I´m trying to understand where did the gaussian expression ds^2=du^2+2dudv+dv^2 comes from.?

Question

This expression is supossed to give the distance between two points over gaussian coordinates (u, v). I know that this is a subject of differenctial geometry but I have not idea on this field.
Still, I like to understand the ideas behind this expression. What's the reasoning that arrive to this expression?

bpiguy: What has to do Pythagora's theorem with the expression ds^2=du^2+2dudv+dv^2. If I´m not wrong this is only the develop of a perfect cuadratic binomian, the (du+dv)^2 expression. I think that here are some ideas that I´m not having account.
On all what you say there is no mention of the gaussian coordinates of the surface.

Answer 1

My earlier answer was completely off the mark. Questioned about that previous answer, I did some research, and found the following in an article on differential geometry, which I'll paraphrase.

A line in a plane can be written in parametric form, x = x(t), y = y(t), so that any point is determined as a function of t.

A curved surface S in 3-space can be written in parametric form, x = x(u,v), y = y(u,v), z = z(u,v), so that any point is determined as a function of u and v. This formulation is due to Gauss.

The Euclidean distance formula between two points (x, y, z) and (x+dx, y+dy, z+dz) is ds^2 = dx^2 + dy^2 + dz^2.

A line in the curved surface S may be described by a function v = f(u).

If the differential elements of the parametric equations for x, y, and z are determined (probably the total differentials, using two partials) as dx, dy, and dz, all of which are functions of u and v, and if these are inserted into the Euclidean distance formula above, the result (which is a function of u and v only) takes the form

ds^2 = Edu^2 + 2Fdudv + Gdv^2

where the coefficients E, F, and G involve only the first derivatives of x, y, and z with respect to u and v.

Now if, in addition, v is a function of u, then the length of a curve in the curved surface S can be determined by integrating an expression of the form ds = g(u)du.

There's more to the article, but this is the idea. In your question, you're interested in the case where E = F = G = 1. My guess is that may be a special (or degenerate) case of the x, y, and z functions (such as the "curved surface" S being a plane or a sphere, for example).

Anyway, this is a lot better than my first answer.