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Math Help - differential geometry problem

  1. #1
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    differential geometry problem

    A catenoid is a surface of revolution. Points in a cateniod, \tilde{p}\\\epsilon\\\\C, can be written parametrically as

    \tilde{x}(\tilde{p})=(c\cosh(\frac{u}{c})\cos(v),c  \cosh(\frac{u}{c})\sin(v),u)

    a. calculate \frac{\partial\tilde{x}}{\partial\\u}\mid_{p} and \frac{\partial\tilde{x}}{\partial\\v}\mid_{p}

    \frac{\partial\tilde{x}}{\partial\\u}\mid_{p}=(\si  nh(\frac{u}{c})\cos(v),\sinh(\frac{u}{c})\sin(v),1  )

    \frac{\partial\tilde{x}}{\partial\\u}\mid_{p}=(-c\cosh(\frac{u}{c})\sin(v),c\\\cosh(\frac{u}{c})\c  os(v),0)


    b. If a metric on C is defined by g(\tilde{X},\tilde{Y}):=^{(3)}g(\tilde{X},\tilde{Y  }) where \tilde{X},\tilde{Y} are vector fields tangent to C, show that the components g'_{ij} of g relative to the basis \{\frac{\partial\tilde{x}}{\partial\\u},\frac{\par  tial\tilde{x}}{\partial\\v}\} for vector fields on C are

    [g'_{ij}]=\left[\begin{array}{cc}\cosh^{2}(\frac{u}{c})&0\\0&c^2\c  osh^{2}(\frac{u}{c})\end{array}\right]

    E=\|\frac{\partial\tilde{x}}{\partial\\u}\|^{2}=(\  sinh^{2}(\frac{u}{c})\cos^{2}(v))+(\sinh^{2}(\frac  {u}{c})\sin^{2}(v))+1^{2}=\cosh^{2}(\frac{u}{c})

    F=\frac{\partial\tilde{x}}{\partial\\u}.\frac{\par  tial\tilde{x}}{\partial\\v}=0? ive tried to do the math, but cant get zero, but all

    G=\|\frac{\partial\tilde{x}}{\partial\\v}\|^{2}=(-c^{2}\cosh^{2}(\frac{u}{c})\sin(v))+(c^{2}\cosh^{2  }(\frac{u}{c})\cos(v))=c^{2}\cosh(\frac{u}{c})

    e=\frac{1}{c}
    f=0
    g=-c

    I cant find out how to find these values properly, coz according to my maths wolfram and other are wrong, they a back to front from what i calculated?

    c.Use the metric of part b. in Gauss's formula to calculate K(\tilde{p})\\\epsilon\\H

    K(\tilde{p})=\frac{eg-f}{EG-F}=\frac{-1}{c^{2}\cosh^{4}(\frac{u}{c})}=\frac{-\\sech^{4}(\frac{u}{c})}{c^{2}}

    d. Using the metric of part b. show that the only non-zero \Gamma^i_{jk} are \Gamma^1_{11}= \Gamma^2_{12}= \Gamma^2_{21}= \frac{\tanh(\frac{u}{c})}{c}) and \Gamma^1_{22} =-c\tanh(\frac{u}{c}). Hence write the geodesic equations on the catenoid. Dont solve them

    \Gamma^1_{11}=\frac{1}{2}g^{11}\{\frac{\partial\\g  _{11}}{\partial\\u}+\frac{\partial\\g_{11}}{\parti  al\\u}-\frac{\partial\\g_{11}}{\partial\\u}\}=\frac{1}{2}  \{\frac{2\cosh(\frac{u}{c})\sinh(\frac{u}{c}}{c}\}  =\frac{\tanh(\frac{u}{c})}{c}

    then form there i have no idea... and i think the other questions are quite wrong...

    e.are the v-coordinate curves geodesic curves? [hint use your geodesic equations from d. above]
    Last edited by fredrick08; May 26th 2010 at 11:51 PM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by fredrick08 View Post
    A catenoid is a surface of revolution. Points in a cateniod, p e C, can be written parametrically as

    x~(p~)=(c*cosh(u/c)cos(v),c*cosh(u/c)sin(v),u)

    a. calculate dx~/du|p and dx~/dv|p

    dx~/du|p=(sinh(u/c)cos(v),sinh(u/c)sin(v),1)

    dx~/dv|p=(-c*cosh(u/c)sin(v),c*cosh(u/v)cos(v),0)

    b. If a metric on C is defined by g(X,Y):=^(3)g(X,Y), where X,Y are vector fields tangent to C, show that the components g'ij of g relative to the basis {dx~/du,dx~/dv} for vector fields on C are

    [g'ij]=[cosh^2(u/c) 0
    0 c^2*cosh^2(u/c)]

    E=||dx~/du||^2=(sinh^2(u/c)cos^2(v))+(sinh^2(u/c)sin^2(v))+1^2=cosh^2(u/c)

    F=dx~/du.dx~/dv=0??? ive tried to do the math, but cant get zero, but all

    G=||dx~/dv||^2=(-c^2*cosh^2(u/c)sin^2(v))+(c^2*cosh^2(u/c)cos(v))=c^2(cos^2(v)+sin^2(v))cosh^2(u/c)=c^2*cosh^2(u/c)

    e=1/c
    f=0
    g=-c

    I cant find out how to find these values properly, coz according to my maths wolfram and other are wrong, they a back to front from what i calculated?

    c.Use the metric of part b. in Gauss's formula to calculate K(p) e H.

    K(p)=eg-f/(EG-F)=-1/(c^2*cosh^4(u/c))=-sech^4(u/c)/c^2

    d. Using the metric of part b. show that the only non-zero \Gamma^i_{jk} are \Gamma^1_{11}= \Gamma^2_{12}= \Gamma^2_{21}=tanh(u/c)/c and \Gamma^1_{22}=-c*tanh(u/c). Hence write the geodesic equations on the catenoid. Dont solve them

    \Gamma^1_{11}=1/2g^11{dg11/du+dg11/gu-dg11/du}=1/2{2*cosh(u/c)sinh(u/c)/c}=tanh(u/c)/c

    then form there i have no idea... and i think the other questions are quite wrong...

    e.are the v-coordinate curves geodesic curves? [hint use your geodesic equations from d. above]
    I have very little experience with Diff. geo and most likely would not be able to help you. That said, I do understand what and what is not acceptable in terms of written exposition. This is almost undreadable. Consider writing it in LaTeX if at all possible.
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  3. #3
    Junior Member
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    ya i know it looks bad... but i have no idea, how to write latex.. i will try


    I did my best to convert it to LATEX, i couldnt find how to put the tildes below the letters.. so they are above.. doesnt make much difference but they are meant to be below

    plz someone hlep me
    Last edited by fredrick08; May 26th 2010 at 11:53 PM.
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