Stability of 4D atoms

[quickfur]

How do you set it up so that there is an outward force on the mass(es) outside the plane of rotation so that it doesn't fall inwards?

An outward force on the masses would not be necessary, just as no outward force is necessary to prevent the Earth from falling into the Sun.

Uhm, yes an outward force is necessary. In the case of the Earth, this is provided by the linear momentum of the Earth which is perpendicular to the gravitional force of the sun. This has to be of the right magnitude in order for the Earth to remain in orbit. Without this, the Earth will fall into the sun.

However, in practice, I know that such systems would be extremely improbable. All the masses would have to have exactly the right mass, and be in exactly the right position at the right time, with the right velocity. It is almost certain that no such system exists in the Universe.

I still don't see how a tetrahedral configuration of rotating bodies in 3D can be in orbital equillibrium. I can't think of any configuration of linear momentum of each body w.r.t. the others that would keep it from falling inwards. There's always at least one body that will have an unbalanced force (no momentum to counteract it) towards the other bodies, which means it won't remain in tetrahedral position w.r.t. to the others.

[jinydu]

Regarding the first topic: The Earth has linear momentum (of course), but linear momentum is not the same thing as force (for one thing, linear momentum is measured in kgms^-1 whereas force is measured in Newtons).

As for the tetrahedron, I'm actually not so sure it will work. Have you had a look at that link?

[wendy]

When one considers the planet rotating in clifford-style, then every part is going around the centre of the circle in a circle. So one is looking at a momentum that is {physics mode="rusty"}

mw2^r momentum of mass element

dr^3 w^2.r momentum of a shell

dr^4 w^2 dr momentum of a shell

{/physics}.

This is a different momentum to one where it has only one axis of turning, or turns on several axies.

I think we're seeing something enitirely different here.

Wendy

[quickfur]

Regarding the first topic: The Earth has linear momentum (of course), but linear momentum is not the same thing as force (for one thing, linear momentum is measured in kgms^-1 whereas force is measured in Newtons).

Sorry, I should've distinguished between the momentum and the centrifugal force that results from it. What I meant is that in order to remain in orbit, the orbiting body must have a velocity perpendicular to the direction of gravitational attraction, so that it balances out the force of gravity that's pulling it inwards. In the case of the tetrahedral configuration, I couldn't think of any possible motion of the orbiting bodies that would give all bodies the required velocities to remain in tetrahedral configuration.

As for the tetrahedron, I'm actually not so sure it will work. Have you had a look at that link?

Yeah I did. I can see how 3 bodies orbiting in a circle (or N bodies orbiting in a circle, for that matter) would be a stable configuration; but I just don't see how that could still be true when there's a 4th body outside the plane defined by the other 3.

[jinydu]

I managed to find four points in 3-space that are mutually equisitant. If I could only find infinitely such quadruplets of points, such that I could join the respective points into a circle, the problem would be solved.

However, progress was hindered because I didn't know how to write the equation of a circle (not a sphere) in 3D.

[quickfur]

I managed to find four points in 3-space that are mutually equisitant. If I could only find infinitely such quadruplets of points, such that I could join the respective points into a circle, the problem would be solved.

Wait, are you just looking for equidistant points in 3-space, or is there some additional criteria? 'cos if you just want N equidistant points, you can just pick the vertices of the regular polyhedra. But that does not have any necessary correspondence with stable orbital systems.

However, progress was hindered because I didn't know how to write the equation of a circle (not a sphere) in 3D.

You can write it as x² + y² = r² and z=0. Note that in this form, you need two criteria since otherwise you get a cylinder. (This is true for all (n-2)-dimensional curves embedded in n-space.) Alternatively, you can write it in parametric form:

<b>r</b>(t) = (r cos t)<b>i</b> + (r sin t)<b>j</b>

where <b>i</b> and <b>j</b> are perpendicular unit vectors lying on the plane of rotation. This form allows you to have arbitrary planes of rotation about the origin, and is simpler to manage than the non-parametric form.

[jinydu]

Wait, are you just looking for equidistant points in 3-space, or is there some additional criteria? 'cos if you just want N equidistant points, you can just pick the vertices of the regular polyhedra. But that does not have any necessary correspondence with stable orbital systems.

I meant "mutually equidistant", not "equidistant from a center point".

You can write it as x² + y² = r² and z=0. Note that in this form, you need two criteria since otherwise you get a cylinder. (This is true for all (n-2)-dimensional curves embedded in n-space.) Alternatively, you can write it in parametric form:

<b>r</b>(t) = (r cos t)<b>i</b> + (r sin t)<b>j</b>

where <b>i</b> and <b>j</b> are perpendicular unit vectors lying on the plane of rotation. This form allows you to have arbitrary planes of rotation about the origin, and is simpler to manage than the non-parametric form.

Ok, I'll be more direct and say:

I'm looking for four circles. Each one is centered on the origin and passes through one of the four points I mentioned in that link. The other constraint is that if the masses move along the circle at constant speed, the distance between any two masses should remain the same at all instants. In order words, I was actually looking for the orbital path of all four masses.

[pat]

The vertexes of a tetrahedron are mutually equidistant. It's impossible (in 3-space) to have more than four-points such that the pairwise distances are all equal.

What you've got is a configuration of four points centered at the origin. What you want is a rigid motion of this configuration (you don't want the distances to change). The only rigid motions of three-space are: translation, reflection through a point, reflection through a line, reflection through a plane, and rotation. The reflections are discrete. So, they wouldn't really work for an orbit. The translation isn't much of an orbit either. And, there's no rotation of three-space where all four of your points sweep out a great circle. In fact, the best you can get is two of your four points sweeping out a great circle (an equator) while the other two sweep out higher lattitude lines. So, this doesn't work as an orbit pattern either.

[jinydu]

So, you don't think a 3D analogue of the (3 mass, 2D, equilateral triangle) system will work?

[quickfur]

So, you don't think a 3D analogue of the (3 mass, 2D, equilateral triangle) system will work?

The problem is that rotation is a planar (2D) phenomenon. A stable 3D orbital system might certainly exist, but you'd have to drop the equidistant requirement.

Although, it does make one wonder if in 4D simultaneous perpendicular rotations might constitute a new kind of motion which might give rise to orbit-like systems. But it would have to be quite different from solar-system-like orbital systems that we're familiar with.

[Rkyeun]

Stable tetrahedral orbit in 3D:
Imagine a black hole. Now imagine that this black hole has a rotation, and angular momentum. It is therefor not a point of singularity shrouded in an event horizon, but a ring. If we continue to add angular momentum without adding much mass to the system, we can spin the hole up until it opens into a torus. Then your tetrahedron can simply fall through the opening, decelerate once it reaches the other side, and begin to fall back towards the hole, again passing through the center...

Planetary orbits in 4D:
Orbital acceleration is square, as opposed to the cubic gravity here. A circular orbit is analogous to the top of the hill here, instead of the bottom of a ditch. If you place a marble atop the hill it is going to roll down either side, either fleeing the sun or colliding into it.
We have four fundamental forces in 3(spatial)D to represent our four dimensions (including time), the Strong, the Electromagnetic, the Weak, and Gravity. These forces are responsible for sustaining things and deciding orbits.
Gravity and Electromagnetic are both shell forces, which spread out cubically, meaning that the orbital acceleration which is constant by comparison, cannot match it. Even the circular orbit will be slightly off center, and from there will accelerate out of alignment either falling into or away from the star.
Good news is that Strong and Weak are not shell forces, their ranges are finite and they become far stronger in this realm where infinitely more stuff can pack into that radius. Fusion and chemisty have about the same energy value, and you can walk on the cold surface of the sun as hydrogen burns into helium. Isotopes are heavier and more stable.
(4D may have a fifth fundamental force to govern its extra dimension. Following the pattern of forces, it would be stronger than the Strong force, could attract and repel, would have infinite range as an inverse cubic, and could potentially be able to locally counteract gravity to make pockets of stable orbit equations.)

[Batman3]

Larry Niven's Ringworld was unstable gravitationally(It would have drifted offcenter more and more even if it did not distort.). Therefore there was a need for a sequal book called "Ringworld Engineers" to adjust the orbit. The 4dRingworld with 4d 1/r^3 gravity might be stable gravitationally. I don't know the math for either 3d or 4d. So I don't know.
A Sphring's orbit would be stable because the center of mass would orbit the star(if there were a star at the center) for the same reason Gauss' law says that the total field lines going through a surface are proportional to the total sources inside. Therefore the whole mass of the spring is equivalent in is action on the star as if it were located at a point. Working backward by one of Newton's'(?) laws(action and reaction(?)) the star acts on the sphring as if the spring was located at a point.
Of course the sphring might become all floppy and become assymmetrical, but that is beside the point.
If the sphring could remain in hyperspherical shape, it would(?) have two rotation vectors to hold it from the center. These might maintain its shape anyway. Or not. I don't know. If it was a rigid shape, the two rotation vectors might have different magnitudes and the inside surface would feel a gravity force outward that would be such that the perfectly hyperpherical sphring would have a force directed towards the circle with the larger vectorial magnitude perpendicular to it. (there are 2 perpendicular circles in any hypersphere, rotatable perp'ly any way as you view it from different angles).
To check this imagine the circle with the smaller ratational speed had zero speed. Then the whole force of the rotation would be perp'r to the axis w/non-zero speed. Cutting a 3d slice around this rotating plane, the objects inside the sphere would slide towards the circle. And the same for the rest of the hypersphere.
For a h.sphere with v=0 in one axis that axis would be unable to resist the inward pull of the star and it would collapse(unless it had implicit stability in its structure.). Thus for a ringworld-like sphring, the speeds of te 2 roation vecors would have to be equal.
(Did Wendy say elsewhere to me that a rotating h.sphere tends to stabilize two rotation vectors to equal each other in magnitude? Or am I wrong? If so, what would be the dynamical history of such a sequence?)

[wendy]

If an object rotates in several modes, the energies of the several modes should become equal by friction.

A planet rotating in four dimensions, would then tend to exhibit friction in the lattitudes where the rotation is non-circular, to the extent of transferring energy from the faster to the slower modes of rotation: for example, the transfer modes of energy might be in the form of earthquakes, tides &c.

The same holds true for the rotational and orbital modes of rotation.

[bo198214]

Oh, I couldnt believe that the orbits in 4d would be unstable.
But indeed it seems to be the case.
To verify this I let Maple solve the respective differential equation
(one mass assumed static in the center) numerically
with some different initial conditions.

The differential equation (system) for dimension d is fairly simple:
dd x(t)/(dt)² = - GM x/|x|^d
where x is the location of the revolving mass.
We can assume that the revolving mass remains in the in (x,y) plane,
so only need to regard x in R².

I simplified to GM=1 and let the initial location always be x(0)=(1,0)
and the initial velocity v₀ perpendicular to the x-Axis (d x / dt )(0) = (0,v₀).

In 2D some funny orbits come out reminding me of algebraic curves.
They look stable though not regular by a simple pattern.

In 3D we have the usual ellipses or hyperbolas. Simple and stable.

The higher dimensions have collapsing or expanding spirals. The circle
orbit is the instable balance between both (as Pat already mentioned).
For d=5 its so instable that the numerical computation leads to a
collapsing spiral after some revolvings where should have been a circle.

The pictures of the orbits can you find here.