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Thread: Curvature With Parallel Surfaces

  1. #1
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    Curvature With Parallel Surfaces

    I have a two part question, one of which I have no clue of how to approach, the other has been giving me a difficult time for the last three days. The questions are as follow:

    Le M be an arbitrairy orientable surface, with $\displaystyle U$ the unit normal vector. Define the following map: $\displaystyle F: M \mapsto \mathhbb{R}^3$ by $\displaystyle F(p)=p+\varepsilon U(p)$

    i) Show that the canonical isomorphisms of $\displaystyle \mathhbb{R}^3$ make $\displaystyle U$ a unit normal on $\displaystyle \overline{M}=F(M)$ for which $\displaystyle \overline{S}(\overline{v})=S(v)$, where $\displaystyle v$ are vectors, and $\displaystyle S(v)$ indicates the shape operator.

    I have no idea what's being asked here.


    ii)Derive the following formulas of the Gaussian and Mean Curvature

    $\displaystyle \overline{K}(F) = \frac{K}{J}$ where $\displaystyle J = (1-\varepsilon k_1)(1-\varepsilon k_2)$ and $\displaystyle k_i$ denote the principal vectors

    Before I begin I managed to deduce from a previous problem that $\displaystyle (\overline{v})\times(\overline{w}) = J(p)v\times w$.

    Now I would imagine that since $\displaystyle K=det(k_1,k_2)$ that I would simply have $\displaystyle F(k_1)\times F(k_2)$ wich would simply give $\displaystyle J(p)k_1 \times k_2$ which isn't the desired result.

    So looking at both RHS and LHS I think I would get:

    $\displaystyle F(k_1)\times F(k_2) = (k_1+\varepsilon k_1) \times (k_2+\varepsilon k_2)= k_1k_2(1-\varepsilon)^2$

    then from the RHS I get :

    $\displaystyle \frac{k_1k_2}{(1-\varepsilon k_1)(1-\varepsilon k_2)}$ which has lead nowhere.

    I'm not looking for the answer, but any hints would be greatly appreciated.
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  2. #2
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    Quote Originally Posted by lllll View Post
    I have a two part question, one of which I have no clue of how to approach, the other has been giving me a difficult time for the last three days. The questions are as follow:

    Le M be an arbitrairy orientable surface, with $\displaystyle U$ the unit normal vector. Define the following map: $\displaystyle F: M \mapsto \mathhbb{R}^3$ by $\displaystyle F(p)=p+\varepsilon U(p)$

    i) Show that the canonical isomorphisms of $\displaystyle \mathhbb{R}^3$ make $\displaystyle U$ a unit normal on $\displaystyle \overline{M}=F(M)$ for which $\displaystyle \overline{S}(\overline{v})=S(v)$, where $\displaystyle v$ are vectors, and $\displaystyle S(v)$ indicates the shape operator.

    I have no idea what's being asked here.
    So F "moves" M a short distance parallel to itself. What are "the canonical isomorphisms of $\displaystyle \mathhbb{R}^3$" and what is "the shape operator"?


    ii)Derive the following formulas of the Gaussian and Mean Curvature

    $\displaystyle \overline{K}(F) = \frac{K}{J}$ where $\displaystyle J = (1-\varepsilon k_1)(1-\varepsilon k_2)$ and $\displaystyle k_i$ denote the principal vectors

    Before I begin I managed to deduce from a previous problem that $\displaystyle (\overline{v})\times(\overline{w}) = J(p)v\times w$.

    Now I would imagine that since $\displaystyle K=det(k_1,k_2)$ that I would simply have $\displaystyle F(k_1)\times F(k_2)$ wich would simply give $\displaystyle J(p)k_1 \times k_2$ which isn't the desired result.

    So looking at both RHS and LHS I think I would get:

    $\displaystyle F(k_1)\times F(k_2) = (k_1+\varepsilon k_1) \times (k_2+\varepsilon k_2)= k_1k_2(1-\varepsilon)^2$

    then from the RHS I get :

    $\displaystyle \frac{k_1k_2}{(1-\varepsilon k_1)(1-\varepsilon k_2)}$ which has lead nowhere.

    I'm not looking for the answer, but any hints would be greatly appreciated.
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  3. #3
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    The first question is to prove that M and $\displaystyle \overline{M}$ have the same normal vectors, that is, $\displaystyle \overline{U} \circ F=U$. Note that the differential of F,
    $\displaystyle d\mathbf{F}(v) = v + \varepsilon \cdot d\mathbf{U}(v)$, where v is any vector tangent to M at p.
    Also note that U is nothing but the Gauss map
    $\displaystyle \mathbf{U} : M \mapsto S^2$
    Thus $\displaystyle -d\mathbf{U}$ is the Weingarten map $\displaystyle \mathbf{S}$.

    Identifying the tangent plane of M at $\displaystyle \mathbf{p}$ and the tangent plane of $\displaystyle S^2$ at $\displaystyle \mathbf{U}(\mathbf{p})$( since they're parallel), we can see that
    $\displaystyle d\mathbf{U}(v)=-\mathbf{S}(v)$ is tangent to M.
    Thus $\displaystyle d\mathbf{F}(v)$ is orthogonal to U(p), we proved that the tangent plane of $\displaystyle \overline{M}$ is orthogonal to U(p).
    We're done for i).
    I'm going to prove ii) in another post.
    Last edited by xxp9; Aug 12th 2010 at 06:04 PM.
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  4. #4
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    To see ii), note that for the two principle directions $\displaystyle v_1$ and $\displaystyle v_2$,
    $\displaystyle \mathbf{S}(v_1) = k_1 \cdot v_1$ and $\displaystyle \mathbf{S}(v_2) = k_2 \cdot v_2$, where $\displaystyle k_1$ and $\displaystyle k_2$ are the principle curvatures.
    So $\displaystyle d\mathbf{F}(v_1) = v_1 - \varepsilon \mathbf{S}(v_1) = (1-\varepsilon k_1)\mathbf{v_1}$
    $\displaystyle d\mathbf{F}(v_2) = (1-\varepsilon k_2)\mathbf{v_2}$
    Since $\displaystyle v_1$ and $\displaystyle v_2$ are orthogonal we can easily deduce that the pull back of the volume element
    $\displaystyle \mathbf{F}^{*}(dV_{\overline{M}}) = (1-\varepsilon k_1)(1-\varepsilon k_2)dV_{M}$, where $\displaystyle dV_{\overline{M}}$ and $\displaystyle dV_M$ are the volume elements of $\displaystyle \overline{M}$ and M, respectively.
    Note that the Gaussian curvature $\displaystyle K=\frac{\mathbf{U}^*(d\sigma)}{dV_M}$, where $\displaystyle d\sigma$ is the volume element of $\displaystyle S^2$ and the similar formula for $\displaystyle \overline{M}$, and $\displaystyle \overline{U} \circ F = U$ we're done.
    Last edited by xxp9; Aug 12th 2010 at 06:06 PM.
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