Author Topic: Gradients on a sphere (Read 1457 times) Tweet Share

/0 · « **on:** October 07, 2010, 02:35:25 pm »

Consider the sphere $S^2$ , $r=constant$ in R^3. Then

$\nabla r = \frac{\partial r}{\partial x}\mathbf{i}+\frac{\partial r}{\partial y}\mathbf{j}+\frac{\partial r}{\partial z}\mathbf{k}=\frac{x}{r}\mathbf{i}+\frac{y}{r}\mathbf{j}+\frac{z}{r}\mathbf{k}=\mathbf{e}_r$

$\nabla \theta=\frac{\partial \theta}{\patrial x}\mathbf{i}+\frac{\partial \theta}{\partial y}\mathbf{j}+\frac{\partial \theta}{\partial z}\mathbf{k}=\frac{\cos\theta\cos\phi}{r}\mathbf{i}+\frac{\cos\theta\sin\phi}{r}\mathbf{j}-\frac{\sin\theta}{r}\mathbf{k}=\frac{1}{r^2}\mathbf{e}_\theta$

$\nabla \phi=\frac{\partial \phi}{\partial x}\mathbf{i}+\frac{\partial \phi}{\partial y}\mathbf{j}+\frac{\partial \phi}{\partial z}\mathbf{k}=-\frac{\sin\phi}{r\sin\theta}\mathbf{i}+\frac{\cos\phi}{r\sin\theta}\mathbf{j}=\frac{1}{r^2\sin^2\theta}\mathbf{e}_\phi$

The first one isn't hard to prove, since you have $r=\sqrt{x^2+y^2+z^2}$ , but how do you do the others? thx

Mao · « **Reply #1 on:** October 09, 2010, 02:47:34 am »

Are those spherical coordinates you are working in?

Regardless,
$x=r \cos \theta \sin \phi$

$dx/dx= dx/dr .dr/dx + dx/d\theta. d\theta/dx +dx/d\phi. d\phi/dx$ from chain rule

$1= blah$

I'm on my phone so I cbf doing latex. But yeah, that's the gist

/0 · « **Reply #2 on:** October 09, 2010, 02:53:09 am »

ooh right, thanks Mao

Mao · « **Reply #3 on:** October 09, 2010, 03:03:38 am »

Ignore what I posted before, I forgot this was multivariate.

I've edited, check again.

/0 · « **Reply #4 on:** October 09, 2010, 03:20:54 am »

yep I got the gist of it =]

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Author Topic: Gradients on a sphere (Read 1457 times) Tweet Share

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Gradients on a sphere

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