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Laplacian in 3-D polar coordinates

Posted 9 years ago

Hello, Could someone tell me how to generate the Laplacian operator in 3-D polar coordinates? Every time I tried to use Laplacian[[r^2], {r, theta, phi}, "polar"], I have used the escape key to generate the angles. it returns with an error message saying that "polar does not define a metric in 3-dimension". Thanks

POSTED BY: Kenneth Wong
2 Replies

See Laplacian please (for a post V 7 treatment). See the defined names of co-ordinate systems CoordinateChartData, e.g. people are used to use the noun Spherical for

3-D polar coordinates

as Hans mentioned already in his answer.

POSTED BY: Udo Krause

In[27]:= << VectorAnalysis`

In[28]:= Laplacian[f[r, th, phi], Spherical[r, th, phi]]

$\frac{\csc (\text{th}) \left(r^2 \sin (\text{th}) f^{(2,0,0)}(r,\text{th},\phi )+2 r \sin (\text{th}) f^{(1,0,0)}(r,\text{th},\phi )+\sin (\text{th}) f^{(0,2,0)}(r,\text{th},\phi )+\cos (\text{th}) f^{(0,1,0)}(r,\text{th},\phi )+\csc (\text{th}) f^{(0,0,2)}(r,\text{th},\phi )\right)}{r^2}$

Regards Hans Dolhaine

POSTED BY: Hans Dolhaine
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