- #1
SNOOTCHIEBOOCHEE
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Sorry i wasnt able to get help in the homework department. figured id try here.
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]
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 that's actually arclength squared but whatever).
But I am 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.
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 that's actually arclength squared but whatever).
But I am 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.