Differential Geometry: Coordinate Patches

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

1. 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]
3. 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.
 

lavinia

Science Advisor
Gold Member
3,076
535
A curve tangent to a coordinate direction only has one metric tensor component that is not zero - I think.
 

Related Threads for: Differential Geometry: Coordinate Patches

  • Posted
Replies
2
Views
2K
Replies
6
Views
9K
Replies
1
Views
2K
  • Posted
Replies
5
Views
2K
Replies
1
Views
6K
  • Posted
Replies
2
Views
2K
  • Posted
Replies
2
Views
2K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top