Integrating the metric in 3-D Spherical coordinates

[GRstudent]

Guys,

I read that integrating the ds gives the arc length along the curved manifold. So in this case, I have a unit sphere and its metric is ds^2=dθ^2+sin(θ)^2*dψ^2. So how to integrate it? What is the solution for S?

Note, it also is known as ds^2=dΩ^2

Thanks!

 

[tiny-tim]

Hi GRstudent!

(try using the X² button just above the Reply box )

… integrating the ds gives the arc length along the curved manifold. So in this case, I have a unit sphere and its metric is ds^{^2}=dθ²+sin(θ)²*dψ². So how to integrate it? What is the solution for S?

integrate ds "dot" the expression for the curve

for example, if the curve is constant ψ, then ds = dθ ("longitude")

if the curve is constant θ, then ds = sinθdψ ("latitude")

 

[clamtrox]

and if the curve is neither, then you need to parametrize, for example by giving θ = θ(t), ψ=ψ(t), write ds² = (something that depends on t) dt² and then integrate!

 

[GRstudent]

I would like to get an integral solution of spherical coordinates metric.

 

[clamtrox]

Did you have a particular path in mind, whose arc length you want to calculate? Let's take an example: Suppose the path is defined by $\phi = 2 \theta$, $\theta \in [0, \pi]$ and the arc length is
$$ds^2 = d\theta^2 + \sin^2 \theta d\phi^2$$
Plugging the curve parametrization into this formula, and taking the square root, you find
$$ds = \sqrt{d\theta^2 + \sin^2 \theta (2 d\theta)^2} = \sqrt{1+4\sin^2\theta} d\theta$$
and to find the arc length, you just integrate
$$s = \int_\gamma ds = \int_0^\pi \sqrt{1+4\sin^2\theta} d\theta$$

 

[GRstudent]

Thanks!