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Re: Which six independent components, 3D Riemann manifold? Date: 1996/09/10
In article <513is0$ul0@r02n01.cac.psu.edu> ale2@psu.edu (ale2) writes:
> Question, if i make the following nine measurements "in" a 3D Riemann > manifold can i then determine the above six quantities at a point P?: > > Pick eight "near" points and form a cube of non-zero volume. Label one > of the points P and consider the three different loops around the three > faces of the cube which have the common point P. For each loop i carry > around an orthogonal triad of gyroscopes and leave behind an identical > set (i think the gyros ensure parallel transport?). Upon return to > point P i compare the two sets of gyroscopes. The > difference between the two triads will in general require three numbers > to quantify.
I'm assuming an orthogonal triad remains orthogonal on its trip around the loop, is this true in general?
Which six independent components, 3D Riemann manifold? Date: 1996/09/10
from Gravitation and Inertia by Ciufolini and Wheeler:
"...on a Riemannian manifold, the total number of independent components of the Riemann tensor is
Question, is the following one of the sets of six independent components of the Riemann tensor for a 3D manifold M?
Question, if i make the following nine measurements "in" a 3D Riemann manifold can i then determine the above six quantities at a point P?:
Pick eight "near" points and form a cube of non-zero volume. Label one of the points P and consider the three different loops around the three faces of the cube which have the common point P. For each loop i carry around an orthogonal triad of gyroscopes and leave behind an identical set (i think the gyros ensure parallel transport?). Upon return to point P i compare the two sets of gyroscopes. The difference between the two triads will in general require three numbers to quantify.
Question, help me see why the nine above numbers reduce to six?
3D Riemann manifolds, parallel transport and related ? Date: 1996/09/11
This is not homework, just trying to learn this stuff from books and you folk. Many apologies if i ask for too much help :^)
If in a N-dimensional Riemannian manifold i parallel transport a N-tuple of vectors around a loop of non-zero area will the N-tuple of vectors in general be rotated? On the surface of a sphere this is true?
Are there a set of loops at every point P of an N-dimensional Riemann manifold that have special status, maybe in the sense that loops with the right orientation rotate vectors in a special way, does the word eigenvector come up? (boy, that was about as vague as a question can get, sorry for that one).
After traveling around a loop will each vector be rotated by the same amount, that is do the N-tuple of vectors retain the same spatial relationship to one another? On the surface of a sphere this seems true?
Similar question, how many numbers does one need in general to quantify the change that the N-tuple of vectors undergoes when parallel transported around a loop of non-zero area? On a surface the path alone determines how much a vector in the surface rotates and not which way the vector initially points?
If one travels on a loop of zero area does the change in the N-tuple of vector necessarily go to zero? On a surface this seems true, but are there manifolds where this might not be true?
On a smooth surface at each point there are a pair of orthogonal directions (like eigenvectors?) which point in the direction of max and min curvature, does a similar thing happen for N-dimensional Riemann manifolds? Is the generalization straight forward?
Thanks for any help, the books i have are pretty intimidating :^(, but they will be less so if i can develop a "feel" for Riemann manifolds.
Three cheers for Bernhard Riemann (b.Sept. 17,1826, Breselenz, Hannover--d. July 20, 1866) Hip Hip Huray!, Hip Hip Huray!, Hip Hip Huray!
Re: 3D Riemann manifolds, parallel transport and related ? Date: 1996/09/13
In article <5175c4$s8e@r02n01.cac.psu.edu> ale2@psu.edu (ale2) writes:
> Three cheers for Bernhard Riemann (b.Sept. 17,1826, Breselenz, > Hannover--d. July 20, 1866) Hip Hip Huray!, Hip Hip Huray!, Hip Hip > Huray! > I should not have forgotten Riemann's teacher, Karl Friedrich Gauss. Three cheers for Karl Gauss, Hip Hip Huray!, Hip Hip Huray!, Hip Hip Huray!
The following is from "Gravitation and Inertia" by Ciufolini and Wheeler:
"The thoughts of the great mathematician Karl Friedrich Gauss about curvature stemmed not from theoretical spheres drawn on paper but from concrete, down-to-Earth measurements. Commissioned by the government in 1827 to make a survey map of the region for miles around Gottingen, he found that the sum of the angles in his largest survey triangle was different from 180 degrees. The deviation from 180 degrees observed by Gauss--almost 15 seconds of arc-- was both inescapable evidence for and a measure of the curvature of the surface of Earth.
To recognize that straight and initially parallel lines on the surface of a sphere can meet was the first step in exploring the idea of curved space. Second came the discovery of Gauss that we do not need to consider a sphere or other two-dimensional surface to be embedded in a three-dimensional space to define its geometry...!"
Touchdown, professor Gauss!
Re: The Quack Attack! (or Sorry but at the quantum level you are 'implicitly' not allowed to see anything.) Date: 1996/09/13
In article <51b8ck$a7n@news.alaska.edu> asswc1@uaa.alaska.edu writes:
> Second their is the wave/particle duality, which is infuriating in itself. > Call the damn thing a wavicle if you have to because it still won't matter > in the long run. The crux of the problem is that people are so connected > to visual stimulus they need a picture, sorry but at the quantum level you > are 'implicitly' not allowed to see anything.
Good mathematicians can probably "see" some stuff that is more weird than QM? I like pictures, i think one can be found for QM but don't bet the farm on it.
> I propose though that a > quantum is a little nastier than that and is really a fractal like > creature.
You don't think we can picture QM and here you go trying to make a picture ?
Plastic straws, kite string, and 3D skeleton manifolds. Date: 1996/09/06 My big black GR book was cursing at me to open it up and get less stupid. So i open to my favorite page in the whole book, page 1168. Its the picture on this page which gives me hope that i will understand the subject soon. The description of figure 42.1 is as follows:
"A 2-geometry with continuously varying curvature can be approximated arbitrarily closely by a polyhedron built of triangles, provided only that the number of triangles is sufficiently great and the size of each sufficiently small. The geometry in each triangle is Euclidean. The curvature of the surface shows up in the amount of deficit angle at each vertex (portion ABCD of polyhedron laid out above on a flat surface)."
While trying to picture a 4D skeleton manifold i remembered we had some plastic drinking straws.....
Time to build some 3D skeleton manifolds. All you need are a bunch of plastic drinking straws, some kite string, and a means of threading the kite string through the straws (i used a piece of wire longer then the straws that had a 180 degree bend at the end to grab the string).
Building 3D skeleton manifolds requires no brain work so the first thing you do is put on some good music. Then run the kite thread through three straws, pull the thread tight and tie a knot. Now build upon this triangle. The idea is to imagine that you have a bunch of equal size tetrahedron and you glue them together, one triangular face to another. In this way you can build many different shapes. Tetrahedrons are the true three dimensional building blocks?
For more fun "triangulate" space as above but use straws of random length where you can. If you triangulate space with random length straws not every straw can be random and still have the structure "close".
Again, something fun for the whole family?
Re: Plastic straws, kite string, and 3D skeleton manifolds. Date: 1996/09/13
In article <50op0a$1bng@r02n01.cac.psu.edu> ale2@psu.edu (ale2) writes:
> > Building 3D skeleton manifolds requires no brain work so the first > thing you do is put on some good music. Then run the kite thread > through three straws, pull the thread tight and tie a knot. Now build > upon this triangle. The idea is to imagine that you have a bunch of > equal size tetrahedron ^^^^^^^^^^^^^^^^^^^^^^
If your "building blocks" are all the same, i think you will not be able to construct large "solids" from small identical tetrahedron, the structure won't "close". But then if you actually build such structures you will figure this out?
> and you glue them together, one triangular face > to another. In this way you can build many different shapes. > Tetrahedrons are the true three dimensional building blocks?
Need metric for simple vibrating system. Date: 1996/09/15
I need the metric, ds^2, for the following "simple" vibrating system. Consider two solid balls of radius R connected by a long (say 10 R) and thin (say R/10) rod. The set-up looks like a weight lifters dumbbell. Now store elastic energy in the system as follows. Grab the two balls and pull them apart putting the rod under tension. Now let the system go and watch it vibrate. Assume the simplest mode of vibration, that is the rod does not buckle.
Many intro texts to General Relativity give the metric for a point mass and the metric for a rotating point mass. I would like to see what kind of terms enter the metric because of stress in solids.
Homework, what form does the metric have right next to the rod but "far" from the two balls, how does tension and compression effect spacetime? Homework, do as above but instead of pulling the balls apart give them a twist putting the rod under torsion and then let the system vibrate, how does torsion effect spacetime.
Re: Help with Euler's Formula. Date: 1996/09/16
In article <51ia1m$31k@math.ucr.edu> baez@math.ucr.edu (john baez) writes:
> The Euler characteristic is important in all dimensions; > it can be defined as b0 - b1 + b2 - b3 + ..., where bn > is the number of n-dimensional polyhedra or "cells" in > your space, which you have chopped into polyhedra. >
Do we get some interesting invariants in dimension N > 2 similar to the Gauss-Bonnet formula for closed surfaces?
Would you care to point us in the right direction for N = 3?
By the way, Schaum's outline series has a nice, cheap book on Differential Geometry which basically cumulates with the Gauss-Bonnet formula. Why do they always stop with dimension N=2, seems like the "good" stuff really starts in higher dimensions.
Re: Help with Euler's Formula. Author: john baez <baez@math.ucr.edu> Date: 1996/09/19Forums:sci.physics, sci.math
In article <51km7q$16ra@r02n01.cac.psu.edu>, ale2 <ale2@psu.edu> wrote: >In article <51ia1m$31k@math.ucr.edu> >baez@math.ucr.edu (john baez) writes:
>> The Euler characteristic is important in all dimensions; >> it can be defined as b0 - b1 + b2 - b3 + ..., where bn >> is the number of n-dimensional polyhedra or "cells" in >> your space, which you have chopped into polyhedra.
>Do we get some interesting invariants in dimension N > 2 similar to the >Gauss-Bonnet formula for closed surfaces?
The Euler characteristic is a topological invariant that makes sense for any topological space whatsoever. For spaces that can be chopped into polyhedra it can be computed as above. (Topologists usually work, not with "polyhedral complexes", but with the more special "simplicial complexes" or the more general "CW complexes, and similar formulas apply for these.) However, you seem to be looking for a nice differential-geometric way to define it for smooth manifolds. There is a generalization of the Gauss-Bonnet formula that works in all dimensions for compact oriented manifolds, and expresses the Euler characteristic as an integral of some expression involving the curvature of a Riemannian metric; this expression is called the Euler class.
>Would you care to point us in the right direction for N = 3?
Once you get past N = 2, you might as well learn it for all N. One place to look is in one of the appendices of Milnor and Stasheff's book "Characteristic Classes". The Euler class is just one of many nice topological invariants of manifolds that can be expressed as an integral over the manifold of an expression involving curvature. These invariants are called characteristic numbers, and the things they are integrals of are called characteristic classes. They are very important in modern physics and geometry and good to learn about.
Milnor and Stasheff's book takes a topological rather than differential-geometric approach except in that appendix. My book "Gauge Fields, Knots and Gravity" has a wee bit about Chern classes, which are in many ways the most important characteristic classes. I take a lowbrow differential-geometric approach.
I don't know the perfect reference. Books on the Atiyah-Singer index theorem are overkill, but a lot of the good stuff can only be found there, it seems. You might check out the two volumes on geometry for physicists by Nash and Sen (the latter one, called something like "Differential geometry and quantum field theory", is only by one of them). You might also try the several-volume series called something like "Modern Differential Geometry" by Dubrovin and Fomenko, which is readable but has some errors. (I'm away from my desk, so for more precise references you might look at "Gauge Fields, Knots and Gravity".)
>By the way, Schaum's outline series has a nice, cheap book on >Differential Geometry which basically cumulates with the Gauss-Bonnet >formula. Why do they always stop with dimension N=2, seems like the >"good" stuff really starts in higher dimensions.
There is a tradition of having introductory courses on differential geometry that concentrate on surfaces, perhaps because these are easy to visualize and this is how the subject got started. However, I find these a waste of time. Physics requires that you understand arbitrary N, or at least N = 4.
Learn about characteristic classes; they're great fun!
Re: Help with Euler's Formula. Date: 1996/09/20
Thanks for your reply!
baez@math.ucr.edu (john baez) writes: > The Euler characteristic is a topological invariant that makes > sense for any topological space whatsoever. For spaces that > can be chopped into polyhedra it can be computed as above. > (Topologists usually work, not with "polyhedral complexes", > but with the more special "simplicial complexes" or the more > general "CW complexes, and similar formulas apply for these.) > However, you seem to be looking for a nice differential-geometric > way to define it for smooth manifolds. There is a generalization > of the Gauss-Bonnet formula that works in all dimensions for > compact oriented manifolds, and expresses the Euler characteristic > as an integral of some expression involving the curvature of > a Riemannian metric; this expression is called the Euler class. > I'm dying with suspense! I can't read and understand these books fast enough!
The total curvature of a one sphere is (1/R)*2Pi*R = 2Pi ? The total curvature of a two sphere is 4Pi? So the total curvature of a three sphere is 6Pi? Probably not :^( . I want to get hold of a closed 3-manifold and start deforming it!
Maybe a related question. If the total curvature of a closed surface which is topologically equivalent to a two sphere is a constant then it seems like if we look at a small patch of the surface while the surface is being deformed there should be some conservation laws for curvature just like the conservation laws for charge? Something like (still stuck in Euclidean space):
Div (Gaussian curvature "flow") + (Gaussian curvature),t = 0 ????
Another possible differential geometry reference (dinosaur), "Encyclopaedia Britannica" 15th ed., volume 7, article, Differential Geometry. I like this reference because at the very least it packs a bunch of buzz words into a very small (5 page) article. Buzz words like: first and second fundamental forms, manifolds and tensor bundles, De Rhan and Hodge theorems, Gauss-Bonnet formula and characteristic classes. If you were really smart you could probably learn the subject from the article? Good for at least an overview of the subject?
I would suggest Encyclopaedia Britannica gather up all the math articles in their encyclopaedia and put them together in a math reference kind of like how Scientific American gathers related articles and republishes them. Good idea? Probably not :^(
Link to thread of next article. Get out your 4-D graph paper, viewing triangulated 4space. Date: 1996/09/30
I'm still stuck in Euclidean space and in the process of trying to escape from here came upon the following, possibly interesting, thing to graph in your head (Your quantum conscious computer, right Jack &^) or on your computer screen (Mathematica software required $^(
Consider 5 points in 4-D Euclidean space such that they define a non-zero 4-volume. Connect line segments between all pairs of points. For simplicity let 4 of the points have the same value, a, of the 4th coordinate. These four points along with their line segments define a tetrahedron, the addition of the 5th point with a different 4th coordinate value, b, defines (proper name?) a 4-dimensional volume.
Now lets make a movie! Consider the above 2-dimensional framework which defines a 4-dimensional volume, and consider the 3-dimensional "slices" of the framework for a fixed value of the 4th coordinate. Say the (min,max) values of the 4th coordinate of our framework are (a,b). Let each frame of the movie be the 3-dimensional slice of the above framework whose 4th coordinate is given by:
a + [(b-a)/N] n
N is the number of frames in the movie and n is an integer that ranges from 0 to N. Now try to picture the movie, with just a little thought i think anyone will see it.
The movie:
For n=0 we see 4 points and 6 line segments which form a tetrahedron,
for n=1 the tetrahedron disappears and we see four dots, and
for increasing n the four points converge for the 5th point. They will all move with the same speed?
For more fun consider a large region of Euclidean 4space from which you select a large number of random points and then triangulate the space using these points. Again take slices to make a movie. What do we see! Points that merge and diverge some "moving" fast and some "moving" slow. I'll give anyone 5$ for the movie %^)
You might ask, what is the point? Now take your large framework and give each segment a color, here's how, if our framework has M segments pick at random M stars from the night sky and assign the surface color of one star to one segment, one star for each line segment. Assume that our random selection of stars has just as many stars that are reddish as are bluish. Let the color represent a one dimensional scaling factor? If a red segment has a length of 2.5 units it really is to represent a length of more then 2.5 units....
Arrrgghh, time to sleep.
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