Differential Geometry: Coordinate Patches

  • #1
Sorry i wasnt able to get help in the hw department. figured id try here.

Homework Statement



For a coordinate patch x: U--->[tex]\Re^{3}[/tex]show that[tex]u^{1}[/tex]is arc length on the [tex]u^{1}[/tex] curves iff [tex]g_{11} \equiv 1[/tex]

The Attempt at a Solution



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

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

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

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

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:

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

Any help appreciated.
 

Answers and Replies

  • #2
lavinia
Science Advisor
Gold Member
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A curve tangent to a coordinate direction only has one metric tensor component that is not zero - I think.
 

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