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in the euclidean plane the distance from the origin to a point is

$s^2 = x^2 + y^2 $

I am reading a paper which say that this could be called an algabraic metric for the plane.

the paper then states that the algebraic metric on the sphere is this:

$ s^2 = \alpha x^2 + \beta y^2 + \gamma xy $

however if we choose a sphere with constant radius R how exactly do we come about finding this expression for the great circle distance, and what are those constants? I would appreciate some pointers as to how to derive this equation and what those constants are. I note that there is no z term- is this because it is eliminated using the equation for a sphere e.g. $ z = \sqrt {R^2 -x^2 -y^2} $ does that make any sense?

NOTE: I am talking about distances on THE SURFACE of a shere

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when i googled great circle distance all i could find was stuff using polar coordinate (actually in the form of latitudes and longitudes. nothing in terms of constants and xs and ys –  Timtam Jun 4 '12 at 4:11
    
Can you cite the paper? Shouldn't a distance be something between 2 points on the sphere? Where is the second? –  draks ... Jun 4 '12 at 5:48
    
Not exactly what you are looking for: Let $p_i = (x_i,y_i,z_i)$, $i= 1,2$ be two points on the surface of the sphere. Then let $\alpha = \arcsin\frac{||p_1 \times p_2||}{R^2}$. The great circle distance is then $\alpha R$. –  copper.hat Jun 4 '12 at 8:02
    
I am assuming we assume one point is an arbitrary origin something like (0,0,R) –  Timtam Jun 4 '12 at 8:52
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1 Answer 1

I think the question involves Riemannian metric on the sphere, rather than metric as two-point distance. It is true that in local coordinates $x,y$ the Riemannian metric on any surface can be written as $ds^2 = \alpha dx^2 + \beta dy^2 + \gamma dx\,dy$ (did you "calcel" $d$ here, by any chance?). The coefficients $\alpha,\beta,\gamma$ depend on the coordinates we use, as well as the point at which we look.

For example, the upper hemisphere can be parametrized via $\vec r(x,y) = (x,y,\sqrt{R^2-x^2-y^2})$. The quadratic form $ds^2$ is calculated from $|\vec r_x dx+ \vec r_y dy|^2$ where the subscripts denote derivatives. I get $$ds^2 = \left(1+\frac{x^2}{R^2-x^2-y^2}\right)\,dx^2 + \left(1+\frac{y^2}{R^2-x^2-y^2}\right)\,dy^2 + \frac{xy}{R^2-x^2-y^2} \,dx\,dy $$

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