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.