spherical geometry, some simple things

by trancefishy
Tags: geometry, simple, spherical, things
trancefishy is offline
Dec23-04, 05:37 PM
P: 75
So, i'm working my way through "Geometry from a Differentiable viewpoint" (or, trying to get through section 1.1, anyways).

right now, it's spherical geometry. so far, a great circle has been defined as the set of points on the sphere that intersect with a plane that intersects the origin of the sphere. The area of a lune, the "sherical pythagorean theorem", and the spherical sine theorem have all been presented with very brief proofs. It has taken me 5 hours of good work to get through 5 pages of this stuff.

the exercise that I'm doing is this : "The sphere of radius one can be coordinatized as teh set of points [tex](x,y,z)[/tex] in [tex]\mathbb{R}^3[/tex] satisfying [tex]x^2 + y^2 + z^2 = 1 [/tex], or as the set of points [tex](1,\psi,\theta)[/tex] in spherical coordinates, with [tex]0 \leq \psi \leq 2\pi [/tex], and [tex]0 \leq \theta \leq \pi [/tex]. In these two coordinate systems, determine the distance along great circles between two arbitrary points on the sphere as a function of the coordinates."

I worked on this thing for quite a while, first in spherical coordinates (they make more sense), then, deciding maybe i should use the [tex]\mathbb{R}^3[/tex] coordinate grid. i got essentially nowhere. finally, i looked at the answer, hoping to see something I missed. I did, kind of, but after working on those, i still don't have any idea how they got the answers, and I feel like i'm missing something essential to doing this stuff.

the Answers are : for [tex]\mathbb{R}^3[/tex], the great-circle distance between A and B (the arbitrary points) = [tex]\arccos(A \cdot B)[/tex], and for spherical coordinates is [tex]\arccos (\cos (\psi_1 - \psi_2)\sin \theta_1 \sin \theta_2 + \cos \theta_1 \cos \theta_2)[/tex]

So, if anyone can tell me maybe what I could read to get myself filled in on this, that would be great. I haven't looked at this in a couple days, so, I'm going to go ahead and continue working on it, and checking up on this, as this book looks super cool, but I want to make sure I have the proper foundations. By the way, i have taken and aced calc I and II, and matrix theory/linear algebra, so that you know what kind of background I have.
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Astronuc is offline
Dec23-04, 09:04 PM
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P: 21,630
The length (s) of an arc on a circle is simply the product of the radius and the inscribed angle [itex]\theta[/itex], so

s = r [itex]\theta[/itex].

And the circumference is just 2[itex]\pi[/itex]r.

On a sphere, the arc length between two points is given by the same relation, and represents the loci of points defined by the intersection of the plane defined by the two vectors and the spherical surface and bound by the two points of interest.

Thus if one has two vectors of radius, r, the length of an arc on the surface of the sphere is given by the product of r and the angle between the vectors, and that angle can be found from

[itex]\vec{A_1}\,\cdotp\,\vec{A_2} = A_1 A_2 cos\, \theta[/itex]

and for r = 1, then [itex] A_1 = A_2 = 1[/itex],

and the angle [itex]\theta[/itex] is simply the arccos of the dot product.

Try to write two vectors in spherical condinates [itex](1, \psi_1, \theta_1 )[/itex] and [itex](1, \psi_2 ,\theta_2)[/itex] and take the dot product on the unit sphere.
dextercioby is offline
Dec24-04, 05:42 AM
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P: 11,863
Let [itex] \vec{A} [/itex] and [itex] \vec{B} [/itex] two arbitrary vectors who share the same origin and that is chosen as the origin of the Oxyz coordinate system.
Consider the scalar product between the 2 vectors:
[tex] \vec{A}\cdot\vec{B} =|\vec{A}||\vec{B}| \cos \theta [/tex] (*)
,where [itex] \theta [/itex] is the angle between the 2 vectors (remember they share the same origin).
Now use the fact that the vectors are somehow related to the orthonormal system of coordinates.
Then u can write:
[tex] \vec{A}=|\vec{A}|\sin \theta_{1}\cos \phi_{1}\vec{i}+|\vec{A}|\sin \theta_{1}\sin \phi_{1}\vec{j}+|\vec{A}|\cos \theta_{1}\vec{k} [/tex]
[tex] \vec{B}=|\vec{B}|\sin \theta_{2}\cos \phi_{2}\vec{i}+|\vec{B}|\sin \theta_{2}\sin \phi_{2}\vec{j}+|\vec{B}|\cos \theta_{2}\vec{k} [/tex]

Make the scalar product and then obtain:
[tex]\vec{A}\cdot{\vec{B}=|\vec{A}||\vec{B}| [\cos \theta_{1}\cos \theta_{2}+\sin \theta_{1}\sin \theta_{2}\cos(\phi_{1}-\phi_{2})] [/tex] (**)

Equate the formulas denoted by (*) and (**),simplify through the product of the moduli of the 2 vectors and finally obtain:

[tex] \cos \theta=\cos \theta_{1}\cos \theta_{2}+\sin \theta_{1}\sin \theta_{2}\cos(\phi_{1}-\phi_{2}) [/tex](***)
,which is exactly the formula u were looking for.
As u can see,my method is structurally similar with the one by Astronuc,but my vectors are not necessarily unit vectors.They're not even of equal modulus.

Problem:Take an atlas and compute the geodesical distance in Km between NewYork and San Francisco.
HINTS:Consider Earth as a sphere.Find the mean radius (probably given in the atlas),use the formula (***).Carefully,though.


PS.Math is beauty...

trancefishy is offline
Dec26-04, 11:56 AM
P: 75

spherical geometry, some simple things

Thanks you very much :-) though i don't get the satisfaction of having done it myself now, but, at least i get it...

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