a few questions about gravity in 4D

[PWrong]

Since we know that orbits don't work in 4D, it's quite likely that a lot of other things don't work. So here's a short list of questions I think we should probably try to answer. If none of these apply in 4D, then there may not be much hope for gravity in 4D.

1. Is gravitational force conservative? It might be hard to find this out without a 4D version of curl, but I'm not sure.

2. Does the shell theorem apply in 4D? That is, can we treat a solid glome as if it's entire mass was concentrated in the center? I've had a go at this, but it doesn't look like it'll work.

3. If not, is there any surface that we can treat in this way?

4. Could we get a stable orbit if there was some kind of damping force affecting the orbiting body, for instance, a cloud of rocks around a planet?
We'd get an equation that would look something like,
r'' +kr' + GM/|r|^3 *r/|r| = 0
but I doubt I could solve that one.

[jinydu]

I can answer the first question:

For a mass M centered on the origin, the gravitational field should be:

E = -GM/(x^2+y^2+z^2+w^2)^(3/2) * (<x,y,z,w>/(x^2+y^2+z^2+w^2)^(1/2))

E = -GM/(x^2+y^2+z^2+w^2)^2 * <x,y,z,w>

The gravitational field is conservative if and only if there exists some scalar function, U(x,y,z,w) such that:

Gradient (U) = -E

In other words, we must have:

-pU/px = -GMx/(x^2+y^2+z^2+w^2)^2

-pU/py = -GMy/(x^2+y^2+z^2+w^2)^2

-pU/pz = -GMz/(x^2+y^2+z^2+w^2)^2

-pU/pw = -GMw/(x^2+y^2+z^2+w^2)^2

where pU/px means the partial derivative of U with respect to x, etc.

It looks like we could find U using a substitution, and then partial integration. But because I'm feeling lazy, I used Mathematica. It is not hard (even without using pen and paper) to confirm, by taking partial derivatives, that one such U is:

U = -GM/2(x^2+y^2+z^2+w^2)

or to put it in a more familiar form:

U = -GM/2r^2

Of course, we're free to add an arbitrary constant. This stems from the fact that potential is really only defined up to a constant; only changes in potential are fully well defined.

So in short, yes an inverse cube gravitatonal force is conservative.

[jinydu]

As for your second question, I don't see what the trouble is. It looks like a straightforward generalization of Gauss' Law.

In our 3D universe, Gauss' Law for gravitation says that for every closed surface:

Surface Integral (E dot dA) = M(in)*k
where E (at a particular point on the surface) is the gravitational field, dA (at a particular point on the surface) is a vector perpendicular to the surface, pointing outwards, with infinitessimal length, M(in) is the mass enclosed by the surface and k is a universal constant that depends only on the units that we're using.

Using the same logic used to derive Gauss' Law in 3D, it shouldn't be hard to say that in 4D, we would have:

Hypersurface Integral (E dot dV) = M(in)*k

Now, suppose our distribution of mass is radially symmetric (and finite in size). Then it makes sense to choose a (4D) hypersphere as our Gaussian surface. Let's choose our hypersphere so that it encloses the entire distribution of mass, so that M(in) = M, the total mass of the distribution.

Since our distribution of mass is radially symmetric, we would expect that the magnitude of the gravitational field is constant over our hypersphere. Thus, we can pull E out of the integral, and we have:

|E| * Hypersurface Integral (dV) = Mk

But the Hypersurface Integral of dV is just the sum over all the infinitessimal surface volumes. This sum is just the total surface volume of the hypersphere, which according to http://mathworld.wolfram.com/Hypersphere.html is 2(pi^2)(r^3), where r is the radius of the hypersphere.

So:

|E| * 2(pi^2)(r^3) = Mk

|E| = M * k/2(pi^2)(r^3)

If we define G = k/2(pi^2), that equation becomes:

|E| = MG/r^3

Since we normally expect gravity to be an attractive force, the gravitational field should point towards the origin. So finally, we get:

E = -GM/r^3 * r(hat)
where r(hat) is a unit vector that points fom the origin to the point in question.

Notice that this does not depend on the distribution itself (so long as the distribution is radially symmetric, and we're outside the entire distribution).

Since a solid glome is radially symmetric, yes, we can treat it as a point mass (which is also radially symmetric), so long as we're talking about points outside the glome.

And for your fourth and final question; since I don't even know enough to understand the proof that standard 4D orbits are unstabe; I'm afraid you'll have to ask someone with more mathematical knowledge.

[wendy]

The equations that we derive from three dimensions are a description of nature, rather than the other way around. Accordingly, we note that curl *does* work in 4d, but it is over three vectors, not two.

On the other hand, 4d has things like the swirl relation or fibulation of space, which may lead to an interesting way of producing the required stabilisation of orbits.

One must also understand that other things come into play as well: for example, the classic Bohr model of atoms are stabalised by quantum effects. It is not at all inconcievable that some kind of spacial quantisation or a rotating rod-like structure might lead to an intense radiadiation of a field perpendicular to the body: this would produce the inverse square law and its attendent stable elliptical orbits, at a level where life might evolve.

The correct approach is not so much to dismiss the theory according to the deeper understandings of 3d, but rather to try and replicate this understanding in 4d. What _are_ the conditions needed to make 4d stable is the real question.

[PWrong]

Thanks for your help Jinydu. I hadn't thought of extending Gauss's Law. I've been trying to replicate Newton's proof of the shell theorem, which uses a single point instead of spherical shells.

Accordingly, we note that curl *does* work in 4d, but it is over three vectors, not two.

I only briefly covered curl in physics, but it takes a single vector field, not two vectors. It's the cross product of "del" and the field, and I think it's a bit like angular momentum.

I doubt there is any simple extension of curl in 4D, because there is a whole plane of possible vectors perpendicular to the plane of rotation.
Divergence should still work, although I haven't formally studied that yet.

I haven't heard of swirl or fibulation, and neither of them turned up on google or mathworld. What are they?

One must also understand that other things come into play as well: for example, the classic Bohr model of atoms are stabalised by quantum effects. It is not at all inconcievable that some kind of spacial quantisation or a rotating rod-like structure might lead to an intense radiadiation of a field perpendicular to the body: this would produce the inverse square law and its attendent stable elliptical orbits, at a level where life might evolve.

The correct approach is not so much to dismiss the theory according to the deeper understandings of 3d, but rather to try and replicate this understanding in 4d. What _are_ the conditions needed to make 4d stable is the real question.

You're right, a 4D universe would be more stable if we imposed certain conditions. For instance, planets could orbit around a spheridrical sun. But one of the interesting features of 4D is that most forces will have inverse cube laws. There's no point in fighting it by removing all the hyperspheres.
You might as well confine all matter and all fields to a particular realm.

Imagine if Fred from 2D said "the inverse square law is too hard for me, so I'll just pretend the 3D universe doesn't contain any spheres."

[pat]

What _are_ the conditions needed to make 4d stable is the real question.

So, you're advocating a sort of weak anthropic principle... if radiant gravity doesn't work, then there wouldn't be a 4-D universe with radiant gravity. So, we shouldn't bother with it. Rather, we should ask, "what would there be instead?"

Or, did I miss your point?

[wendy]

I assume so. The approach is more akin to:

1. Assume it works.

2. Let it run a bit

3. See why it works

The bit that many people find hard is "let it run a bit", because this requires that one has a clue on how 4d works. Most people, me included don't, but i learn from a spot of 304.8 mm. kitbashing.

Clifford rotations were explored by looking at the night sky, and from there, a sun gives the notion of how a 4d planet could be inclined (and seasons work).

After all, gravity has been around for a million years, but the notion of a radiant field being both local and global is due to newton. it quite might be the case that the radiant field might not affect planets as much as some other kind of local planet-form thing, that has a inverse square law.

i really don't know. i have a bachelor degree in applied physics, and i do read a lot of maths, but i tend to look at things at a more unusual level (ie what does it take to convert to base 120).

So it's not that i make outrageous statements of physics as such: i am aware of lots of issues, but find that the assumption of these issues as working in 4d is unwarranted.

So on the other hand, if the current crop does not work, let's find a crop that does work.

So þe weak anthropologicial principle is more likely to find life-conditions, since its working premise is the Riki Tiki Tava principle (let's run outside and find out).

[PWrong]

1. Assume it works.

2. Let it run a bit

3. See why it works

How can you run a hypothetical system just by assuming it works?

it quite might be the case that the radiant field might not affect planets as much as some other kind of local planet-form thing, that has a inverse square law.

What I don't understand is why we would want an inverse square law in 4D. You're just turning an interesting 4D problem into a trivial 3D problem. The inverse cube law is much more natural for a 4D universe, and it doesn't necessarily mean a loss of stability. I've already suggested one simple idea that might work, a drag force on the orbiting body.

[jinydu]

I just had an idea: How about combining extra time derivatives and the inverse-cube law together? Maybe that would make things work.

But of course, in order to even begin such an attempt, a lot more would have to be done with 'F = mj' kinematics first.

[wendy]

You can assume that the thing works, and let it happen. It's not hard. It's a matter of colouring in the dots. That is, the certian implications of four dimensions become self evident when you go there. But just because you turn up there, it does not mean that you understand the underlying physics.

The thing with the inverse square law is that it, and it alone, allow stable elliptical orbits. That is, it allows a planet to move from circular orbit to something else, without falling into the sun, or shooting off.

Of course, i am aware that the radiant laws suggest an inverse-cube law. But the inverse cube law is not even obeyed even in our world: we have the likes of acretion-disks, and power pulsars, which radiate things over linear space without diminishing.

The inverse cube law is more natural in 4d, based on the assumption that the radiant field exits the source with glomar symmetry. If this is not the case, then the inverse-cube law does not need to apply.

Failing nature to offer a large-scale way to stabalise orbits, then we need to look at this from different angles: either abandon the planet/sun idiom, suggest that the sun has some non-glomar feature (ie so planets lie in an inverse-square law), allow something else to create orbits, or assume some large-scale quantisation of the orbits.

None the same, one can explore 4d as easily as a foreign city, where the sun and the moon and the stars rise and set differently then they do in the known world.

And for this, one assumes the foreign world is not different to ours except it is 4d.

[PWrong]

I've just written a program in blitz basic that simulates gravity with an "inverse n'th law". I don't know of any way to get the program on the net, (at least not without actually buying the program), but it should be pretty easy to do in any programming language.

It seems like not all 4D orbits are completely unstable. You can't get an elliptical orbit, but sometimes a planet will spiral in and out repeatedly.
Also, adding a drag force does seem to make everything more stable, although I'm not sure yet.

[jinydu]

Another thing that we might want to keep in mind is that stability deals with what happens as t ---> infinity.

What if the planet was eventually fated to crash into the star, but this wouldn't happen for say, 100 trillion years? In that case, the lack of stability might not be a problem.

[PWrong]

I think I've just found a problem with your second proof.

Since we normally expect gravity to be an attractive force, the gravitational field should point towards the origin.

You can't assume that, it's exactly what you're trying to prove.
The field due to each point in the sphere is directed to that point. But the test particle is closer to some points than others, so you can't assume that they all cancel out to direct towards the origin.

I have a feeling the only way to prove it is by extending Newton's shell theorem. This page has one proof of the theorem. There is also a similar proof that uses potential energy instead of force.
http://teacher.pas.rochester.edu/phy121/LectureNotes/Chapter14/Chapter14.html

I've tried both ways, but I get different answers each time. The integral doesn't cancel out nicely with either method. [/quote]

[Batman3]

Does anyone know the nature of 4d rotations? I know there are 2 directions available using 6 numbers to describe them(the other 2 disappear because the 2 directions ar at right angles to each other(?)).
How do these directions interact? Can you take 2 numbers of one and exchange them wth 2 of the other? No? What about 4d asteroids precessing in their rotations or tumbling?
Also what keeps the 6*10^24 kg of the earth from collapsing to a point? EM forces aren't strong enough. Do we just assume the core is incompressible? QMechanics(of a sort)?

[wendy]

The simple rotations in 4d correspond to points on a bi-glomohedric pyramid. This is a 5d "surface" in 7D, formed by the pyramid product of two 3d spheres.

Any possible rotation corresponds to a point on this surface. The nature of things is that the point would tend to move to one of the bases of the pyramid, where one would have "clifford-rotation", every point going around the centre at the same rate.

[Batman3]

What do "bi-glomohedric ","pyramid","pyramid product","clifford rotation" mean? Do you mean a 6d 'surface' in 7d?(You said 5d surface which does not seem to make sense). What do you mean that the point would move to one of the bases of the pyramid? How many bases are there? What happens at a base? Is there more than one point per base? When you say every point goes around the centre at the same rate do you mean that both 4d rotation 'vectors' have the same magnitude? If these are "simple rotations" are there "complex rotations"? Again , what about precessions? Can I visualize this?

[wendy]

Surface is given in quotes, because most people think of these as surfaces. The technical term is 'manifold', but i think that of a crat-thing, so i use 'fabric' (as in the fabric of space). 1-fabric = button, 2-fabric = thread, etc.

The function rss() is 'root-sum-square', ie sqrt(x*x + y*y + ...).

A glomohedrix is a globe-shaped 2d-fabric. This corresponds to the surface of a 3d sphere: ie rss(x1, x2, x3).

A pyramid is a product of bases, which is 1 at a given base, and tapers to zero at the "apex". One can visualise this in the plane x+y+z=1, applied to the sizes of perpendicular bases X, Y, Z. At any given point in the plane represents a prism xX, yY, zZ, so one vanishes to the other.

A bi-glomohedrix puramid then is a 5-fabric (ie a petix), in seven dimensions, represented by the intersection of two equations:

rss(x1, x2, x3) = 1-x4 ; 1+x4 = rss(x5, x6, x7)

Each equal sign represents a loss of dimension, so it is only a 5d fabric. (petix)

The middle section of this is when x4 = 0, corresponds to the tetrix (4-fabric) in 6D, rss(x1,x2,x3)=1, rss(x4,x5,x6)=1 which is a 'ridge' of a bispheric prism.

Any valid rotation corresponds to a point on this petix.

If we set up a single rotation, in wheel-fashion [fixed axle], it corresponds to x4 = 0. If the rotations in two orthogonal axies (eg wx, yz) are the same speed, then it maps onto one of the bases (ie x4=1, x4=-1), and every point goes around the centre in a proper axis.

For rotations at different speeds, we have x4 as the ratio of speeds, and the point corresponding to the sense of the rotation.

These are 'simple' or 'inertial' rotations, because they do not suppose the action of an external force (eg tumbling).

For a balanced object, like a planet, the tendency is for x4 to move from zero [balance of energy], and tidal friction will eventually set x4 = 1 or x4 = -1. This rotation makes a "clifford rotation", where every point moves in "clifford parallels". It is also known as Hopf Fibulation.

W

[Batman3]

Wendy, are you saying that all 4d rotations can be described by 2 perpendicular "vectors":2 perpendicular directions with 2 amplitudes each?
If so can one exchange xy - zw to xz - yw or xw - yz while maintaining the same meaning of rotation? In 3d you can't exchange xy -z for xz -y or yz -x while keeping the same axis and magnitude of rotation. Can you do this in 4d?

I suppose rotating coordinate frames is done by multiplying vectors by 4x4 matrices of a certain type. What type?
The eq'n |x|=1 describes 2 points in 1d
|x|+|y|=1 describes a square in 2d
|x|+|y|+|z|=1 describes an octahedron in 3d
|x|+|y|+|z|+|w|=1 describes what? in 4d


Also, regarding impirical evidence in 4d, I presume this can be done by keeping the Biblical Mosaic commandments from 3d as applied to 4d. This would have to be progressive, I think.

[wendy]

The phase-space of four-dimensional rotation, is a biglomohedric pyramid. This is a 5d manifold [petix] in 7D.

In four dimensions, it amounts to the following elements:

You can have one rotation, in any platohedrix [flat-2d-cloth], and an entirely different and independent rotation in the orthogonal platohedrix. The net effect is that any point not on either of these, will follow a orbit confined to a torus, being the surhedron in a bi-circular prism [duocylinder]. This is represented as a general point on the sloping bit of the pyramid: the bit between the bases.

If one makes the two rotations equal in magnitude, then the torii essentially disappear, because everything goes around the centre, and one can regard any track as the generator.

For a planet so rotating, one can plot the risings of stars on a glomohedrix [globe-shaped 2-cloth], or 3d-sphere-surface. Every point on this surface represents a complete circle in the sky, and a complete circle on the ground. Every town on the circle will have the same stars as zenith.

By doubling the distances of the rising of stars (from the most east point), one gets a sphere where the poles are East-most point [place where the zenith-stars rise], and the line between rising and setting. Regardless of where you are on the world, you will come up with the same sphere, except that the east-point and the equator will be different.

This is a glomohedrix, being one of the bases of the pyramid.

You can then take any great circle on the 4d sphere, and its orthogonal, and slow down the orthogonal, until it stops and reverses. This produces a trace in the pyramid, that runs from one base to the other. The other base is identical to the first, except the direction-sense spins the other way [ie left - corkscrew vs right]. Every great circle is left-parallel to exactly one right-parallel circles, unless it is in the right-parallel circles.

The net effect is that every great circle corresponds to a coordinate (x,y) where x,y represents a left- and right- rising-sphere (ie glomohedrix). The complete phase-rotation space is then easy to derive, because there must be a fast and slow rotation, and that there is context (left-vs-right) and also direction. All together, these correspond to every point of a bi-glomohedric pyramid.

W

[Batman3]

I heard that in 3d the Polhod rolls without slipping on the Herpolhod in the Invariable Plane.
--Dynamics

[PWrong]

I have a feeling the only way to prove it is by extending Newton's shell theorem. This page has one proof of the theorem. There is also a similar proof that uses potential energy instead of force.
http://teacher.pas.rochester.edu/phy121 ... ter14.html

I've tried both ways, but I get different answers each time. The integral doesn't cancel out nicely with either method.

I finally managed to prove it. The integral is
U = - 2GMm/pi * Int [sin² x / (r^2 + R^2 - 2Rr cos x) dx]

from 0 to pi, where R is the radius of the shell, r is the distance to the point mass, and x is the angle.

The answer simplifies to
U = -GMm/2r^2 - GMm/2R^2
I thought this was wrong at first, but it's just got an extra constant.
F = - dU/dr = GMm/r^3

So fortunately, you can treat a glome as a point mass.

[thigle]

pw: could you please expand a bit on what you mean by 'point mass' ?

batman3: what is polhod and herpolpod ?

[PWrong]

Well, what this means is that the radius of a spherical object doesn't make any difference to it's gravitational pull, only the mass is important.
For instance, if you compressed the earth into the size of a tennis ball, without changing it's mass, it wouldn't affect the moon's orbit.

[Batman3]

p space is the space of the angular velocity unit vectors.
The INERTIA ELLIPSOID has in p space lengths of I1, I2, I3 (the princple moments of inertia of a rigid body such as an asteroid), w/ dimensions of kg meters^2.
The INVARIABLE PLANE is defined bya plane perpendicular to L , the angular momenum vector. (There is no torque allowed for this case).
The ellipsoid's origin is always a fixed distance in terms of the I1, I2 and I3 away from the plane and so can be said to always touch it.
The POLHODE is the path 'traced out' on the ellipsoid by the point of contact with the plane.
The HERPOLHODE is the path traced out on the plane.

The relationship between the angular velocity, the angular momntum, the moments of inertia determine the precession (etc.?) of the rigid body.

This translates into the statement, "The Polhode rolls w/out slipping on the Herpolhode lying in the Invariable Plane." Appearantly it is easier to do the physics from this perspective than by brute intellectual force.

SOURCE: condensed from CLASSICAL MECHANICS by Herbert Goldstein. copyright 1980 by Addison-Wesley Pub. Comp. Inc.
pages 202-207

[jinydu]

Well, what this means is that the radius of a spherical object doesn't make any difference to it's gravitational pull, only the mass is important.
For instance, if you compressed the earth into the size of a tennis ball, without changing it's mass, it wouldn't affect the moon's orbit.

Just in case this isn't clear: What PWrong said applies when you are looking at points outside of the sphere. If you look at points inside the sphere, the gravitational field is indeed different from that of a point mass.

In general, it can be shown using Gauss' Law that any spherically symmetric object can be treated as a point mass, when you are looking outside of the object.

[thigle]

how is then 'looking at points outside of the sphere' formally different from 'looking at points inside the sphere' ? or you mean surface/volume distinction ?

[PWrong]

how is then 'looking at points outside of the sphere' formally different from 'looking at points inside the sphere' ? or you mean surface/volume distinction ?

The moon is outside the earth, so it can pretend the earth is a point. I live several kilometers below the earth's surface , so gravity is different for me.

Actually, if you're inside the sphere, everything outside cancels out. So if you're just talking about a hollow shell, then there's no force at all.

If you're talking about being inside a solid ball, then you can ignore everything farther from the centre than you are, and only consider what's closer than you. So if you were underground at a distance r from the centre of the earth, then you could pretend that the earth was a sphere of radius exactly r.

The mass of this part of the earth is density*4/3 pi r^3. If we let density=1, the force is F = -GMm/r^2 = -4/3Gm pi r.
Now let k = 4/3Gm pi, so a = -kr

This is an equation for simple harmonic motion. This means that if you dug a tunnel through the earth and jumped in, then you would fall all the way to the other side, then fall back again, and continue endlessly (if you ignore air resistance and other things that ruin all my fun).

Incidentally, the exact same thing happens in 4D. I'm trying to use this in my gravity program. If you put the whole system inside a big uniform dust cloud that has it's own gravity, it seems to stabilise things a bit.

[thigle]

thanx, it's clear now. good luck with your programming. (I always envied people who can program :twisted: ) :wink:

[wendy]

i was under the understanding that the reduction of a spherical body to a point at the centre is valid only in one and three dimensions. In other dimensions, it's wonkier than a metric ruler.

Wendy

[PWrong]

Well, I tried looking up a proof of it, but I couldn't find one, so I had to do it myself. You can try integrating it yourself if you don't think it's right.