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Re: [Q:] Three Generations of Matter. Date: 1996/06/07
In article <31b4b999.139089972@aklobs.org.nz> rtomes@kcbbs.gen.nz (Ray Tomes) writes:
.... there is no reason to believe that > there are only 3 families.
There is if you have the silly idea that the number of families is related to the number of space dimensions :)
Re: most of you know this few of you care? The surface x^4+y^4+z^4=1 Date: 1996/11/15
In article <1996Nov14.221903.29402@lafn.org> ba137@lafn.org (Brian Hutchings) writes: > > In a previous article, ale2@psu.edu (ale2) says: > > so, *how* is it rounded?... assuming (having no computer loaded > with *that kind* of software) that it has no discontinuities, at all, and > no flat areas; eh?
The post started because i thought enough readers (1?, 0, -1?) of sci.math did not know of the simple surface in question. I saw the surface in:
"Modern Differential Geometry of Curves and Surfaces" by Alfred Gray. And i like to plug books i feel are above the ordinary. How one goes about proving the surface is smooth and in fact is a "rounded cube" is not my interest.
So... I'm not sure what your getting at...this is not rocket science. > > >The equation represents the surface of the rounded cube, but its real > > >he: scribbles the equation x^4+y^4+z^4=1 on a drink napkin... > -- > You *don't* have to be a rocket scientist. (College Career Counselor > to me, again <sob>) > > There is no dimension without time. --RBF (Synergetics, 527.01)
Tinker-Toy Spin, a holiday gift idea. Date: 1996/12/24
Tinker-Toy Spin? Consider the function of time, t, and the cylindrical coordinate phi, exp(-i[w*t+(N/2)*phi])
where w is a constant and N is an integer. Evaluate this function at some point (t,rho,phi,z). Now ask yourself the question:
Which direction should i travel such that the phase of the above function changes most rapidly (remember the function depends on space AND time)? Then determine the actual rate of change. Do this for several representative points.
For fun, consider how the phase, (-i[w*t+(N/2)*phi]), changes near the z-axis in different directions.
Now consider this function in another rest frame!
The perfect gift for the person who collects functions?
Re: Experiments to determine the speed of gravity propagation? Date: 1996/08/02
When we measure the gravitational "magnetic" effect of the earth we will be able to determine the "speed" of gravitational radiation just as the permeability and permittivity of space, which determine the speed of light, can be measured using only the effects of virtual photons between currents and static charges?
Why do we need trigonometry. A simple example for children. Date: 1996/08/02
I routinely torture my ten year old daughter by trying to teach her math that is easy but that she will likely only need in high school. The latest attempt was to teach her some trigonometry. Things were going pretty bad until i thought of the following example that brought the need of trig to life.
Go in your back yard or some park and show your children how one can measure distances without rulers (one must make one measurement though) but by measuring angles! One picks two spots in the park from which to measure angles, and measure the distance between these two places. Using some soda straws and a cheap protractor and some imagination one can then measure angles between some object you wish to measure the distance to and the other reference point. This measurement is then repeated at the other reference point. Then using some trigonometry one can determine the distance from the object in question to the reference points.
Any ten year old with enough patience could learn to do the necessary calculations, and for some i think it would be fun!
Steel balls, 5-min epoxy, and loads of fun making "crystals". Date: 1996/09/05
While i should be trying to understand manifolds (actually the thought came while trying to understand GR) instead i take time to talk of a little harmless fun i just had, and if you want to have the same get together the following:
(note, clean the packing oil off the balls before you epoxy them together)
Now make some "crystals". The easiest to make and maybe the most beautiful is the face-centered cubic? Squeeze out a very very small amount of epoxy onto a piece of paper and mix with the end of a sharp pencil. Using the tip of the pencil to apply the epoxy, glue one ball to a piece of paper which rests on a flat table. Now quickly and neatly epoxy more balls into a close packed triangular formation. Epoxy ball to ball and not to the paper. An equilateral triangle of 5 balls per side for a total of 15 balls is good for starters. Make sure the balls are as close packed as possible before the epoxy starts to set. Let harden.
Now epoxy a second triangular layer of 10 balls on top of the first, and let harden. Make sure you put epoxy in the right places.
Continue in this way until you have a tetrahedron shaped stack of balls.
Now admire your work and make more crystals or wonder why you wasted your time and money so. I am interested in the visual so i guess it was worth it. This might be a good thing for kids to do but adults would probably have better things to do with their free time like watch TV commercials or argue with A. Plutonium ;^)
Walk up the surface z=phi, and slide down z= phi + rho^2 Date: 1996/07/14
Get out your 3 dimensional graph paper (or your Mathematica software) and graph the following surfaces given in cylindrical coordinates (rho,phi,z):
Let there be a gravitational force in the negative z direction. Let the surfaces be of smooth metal so you can climb up the steep parts with sneakers but then slide down on a frictionless mat.
Climb up the first surface by keeping a constant distance from the z axis. Young people would take the steep climb up the surface in a spiral of small constant rho and older people would take the gentle climb up in a spiral of large constant rho.
Now keeping your z coordinate constant move in the negative rho direction. At some point the surface will get so steep you will slip, and slide on your frictionless butt. Yeeeeeeeee!
The second surface is more fun to slide down, it has banked turns!
For the mathematically inclined determine:
Bonus question. Do as above but for the surface z = phi + sin(rho + a) where a is a constant. Determine if and when the puck leaves this surface.
For the mathematically gifted determine what similar? kind of slide 4+1 dimensional people might use.)
On a different note, consider the first surface given above, z = phi. Let z represent the time axis. Consider the family of surfaces z = phi + a where again a is a constant. Take the surface normal to represent "flow" of something. At large rho the surface normal is in the time direction but closer to the t axis the surface normal has an increasing component in the spacelike direction.
(Neat surfaces.)
Re: Surfaces z=phi, z=2phi; unit normal; invariants? Date: 1996/07/18
Should of also suggested that one check out how the "surface" t = a*phi looks in another, moving coordinate system.
Surfaces z=phi, z=2phi; unit normal; invariants? Date: 1996/07/18
Consider the following surface as given in cylindrical coordinates:
where a is some constant.
Consider the unit normal to this surface. The surface unit normal has components?:
and
Check that the normal is (normal and) mostly in the z direction for large rho and the normal is mostly in the negative phi direction for small rho.
Note, the slope of the surface goes as a/r, cool surface?
The spacing, s, between sheets is given by:
Check that this goes to 2*Pi*a for large rho.
Can one find any interesting invariants for (1) ? Here goes?....
Consider one "sheet" of surface (1), that is let phi have a range of only 0 to 2*Pi. Associated with this sheet is a "vector field" F. Let the strength of the field go as 1/s. F has components:
Note F_t depends only on the spacing (like energy?) and F_phi is independent of the spacing (like spin?)
Some of the math might be bad but i think the main idea, what ever that was, is still good? Hope this was worth reading?
While your still there, are you still there?
Consider the following operations on z = a*phi. Make a half-plane cut starting at the z axis and parallel to the z axis. The surface z = a*phi will now consist of many disjoint sheets. Now consider the ways one reconnect these sheets and make the surface whole again.
One way of reconstruction produces an infinite family of parallel surfaces (spin zero?). Another way produces a two parallel, interwoven surfaces (spin 1?). Neat but probably irrelevant stuff, where is the top of this damn mountain :(
Re: Surfaces z=phi, z=2phi; unit normal; invariants? Date: 1996/07/19
In article <4slquq$252p@hearst.cac.psu.edu> ale2 writes: > > Note, the slope of the surface goes as a/r, cool surface? > Should be, ...the slope of the surface in the phi direction goes as a/r, ...
Re: Surfaces z=phi, z=2phi; unit normal; invariants? Date: 1996/07/19
In article <4slquq$252p@hearst.cac.psu.edu> ale2 writes: > >F_t = 1/2*Pi*a >F_phi = 1/2*Pi*rho > Should be:
Name of this surface in cyl. coordinates, z = phi*rho^2 ? Date: 1996/11/09
The surface z = - phi (in cylindrical coordinates) has a "dual" (in some sense, see below) surface, the surface z = phi*rho^2 ?
Consider the surface z = -phi that is the union of a bunch of lines, those lines being helices of radius rho (0 < rho < oo) and slope (in the phi direction) of -1/rho.
Consider the surface z = phi*rho^2 that is the union of a bunch of lines, those lines being helices of radius rho (0 < rho < oo) and slope (in the phi direction) of rho.
Graph both surfaces together in your head or on a computer screen.
Answer this question please. For some fixed value rho, call it alpha, look only at those points of both surfaces that have the same value for rho, alpha. Are we are looking at two helices which will always intersect at 90 degrees ?
The surface z = -phi is called a Left Conoid (Diff Geo, Lipschutz). If the surface z = phi*r^2 does not have a name can i give it one?
How about calling it Conoids Friend,
as i feel this fits in a way?
What would you call this surface? Its proper name, right?
This is where some graphing software would come in handy?
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