Deriving the area of a spherical triangle from the metric

  • #1
1,872
147
The metric for 2-sphere is $$ds^2 = dr^2 + R^2sin(r/R)d\theta^2$$

Is there an equation to describe the area of an triangle by using metric.

Note: I know the formulation by using the angles but I am asking for an equation by using only the metric.
 

Answers and Replies

  • #2
55
13
Hello,
I think that the area of a triangle in metric would be the same as in imperial. It is just a ratio really. 1/2 B * H
would be the same for metric right?

unless you take the sector and neglect the segment.; Triangle = sector - segment. You could probably derive this in radians easily enough.

jeff
 
  • #3
1,872
147
Hello,
I think that the area of a triangle in metric would be the same as in imperial. It is just a ratio really. 1/2 B * H
would be the same for metric right?

unless you take the sector and neglect the segment.; Triangle = sector - segment. You could probably derive this in radians easily enough.

jeff
I did not understand what you are trying to mean.
 
  • #5
1,872
147
Well kind of. But I understand the problem ...so I guess theres no need
 
  • #6
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,652
Is there an equation to describe the area of an triangle by using metric.
It is unclear what you mean. You obviously need some information about the triangle itself. Which information about the triangle do you want to use? Or are you after a more general integral expression for the area of a part of the sphere?

I think that the area of a triangle in metric would be the same as in imperial.
This is not about units, it is about the metric tensor used in differential geometry as applied to a curved surface.
 
  • #7
1,872
147
1572603145456.png


It is unclear what you mean. You obviously need some information about the triangle itself
Yes I realized that after some time later. First I thought that from the metric we can define a general area equation for any spherical triangle. However as you mentioned thats possible.. Lets suppose we have a triangle like this. N is the north pole and it makes an angle of A. From the metric I find that the angle can be expressed as

$$Area = \int_0^A \int_0^r Rsin(r/R)drd\theta$$
 
Last edited:
  • #8
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,652
Your triangle is not well defined, you need to specify three numbers to completely specify the triangle.
 
  • #9
1,872
147
Your triangle is not well defined, you need to specify three numbers to completely specify the triangle.
Well yes sorry the other length is also r. So the bottom length becomes ##ARsin(r/R)dr## (from ##ds = Rsin(r/R)d\theta)##

Actually the area is $$Area = \int_0^rARsin(r/R)dr$$ but this is also equal to $$Area = \int_0^A \int_0^r Rsin(r/R)drd\theta$$
 
  • #10
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,652
That is not a triangle. A triangle has geodesics as its sides and the line with the bottom length you have specified is not a geodesic unless ##r## is the distance from the pole to the equator.
 
  • #11
1,872
147
That is not a triangle. A triangle has geodesics as its sides and the line with the bottom length you have specified is not a geodesic unless ##r## is the distance from the pole to the equator.
Hmm I know that ##r ∈[0, \pi R/2)##. But thats confusing. I mean

1572605300894.png


$$ds^2 = dr^2 + R^2sin(r/R)^2d\theta^2$$ so for ##h##, ##dr = 0 (?)##.

I see you are meaning that if ##r\ne \pi R/2## then ##dr\ne0##

If we know the value of the A is it possible to find the h from sherical trigonometry (without knowing the value of ##r##)? I guess metric is useless to find the ##h## then
 
  • #12
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,652
Unless you know three parameters describing the triangle, it is impossible to find the others. This is true also in Euclidean space. If you do not know r, it is impossible to find h just based on the angle.
 
  • #13
1,872
147
Hmm I see. I ll try to look something else for solution. Thanks then
 
  • #14
1,872
147
line with the bottom length you have specified is not a geodesic unless rrr is the distance from the pole to the equator.
I ma kind of having trouble to understand this. At post 11 $$h = ARsin(r/R)$$ ? I mean it has to be that way..
 
  • #15
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,652
I ma kind of having trouble to understand this. At post 11 $$h = ARsin(r/R)$$ ? I mean it has to be that way..
Again, this is not a triangle. It is a circle sector.
 
  • #16
1,872
147
Again, this is not a triangle. It is a circle sector.
Okay.. umm then how can I find the area of that thing. What should I do ? Is it better maybe to use spherical coordinates ?
 
  • #17
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,652
Of the circle sector or the triangle?
 
  • #19
DEvens
Education Advisor
Gold Member
1,203
457
The metric for 2-sphere is $$ds^2 = dr^2 + R^2sin(r/R)d\theta^2$$
Here, by ##r## do you mean ##R \theta##? Then your metric would be the more familiar form.
$$ds^2 = R^2 ( d^2 \theta + \sin^2(\theta ) d^2 \phi )$$

And ##\theta## would have range 0 to ## \pi## and ##\phi## would have range 0 to ##2 \pi##. Then it all becomes very familiar surface-of-a-sphere type geometry. Is that it?
 
  • Like
Likes Arman777
  • #20
1,872
147
Well that could have helped I guess, yes.
But I used the spherical trigonometry to find the area of the triangle.. But the metric you wrote also seems well for my problem as well. Sadly I already handed out the paper...
 

Related Threads on Deriving the area of a spherical triangle from the metric

  • Last Post
Replies
2
Views
3K
Replies
3
Views
5K
Replies
3
Views
2K
  • Last Post
Replies
1
Views
2K
Replies
6
Views
1K
Replies
2
Views
9K
  • Last Post
2
Replies
36
Views
16K
  • Last Post
Replies
4
Views
5K
Replies
20
Views
3K
Top