differential geometry problem

May 2009
45
1
A catenoid is a surface of revolution. Points in a cateniod, \(\displaystyle \tilde{p}\\\epsilon\\\\C\), can be written parametrically as

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

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

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

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


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

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

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

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

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

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

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 \(\displaystyle K(\tilde{p})\\\epsilon\\H\)

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

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

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

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]
 
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Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
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 \(\displaystyle \Gamma^i_{jk}\) are \(\displaystyle \Gamma^1_{11}\)=\(\displaystyle \Gamma^2_{12}\)=\(\displaystyle \Gamma^2_{21}\)=tanh(u/c)/c and \(\displaystyle \Gamma^1_{22}\)=-c*tanh(u/c). Hence write the geodesic equations on the catenoid. Dont solve them

\(\displaystyle \Gamma^1_{11}\)=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.
 
May 2009
45
1
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
 
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