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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*}

Tue, 2007-02-13 22:15 | eric