derivation of the Laplacian from rectangular to spherical coordinates

We begin by recognizing the familiar conversion from rectangular to spherical coordinates (note that $\varphi$ is used to denote the azimuthal angle, whereas $\theta$ is used to denote the polar angle)

$x=r\sin (\theta )\cos (\varphi ),y=r\sin (\theta )\sin (\varphi ),z=r\cos (\theta ),$ (1)

and conversely from spherical to rectangular coordinates

$r=\sqrt{x^2+y^2+z^2},\theta =\arccos \left({\displaystyle \frac{z}{r}}\right),\varphi =\arctan \left({\displaystyle \frac{y}{x}}\right).$ (2)

Now, we know that the Laplacian in rectangular coordinates is defined in the following way

${\nabla }^{2}f$ $=$ ${\displaystyle \frac{{\partial }^{2}f}{\partial x^2}}+{\displaystyle \frac{{\partial }^{2}f}{\partial y^2}}+{\displaystyle \frac{{\partial }^{2}f}{\partial z^2}}.$ (3)

We also know that the partial derivatives in rectangular coordinates can be expanded in the following way by using the chain rule

${\displaystyle \frac{\partial f}{\partial x}}$ $=$ ${\displaystyle \frac{\partial f}{\partial r}}{\displaystyle \frac{\partial r}{\partial x}}+{\displaystyle \frac{\partial f}{\partial \theta }}{\displaystyle \frac{\partial \theta }{\partial x}}+{\displaystyle \frac{\partial f}{\partial \varphi }}{\displaystyle \frac{\partial \varphi }{\partial x}},$ (4)

${\displaystyle \frac{\partial f}{\partial y}}$ $=$ ${\displaystyle \frac{\partial f}{\partial r}}{\displaystyle \frac{\partial r}{\partial y}}+{\displaystyle \frac{\partial f}{\partial \theta }}{\displaystyle \frac{\partial \theta }{\partial y}}+{\displaystyle \frac{\partial f}{\partial \varphi }}{\displaystyle \frac{\partial \varphi }{\partial y}},$ (5)

${\displaystyle \frac{\partial f}{\partial z}}$ $=$ ${\displaystyle \frac{\partial f}{\partial r}}{\displaystyle \frac{\partial r}{\partial z}}+{\displaystyle \frac{\partial f}{\partial \theta }}{\displaystyle \frac{\partial \theta }{\partial z}}+{\displaystyle \frac{\partial f}{\partial \varphi }}{\displaystyle \frac{\partial \varphi }{\partial z}}.$ (6)

The next step is to convert the right-hand side of each of the above three equations so that it only has partial derivatives in terms of $r$, $\theta$ and $\varphi$. We can do this by substituting the following values (which are easily derived from (2)) in their respective places in the above three equations

${\displaystyle \frac{\partial r}{\partial x}}=\sin (\theta )\cos (\varphi ),{\displaystyle \frac{\partial \theta }{\partial x}}={\displaystyle \frac{1}{r}}\cos (\theta )\cos (\varphi ),{\displaystyle \frac{\partial \varphi }{\partial x}}=-{\displaystyle \frac{1}{r}}{\displaystyle \frac{\sin (\varphi )}{\sin (\theta )}},$

${\displaystyle \frac{\partial r}{\partial y}}=\sin (\theta )\sin (\varphi ),{\displaystyle \frac{\partial \theta }{\partial y}}={\displaystyle \frac{1}{r}}\cos (\theta )\sin (\varphi ),{\displaystyle \frac{\partial \varphi }{\partial y}}={\displaystyle \frac{1}{r}}{\displaystyle \frac{\cos (\varphi )}{\sin (\theta )}},$

${\displaystyle \frac{\partial r}{\partial z}}=\cos (\theta ),{\displaystyle \frac{\partial \theta }{\partial z}}=-{\displaystyle \frac{1}{r}}\sin (\theta ),{\displaystyle \frac{\partial \varphi }{\partial z}}=0.$ (7)

After the substitution, equation (4) looks like the following

${\displaystyle \frac{\partial f}{\partial x}}$ $=$ $\sin (\theta )\cos (\varphi ){\displaystyle \frac{\partial f}{\partial r}}+{\displaystyle \frac{1}{r}}\cos (\theta )\cos (\varphi ){\displaystyle \frac{\partial f}{\partial \theta }}-{\displaystyle \frac{1}{r}}{\displaystyle \frac{\sin (\varphi )}{\sin (\theta )}}{\displaystyle \frac{\partial f}{\partial \varphi }}.$ (8)

Assuming that $f$ is a sufficiently differentiable function, we can replace $f$ by $\frac{\partial f}{\partial x}$ in the above equation and arrive at the following

${\displaystyle \frac{{\partial }^{2}f}{\partial x^2}}$ $=$ $\sin (\theta )\cos (\varphi ){\displaystyle \frac{\partial }{\partial r}}\left[{\displaystyle \frac{\partial f}{\partial x}}\right]+{\displaystyle \frac{1}{r}}\cos (\theta )\cos (\varphi ){\displaystyle \frac{\partial }{\partial \theta }}\left[{\displaystyle \frac{\partial f}{\partial x}}\right]-{\displaystyle \frac{1}{r}}{\displaystyle \frac{\sin (\varphi )}{\sin (\theta )}}{\displaystyle \frac{\partial }{\partial \varphi }}\left[{\displaystyle \frac{\partial f}{\partial x}}\right].$ (9)

Now the trick is to substitute equation (8) into equation (9) in order to eliminate any partial derivatives with respect to $x$. The result is the following equation

${\displaystyle \frac{{\partial }^{2}f}{\partial x^2}}$ $=$ $\sin (\theta )\cos (\varphi ){\displaystyle \frac{\partial }{\partial r}}\left[\sin (\theta )\cos (\varphi ){\displaystyle \frac{\partial f}{\partial r}}+{\displaystyle \frac{1}{r}}\cos (\theta )\cos (\varphi ){\displaystyle \frac{\partial f}{\partial \theta }}-{\displaystyle \frac{1}{r}}{\displaystyle \frac{\sin (\varphi )}{\sin (\theta )}}{\displaystyle \frac{\partial f}{\partial \varphi }}\right]+$

${\displaystyle \frac{1}{r}}\cos (\theta )\cos (\varphi ){\displaystyle \frac{\partial }{\partial \theta }}\left[\sin (\theta )\cos (\varphi ){\displaystyle \frac{\partial f}{\partial r}}+{\displaystyle \frac{1}{r}}\cos (\theta )\cos (\varphi ){\displaystyle \frac{\partial f}{\partial \theta }}-{\displaystyle \frac{1}{r}}{\displaystyle \frac{\sin (\varphi )}{\sin (\theta )}}{\displaystyle \frac{\partial f}{\partial \varphi }}\right]-$

${\displaystyle \frac{1}{r}}{\displaystyle \frac{\sin (\varphi )}{\sin (\theta )}}{\displaystyle \frac{\partial }{\partial \varphi }}\left[\sin (\theta )\cos (\varphi ){\displaystyle \frac{\partial f}{\partial r}}+{\displaystyle \frac{1}{r}}\cos (\theta )\cos (\varphi ){\displaystyle \frac{\partial f}{\partial \theta }}-{\displaystyle \frac{1}{r}}{\displaystyle \frac{\sin (\varphi )}{\sin (\theta )}}{\displaystyle \frac{\partial f}{\partial \varphi }}\right].$

In the hopes of simplifying the above equation, we operate the derivates on the operands and get

${\displaystyle \frac{{\partial }^{2}f}{\partial x^2}}$ $=$ $\sin (\theta )\cos (\varphi )[\sin (\theta )\cos (\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial r^2}}-{\displaystyle \frac{1}{r^2}}\cos (\theta )\cos (\varphi ){\displaystyle \frac{\partial f}{\partial \theta }}+{\displaystyle \frac{1}{r}}\cos (\theta )\cos (\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \theta }}+$

${\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\sin (\varphi )}{\sin (\theta )}}{\displaystyle \frac{\partial f}{\partial \varphi }}-{\displaystyle \frac{1}{r}}{\displaystyle \frac{\sin (\varphi )}{\sin (\theta )}}{\displaystyle \frac{{\partial }^{2}f}{\partial \varphi \partial r}}]+{\displaystyle \frac{1}{r}}\cos (\theta )\cos (\varphi )[\cos (\theta )\cos (\varphi ){\displaystyle \frac{\partial f}{\partial r}}+$

$\sin (\theta )\cos (\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \theta }}-{\displaystyle \frac{1}{r}}\sin (\theta )\cos (\varphi ){\displaystyle \frac{\partial f}{\partial \theta }}+{\displaystyle \frac{1}{r}}\cos (\theta )\cos (\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial {\theta }^2}}+{\displaystyle \frac{1}{r}}{\displaystyle \frac{\sin (\varphi )\cos (\theta )}{{\sin }^2(\theta )}}{\displaystyle \frac{\partial f}{\partial \varphi }}-$

${\displaystyle \frac{1}{r}}{\displaystyle \frac{\sin (\varphi )}{\sin (\theta )}}{\displaystyle \frac{{\partial }^{2}f}{\partial \theta \partial \varphi }}]-{\displaystyle \frac{1}{r}}{\displaystyle \frac{\sin (\varphi )}{\sin (\theta )}}[-\sin (\theta )\sin (\varphi ){\displaystyle \frac{\partial f}{\partial r}}+\sin (\theta )\cos (\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \varphi }}-$

${\displaystyle \frac{1}{r}}\cos (\theta )\sin (\varphi ){\displaystyle \frac{\partial f}{\partial \theta }}+{\displaystyle \frac{1}{r}}\cos (\theta )\cos (\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial \theta \partial \varphi }}-{\displaystyle \frac{1}{r}}{\displaystyle \frac{\cos (\varphi )}{\sin (\theta )}}{\displaystyle \frac{\partial f}{\partial \varphi }}-{\displaystyle \frac{1}{r}}{\displaystyle \frac{\sin (\varphi )}{\sin (\theta )}}{\displaystyle \frac{{\partial }^{2}f}{\partial {\varphi }^2}}].$

After further simplifying the above equation, we arrive at the following form

${\displaystyle \frac{{\partial }^{2}f}{\partial x^2}}$ $=$ $[{\sin }^2(\theta ){\cos }^2(\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial r^2}}-{\displaystyle \frac{1}{r^2}}\cos (\theta )\sin (\theta ){\cos }^2(\varphi ){\displaystyle \frac{\partial f}{\partial \theta }}+{\displaystyle \frac{1}{r}}\cos (\theta )\sin (\theta ){\cos }^2(\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \theta }}+$

${\displaystyle \frac{1}{r^2}}\sin (\varphi )\cos (\varphi ){\displaystyle \frac{\partial f}{\partial \varphi }}-{\displaystyle \frac{1}{r}}\sin (\varphi )\cos (\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial \varphi \partial r}}]+[{\displaystyle \frac{1}{r}}{\cos }^2(\theta ){\cos }^2(\varphi ){\displaystyle \frac{\partial f}{\partial r}}+$

${\displaystyle \frac{1}{r}}\sin (\theta )\cos (\theta ){\cos }^2(\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \theta }}-{\displaystyle \frac{1}{r^2}}\sin (\theta )\cos (\theta ){\cos }^2(\varphi ){\displaystyle \frac{\partial f}{\partial \theta }}+{\displaystyle \frac{1}{r^2}}{\cos }^2(\theta ){\cos }^2(\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial {\theta }^2}}+$

${\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\sin (\varphi )\cos (\varphi ){\cos }^2(\theta )}{{\sin }^2(\theta )}}{\displaystyle \frac{\partial f}{\partial \varphi }}-{\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\cos (\theta )\sin (\varphi )\cos (\varphi )}{\sin (\theta )}}{\displaystyle \frac{{\partial }^{2}f}{\partial \theta \partial \varphi }}]+[{\displaystyle \frac{1}{r}}{\sin }^2(\varphi ){\displaystyle \frac{\partial f}{\partial r}}-$

${\displaystyle \frac{1}{r}}\sin (\varphi )\cos (\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \varphi }}+{\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\cos (\theta ){\sin }^2(\varphi )}{\sin (\theta )}}{\displaystyle \frac{\partial f}{\partial \theta }}-{\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\cos (\theta )\cos (\varphi )\sin (\varphi )}{\sin (\theta )}}{\displaystyle \frac{{\partial }^{2}f}{\partial \theta \partial \varphi }}+$

${\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\sin (\varphi )\cos (\varphi )}{{\sin }^2(\theta )}}{\displaystyle \frac{\partial f}{\partial \varphi }}+{\displaystyle \frac{1}{r^2}}{\displaystyle \frac{{\sin }^2(\varphi )}{{\sin }^2(\theta )}}{\displaystyle \frac{{\partial }^{2}f}{\partial {\varphi }^2}}].$

Notice that we have derived the first term of the right-hand side of equation (3) (i.e. $\frac{{\partial }^{2}f}{\partial x^2}$) in terms of spherical coordinates. We now have to do a similar arduous derivation for the rest of the two terms (i.e. $\frac{{\partial }^{2}f}{\partial y^2}$ and $\frac{{\partial }^{2}f}{\partial z^2}$). Lets do it!

After we substitute the values of (7) into equation (5) we get

${\displaystyle \frac{\partial f}{\partial y}}$ $=$ $\sin (\theta )\sin (\varphi ){\displaystyle \frac{\partial f}{\partial r}}+{\displaystyle \frac{1}{r}}\cos (\theta )\sin (\varphi ){\displaystyle \frac{\partial f}{\partial \theta }}+{\displaystyle \frac{1}{r}}{\displaystyle \frac{\cos (\varphi )}{\sin (\theta )}}{\displaystyle \frac{\partial f}{\partial \varphi }}.$ (10)

Again, assuming that $f$ is a sufficiently differentiable function, we can replace $f$ by $\frac{\partial f}{\partial y}$ in the above equation and arrive at the following

${\displaystyle \frac{{\partial }^{2}f}{\partial y^2}}$ $=$ $\sin (\theta )\sin (\varphi ){\displaystyle \frac{\partial }{\partial r}}\left[{\displaystyle \frac{\partial f}{\partial y}}\right]+{\displaystyle \frac{1}{r}}\cos (\theta )\sin (\varphi ){\displaystyle \frac{\partial }{\partial \theta }}\left[{\displaystyle \frac{\partial f}{\partial y}}\right]+{\displaystyle \frac{1}{r}}{\displaystyle \frac{\cos (\varphi )}{\sin (\theta )}}{\displaystyle \frac{\partial }{\partial \varphi }}\left[{\displaystyle \frac{\partial f}{\partial y}}\right].$ (11)

Now we substitute equation (10) into equation (11) in order to eliminate any partial derivatives with respect to $y$. The result is the following

${\displaystyle \frac{{\partial }^{2}f}{\partial y^2}}$ $=$ $\sin (\theta )\sin (\varphi ){\displaystyle \frac{\partial }{\partial r}}\left[\sin (\theta )\sin (\varphi ){\displaystyle \frac{\partial f}{\partial r}}+{\displaystyle \frac{1}{r}}\cos (\theta )\sin (\varphi ){\displaystyle \frac{\partial f}{\partial \theta }}+{\displaystyle \frac{1}{r}}{\displaystyle \frac{\cos (\varphi )}{\sin (\theta )}}{\displaystyle \frac{\partial f}{\partial \varphi }}\right]+$

${\displaystyle \frac{1}{r}}\cos (\theta )\sin (\varphi ){\displaystyle \frac{\partial }{\partial \theta }}\left[\sin (\theta )\sin (\varphi ){\displaystyle \frac{\partial f}{\partial r}}+{\displaystyle \frac{1}{r}}\cos (\theta )\sin (\varphi ){\displaystyle \frac{\partial f}{\partial \theta }}+{\displaystyle \frac{1}{r}}{\displaystyle \frac{\cos (\varphi )}{\sin (\theta )}}{\displaystyle \frac{\partial f}{\partial \varphi }}\right]+$

${\displaystyle \frac{1}{r}}{\displaystyle \frac{\cos (\varphi )}{\sin (\theta )}}{\displaystyle \frac{\partial }{\partial \varphi }}\left[\sin (\theta )\sin (\varphi ){\displaystyle \frac{\partial f}{\partial r}}+{\displaystyle \frac{1}{r}}\cos (\theta )\sin (\varphi ){\displaystyle \frac{\partial f}{\partial \theta }}+{\displaystyle \frac{1}{r}}{\displaystyle \frac{\cos (\varphi )}{\sin (\theta )}}{\displaystyle \frac{\partial f}{\partial \varphi }}\right].$

Now we operate the operators and get

${\displaystyle \frac{{\partial }^{2}f}{\partial y^2}}$ $=$ $\sin (\theta )\sin (\varphi )[\sin (\theta )\sin (\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial r^2}}-{\displaystyle \frac{1}{r^2}}\cos (\theta )\sin (\varphi ){\displaystyle \frac{\partial f}{\partial \theta }}+{\displaystyle \frac{1}{r}}\cos (\theta )\sin (\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \theta }}-$

${\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\cos (\varphi )}{\sin (\theta )}}{\displaystyle \frac{\partial f}{\partial \varphi }}+{\displaystyle \frac{1}{r}}{\displaystyle \frac{\cos (\varphi )}{\sin (\theta )}}{\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \varphi }}]+{\displaystyle \frac{1}{r}}\cos (\theta )\sin (\varphi )[\sin (\theta )\sin (\varphi ){\displaystyle \frac{\partial f}{\partial r\partial \theta }}+$

$\cos (\theta )\sin (\varphi ){\displaystyle \frac{\partial f}{\partial r}}-{\displaystyle \frac{1}{r}}\sin (\theta )\sin (\varphi ){\displaystyle \frac{\partial f}{\partial \theta }}+{\displaystyle \frac{1}{r}}\cos (\theta )\sin (\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial {\theta }^2}}-{\displaystyle \frac{1}{r}}{\displaystyle \frac{\cos (\varphi )cos(\theta )}{{\sin }^2(\theta )}}{\displaystyle \frac{\partial f}{\partial \varphi }}+$

${\displaystyle \frac{1}{r}}{\displaystyle \frac{\cos (\varphi )}{\sin (\theta )}}{\displaystyle \frac{{\partial }^{2}f}{\partial \theta \partial \varphi }}]+{\displaystyle \frac{1}{r}}{\displaystyle \frac{\cos (\varphi )}{\sin (\theta )}}[\sin (\theta )\cos (\varphi ){\displaystyle \frac{\partial f}{\partial r}}+\sin (\theta )\sin (\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \varphi }}+$

${\displaystyle \frac{1}{r}}\cos (\theta )\cos (\varphi ){\displaystyle \frac{\partial f}{\partial \theta }}+{\displaystyle \frac{1}{r}}\cos (\theta )\sin (\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial \theta \partial \varphi }}-{\displaystyle \frac{1}{r}}{\displaystyle \frac{\sin (\varphi )}{\sin (\theta )}}{\displaystyle \frac{\partial f}{\partial \varphi }}+{\displaystyle \frac{1}{r}}{\displaystyle \frac{\cos (\varphi )}{\sin (\theta )}}{\displaystyle \frac{{\partial }^{2}f}{\partial {\varphi }^2}}],$

and after some simplifications

${\displaystyle \frac{{\partial }^{2}f}{\partial y^2}}$ $=$ $[{\sin }^2(\theta ){\sin }^2(\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial r^2}}-{\displaystyle \frac{1}{r^2}}\sin (\theta )\cos (\theta ){\sin }^2(\varphi ){\displaystyle \frac{\partial f}{\partial \theta }}+{\displaystyle \frac{1}{r}}\sin (\theta )\cos (\theta ){\sin }^2(\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \theta }}-$

${\displaystyle \frac{1}{r^2}}\sin (\varphi )\cos (\varphi ){\displaystyle \frac{\partial f}{\partial \varphi }}+{\displaystyle \frac{1}{r}}\sin (\varphi )\cos (\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \varphi }}]+[{\displaystyle \frac{1}{r}}\cos (\theta )\sin (\theta ){\sin }^2(\varphi ){\displaystyle \frac{\partial f}{\partial r\partial \theta }}+$

${\displaystyle \frac{1}{r}}{\cos }^2(\theta ){\sin }^2(\varphi ){\displaystyle \frac{\partial f}{\partial r}}-{\displaystyle \frac{1}{r^2}}\sin (\theta )\cos (\theta ){\sin }^2(\varphi ){\displaystyle \frac{\partial f}{\partial \theta }}+{\displaystyle \frac{1}{r^2}}{\cos }^2(\theta ){\sin }^2(\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial {\theta }^2}}-$

${\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\sin (\varphi )\cos (\varphi ){\cos }^2(\theta )}{{\sin }^2(\theta )}}{\displaystyle \frac{\partial f}{\partial \varphi }}+{\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\cos (\theta )\sin (\varphi )\cos (\varphi )}{\sin (\theta )}}{\displaystyle \frac{{\partial }^{2}f}{\partial \theta \partial \varphi }}]+[{\displaystyle \frac{1}{r}}{\cos }^2(\varphi ){\displaystyle \frac{\partial f}{\partial r}}+$

${\displaystyle \frac{1}{r}}\cos (\varphi )\sin (\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \varphi }}+{\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\cos (\theta ){\cos }^2(\varphi )}{\sin (\theta )}}{\displaystyle \frac{\partial f}{\partial \theta }}+{\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\cos (\theta )\cos (\varphi )\sin (\varphi )}{\sin (\theta )}}{\displaystyle \frac{{\partial }^{2}f}{\partial \theta \partial \varphi }}-$

${\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\sin (\varphi )\cos (\varphi )}{{\sin }^2(\theta )}}{\displaystyle \frac{\partial f}{\partial \varphi }}+{\displaystyle \frac{1}{r^2}}{\displaystyle \frac{{\cos }^2(\varphi )}{{\sin }^2(\theta )}}{\displaystyle \frac{{\partial }^{2}f}{\partial {\varphi }^2}}].$

Now its time to derive $\frac{{\partial }^{2}f}{\partial z^2}$. After our substitution of value in (7) into equation (6) we get

${\displaystyle \frac{\partial f}{\partial z}}$ $=$ $\cos (\theta ){\displaystyle \frac{\partial f}{\partial r}}-{\displaystyle \frac{1}{r}}\sin (\theta ){\displaystyle \frac{\partial f}{\partial \theta }}.$ (12)

Once more, assuming that $f$ is a sufficiently differentiable function, we can replace $f$ by $\frac{\partial f}{\partial z}$ in the above equation which gives us the following

${\displaystyle \frac{{\partial }^{2}f}{\partial z^2}}$ $=$ $\cos (\theta ){\displaystyle \frac{\partial }{\partial r}}\left[{\displaystyle \frac{\partial f}{\partial z}}\right]-{\displaystyle \frac{1}{r}}\sin (\theta ){\displaystyle \frac{\partial }{\partial \theta }}\left[{\displaystyle \frac{\partial f}{\partial z}}\right].$ (13)

Now we substitute equation (12) into equation (13) in order to eliminate any partial derivatives with respect to $z$ and we arrive at

${\displaystyle \frac{{\partial }^{2}f}{\partial z^2}}$ $=$ $\cos (\theta ){\displaystyle \frac{\partial }{\partial r}}\left[\cos (\theta ){\displaystyle \frac{\partial f}{\partial r}}-{\displaystyle \frac{1}{r}}\sin (\theta ){\displaystyle \frac{\partial f}{\partial \theta }}\right]-$

${\displaystyle \frac{1}{r}}\sin (\theta ){\displaystyle \frac{\partial }{\partial \theta }}\left[\cos (\theta ){\displaystyle \frac{\partial f}{\partial r}}-{\displaystyle \frac{1}{r}}\sin (\theta ){\displaystyle \frac{\partial f}{\partial \theta }}\right].$

After operating the operators we get

${\displaystyle \frac{{\partial }^{2}f}{\partial z^2}}$ $=$ $\cos (\theta )\left[\cos (\theta ){\displaystyle \frac{{\partial }^{2}f}{\partial r^2}}+{\displaystyle \frac{1}{r^2}}\sin (\theta ){\displaystyle \frac{\partial f}{\partial \theta }}-{\displaystyle \frac{1}{r}}\sin (\theta ){\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \theta }}\right]-$

${\displaystyle \frac{1}{r}}\sin (\theta )[-\sin (\theta ){\displaystyle \frac{\partial f}{\partial r}}+\cos (\theta ){\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \theta }}-{\displaystyle \frac{1}{r}}\cos (\theta ){\displaystyle \frac{\partial f}{\partial \theta }}-$

${\displaystyle \frac{1}{r}}\sin (\theta ){\displaystyle \frac{{\partial }^{2}f}{\partial {\theta }^2}}],$

and then simplifying

${\displaystyle \frac{{\partial }^{2}f}{\partial z^2}}$ $=$ $[{\cos }^2(\theta ){\displaystyle \frac{{\partial }^{2}f}{\partial r^2}}+{\displaystyle \frac{1}{r^2}}\cos (\theta )\sin (\theta ){\displaystyle \frac{\partial f}{\partial \theta }}-{\displaystyle \frac{1}{r}}\cos (\theta )\sin (\theta ){\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \theta }}]+[{\displaystyle \frac{1}{r}}{\sin }^2(\theta ){\displaystyle \frac{\partial f}{\partial r}}-$

${\displaystyle \frac{1}{r}}\sin (\theta )\cos (\theta ){\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \theta }}+{\displaystyle \frac{1}{r^2}}\sin (\theta )\cos (\theta ){\displaystyle \frac{\partial f}{\partial \theta }}+{\displaystyle \frac{1}{r^2}}{\sin }^2(\theta ){\displaystyle \frac{{\partial }^{2}f}{\partial {\theta }^2}}].$

Now that we have all three terms of the right hand side of equation (3)(i.e. $\frac{{\partial }^{2}f}{\partial x^2}$, $\frac{{\partial }^{2}f}{\partial y^2}$ and $\frac{{\partial }^{2}f}{\partial z^2}$), we add them all together (because of equation (3)) to get the laplacian in terms of $r$, $\theta$ and $\varphi$

${\nabla }^{2}f$ $=$ $[{\sin }^2(\theta ){\cos }^2(\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial r^2}}-{\displaystyle \frac{1}{r^2}}\cos (\theta )\sin (\theta ){\cos }^2(\varphi ){\displaystyle \frac{\partial f}{\partial \theta }}+{\displaystyle \frac{1}{r}}\cos (\theta )\sin (\theta ){\cos }^2(\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \theta }}+$

${\displaystyle \frac{1}{r^2}}\sin (\varphi )\cos (\varphi ){\displaystyle \frac{\partial f}{\partial \varphi }}-{\displaystyle \frac{1}{r}}\sin (\varphi )\cos (\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial \varphi \partial r}}]+[{\displaystyle \frac{1}{r}}{\cos }^2(\theta ){\cos }^2(\varphi ){\displaystyle \frac{\partial f}{\partial r}}+$

${\displaystyle \frac{1}{r}}\sin (\theta )\cos (\theta ){\cos }^2(\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \theta }}-{\displaystyle \frac{1}{r^2}}\sin (\theta )\cos (\theta ){\cos }^2(\varphi ){\displaystyle \frac{\partial f}{\partial \theta }}+{\displaystyle \frac{1}{r^2}}{\cos }^2(\theta ){\cos }^2(\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial {\theta }^2}}+$

${\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\sin (\varphi )\cos (\varphi ){\cos }^2(\theta )}{{\sin }^2(\theta )}}{\displaystyle \frac{\partial f}{\partial \varphi }}-{\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\cos (\theta )\sin (\varphi )\cos (\varphi )}{\sin (\theta )}}{\displaystyle \frac{{\partial }^{2}f}{\partial \theta \partial \varphi }}]+[{\displaystyle \frac{1}{r}}{\sin }^2(\varphi ){\displaystyle \frac{\partial f}{\partial r}}-$

${\displaystyle \frac{1}{r}}\sin (\varphi )\cos (\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \varphi }}+{\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\cos (\theta ){\sin }^2(\varphi )}{\sin (\theta )}}{\displaystyle \frac{\partial f}{\partial \theta }}-{\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\cos (\theta )\cos (\varphi )\sin (\varphi )}{\sin (\theta )}}{\displaystyle \frac{{\partial }^{2}f}{\partial \theta \partial \varphi }}+$

${\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\sin (\varphi )\cos (\varphi )}{{\sin }^2(\theta )}}{\displaystyle \frac{\partial f}{\partial \varphi }}+{\displaystyle \frac{1}{r^2}}{\displaystyle \frac{{\sin }^2(\varphi )}{{\sin }^2(\theta )}}{\displaystyle \frac{{\partial }^{2}f}{\partial {\varphi }^2}}]+[{\sin }^2(\theta ){\sin }^2(\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial r^2}}-$

${\displaystyle \frac{1}{r^2}}\sin (\theta )\cos (\theta ){\sin }^2(\varphi ){\displaystyle \frac{\partial f}{\partial \theta }}+{\displaystyle \frac{1}{r}}\sin (\theta )\cos (\theta ){\sin }^2(\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \theta }}-{\displaystyle \frac{1}{r^2}}\sin (\varphi )\cos (\varphi ){\displaystyle \frac{\partial f}{\partial \varphi }}+$

${\displaystyle \frac{1}{r}}\sin (\varphi )\cos (\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \varphi }}]+[{\displaystyle \frac{1}{r}}\cos (\theta )\sin (\theta ){\sin }^2(\varphi ){\displaystyle \frac{\partial f}{\partial r\partial \theta }}+{\displaystyle \frac{1}{r}}{\cos }^2(\theta ){\sin }^2(\varphi ){\displaystyle \frac{\partial f}{\partial r}}-$

${\displaystyle \frac{1}{r^2}}\sin (\theta )\cos (\theta ){\sin }^2(\varphi ){\displaystyle \frac{\partial f}{\partial \theta }}+{\displaystyle \frac{1}{r^2}}{\cos }^2(\theta ){\sin }^2(\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial {\theta }^2}}-{\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\sin (\varphi )\cos (\varphi ){\cos }^2(\theta )}{{\sin }^2(\theta )}}{\displaystyle \frac{\partial f}{\partial \varphi }}+$

${\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\cos (\theta )\sin (\varphi )\cos (\varphi )}{\sin (\theta )}}{\displaystyle \frac{{\partial }^{2}f}{\partial \theta \partial \varphi }}]+[{\displaystyle \frac{1}{r}}{\cos }^2(\varphi ){\displaystyle \frac{\partial f}{\partial r}}+{\displaystyle \frac{1}{r}}\cos (\varphi )\sin (\varphi ){\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \varphi }}+$

${\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\cos (\theta ){\cos }^2(\varphi )}{\sin (\theta )}}{\displaystyle \frac{\partial f}{\partial \theta }}+{\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\cos (\theta )\cos (\varphi )\sin (\varphi )}{\sin (\theta )}}{\displaystyle \frac{{\partial }^{2}f}{\partial \theta \partial \varphi }}-{\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\sin (\varphi )\cos (\varphi )}{{\sin }^2(\theta )}}{\displaystyle \frac{\partial f}{\partial \varphi }}+$

${\displaystyle \frac{1}{r^2}}{\displaystyle \frac{{\cos }^2(\varphi )}{{\sin }^2(\theta )}}{\displaystyle \frac{{\partial }^{2}f}{\partial {\varphi }^2}}]+[{\cos }^2(\theta ){\displaystyle \frac{{\partial }^{2}f}{\partial r^2}}+{\displaystyle \frac{1}{r^2}}\cos (\theta )\sin (\theta ){\displaystyle \frac{\partial f}{\partial \theta }}-{\displaystyle \frac{1}{r}}\cos (\theta )\sin (\theta ){\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \theta }}]+$

$\left[{\displaystyle \frac{1}{r}}{\sin }^2(\theta ){\displaystyle \frac{\partial f}{\partial r}}-{\displaystyle \frac{1}{r}}\sin (\theta )\cos (\theta ){\displaystyle \frac{{\partial }^{2}f}{\partial r\partial \theta }}+{\displaystyle \frac{1}{r^2}}\sin (\theta )\cos (\theta ){\displaystyle \frac{\partial f}{\partial \theta }}+{\displaystyle \frac{1}{r^2}}{\sin }^2(\theta ){\displaystyle \frac{{\partial }^{2}f}{\partial {\theta }^2}}\right].$

It may be hard to believe but the truth is that the above expression, after some miraculous simplifications of course, reduces to the following succinct form and we finally arrive at the Laplacian in spherical coordinates!

${\nabla }^{2}f$ $=$ ${\displaystyle \frac{{\partial }^{2}f}{\partial r^2}}+{\displaystyle \frac{1}{r^2}}{\displaystyle \frac{{\partial }^{2}f}{\partial {\theta }^2}}+{\displaystyle \frac{1}{r^2}}{\displaystyle \frac{1}{{\sin }^2(\theta )}}{\displaystyle \frac{{\partial }^{2}f}{\partial {\varphi }^2}}+{\displaystyle \frac{2}{r}}{\displaystyle \frac{\partial f}{\partial r}}+{\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\cos (\theta )}{\sin (\theta )}}{\displaystyle \frac{\partial f}{\partial \theta }}.$ (14)

We can write the Laplacian in an even more compact form as

${\nabla }^{2}f$ $=$ ${\displaystyle \frac{1}{r^2}}{\displaystyle \frac{\partial }{\partial r}}\left[r^2{\displaystyle \frac{\partial f}{\partial r}}\right]+{\displaystyle \frac{1}{r^2\sin (\theta )}}{\displaystyle \frac{\partial }{\partial \theta }}\left[\sin (\theta ){\displaystyle \frac{\partial f}{\partial \theta }}\right]+{\displaystyle \frac{1}{r^2{\sin }^2(\theta )}}{\displaystyle \frac{{\partial }^{2}f}{\partial {\varphi }^2}}.$ (15)

Title	derivation of the Laplacian from rectangular to spherical coordinates
Canonical name	DerivationOfTheLaplacianFromRectangularToSphericalCoordinates
Date of creation	2013-03-22 17:04:57
Last modified on	2013-03-22 17:04:57
Owner	swapnizzle (13346)
Last modified by	swapnizzle (13346)
Numerical id	11
Author	swapnizzle (13346)
Entry type	Topic
Classification	msc 53A45