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#1 2010-01-26 07:24:21

almost there
Member
Registered: 2009-11-11
Posts: 21

osculating circles and spherical curves

I have to show:

A curve, c(s), is spherical if and only if there is some fixed point, a, located in every normal plane of c(s).

Of course, this fixed point must be the center of the sphere, but I am having a hard time showing it. I want to say that for all s, the normal vector, N(s), points directly at the center and since the curve is parametrized by arclength, the center is actually at the terminal point of N(s), but I cannot formalize this. The only things I can think to work with are the Frenet equations and the fact that the center of the osculating sphere for any point on a spherical curve is constant.

Please help! I know this is probably not a very difficult problem, but I am very uncomfortable in my differential geometry course and feel way over my head. Thanks in advance.

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#2 2010-01-26 17:05:14

almost there
Member
Registered: 2009-11-11
Posts: 21

Re: osculating circles and spherical curves

I think I got one direction (that c(s) spherical implies the center is in the normal plane for every normal plane), but haven't figured the other direction out.

Let's say I wanted to show that if there is some fixed point in every normal plane of a curve, then the curve must be spherical (lie on the surface of a sphere).

I did not mention this before, but the curve is in R^3.

I know the normal plane is the span of the normal and binormal vectors. So in general to say that some point a_0 is in the normal plane to some curve, c,  at some point s_0 would mean that there exist m_1 and m_2 such that

But by assumption a_0 is in the normal plane for ALL s (is in every normal plane of the curve c(s) ).

Before I continue, I'd like to get someone to put a check on my intuition please:

Does a_0 being in every normal plane mean that for each s there exist m_1 and m_2 as above, or do the same m_1 and m_2 work for every point on the curve? I think that the same m_1 and m_2 should work but am uncertain.

Thanks in advance.

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