Power Series for Volume of Balls in Riemannian Manifold


by tornado28
Tags: differential, differentiation, geometry, power series, riemannian
tornado28
tornado28 is offline
#1
Dec7-12, 02:10 PM
P: 14
I'm trying to work out the following problem: Find the first two terms of the power series expansion for the volume of a ball of radius r centered at p in a Riemannian Manifold, M with dimension n. We are given that

[tex] Vol(B_r(p)) = \int_S \int_0^r \det(d(exp_p)_{tv})t^{n-1}\mathrm{d}t \mathrm{d}v[/tex]

(I'm ignoring the cut locus distance because we are only interested in small r anyway.) I assume that [itex] S = S^{n-1} = \{v \in T_pM : |v| = 1 \} [/itex]. Clearly the constant term is 0, because a ball of radius 0 has no area in any manifold, so I think that they are really asking for the first two non-zero terms. To find the next term we need to calculate the derivative of [itex] Vol(B_r(p)) [/itex] wrt r.

[tex] \frac{\mathrm{d}}{\mathrm{d}r} Vol(B_r(p)) =
\frac{\mathrm{d}}{\mathrm{d}r}\int_S \int_0^r \det(d(exp_p)_{tv})t^{n-1}\mathrm{d}t \mathrm{d}v = \int_S \frac{\mathrm{d}}{\mathrm{d}r} \left(\int_0^r \det(d(exp_p)_{tv})t^{n-1}\mathrm{d}t\right) \mathrm{d}v = \int_S \det(d(exp_p)_{rv})r^{n-1} \mathrm{d}v[/tex]

I can pass the derivative under the integral sign because S is compact and the inner integral is a continuous function of v. (Right?) So the second term is 0 also, unless n=1. Here's where I run into trouble. I'm not sure how to differentiate [itex]\det(\mathrm{d}(exp_p)_{rv})[/itex]. I do know that [itex]\mathrm{d}(exp_p)_{tv} \cdot tw = J(t)[/itex] where [itex]J(t)[/itex] is the Jacobi Field along [itex]\gamma_v[/itex] defined by [itex]J(0)=0[/itex] and [itex]J'(0)=w[/itex]. My guess is that the solution uses this along with some linear algebra trick to find the derivative.

Thanks in advance for your help!
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Vargo
Vargo is offline
#2
Dec7-12, 09:27 PM
P: 350
My notation might be a bit rusty, but the differential of the exponential at 0 is the identity. So d(exp)(rv)= I + O(r).

So the first non zero term will come from that I, which has determinant 1. You integrate over v to get the usual surface area of a (Euclidean) sphere. Since you are looking at the derivative of V, the antiderivative gives the volume of the euclidean ball.

To get the second term, you'll need the next term in the differential of that exponential. It will have to involve curvature somehow. From your formula, it looks like you might take J(w,t)/t and find the second non zero term in its expansion. (The first non zero term is w, which is why d(exp)(0) = I.
tornado28
tornado28 is offline
#3
Dec9-12, 11:15 PM
P: 14
Thanks. That gets me the first term, and at least gives me an educated guess for the second term. I'm guessing that [itex]\frac{d}{dr} \left( d(\exp_p)_{rv} \right) = ric_p(v)[/itex]. When you integrate this over the whole ball you end up getting the scalar curvature which is what was supposed to happen.

Vargo
Vargo is offline
#4
Dec10-12, 08:19 AM
P: 350

Power Series for Volume of Balls in Riemannian Manifold


Since you mentioned the Ricci tensor, I looked it up on Wikipedia. They have formulas there that look relevant. In particular, under the heading "direct geometric meaning", they have the first two nonzero terms in the Taylor expansion of the volume element in geodesic normal coordinates.

http://en.wikipedia.org/wiki/Ricci_curvature


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