My Math Forum Finding the Radial Shadow that a manifold casts on a sphere centered at $(a,b,c)$

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 January 28th, 2017, 06:37 PM #1 Newbie   Joined: Jan 2017 From: Toronto Posts: 2 Thanks: 0 Finding the Radial Shadow that a manifold casts on a sphere centered at $(a,b,c)$ Hello everyone, I am new to this forum and as such, I extend my warmest greetings. I have a question about this problem and I want to understand it, but I'm not sure if my logic is solid. The question is: Let M be compact, connected, oriented surface with boundary in $\mathbb{R}^3$, and $(a,b,c)$ not in M. Define:$$\Omega_{a,b,c} = \int_{M}\omega_{a,b,c}$$ where:$$\omega_{a,b,c} = \frac{(x-a)dy\wedge dz+(y-b)dz\wedge dx+(z-c)dx\wedge dy}{[(x-a)^2+(y-b)^2+(z-c)^2]^{3/2}}$$Show there is a path $j(t)$ crossing $M$ transversally, such that if $\mu$ is an orientation form on $M$ and $j'(t)\wedge \mu$ is orientation form on $\mathbb{R}^3$ (for $j'(t)$, using the metric to identify tangent and co-tangent bundles), then: $$\lim_{t_{1}\rightarrow0^{-}, t_2\rightarrow0^{+}}\Omega(j(t_2)-j(t_1))=-4\pi$$ The question ends here. Now reading a solution manual to Spivak's Calculus on Manifolds, I found that if $M$ is a boundary surface in $\mathbb{R}^3$ of a manifold $N$, then the integral on $M$ of $\omega_{a,b,c}$ is also $-4\pi$. Now, the way I think I should approach this is by letting $N=M\cup M'$, where $M'$ is a 2-dimensional Manifold. Then, let $t_2 = (a,b,c)$ and $t_1 = (a',b',c')$. Then $t_2\notin N$, $t_1 \in N-\partial N$. Then: $$\Omega(j(t_2)) +\int_{M'}\omega_{a,b,c}=0,\Omega(j(t_1))+\int_{M' }\omega_{a,b,c}=-4\pi$$ Subtracting these two and taking the limit as $t_2\rightarrow0^{+}, t_1\rightarrow0^{-}$ gives the desired limit. Is this a fine answer for this question? Or am I missing something that could be included in this solution? Thanks for your time. Last edited by mathandeconstudent; January 28th, 2017 at 06:49 PM.

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