# Differential Geometry: Coordinate Patches

#### SNOOTCHIEBOOCHEE

Sorry i wasnt able to get help in the hw department. figured id try here.

1. Homework Statement

For a coordinate patch x: U--->$$\Re^{3}$$show that$$u^{1}$$is arc length on the $$u^{1}$$ curves iff $$g_{11} \equiv 1$$
3. The Attempt at a Solution

So i know arc legth of a curve $$\alpha (t) = \frac{ds}{dt} = \sum g_{ij} \frac {d\alpha^{i}}{dt} \frac {d\alpha^{j}}{dt}$$ (well thats actually arclength squared but whatever).

But im not sure how to write this for just a $$u^{1}$$ curve. A $$u^{1}$$ curve throught the point P= x(a,b) is $$\alpha(u^{1})= x(u^{1},b)$$

But i have no idea how to find this arclength applies to u^1 curves.

Furthermore i know some stuff about our metric $$g_{ij}(u^{1}, u^{2})= <x_{i}(u^{1}, u^{2}), x_{j}(u^{1}, u^{2})$$

But i do not know how to use that to show that u^1 must be arclength but here is what i have so far:

$$g_{11}(u^{1}, b)= <x_{1}(u^{1}, u^{2}), x_{2}(u^{1}, u^{2})>$$ We know that $$x_{1}= (1,0)$$ and that is as far as i got :/

Any help appreciated.

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#### lavinia

Science Advisor
Gold Member
A curve tangent to a coordinate direction only has one metric tensor component that is not zero - I think.

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