Torus geometry & gravity.

A long time ago I spent some time analyzing the torus (doughnut or bagel).

I managed to develop (and then forget) differential equations for geodesics (shortest distance curves), but never managed to develop an analytical expression for the gravitational field.

Can anybody provide differential equations for geodesics? I would like to avoid redoing the work. I would also be interested in equations for geodesics on a spheroid with three unequal axes, which I think are more difficult to develop than equations for the torus.

I do not think it is possible to develop an analytical expression for the gravitational field, but would like to know if anybody thinks otherwise. Actually, I think that an expression for the field (or the potential) due to mass uniformly distributed on a circle would be enough to solve this problem.

Has anybody here have any good ideas on the above?

gravity = 9.8

Just out of morbid curiosity, what are you doing this for Guv?

The torus is an interesting object. I would like to investigate it as a play project.

There is a property called total or Gaussian curvature. The torus has regions of positive, negative, and zero total curvature, making it an example of all three possibilities. BTW: did you know that there is a surface called a pseudo sphere which has constant total curvature which is negative?

If you think of a torus as being defined by two radii (R and r), it can be viewed as a sphere in one limiting case (R = 0), a circle for r = 0, and a cylinder as R grows without bound. If R = r or R < r, it is a surface with one or two dimple like points. Note: R is the radius of a circle on which circles of radius r are centered. One often refers to various circles on the torus as the outer equator (Radius R + r), the inner equator (R - r), the polar circles (like north & south poles on a sphere, Radius = R), the circles of constant longitude (Radius = r), and the circles of constant latitude.

The torus has interesting geodesics (shortest distance curves). Some are circles: The outer & inner equators, and the circles of constant longitude. Some are similar to sine waves, and some are similar to helices. There is one that is asymptotic to the inner equator from both sides.

A problem I always wanted to solve was the determination of the shortest distance curve from a point on the outer equator to a point 180 degrees away on the outer equator.

One type of shortest distance curve between points on the outer equator is like a sine wave starting at a point on the outer equator, rising above it, returning to cross it, continuing below it, and returning to start another phase of the wave. Another type of shortest distance curve is like a helix: It starts at the outer equator, moves toward and crosses the north (or south) polar circle, crosses the inner equator, and returns to the outer equator after crossing the other polar circle.

I am almost certain that it can be proven that for nearby points on the outer equator, the outer equator itself is the shortest distance curve. I think it is fairly obvious (and can be proven) that for points further apart than some minimum difference in longitude, a sine wave like curve is the shortest distance. For more distant points (separated by maybe 150-180 degrees of longitude), it is not known (to me) if the sine wave like curves are the shortest. Perhaps a helix like curve might be shorter.

If I redeveloped the differential equations, I would investigate the torus a bit, just as a fun project. One of the investigations would compare sine wave like and helix like geodesics.

The gravitational field around a torus is mildly interesting. There is zero gravity at the center, but it is not a stable position. Move slightly off center, and you are pulled (slowly at first) toward the surface of the torus. If an intelligent species lived on a torus, athletic competitions might be a bit weird. At the very least, some world records would have to specify the latitude. You could jump highest and lift the most weight at the inner equator, while your performance would be worst at the outer equator. In sports like basketball and golf, different types of athletes might be the more successful at the extreme locations.

you're first post at 1 in the morning - ouch!

i get what you're saying guv and yep it is a pretty interesting solid. unfortunately geodesics aren't in p1, p2, p3, m1, m2, s1, s2 or s3 which are the modules i've done at college. perhaps the next few will help!...

you obviously have some experience with geodesics and problems of this type - what's the general approach for a solution?

DavidHooper: There is a mathematical discipline called Differential Geometry, which applies calculus to geometry. Following are some thoughts which can be understood by a person with very little back ground knowledge in this discipline.

Differential Geometry is the mathematics of Relativity theory, which views the laws of physics as static objects in a 4D space. For example: The path of an object is viewed as a stationary curve in 4D space, instead of an object moving along a 3D curve. BTW: The 4D curve is called a World Line. Some of the concepts of relativity seem easier to deal with as geometry, and certain insights were discovered by thinking of the geometry. For example, the world lines of objects moving in a gravitational field are geodesics or shortest distance curves.

Now back to more mundane applications. The equations of a Torus are as follows.

Code:

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X = [ R + r * cos(Latitude) ] * cos(Longitude)
Y = [ R + r * cos(Latitude) ] * sin(Longitude)
Z = r * sin(Latitude)

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Where R > r > 0 results in an ordinary Torus. Latitude and Longitude are angles, which must be in radians for certain formulae to be correct.

Note that for R = 0, the equations represent a sphere of radius r, while for r = 0, you have a circle. As R grows without bound, the torus geometry approaches that of a cylinder.

After some theory and algebraic manipulation, it can be shown that the following is a formula for the distance (arc length) along a curve on the surface of the torus.

Code:

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ds^2 = [ R + r * cos(V) ]^2 * dU^2 + r^2 * dV^2

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ds is an infinitesimal distance along a curve. U and V are Longitude and Latitude, respectively (I get writer’s cramp spelling them out too many times). dU and dV are infinitesimal changes in Longitude and Latitude, respectively. Note that for R = 0, the formula gives the distance between two points on a sphere. ds^2 = dX^2 + dY^2 is the corresponding equation for arc length along a curve in a plane.

Integration of arc length (often numerically) is usually required to determine the distance between two points along a curve on a surface. One approach to determining geodesics is to find curves for which the arc length is a minimum. Another approach takes advantage of a different property of a geodesic.

The straight line from a point on a geodesic to the instantaneous center of curvature is always perpendicular to the surface. On a sphere, this is easy to visualize. The center of curvature for all great circles is the center of the sphere. A line from the center of the sphere to any point on the surface is perpendicular to the sphere. Note that this is not true for a circle of constant latitude. For a circle of constant latitude, the center of curvature is on the diameter through the North & South Poles. Except for the equator, a line from the polar axis to the surface is not perpendicular to the sphere, indicating that the constant latitude circles are not geodesics.

So to determine differential (or ordinary) equations for geodesics, you either work on minimizing the arc length equation or develop equations from the perpendicular property.

For numerical approximations, ds, dU, & dV can be small finite amounts. There is a reason for shortest distance curves being referred to as geodesics. A geodesic is always the shortest distance between two nearby points on the geodesic, but it might not be the shortest distance between more distant points on the curve. On a simple surface like a plane, there is one and only one geodesic between two points (a straight line), and it is the shortest distance.

On some surfaces, there can be many geodesics between two points, with some being longer than others. For example, consider two points vertically aligned on a cylinder. The shortest distance geodesic between them is a straight line parallel to the axis of the cylinder. There is another geodesic connecting the two points, which spirals around the cylinder once. There is another which spirals twice, et cetera. If you sliced the cylinder vertically and flattened it out, the shortest distance line would be a vertical line. The other geodesics would be slanted straight lines interrupted by the slice made in the cylinder. The fact that the spirals become broken straight lines, indicates that they are shortest distance curves for nearby points.

A cylinder is geometrically identical to a plane for small geometric shapes. The interior angles of a triangle add up to 180 degrees. The Pythagorean theorem is valid. This is precisely true, not approximately true as it is for small regions of a huge sphere. One of the investigations of differential geometry is determining surfaces for which Euclidean geometry is valid. It is not that easy to prove, but it seems obvious that a surface which can be slit and flattened out without wrinkling or tearing is a Euclidean surface. The cone is another example. I am pretty sure that there are some weird surfaces with this property.

Total or Gaussian curvature is an interesting property. Imagine erecting a perpendicular to a surface and passing a plane thought the perpendicular. Consider rotating the plane, using the perpendicular as the axis of rotation. At each position, the intersection of the plane and the surface is a curve. One of these intersection curves has a maximum curvature, and one has a minimum curvature. The product of the maximum and minimum curvatures is called the total or Gaussian curvature for the surface.

For the sphere, the total curvature is obviously constant. For an ellipsoid, the total curvature varies as you move from one point on the surface to another. For a cone or a cylinder, the total curvature is always zero, because one of the intersection curves is always a straight line parallel to the axis of the cylinder. Euclidean surfaces always have zero total curvature.

Some surfaces have points where the total curvature is negative. This occurs when the center of curvature is on opposite sides of the surface. The inner part of a torus has negative curvature. There is a surface called a pseudo sphere whose total curvature is constant and negative. It is difficult to describe in words. It looks like a simple single tone trumpet-like wind instrument The Positive curvature grows without bound and the negative curvature approaches zero as a point moves toward the mouthpiece of the trumpet. I think the surface becomes weird if you move too far in the opposite direction.. The following site shows one.
http://www.paddle.mb.ca/Students/Pseudosphere.html

Surfaces with constant total curvature have an interesting property in common. If you make a copy of one piece of the surface, you can slide it around the entire surface without wrinkling or tearing it. For the sphere (constant positive total curvature) and the plane (constant zero total curvature), this property is obvious. For the cylinder and cone (also constant zero total curvature), it is not as obvious, but is easily visualized when somebody points it out. For the pseudosphere (constant negative total curvature), it does not seem true, but it has been proven.

Quote:

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Running minimizes the time spent in the rain. This answer might be correct for any rainfall angle.

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This, from The Great Rain Debate, could be an application of geodesics?? anyway, i get what ure saying - thankx for writing all that up!!

For any point x from the torus, you could draw a plane trough it and the torus center which would divide the torus in two. Both sides horizontal component (perpendicular to the plane) of gravity field would take out each other.

Then you could say the torus contains of infinitely thin sloping cylinders, calculate the gravity center of one, they should all be on a circle with center in the center of the torus and perpendicular to the plane.

Then you integrate the effect of all these sloping cylinders on point x.

Not sure if this is good idea, but it's the only thing i could think of :p

Kedaman: A long time ago, I tired to determine the gravitational field around a torus and decided that it could only be determined by numerical integration. Since you would have to do evaluate a 3D integral for each point, it is not a practical solution. It could be used to develop some insight, but is useless otherwise.

I think the best approach is to try to determine the field due to mass uniformly distributed on a circle. If an analytical equation could be determined for the circle, I think you could use it to develop an equation for the torus.

The applicable mathematics is not one of my strong points, so I am not completely convinced that the problem has no analytical solution.

I often wondered if some approximating function could be developed. Perhaps use numerical integration to calculate values for several hundred (or several thousand) points and use some curve fitting technique to determine an approximatiing function.

Quote:

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The applicable mathematics is not one of my strong points, so I am not completely convinced that the problem has no analytical solution.

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:( i had probs in Grade 10 calculus...

I think after seeing this thread, ill just make my way over to the left, and out the "Maths Forum" doors. :D

Ph34R
Don't worry too much, there will be stuff in between that helps you understand what comes further on :)

Guv
Okay, I think I have no chance of going into numerical integration, I had a chance to take the 4'th math course, which is about numerical methods, but I didn't since it was optional and it requires math 3 which i skipped last year. I guess I'll get a new chance each year but I don't know it is worth it.

keda: what's happened to your member status?! aka nugster?

guv: let us know the answer if you find it

Ph34R: *chuckle*

Kedaman & Others: Numerical integration sounds formidable, but is not as difficult as it sounds. At least most of the practical stuff is not tuff.

If you have y = Function(x) and want to find the integral of the function numerically, do the following.

• Imagine making an XY-Graph of the function.
  
• The integral of the function is the area under the curve represented by your graph.
  
• To do the integration numerically, you imagine making a lot of vertical slices, all the same width (Call the width Xwide).
  
• You have to calculate the area of each slice and add them all up.
  
• The Y values are the height of the slices.
  
• The X values are InitialX + Xwide, InitialX + 2*Xwide, et cetera.
  
• It is not a bad approximation to think of each slice being a trapezoid. Use two Y values for the base and top and the Xwide value for the height of each trapezoid. Area is Xwide times average of the two Y values.

If you pick a small value for Xwide, you have a good numerical approximation for the integeral.

You merely have to know the formula for area of a trapezoid, and be able to evaluate the function. That is not so tuff.

Numerical solution of simple differential equation, which assumes you know a derivative but not the function.

• The unknown function defines a curve, and the known derivative defines the tangent at each point on the curve.
  
• Assume you know one point LastY = Function(LastX). Id est: You know one point on the curve.
  
• You can approximate another point by NextY = LastY + Derivative(LastX)*DeltaX and NextX = LastX + DeltaX
  
• Use NextX & NextY to calculate another point.
  
• Keep it up until you have computed enough points to make you happy.

The above approximates the curve by assuming that moving a small distance along the tangent is a good approximation to the next point on the curve. This approximation is very good if DeltaX is very small and the curve does not have extreme curvature.

Now you know haw to numerically solve a simple differential equation.

Does any of the above seem difficult? There are various complications which can make it tuff. There are slightly more complicated methods which improve the accuracy of the solutions.

The the concepts are pretty much described by the above, which is all you need to know to solve a vast range of problems numerically.

Perhaps when you've mastered the torus and you're feeling ambitious, you can look at the torus knot? That sounds pretty complicated.
http://www.vbforums.com/attachment.php?s=&postid=575430

That image looks a whole lot like an IOTD I saw on Flipcode a while back =).

Z.