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Metric Tensor on a Mobius Strip?

  1. Jun 1, 2013 #1
    I was bored, so I tried to do something to occupy myself. I started going through withdrawal, so I finally just gave in and tried to do some math. Three months of no school is going to be painful. I think I have problems. MATH problems. :tongue:

    Atrocious comedy aside, Spivak provides a parametric equation for a Mobius strip on page 10 of A Comprehensive Introduction to Differential Geometry (Vol. 1, 3rd Edition):

    $$f(t,\theta)=\left(\cos\theta\left(2+t\cos\frac{\theta}{2}\right), \, \sin\theta\left(2+t\sin\frac{\theta}{2}\right), \, t\sin\frac{\theta}{2}\right)$$

    I decided to try defining an induced metric tensor on the Mobius strip. I found the Jacobian,

    $$\mathfrak{J}=\begin{bmatrix} \cos\frac{\theta}{2}\cos\theta & -2\sin\theta-t\left(\sin\theta\cos\frac{\theta}{2}+\frac{1}{2}\cos\theta\sin \frac{\theta}{2}\right) \\ \sin\frac{\theta}{2}\sin{\theta} & 2\cos{\theta}+t\left(\cos\theta\sin\frac{\theta}{2}+\frac{1}{2}\sin \theta\cos\frac{\theta}{2}\right) \\ \sin\frac{\theta}{2} & \frac{t}{2}\cos\frac{\theta}{2}\end{bmatrix}$$

    I then multiply the transpose of the Jacobian by the Jacobian. So, with much simplification, I get that the metric tensor would be

    $$\mathfrak{g}=\begin{bmatrix}1-\frac{1}{4}\cos(\theta)+\frac{1}{4}\cos{3\theta} & \sin{2\theta}\left(\sin{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}+t\left(\frac{\sin^2{\frac{\theta}{2}}}{2}-\frac{3\cos{\theta}}{8}-\frac{1}{4}\right)\right)+\frac{t}{4}\sin^3{\theta} \\ \sin{2\theta}\left(\sin{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}+t\left(\frac{\sin^2{\frac{\theta}{2}}}{2}-\frac{3\cos{\theta}}{8}-\frac{1}{4}\right)\right)+\frac{t}{4}\sin^3{\theta} & \left(\frac{t^2}{16}\cos{\theta}-\frac{9t^2}{16}\cos{3\theta}+\frac{3 t^2}{4}+2 t \sin{\frac{\theta}{2}}-\frac{t}{2}\sin{\frac{3\theta}{2}}+\frac{3t}{2}\sin{\frac{5\theta}{2}}+2 t\cos{\frac{\theta}{2}}-\frac{t}{2}\cos{\frac{3\theta}{2}}-\frac{3t}{2}\cos{\frac{5\theta}{2}}+4\right)\end{bmatrix}$$

    which is ugly. Is this right?
     
  2. jcsd
  3. Jun 1, 2013 #2

    WannabeNewton

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    Jesus Christ man! Well I'm stuck at home for 3 months without school as well so I feel your pain. Did you do this by hand? If so you can just check using wolfram alpha to see if your answer is correct because your formulas (and Spivak's parametric equation) are fine so if anything went wrong it would be in the computations.

    If you're bored and wanna tackle a new math subject then hit me up. I'm bored too and would love to learn some new math.
     
  4. Jun 2, 2013 #3
    I didn't do it all by hand, since I cheated and had a list of trig identities open on my browser. :frown:

    Thanks, though. I wasn't sure if I was doing the process right. I'm new to Differential Geometry, so I intend to keep doing silly things like this until I get comfortable, after which I can start doing more serious stuff. :tongue:
     
  5. Jun 2, 2013 #4

    WannabeNewton

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    CHEATER! Lol jk. But yes you can very well calculate the Jacobian and left multiply by its transpose to get the metric tensor when you have a parametrization given. I must say if you have the endurance to do computations like that, you must be something other than human :)
     
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