somewhere near the beginning.

A cool proof of what exactly?

Filed under: Mathematics — Alex @ 2:36 pm 4/29/2005

Today, we had a review in preparation for our test on Monday on differential forms, and I saw the first concrete example of the power they give you. Now I’m even more enamored of them. Here’s the example:

Let \displaystyle \eta = \frac{x dy - y dx}{x^2 + y^2} in \R^2\backslash\{\vec{0}\}.

  1. prove: d \eta = 0
  2. Let \gamma : [0, 2\pi] \rightarrow S^2 be defined by t \mapsto (r\cos t, r \sin t). Compute \int_\gamma \eta .
  3. Let \Gamma be any C^{\prime\prime} curve in \R^2\backslash\{\vec{0}\} from [0,2\pi] with \Gamma(0) = \Gamma(2\pi) such that for all t \in [0,2\pi], the interval [\gamma(t), \Gamma(t)] does not contain \vec{0}. Prove \int_{\Gamma} \eta = \int_{\gamma} \eta.
  4. Awesome application: \displaystyle \int_0^{2\pi} \frac{ab}{a^2\cos^2 t + b^2 \sin^2 t} dt = 2 \pi.

All except for the second to last are pretty easy. Once you see how to prove the second to last, it’s not only easy, but enlightening. So here’s the proof:

Define a 2-surface \Phi : [0, 2\pi] \times [0,1] \rightarrow \R^2\backslash\{\vec{0}\} by (u,v) \rightarrow (1-u)\Gamma(t) + u \gamma(t). Notice by the special condition, the image of \Phi does not contain \vec{0}. \partial \Phi consists of 4 pieces:

  1. \Gamma : [0, 2\pi] \ni t \mapsto \Gamma(t).
  2. \Theta: [0,1] \ni u \mapsto (1-u)\Gamma(2\pi) + u \gamma(2\pi).
  3. \gamma: [0, 2\pi] \ni t \mapsto \gamma(t) .
  4. \Theta: [0,1] \ni u \mapsto (1-u)\Gamma(0) + u \gamma(0) = \Gamma(2\pi) + u \gamma(2\pi).

Applying Stokes’ Theorem: 0 = \displaystyle \int_\Phi d\eta = \int_{\partial \Phi} \eta = \left( \int_\Gamma + \int_\Theta - \int_\gamma - \int_\Theta ) \eta \Rightarrow \int_\Gamma \eta = \int_\gamma \eta.

Looking at this now, it seems like something similar to this holds for any 1-form \eta with d\eta = 0. … And now after thinking about it some more, that looks like a conservative vector field– or whatever it is that gravity is.

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