbiting cases).

Once you throw in a 1/x^3 factor, though, it's totally a different ballgame, 'cos now you're dealing with cubic (i.e., x^3) curves, which are of a fundamentally different sort: almost all cubic curves are aperiodic.

Yes, thanks. You could tell I was cautious to say otherwise because past experience has taught me that you know and investigate your stuff well. Also, particular things you said were ringing alarm bells.
When I first compared the 1/x^2 and 1/x^3 graphs separately I thought they looked fairly similar but when I looked at them superimposed I could see that the former had a more gradual curve and the later look more like a corner.
This probably parallels what you were saying originally; similar to along the lines of it being easier to balance a ball on a dome than it is to balance a ball on the corner of a rounded rectangle. ie.
Image

It is sad and I can't immediately think of a solution. It does seem to show - along with other things - how much of a Goldilocks zone we appear to live in. Its quite amazing.
You would need some other dynamics to overcome this it would seem; if it is at all possible to create a Goldilocks zone in a 4D universe?
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Re: 4D orbits again

Postby quickfur » Mon Sep 16, 2013 2:59 pm

gonegahgah wrote:
quickfur wrote:Yep, now you're starting to understand why 4D orbits are so unlikely. :)

Not to mention, there is a fundamental connection between 1/x^2 gravity and elliptical orbits, in the sense that the ^2 exponent connects it with conic sections (which are curves of the form Ax^2 + By^2 + Cy = D). An ellipse, for example, has an equation of the form (Ax)^2 + (By)^2 = C. The quadratic nature of these equations link them to circles and ellipses (also parabolas and hyperbolas, but those are the non-orbiting cases).

Once you throw in a 1/x^3 factor, though, it's totally a different ballgame, 'cos now you're dealing with cubic (i.e., x^3) curves, which are of a fundamentally different sort: almost all cubic curves are aperiodic.

Yes, thanks. You could tell I was cautious to say otherwise because past experience has taught me that you know and investigate your stuff well. Also, particular things you said were ringing alarm bells.
When I first compared the 1/x^2 and 1/x^3 graphs separately I thought they looked fairly similar but when I looked at them superimposed I could see that the former had a more gradual curve and the later look more like a corner.
This probably parallels what you were saying originally; similar to along the lines of it being easier to balance a ball on a dome than it is to balance a ball on the corner of a rounded rectangle. ie.
Image

It is sad and I can't immediately think of a solution. It does seem to show - along with other things - how much of a Goldilocks zone we appear to live in. Its quite amazing.
You would need some other dynamics to overcome this it would seem; if it is at all possible to create a Goldilocks zone in a 4D universe?

Yep, it's quite amazing indeed.

OTOH, as far as a 4D universe is concerned, the way I see it is that it would be of a fundamentally different organization than anything we've ever experienced or imagined. Dimensional analogy is useful for helping us come to grips with how 4D space works, but it is by no means anything even remotely resembling what a "true native" 4D universe would look like. If a 4D universe actually exists, and if 4D living beings actually exist, I'd expect that they'd be so foreign to our concept that we wouldn't even recognize them as living beings in the first place. This is where realism gets in the way of fun, I guess. I'm perfectly happy with postulating unlikely hypothetical circular planetary orbits, since it's more fun that way (we get to actually understand what we're talking about, for one thing), rather than trying to grapple with a "true" 4D world that is beyond our ability to comprehend. Ultimately, it all boils down to using analogies of familiar objects to help us grok the nature of 4D space. It's totally another kettle o' fish to talk about native self-sustaining systems that arise naturally in a 4D environment. We can't even answer such things about our own world (except what we observe externally), let alone a 4D one!
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Re: 4D orbits again

Postby gonegahgah » Sat Sep 21, 2013 8:39 am

quickfur wrote:OTOH, as far as a 4D universe is concerned, the way I see it is that it would be of a fundamentally different organization than anything we've ever experienced or imagined. Dimensional analogy is useful for helping us come to grips with how 4D space works, but it is by no means anything even remotely resembling what a "true native" 4D universe would look like. If a 4D universe actually exists, and if 4D living beings actually exist, I'd expect that they'd be so foreign to our concept that we wouldn't even recognize them as living beings in the first place. This is where realism gets in the way of fun, I guess. I'm perfectly happy with postulating unlikely hypothetical circular planetary orbits, since it's more fun that way (we get to actually understand what we're talking about, for one thing), rather than trying to grapple with a "true" 4D world that is beyond our ability to comprehend. Ultimately, it all boils down to using analogies of familiar objects to help us grok the nature of 4D space. It's totally another kettle o' fish to talk about native self-sustaining systems that arise naturally in a 4D environment. We can't even answer such things about our own world (except what we observe externally), let alone a 4D one!

I'm a big believer in the effects of relative spin and also of the battle between gravity and pressure, and that these are relevant to explaining things in our world.
I believe science ignores these aspects because they are difficult; especially spin.
I'm hoping this could hopefully lead to some 4D solutions maybe; but I'm just guessing of course.

The interesting thing about spin is that it mitigates the effects of gravity.
If a planet were able to spin fast enough it would break apart because the spin would effectively cancel out the effect of the gravity.
Some recognition is given to spin at planet and larger scales. It is known in science as frame dragging.
If you shot a series of spinning planets passed an asteroid you would have to not only take gravity into account but this frame dragging as well.

As we've discussed a 4D world is interesting in that it can allow two simultaneous axis of spin at once.
I wonder what effect this would have on the shape of planets?
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Re: 4D orbits again

Postby quickfur » Mon Sep 23, 2013 11:45 pm

gonegahgah wrote:[...]
As we've discussed a 4D world is interesting in that it can allow two simultaneous axis of spin at once.
I wonder what effect this would have on the shape of planets?

That's an interesting question! In 3D, planets usually have some amount of spin, and planets with a large amount of spin tends to "bulge" at their equators due to centrifugal effects.

A 4D planet would, if Wendy is right, eventually settle into a double rotation where the two orthogonal rotations have the same rate. If this rate of rotation is very high, then we'd expect the planet to be deformed along two great circles, and have a shape similar to a rounded version of what Wendy calls a bicircular tegum. Basically, it's a deformation of the 4D sphere where it bulges along two orthogonal great circles, with a toroidal region in between.

And assuming the weather model I've described elsewhere, this toroidal region corresponds with the 4D equivalent of the temperate zones, and the two bulging great circles with the tropics and the zone of perpetual twilight. Hmm. Interesting.

Anyway, I'm not sure how all of this relates to planetary systems, because this kind of double rotation requires a solid spherical body, not isolated orbiting objects. There just doesn't seem to be any way to get a 4D analogue of a solar system that lasts long enough to be interesting.

OTOH, the thought just occurred to me that perhaps it's possible for large 4D brown dwarfs to generate enough internal heat to sustain life, so that instead of drawing energy from a sun, the inhabitants would draw energy from the hot interior of the brown dwarf. But then it wouldn't have interesting weather patterns analogous to Earth, and it wouldn't have seasons and day/night cycles. So it'd be pretty boring. :P
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Re: 4D orbits again

Postby Polyhedron Dude » Tue Sep 24, 2013 6:34 am

If the two rotations have the same rate, something amazing happens, every point on its surface would trace a great circle as it spins. The glome is now "swirling" - AKA Clifford rotation. The two equators would actually vanish and blend in as two circles in a continuum of equators that swirl around each other, this is called Hopf fibration. Every point on the glome would actually be on an equator. If the rate of rotation was very fast, the glome would simply expand.
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Re: 4D orbits again

Postby wendy » Tue Sep 24, 2013 8:13 am

The transfer of energy between modes of rotation and vibration are quite well understood: the branch of thermodynamics deals with it at atomic levels, and the effect of tidal drag etc is how it happens at a planetary scale. A planet in four dimensions has two independent rotations, which would, for a fairly spherically symmetrical planet, settle down into a clifford-style rotation, where every point on the surface goes around the centre, since this equalises the energies in the different modes of rotation.

Supposing that we have a means for stable orbits, it is possible for a planet in 4d to have seasons etc, even though there is no solistace etc. Because every part of the planet goes around the centre, they see the sun for 12 hours every day, no longest-day shortest-day stuff. But seasons are given by more than solstaces, but also cumulations of the sun: how high does it get?

In 3d, if you divide the sphere by its rotation, you get the gimble or half-cirlce that is used to hold the globe. This goes from +90 to -90, with 0 as the equator or great circle.

In 4d, the division of the glome by its rotation, gives a sphere. Each point of this sphere represents all the places that see the same star as its zenith: a hedrix through the centre. Any diametricly opposite pair of points represent a valid pair of axies for defining the rotation. Each line of lattitude on this sphere represents a 'torus' or bi-circular prism or duo-cylinder. Unfolded this gives a rectangle of sides cos(theta), sin(theta), the normal rotations going diagonally bottom left to top right, eg, and 'toric parallels' to it.

If the motion of the sun in the sky is not one of these lines, it will cross a torus in the opposite direction (top-left to bottom right), and on the sphere you will get a 'tropics' lattitude where it is possible for the sun to be overhead. Where in 3d, the sun varies between the tropics of capricorn and cancer, crossing the equator, in 4d, the sun stays over the tropics, and moves through all seasons, so there's a tropic of leo and a tropic of taurus etc.

If we want to find out how the climate is, we suppose that this circle of tropics is 23.5 degrees, where 0 degrees is one pole, and 90 degrees is the the opposite pole: angles are all doubled, so our tropics at 23.5 degrees, becomes 47 degrees north of the SP, or 43 degrees S. The northern artics is 43 N, so all of the temperate zone lies between 43S and 43N, ie a 4d planet tilted at the same tilt as our world is prodominately temperate.

The cumulation of the sun is the same distance as the distance from your local dot on this and where the sun is. So if the sun is on the same 'longitude' as you are, it's mid summer, and if it's opposite, then it's mid winter. In place of having the seasons, and the seasons displaced by 6 months in the N hemisphere, you get 'season-zones' like time zones. So just as you can be '15 hours behind X' time-wise, you can be '3 months ahead of X' by the season-zone.

In total, it is possible to calculate three points for a given locality, like you do in 3d (lattitude = climate, longitude = day-cycle, + 2 season-phases N/S). In 4d, you get climate-line (hot to cold), and year-phase (winter in december, vs winter in april), and day-cycle (noon at 9.00 pm GMT or noon at 2 am GMT.
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Re: 4D orbits again

Postby Polyhedron Dude » Tue Sep 24, 2013 9:57 am

wendy wrote:In total, it is possible to calculate three points for a given locality, like you do in 3d (lattitude = climate, longitude = day-cycle, + 2 season-phases N/S). In 4d, you get climate-line (hot to cold), and year-phase (winter in december, vs winter in april), and day-cycle (noon at 9.00 pm GMT or noon at 2 am GMT.


I like to call the 4-D directions lattitude (climate, measures north and south), longitude (day cycle, measures east and west), and laptitude (year-phase, measures marp and garp).
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Re: 4D orbits again

Postby quickfur » Tue Sep 24, 2013 5:59 pm

Polyhedron Dude wrote:If the two rotations have the same rate, something amazing happens, every point on its surface would trace a great circle as it spins. The glome is now "swirling" - AKA Clifford rotation. The two equators would actually vanish and blend in as two circles in a continuum of equators that swirl around each other, this is called Hopf fibration. Every point on the glome would actually be on an equator. If the rate of rotation was very fast, the glome would simply expand.

You are right, I didn't think of it this way before. But it totally makes sense because of the way the rotations work out. Amazing indeed! So there is no distinction between the two "equators" and the rest of the planet's surface.

So you could say that the planet's orientation doesn't even matter anymore, what determines the seasons is basically which fibre bundle (i.e., toroidal region) is most illuminated by the sun. In my previous analysis, I assumed that one of the equators would coincide with the planet's orbital plane (well, at least, in the simplest case). This would then give rise to a toroidal region that's analogous to our tropical regions -- they get the most sunshine, have full day/night cycles with equal day/night lengths; an orthogonal toroial region that is in perpetual twilight -- the sun never rises above the horizon, but circles around it as the planet rotates; and another toroidal region in between these two which behave like the temperate zones.

Now, in a slightly more realistic case (as realistic as it can get if we assume a perfect circular orbit), the planet's rotation would be at a slight angle to the orbital plane, so the tropical region would vary slightly over the year, but more-or-less remain at almost equal day/night lengths, and one would expect that temperatures remain more or less constant throughout the year. The perpetual twilight region (the "arctic circle"? :lol: Though "circle" here is used in a different sense. Maybe "arctic zone"?) would undergo half-year day and half-day night, just like on our Earth above the arctic circle. The temperate zones would then undergo yearly seasons. The interesting thing, though, is that the arctic zone in the 4D case isn't confined to a small area around the poles -- there are no poles in a planet undergoing Clifford rotation -- but covers an area just as wide as the tropical zone! So a rather significant proportion of the planet's surface would have arctic climate. (In higher dimensions, if we carry this analogy further, the arctic region would actually cover the majority of the planet's surface -- the higher the dimension you go, the more frigid the planet becomes; the warm tropical regions would just be confined to a narrow band.)
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Re: 4D orbits again

Postby wendy » Wed Sep 25, 2013 7:50 am

Let's look at clifford rotations from complex numbers and CE2.

CE2 is simply algebraic euclidean geometry, done with complex numbers. So you use your ordinary rules, like that of a line as "y = ax+b", but all of the numbers are complex. y=b when x=0, so if we are going to set b=0.

The equation y=ax tells us that there are infinite argand diagrams, one for every complex number 'a', that contain the point (0,0). In 4d, this means that you can pass an infinite number of hedrixes (2d spaces), through a point where the origion is the sole common point to all planes. Drawing a glome, centred on the origion means that the surface contains a mob of great circles that don't intersect, and that there is a great circle for every point. These make 'clifford parallels'.

Put w = rotation speed, and t = time, we write y.wt = ax.wt, which means that you can rotate this glome at a constant, consistant rate, ie w=w' as t=t', such that y/x = a (ie the points never leave the same circle as they rotate). This is 'clifford rotation'.

A circle, not part of the fibulation, will contain points belonging to several parallels. It crosses each parallel at the same angle, and it is eqidistant, and in the same direction as a singular parallel at the smae distance from the parallels. On the glome, if you draw a line perpendicular to both the parallel and the circle, all of these lines will cross at the common 'polar' circle.
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Re: 4D orbits again

Postby wendy » Thu Sep 26, 2013 6:35 am

The percentage of the planets given to the tropics, temperate and polar are as follows

3D tropics = 39.9%, temperate = 51.8%. polar = 8.3%
4D tropics = 15.9%, temperate = 68.2%, polar = 15.9%

The formula is for an tilt angle of þ, gives

3d tropics = sin(þ); temperate = cos(þ)-sin(þ), polar = 1-cos(þ)
4d temperate = sin(þ); tropics = polar = (1-sin(þ))/2,
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Re: 4D orbits again

Postby Polyhedron Dude » Thu Sep 26, 2013 12:36 pm

wendy wrote:The percentage of the planets given to the tropics, temperate and polar are as follows

3D tropics = 39.9%, temperate = 51.8%. polar = 8.3%
4D tropics = 15.9%, temperate = 68.2%, polar = 15.9%

The formula is for an tilt angle of þ, gives

3d tropics = sin(þ); temperate = cos(þ)-sin(þ), polar = 1-cos(þ)
4d temperate = sin(þ); tropics = polar = (1-sin(þ))/2,


This is quite interesting, any chance you've calculated formulas for 5-D and above? I suspect that the tropics percentage would decrease, the polar would increase, and the temperate would reach a maximum and then decrease.
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Re: 4D orbits again

Postby quickfur » Thu Sep 26, 2013 5:37 pm

Polyhedron Dude wrote:
wendy wrote:The percentage of the planets given to the tropics, temperate and polar are as follows

3D tropics = 39.9%, temperate = 51.8%. polar = 8.3%
4D tropics = 15.9%, temperate = 68.2%, polar = 15.9%

The formula is for an tilt angle of þ, gives

3d tropics = sin(þ); temperate = cos(þ)-sin(þ), polar = 1-cos(þ)
4d temperate = sin(þ); tropics = polar = (1-sin(þ))/2,


This is quite interesting, any chance you've calculated formulas for 5-D and above? I suspect that the tropics percentage would decrease, the polar would increase, and the temperate would reach a maximum and then decrease.

That's my suspicion too, though I have not rigorously checked it. I'm expecting that in a high-enough dimension, only a very narrow band of the planet would actually get significant amounts of sunlight, and most of the planet would be cold and in perpetual twilight (i.e. the sun never passes overhead).

It seems that solar systems that can sustain life are a peculiar feature of 3D, not just in the sense that only 3D and 2D orbits are stable, but also that an orbit-based system in higher dimensions would not provide the same kind of seasons/warm-cold cycles that they do in 3D. It almost suggests that if life is to exist in higher dimensions, it would require far more exotic living environments than we can infer just by dimensional analogy.
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Re: 4D orbits again

Postby wendy » Fri Sep 27, 2013 7:44 am

Seasons are supported in every dimension. It's not hard to see how it works, if you look at the sky, rather than the planets.

The sun plods along its circle in the sky, which we shall call the zodiac. This is a great arrow, the sun moves along it once a year. The zenith point is the point directly overhead, it basically follows some kind of circle in the sky too, once a day. Usually, the zodiac is not one of these zenith circles.

The cumulation of the sun is the same as the angular difference between the current location of the zenith and sun. If this is greater than 90 degrees, the sun is below the horizon. In practice, the sun is essentially stationary, and you get a sinosoid that crosses the horizon, marking day and night. Because the sun moves on its circle, the closest point the sun and zenith gets from day to day, changes also sinosoidally, and it is this that generates the seasons.

All the same, we see that in six dimensions, regardless of the orientations of these circles, there must be at least one circle that experiences neither day and night, or seasons. Instead, the sun moves around a torid shape thing on the middle of the horizon, and a sundail can tell you not only the time of day, but what day it is.

Calculating the portions of the tropics in 5 and higher dimensions is still a bit hard, but this is the current thinking. It has a new measure that does not exist in 4d and lower, as areas that are entirely free of these things.

1. Put tropic circle = T, artic circle = A, so tropics = T, temperate = A-T, and polar = 1-A.

2. The working model for higher dimensions, is that the climata is a bright centre in a dark ring, so you get T = t^m, A=a^m, where m=n-3. The -3 bit is because three dimensions are already used by the seasons, the time of day, and the radius.

3. It's highly unlikely that 'tilt' affects things, so this division must be something that exists in higher than four dimensions.

4. For a sphere of diameter \sqrt{n}, one has most of the volume lying between +1 and -1, which means that most of the light that is falling at any instance is likewise very low in the sky.

It does not prevent, for example, a planet in eight dimensions having two different suns in orthogonal orbits, and even two different year lengths. Good lord.

If a is something like (n-1)/n, then a^m is something like 1/e.

There's something in the new scientist of late, about brown dwarfs being warm enough to have weather and clouds and such. If these things come up from down under, you probably don't need orbits and that sort of thing. Seasons could still be effected by having something sloshing around inside.
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Re: 4D orbits again

Postby quickfur » Fri Sep 27, 2013 4:42 pm

wendy wrote:[...]
It does not prevent, for example, a planet in eight dimensions having two different suns in orthogonal orbits, and even two different year lengths. Good lord.

Now, there's an interesting thought. What about geocentric models for 4D and higher?

If a is something like (n-1)/n, then a^m is something like 1/e.

There's something in the new scientist of late, about brown dwarfs being warm enough to have weather and clouds and such. If these things come up from down under, you probably don't need orbits and that sort of thing. Seasons could still be effected by having something sloshing around inside.

It would result in a remarkably different surface environment, though. For one thing, unless the brown dwarf itself is generating light (unlikely, since for that to happen, it would probably be too hot to live on), the sky would be mostly dark. And in 4D, where orbits are inherently unstable, you wouldn't have galaxies and such either. So the sky would be almost completely black. The ground would be warm, and would be the source of energy, so there wouldn't be trees; rather, plants would simply have minimal surface bodies and send their roots as deep down as possible in order to draw from the heat source below. Ecosystems would work in fundamentally different ways from what we know (which does make it so much more interesting, but also harder to comprehend).
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