Metric Tensor on a Mobius Strip?

In summary, the conversation revolves around the speaker being bored and trying to occupy themselves with math, specifically tackling a problem involving a Mobius strip and differential geometry. The speaker also mentions being stuck at home for three months without school and offers to collaborate with the other person on a new math subject. They also discuss using Wolfram Alpha to check their computations and the speaker admits to using a list of trig identities while working on the problem.
  • #1
Mandelbroth
611
24
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?
 
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  • #2
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 want to tackle a new math subject then hit me up. I'm bored too and would love to learn some new math.
 
  • #3
WannabeNewton said:
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 want to tackle a new math subject then hit me up. I'm bored too and would love to learn some new math.
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:
 
  • #4
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 :)
 
  • #5


I would say that your attempt to define an induced metric tensor on a Mobius strip is commendable. However, there are a few things to consider before determining if your calculations are correct.

Firstly, the Mobius strip is a non-orientable surface, meaning that it does not have a consistent "top" or "bottom" side. This can lead to some complications when defining a metric tensor, as it relies on a consistent orientation of the surface. It may be helpful to consider the Mobius strip as a two-sided surface, with the orientation changing at the "edge."

Secondly, the parametric equation you have chosen may not be the most suitable for defining a metric tensor. It would be beneficial to consider alternative parametrizations and see if they yield a simpler or more elegant metric tensor.

Lastly, it is important to check your calculations and make sure that they are consistent with the properties of a metric tensor. For example, the metric tensor should be symmetric, and the determinant should be non-zero.

In conclusion, while your attempt to define a metric tensor on a Mobius strip is a good exercise in differential geometry, it may require further refinement and verification before it can be considered a valid solution. Keep exploring and experimenting, and don't be afraid to seek guidance from more experienced mathematicians or scientists.
 

1. What is a metric tensor on a Mobius Strip?

A metric tensor on a Mobius Strip is a mathematical object that describes the distance and angles between points on a Mobius Strip, which is a one-sided, non-orientable surface. It is used in the field of differential geometry to study the properties of curved surfaces.

2. How is a metric tensor on a Mobius Strip different from a metric tensor on a flat surface?

A metric tensor on a Mobius Strip is different from a metric tensor on a flat surface because the Mobius Strip has intrinsic curvature, while a flat surface has no intrinsic curvature. This means that the distances and angles on a Mobius Strip will change depending on where they are measured, while on a flat surface they will remain the same.

3. What information does a metric tensor on a Mobius Strip provide?

A metric tensor on a Mobius Strip provides information about the intrinsic geometry of the surface, such as the lengths of curves, angles between curves, and the curvature of the surface at a given point. It can also be used to calculate the shortest distance between two points on the surface.

4. How is a metric tensor on a Mobius Strip calculated?

A metric tensor on a Mobius Strip is calculated using a set of basis vectors that describe how the surface is curved. These basis vectors are then used to calculate the metric tensor components, which are a set of numbers that represent the distance and angles between points on the surface.

5. What are the applications of a metric tensor on a Mobius Strip?

A metric tensor on a Mobius Strip has various applications in the fields of mathematics, physics, and engineering. It is used to study the properties of curved surfaces, such as the behavior of light and particles on these surfaces. It is also used in the development of mathematical models for physical systems, such as the motion of celestial objects or the behavior of fluids.

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