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:(adsbygoogle = window.adsbygoogle || []).push({});

Atrocious comedy aside, Spivak provides a parametric equation for a Mobius strip on page 10 ofA 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?

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# Metric Tensor on a Mobius Strip?

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