Circumference of a circle on a spherical surface

[physlosopher]

I was just reading an intro text about GR, which considers the circumference of a circle on a sphere of radius R as an example of intrinsic curvature - the thought being that you know you're on a 2D curved surface because the circle's circumference will be less than $2\pi r$. They draw a circle on the sphere around the z-axis; in spherical polar coordinates the locus of points with constant radius (distance from the origin) and polar angle. Then to calculate the circumference of the circle, they integrate over the azimuthal angle:

$$\int_0^{2\pi} R sin(\theta)\, d\phi = 2\pi R sin(\theta).$$

This is meant to illustrate that the surface is curved, because the circumference is less than $2\pi R$. But isn't this the wrong R? The R in the integral is the distance from the origin of the coordinate system; the radius of the circle that's meant to illustrate the curvature would have to be the distance from the center of the circle as measured on the surface of the sphere itself - so the point where the sphere intersects the z-axis. (Which should be $R\theta$, no?) Doesn't the integral (without comparing it to $R\theta$) just illustrate that the circle is smaller by $sin(\theta)$ than the circle lying in the xy-plane, rather than illustrating curvature? Is this a flaw in the example, or am I overlooking something?

 

[fresh_42]

I was just reading an intro text about GR, which considers the circumference of a circle on a sphere of radius R as an example of intrinsic curvature - the thought being that you know you're on a 2D curved surface because the circle's circumference will be less than $2\pi r$.

Could you elaborate this? A circle on a sphere is still a circle, i.e. I can cut the sphere in the plane of the circle and get a circle in a plane. No curvature beside the curvature of the circle, and no change in circumference.

 

[physlosopher]

Could you elaborate this? A circle on a sphere is still a circle, i.e. I can cut the sphere in the plane of the circle and get a circle in a plane. No curvature beside the curvature of the circle, and no change in circumference.

My understanding is that if you live on the surface of the sphere (2D manifold?) then a circle drawn on the sphere will indeed be a circle, but its circumference will be less than $2\pi r$ where $r$ is measured from the center of the circle within the manifold. So the center of the circle as seen from someone restricted to the sphere's surface is also on the sphere. It's in the same spirit as examples where triangles can have three right angles if they're restricted to living on a sphere. This is all presented as a discussion of the intrinsic curvature of the sphere.

Sorry if that didn't make sense, and I hope I'm using all of this language correctly - just started looking at this stuff!

 

[Orodruin]

Yes, the radius of the circle is $R\theta$, not $R$.

Note that this is an effect of non-zero curvature, not the definition of non-zero curvature. It is more connected to the Ricci scalar than to the curvature tensor.

Note that you recover the usual relationship between radius and circumference when $\theta \ll 1$ as the circle’s radius becomes much smaller than the radius of curvature.

 

[Orodruin]

Could you elaborate this? A circle on a sphere is still a circle, i.e. I can cut the sphere in the plane of the circle and get a circle in a plane. No curvature beside the curvature of the circle, and no change in circumference.

There is no ”plane of the circle”. You are thinking of the sphere as embedded in $\mathbb R^3$. Here the sphere is considered as a manifold of its own without any reference to an embedding space being necessary (although with the metric implied by the standard embedding). A ”circle” of radius r is the set of points located a distance r away from s point on the manifold referred to as the center of the circle. It can be constructed by following the geodesics from the center for a distance r.

 

[fresh_42]

There is no ”plane of the circle”.

Now that he explained it, I know. If you measure the radius longer than it is, then the circumference becomes smaller, of course. As mathematician I first thought circle = conic section. The sphere is a bad example for embedded or not embedded, since there is not much of a difference.

 

[physlosopher]

Yes, the radius of the circle is RθRθR\theta, not RRR.

Good to hear, that's what I thought. Thanks!

Note that you recover the usual relationship between radius and circumference when θ≪1θ≪1\theta \ll 1 as the circle’s radius becomes much smaller than the radius of curvature.

Makes sense - is this related to the fact that the space is locally flat?

 

[Orodruin]

Makes sense - is this related to the fact that the space is locally flat?

Local flatness is a property of an embedding of a manifold into a higher-dimensional manifold. If you forget about the embedding space it is no longer a concept. It has more to do with the fact that parallel transport around an infinitesimal loop changes a vector by an amount that is proportional to the area of that loop. When you shrink the loop this goes to zero.