How to derive the spherical coordinate form for Laplacian

[kakarotyjn]

Homework Statement

\Delta f = \frac{1}{{r^2 }}\frac{\partial }{{\partial r}}\left( {r^2 \frac{{\partial f}}{{\partial r}}} \right) + \frac{1}{{r^2 \sin \phi }}\frac{\partial }{{\partial \phi }}\left( {\sin \phi \frac{{\partial f}}{{\partial \phi }}} \right) + \frac{1}{{r^2 \sin ^2 \phi }}\frac{{\partial ^2 f}}{{\partial \theta ^2 }}

Homework Equations

\nabla ^2 = \frac{{\partial ^2 }}{{\partial x^2 }} + \frac{{\partial ^2 }}{{\partial y^2 }} + \frac{{\partial ^2 }}{{\partial y^2 }}

The Attempt at a Solution

 

[fzero]

What have you attempted so far?

 

[kakarotyjn]

I have just attmpted to write the coordinate change:x=r sin(\theta)cos(\phi) y=rsin(\theta)sin(\phi) z=rcos(\theta)

I don't know what to do next.

 

[HallsofIvy]

Just use the chain rule, repeatedly. Nothing very deep but labor intensive and tedious!

\frac{\partial f}{\partial x}= \frac{\partial f}{\partial r}\frac{\partial r}{\partial x}+ \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial x}+ \frac{\partial f}{\partial \phi}\frac{\partial \phi}{\partial x}

You will need the coordinate equations, "the other way around":
r= (x^2+ y^2+ z^2)^{1/2}
\phi= arctan(\frac{y}{x})
\theta= arctan(\frac{\sqrt{x^2+ y^2}}{z}
and you will want to write the derivatives in terms of the polar coordinates:
\frac{d\phi}{dx}= \frac{1}{1+ \frac{y^2}{x^2}}\left(-\frac{y}{x^2}\right)
and, multiplying numerator and denominator by x^2,
= \frac{-y}{x^2+ y^2}= -\frac{r sin(\phi)}{r^2 sin^2(\theta)}= -\frac{sin(\phi)}{r sin^2(\theta)}.

If you haven't done it already, I recommend doing polar or cylindrical coordinates first to get the idea.

 

[kakarotyjn]

Just use the chain rule, repeatedly. Nothing very deep but labor intensive and tedious!

\frac{\partial f}{\partial x}= \frac{\partial f}{\partial r}\frac{\partial r}{\partial x}+ \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial x}+ \frac{\partial f}{\partial \phi}\frac{\partial \phi}{\partial x}

You will need the coordinate equations, "the other way around":
r= (x^2+ y^2+ z^2)^{1/2}
\phi= arctan(\frac{y}{x})
\theta= arctan(\frac{\sqrt{x^2+ y^2}}{z}
and you will want to write the derivatives in terms of the polar coordinates:
\frac{d\phi}{dx}= \frac{1}{1+ \frac{y^2}{x^2}}\left(-\frac{y}{x^2}\right)
and, multiplying numerator and denominator by x^2,
= \frac{-y}{x^2+ y^2}= -\frac{r sin(\phi)}{r^2 sin^2(\theta)}= -\frac{sin(\phi)}{r sin^2(\theta)}.

If you haven't done it already, I recommend doing polar or cylindrical coordinates first to get the idea.

Thank you very much,I see