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HW #5: Spherical Coordinates, Duex

This work is borrowed gratefully from Flanders 4.1.

Position in $ \Re^3 $ is given in terms of spherical parameters as:

\[ \vec{r} = &(r\sin{\phi}\cos{\theta}, r\sin{\phi}\sin{\theta}, r\cos{\phi}) \]

with basis $ \{\hat{i}, \hat{j}, \hat{k}\} $. Applying the exterior derivative operator gives:

\begin{align*} d\vec{r} = &(\sin{\phi}\cos{\theta}, \sin{\phi}\sin{\theta}, \cos{\phi})dr \\<br /> &+ (r\cos{\phi}\cos{\theta}, r\cos{\phi}\sin{\theta}, -r\sin{\phi})d\phi \\<br /> &+ (-r\sin{\phi}\sin{\theta}, r\sin{\phi}\cos{\theta}, 0)d\theta \\<br /> = &dr\hat{r} + rd\phi\hat{\phi} + rd\theta\hat{\theta} \\<br /> \end{align*}

with

\begin{align*} \hat{r} &= (\sin{\phi}\cos{\theta}, \sin{\phi}\sin{\theta}, \cos{\phi}) \\<br /> \hat{\phi} &= (\cos{\phi}\cos{\theta}, \cos{\phi}\sin{\theta}, -\sin{\phi}) \\<br /> \hat{\theta} &= (-\sin{\phi}\sin{\theta}, \sin{\phi}\cos{\theta}, 0) \\<br /> \end{align*}

Therefore, (answer part (a))

\begin{align*} d\hat{r} = &(0,0,0)dr \\<br /> &+ (\cos{\phi}\cos{\theta}, \cos{\phi}\sin{\theta}, -\sin{\phi})d\phi \\<br /> &+ (-\sin{\phi}\sin{\theta}, \sin{\phi}\cos{\theta}, 0)d\theta \\<br /> = &\hat{\phi}d\phi + \hat{\theta}d\theta \\<br /> d\hat{\phi} = &(0,0,0)dr \\<br /> &+ (-\sin{\phi}\cos{\theta}, -\sin{\phi}\sin{\theta}, -\cos{\phi})d\phi \\<br /> &+ (-\cos{\phi}\sin{\theta}, \cos{\phi}\cos{\theta}, 0)d\theta \\<br /> d\hat{\theta} = &(0,0,0)dr \\<br /> &+ (-\cos{\phi}\sin{\theta}, \cos{\phi}\cos{\theta}, 0)d\phi \\<br /> &+ (-\sin{\phi}\cos{\theta}, -\sin{\phi}\sin{\theta}, 0)d\theta \\<br /> \end{align*}

HW #3

  1. Hodge Dual in Minkowski Space

    Minkowski 4-space has orthonormal, oriented basis $ \{dx,dy,dz,dt\} $ with volume element (choice of orientation) $ \omega=dx\wedge dy \wedge dz \wedge dt $ and inner product $ g(,) $ defined by

    \begin{align*}<br /> g(dx,dx)&=1 \\<br /> g(dy,dy)&=1 \\<br /> g(dz,dz)&=1 \\<br /> g(dt,dt)&=-1 \\<br /> g(d\alpha_i,d\alpha_{j\ne i})&=0 \\<br /> \end{align*}

    1. Determine the Hodge dual operator * on all forms by computing its action on basis forms at each rank.
      In general, the dual can be computed as (Equation 20, http://oregonstate.edu/~drayt/Courses/MTH434/2007/dual.pdf):

      \[ \ast(\sigma^1\ldots\sigma^p)=g(\sigma^1,\sigma^1)\ldots g(\sigma^p,\sigma^p)\sigma^{p+1}\wedge\ldots\wedge\sigma^n \]

      where $ \sigma^1\ldots\sigma^n=\omega $ is the volume element.
      In particular:

      \begin{align*}<br /> \ast 1=\omega<br /> \end{align*}
      \begin{align*}<br /> \ast dx &=dy \wedge dz \wedge dt \\<br /> \ast dy &=(-dx \wedge dz \wedge dt) \\<br /> \ast dz &=dx \wedge dy \wedge dt \\<br /> \ast dt &=(-1)(-dx \wedge dy \wedge dz) \\<br /> \end{align*}
      \begin{align*}<br /> \ast dx\wedge dy &=dz \wedge dt) \\<br /> \ast dx\wedge dz &=(-dy \wedge dt) \\<br /> \ast dx\wedge dt &=(-1)(-dy \wedge dz) \\<br /> \ast dy\wedge dz &=dx \wedge dt) \\<br /> \ast dy\wedge dt &=(-1)(-dx \wedge dz) \\<br /> \ast dz\wedge dt &=(-1)(dx \wedge dy) \\<br /> \end{align*}
      \begin{align*}<br /> \ast dx \wedge dy \wedge dz &=dt \\<br /> \ast dx \wedge dy \wedge dt &=(-1)(-dt) \\<br /> \ast dx \wedge dz \wedge dt &=(-1)dt \\<br /> \ast dy \wedge dz \wedge dt &=(-1)(-dt) \\<br /> \end{align*}
      \begin{align*}<br /> \ast \omega = -1<br /> \end{align*}
    2. How does your answer change if the opposite orientation is chosen, namely $ \omega = dt\wedge dx \wedge dy \wedge dz $?
      The sign of each dual would change, due to the additional odd number (3) of transpositions to the volume element. The sign of the number of transpositions required to transform one volume element into another is an invariant (independent of a particular permutation).

  2. Spherical Coordinates

    Spherical coords in $ E^3 $ with $ \omega = r^2\sin{<br /> \theta}dr\wedge d\theta \wedge d\phi $

    1. Calculate the action of * on the basis elements of each rank.
    2. Compute the dot- and cross-products of 2 arbitrary "vector fields" (1-forms) using the expressions:
      \begin{align*}<br /> \alpha\cdot\beta&=\ast(\alpha\wedge\ast\beta) \\<br /> \alpha\times\beta&=\ast(\alpha\wedge\beta) \\<br /> \end{align*}

HW #4: Orthogonal Coordinates

Paraboloidal coordinates given as:

\begin{align*}<br /> x &= uv\sin{\phi} \\<br /> y &= uv\cos{\phi} \\<br /> z &= \frac{1}{2}(u^2-v^2) \\<br /> \end{align*}

We express the Euclidean differential line element in terms of the derivatives (?) of our new coordinate basis:

\begin{align*}<br /> dx &= v\sin{\phi}du + u\sin{\phi}dv + uv\cos{\phi}d\phi \\<br /> dy &= v\cos{\phi}du + u\cos{\phi}dv - uv\sin{\phi}d\phi \\<br /> dz &= u du -v dv \\<br /> ds^2 &= dx^2+dy^2+dz^2 \\<br /> &= (u^2+v^2)du^2 + (u^2+v^2)dv^2 + u^2v^2d\phi^2\\<br /> &= (\sqrt{u^2+v^2}du)^2 + (\sqrt{u^2+v^2}dv)^2 + (uvd\phi)^2\\<br /> \end{align*}

$ \{\sqrt{u^2+v^2}du, \sqrt{u^2+v^2}dv, uvd\phi\} $ is an orthonormal basis for $ \bigwedge^1\Re $ (the square of the line element is the product of the squares of the basis elements). The change-of-basis matrix from $ \{dx,dy,dz\} $ to our new basis thus has determinate one. To save typing we will make the substitution $ \sqrt{u^2+v^2}=\eta $. We choose the oriented volume element:

\[ \omega=\eta du\wedge\eta dv\wedge uvd\phi. \]
  1. Compute the gradient of an arbitrary function $ f $by the expression

    \begin{align*} \nabla f &=df \\<br /> df &= f_udu+f_vdv+f_\phi d\phi \end{align*}
  2. Compute the Laplacian of $ f $ by the expression
    \begin{align*}<br /> \Delta f &= \nabla\cdot\nabla f = \ast d\ast df \\<br /> &=\ast d(\frac{f_u}{\eta}\ast(\eta du) + \frac{f_v}{\eta}\ast(\eta dv) + \frac{f_\phi}{uv}\ast{uv d\phi}) \\<br /> &=\ast d(f_u uvdv\wedge d\phi + f_v uvdu\wedge d\phi + \frac{f_\phi}{uv}\eta^2 dudv) \\<br /> &=(vf_u+uvf_{uu}- uf_v-uvf_{uu}+\frac{f_{\phi\phi}}{uv}\eta^2)\ast(dudvd\phi) \\<br /> &=(vf_u+uvf_{uu}- uf_v-uvf_{uu}+\frac{f_{\phi\phi}}{uv}\eta^2) \\<br /> \end{align*}

Homework #1: Decomposable Forms

  1. Let $ \vec{u} $ be an ordinary vector in $ \Re^3 $, so that
    \[ \vec{u} = A\hat{i} + B\hat{j} + C\hat{k} \]

    Find $ \vec{v} $ and $ \vec{w} $ such that

    \[ \vec{u} = \vec{v} \times \vec{w} \]

    We observe that the cross product is an alternating linear function, uniquely determined by its action on the basis,

    \begin{align*}<br /> \hat{j} \times \hat{k} &= \hat{i} \\<br /> \hat{k} \times \hat{i} &= \hat{j} \\<br /> \hat{i} \times \hat{j} &= \hat{k}<br /> \end{align*}

    and also that $ \times $ is antisymmetric and distributive. Bring this all together, we can express $ \vec{u} $ as a product of two vectors:

    \begin{align*}<br /> \vec{u} &= A\hat{j} \times \hat{k} + B\hat{k} \times \hat{i} + C\hat{i} \times \hat{j} \\<br /> &= A\hat{j} \times \hat{k} + (B\hat{k} - C\hat{j})\times \hat{i} \\<br /> &= (A + B) \hat{j} \times \hat{k} + B \hat{k} \times \hat{j} + (B\hat{k} - C\hat{j})\times \hat{i} \\<br /> &= -C \frac{(A + B)}{-C} \hat{j} \times \hat{k} + B \hat{k} \times \hat{j} + (B\hat{k} - C\hat{j})\times \hat{i} \\<br /> &= (B\hat{k} - C\hat{j})\times (\hat{i} + \hat{j} - \frac{(A + B)}{C}\hat{k}) \end{align*}

    Notice that if $ C=0 $, the answer is undefined. One, of A,B,C must be nonzero, however, in which case a symmetric argument will suffice. $ \square $