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Suppose the unit sphere in ℝ3 has coordinates (ρ, η) with ρ as the "co-latitude" angle (measured from positive z-axis) and η as the "longitude" angle measured from positive x-axis in the xy plane. I am given to understand that the metric tensor is

$g = \left[\matrix{1 & 0 \\ 0 & \sin^2\rho}\right]$

and I am further told that this induces a distance metric on the unit sphere.

How can I obtain a distance metric d(x, y) from this? I have seen several definitions similar to this one (see the note), but I am unsure how to actually reduce this to a usable form.

Disclaimer: Posted by an engineer in over his head.

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Given a metric tensor with corresponding matrix $g_{jk}$ and (tangent) vectors $v = v^\ell e_\ell$ and $w = w^m e_m$ (using the Einstein convention in which paired upper and lower indices imply a summation), one has essentially by definition that $g(v,w) = g_{jk}v^j w^k$. For the case you are concerned with, the vectors $v$ and $w$ are the same, and are in fact the tangent vectors of a geodesic. I hope that the rest will be clear. In your case this is just giving arclengths of (portions of) great circles, which probably doesn't require this machinery. – Steve Huntsman Jun 23 '10 at 21:11
Also, you would probably find this better suited to another site mentioned in the FAQ. – Steve Huntsman Jun 23 '10 at 21:12
Essentially, you do not understand the definition. I would suggest to ask any geometer around --- it is very easy to explain by "talking", but by "writing" it is hard to add anything to the definition... – Anton Petrunin Jun 23 '10 at 21:12
You may also be having a hard time with the expressions $\partial/\partial x^j$, for which see (e.g.) – Steve Huntsman Jun 23 '10 at 21:18
up vote 5 down vote accepted

Looking at the other answers and comments posted so far, I feel compelled to add a different answer. Only the advice by Anton Petrunin makes sense to me. You really should find a helpful mathematician and discuss your question in person. For one thing, if it were me, I would start by asking you what you need this for. Your answer would guide me in how to proceed.

Anyway, here are some thoughts (I'm not providing details until you say more about what you want):

First, if all you need is the distance between two points on the unit sphere, that is very easy to compute using the dot product of the two points. It's the same as in the plane (which isn't surprising since any two points on the unit sphere lie in a plane containing the origin).

Second, if you also want to be able to compute the length of a curve that lies in the unit sphere, then that is also easy, because it is the same as the length of the curve viewed as a curve in $R^3$, so you use the same arclength formula as you would for any curve in 3-dimensional space.

But if you need the "distance metric" for something more than this, you should give us some more details and that will help us help you better. The others are trying to explain to you what a Riemannian metric is and how to use it. You might need this, but I doubt it.

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let $(M,g)$ be a Riemannian Manifold, let $ \gamma : [a,b] \rightarrow M$ be a piecewise smooth curve and let $\Omega : M \rightarrow \mathbb{R}^{n}$ be a coordinate chart. The length of $\gamma$ on the manifold is given by

$$L_{g}(\gamma) = \int_{a}^{b} | \gamma'(t)|_{g} dt.$$

$\gamma'(t)$ is an element of the tangent space to M at $\gamma(t)$. In local coordinates it is given by the equation

$$\gamma'(t) = (\gamma^{i})'(t) \frac{\partial}{\partial x^{i}} | _{\gamma(t)}$$

where $\gamma^{i}$ is $\gamma$ composed with the coordinate chart composed with the ith projection map. By definition

$$| \gamma'(t)|_{g} = (g_{\gamma(t)}(\gamma'(t),\gamma'(t)))^{\frac{1}{2}} =[ g_{ij} (\gamma^{i})'(t) (\gamma^{j})'(t) ]^{\frac{1}{2}} $$

Where $g_{ij}$ is the $(i,j)$ component of the metric tensor. You then just plug this into the integral and evaluate (with a computer I hope).

Now to calculate the distance between two points $a$ and $b$ on $M$, just calculate the length of a geodesic beginning at $a$ and ending at $b$. (This can be done on a sphere as geodesics are great circles. I am not so sure about an arbitrary riemannian manifold though...)

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