Chapter 44Coordinate Transformations and the Laplacian

Derivation (1): Computing $r_x,\theta_x,r_{xx},\theta_{xx}$

We differentiate $r=\sqrt{x^2+y^2}$ in Eq. \eqref{eq:44-polar-def} partially with respect to $x$, regarding $y$ as a constant. Differentiating $\sqrt{\ }$ is an instance of differentiating a composite function ($\left(\sqrt{f}\right)'=f'/(2\sqrt f)$), so

$$ r_x=\pdiff rx=\pdiff{}{x}\sqrt{x^2+y^2}=\frac{2x}{2\sqrt{x^2+y^2}}=\frac{x}{\sqrt{x^2+y^2}}=\frac xr $$

Here the last equality simply replaces the denominator by $r$ using $r=\sqrt{x^2+y^2}$. Further, substituting $x=r\cos\theta$ (Eq. \eqref{eq:44-polar-def}), we obtain

\begin{equation} r_x=\frac{r\cos\theta}{r}=\cos\theta \label{eq:44-rx} \end{equation}

Next we differentiate $\theta=\tan^{-1}(y/x)$ partially with respect to $x$. Using the derivative formula for the arctangent, $\left(\tan^{-1}f\right)'=f'/(1+f^2)$ (here $f=y/x$ and $f'=\partial f/\partial x=-y/x^2$; we fixed $y$ and used the quotient rule), we get

$$ \theta_x=\pdiff{}{x}\tan^{-1}\frac yx=\left(-\frac{y}{x^2}\right)\cdot\frac{1}{1+\left(\dfrac yx\right)^2}=-\frac{y}{x^2+y^2} $$

(In the last equality, we multiplied $-y/x^2$ by $x^2/(x^2+y^2)$, the reciprocal of $1+(y/x)^2=(x^2+y^2)/x^2$, and canceled $x^2$.) Substituting $x^2+y^2=r^2$ and $y=r\sin\theta$, we obtain

\begin{equation} \theta_x=-\frac{r\sin\theta}{r^2}=-\frac{\sin\theta}{r} \label{eq:44-thetax} \end{equation}

Next we find the second-order derivatives $r_{xx},\theta_{xx}$. We have $r_{xx}=\partial r_x/\partial x$, but $r_x=\cos\theta$ in Eq. \eqref{eq:44-rx} is not written directly as a function of $x$ (it goes through $\theta$), so here too the chain rule is needed. To differentiate $\cos\theta$ with respect to $x$, we note that $\theta$ is a function of $x$ and use the formula for differentiating a composite function:

\begin{equation} r_{xx}=\pdiff{}{x}(\cos\theta)=\pdiff{}{\theta}(\cos\theta)\cdot\pdiff{\theta}{x}=(-\sin\theta)\cdot\left(-\frac{\sin\theta}{r}\right)=\frac{\sin^2\theta}{r} \label{eq:44-rxx} \end{equation}

(We substituted $\theta_x=-\sin\theta/r$ from Eq. \eqref{eq:44-thetax}.) Similarly, for $\theta_{xx}=\partial\theta_x/\partial x$, this time $\theta_x=-\sin\theta/r$ in Eq. \eqref{eq:44-thetax} contains both $\theta$ and $r$, so we use the product rule and the chain rule at the same time:

$$ \theta_{xx}=\pdiff{}{x}\left(-\frac{\sin\theta}{r}\right) =\pdiff{}{r}\!\left(-\frac{\sin\theta}{r}\right)\!\pdiff rx+\pdiff{}{\theta}\!\left(-\frac{\sin\theta}{r}\right)\!\pdiff{\theta}{x} =\frac{\sin\theta}{r^2}\cdot\cos\theta+\left(-\frac{\cos\theta}{r}\right)\!\left(-\frac{\sin\theta}{r}\right) $$

(For the first term, we differentiated $-\sin\theta/r$ partially with respect to $r$ (treating $\theta$ as a constant) to get $\sin\theta/r^2$, and multiplied this by $r_x=\cos\theta$. For the second term, we differentiated $-\sin\theta/r$ partially with respect to $\theta$ (treating $r$ as a constant) to get $-\cos\theta/r$, and multiplied this by $\theta_x=-\sin\theta/r$.) Both terms have the form $\sin\theta\cos\theta/r^2$, so they combine into

\begin{equation} \theta_{xx}=\frac{\sin\theta\cos\theta}{r^2}+\frac{\sin\theta\cos\theta}{r^2}=\frac{2\sin\theta\cos\theta}{r^2} \label{eq:44-thetaxx} \end{equation}

This gives us all the first- and second-order partial derivatives with respect to $x$.

Derivation (2): Computing $r_y,\theta_y,r_{yy},\theta_{yy}$ and $u_{yy}$

This time we fix $x$ and differentiate partially with respect to $y$. From $r=\sqrt{x^2+y^2}$ we get

\begin{equation} r_y=\frac{y}{\sqrt{x^2+y^2}}=\frac yr=\frac{r\sin\theta}{r}=\sin\theta \label{eq:44-ry} \end{equation}

(The flow of the computation is the same as for $r_x$; we simply substituted $y=r\sin\theta$.) When we differentiate $\theta=\tan^{-1}(y/x)$ partially with respect to $y$, this time $\partial(y/x)/\partial y=1/x$ (with $x$ fixed), so we get

\begin{equation} \theta_y=\frac{1}{x}\cdot\frac{1}{1+\left(\dfrac yx\right)^2}=\frac{x}{x^2+y^2}=\frac{r\cos\theta}{r^2}=\frac{\cos\theta}{r} \label{eq:44-thetay} \end{equation}

The second-order derivatives are found by the same procedure as in the case of $x$:

\begin{equation} r_{yy}=\pdiff{}{\theta}(\sin\theta)\cdot\theta_y=\cos\theta\cdot\frac{\cos\theta}{r}=\frac{\cos^2\theta}{r} \label{eq:44-ryy} \end{equation} \begin{equation} \theta_{yy}=\pdiff{}{r}\!\left(\frac{\cos\theta}{r}\right)\!r_y+\pdiff{}{\theta}\!\left(\frac{\cos\theta}{r}\right)\!\theta_y =-\frac{\cos\theta}{r^2}\cdot\sin\theta-\frac{\sin\theta}{r}\cdot\frac{\cos\theta}{r}=-\frac{2\sin\theta\cos\theta}{r^2} \label{eq:44-thetayy} \end{equation}

(Note that the sign is exactly opposite to that of $\theta_{xx}$.) Using these and repeating exactly the same procedure as for Eqs. \eqref{eq:44-uxx-setup}–\eqref{eq:44-uxx-final} (we only have to read $x\to y$), we get

$$ u_y=u_r r_y+u_\theta\theta_y,\qquad u_{yy}=\bigl\{(u_{rr}r_y+u_{r\theta}\theta_y)r_y+u_r r_{yy}\bigr\}+\bigl\{(u_{\theta r}r_y+u_{\theta\theta}\theta_y)\theta_y+u_\theta\theta_{yy}\bigr\} $$

Substituting Eqs. \eqref{eq:44-ry}–\eqref{eq:44-thetayy} into these, and combining $u_{r\theta}=u_{\theta r}$ in the same way by Schwarz's theorem, we find that

\begin{equation} \boxed{u_{yy}=\sin^2\theta\,u_{rr}+\frac{2\sin\theta\cos\theta}{r}u_{r\theta}+\frac{\cos^2\theta}{r}u_r+\frac{\cos^2\theta}{r^2}u_{\theta\theta}-\frac{2\sin\theta\cos\theta}{r^2}u_\theta} \label{eq:44-uyy-final} \end{equation}

holds.

Derivation (3): Completing the computation of $u_{xx}+u_{yy}$

Let us write Eq. \eqref{eq:44-uxx-final} and Eq. \eqref{eq:44-uyy-final} side by side and add the two equations.

$$ \begin{aligned} u_{xx}&=\cos^2\theta\,u_{rr}-\frac{2\sin\theta\cos\theta}{r}u_{r\theta}+\frac{\sin^2\theta}{r}u_r+\frac{\sin^2\theta}{r^2}u_{\theta\theta}+\frac{2\sin\theta\cos\theta}{r^2}u_\theta\\[4pt] u_{yy}&=\sin^2\theta\,u_{rr}+\frac{2\sin\theta\cos\theta}{r}u_{r\theta}+\frac{\cos^2\theta}{r}u_r+\frac{\cos^2\theta}{r^2}u_{\theta\theta}-\frac{2\sin\theta\cos\theta}{r^2}u_\theta \end{aligned} $$

In these two equations, the coefficients of $u_{r\theta}$ ($\mp2\sin\theta\cos\theta/r$) and the coefficients of the last term $u_\theta$ ($\pm2\sin\theta\cos\theta/r^2$) have the same magnitude and opposite signs. That is, when we add them, these two pairs cancel exactly and disappear (in the original notes as well, a slanted check mark indicating that they “cancel” is actually written on these two pairs of terms). All that remains are the coefficient of $u_{rr}$, $\cos^2\theta+\sin^2\theta=1$, the coefficient of $u_r$, $\dfrac{\sin^2\theta}{r}+\dfrac{\cos^2\theta}{r}=\dfrac1r$, and the coefficient of $u_{\theta\theta}$, $\dfrac{\sin^2\theta}{r^2}+\dfrac{\cos^2\theta}{r^2}=\dfrac1{r^2}$ (in each case we used the basic trigonometric identity $\sin^2\theta+\cos^2\theta=1$), and so the sum collapses to

$$ \nabla^2u=u_{xx}+u_{yy}=u_{rr}+\frac1r u_r+\frac{1}{r^2}u_{\theta\theta} $$

an extremely simple form that does not contain $\theta$.

Derivation: The Laplacian in cylindrical coordinates — why can we simply add the $z$ term?

To find the Laplacian in cylindrical coordinates, we do not need to redo the chain rule from the beginning. Looking at Eq. \eqref{eq:44-cyl-def}, we see that $r,\theta$ are determined by $x,y$ alone, while $z$ is equal to $z$ itself (it does not depend on $x,y$ at all). That is, the variable $z$ is completely independent of $x,y,r,\theta$. Therefore, when we differentiate $u(x,y,z)$ partially with respect to $x$ or $y$, $z$ is treated as a mere constant, so the computation of $u_{xx},u_{yy}$ carried out in Section 44.2 holds as it stands (even if $z$ appears in $u$). Conversely, when we differentiate $u$ partially with respect to $z$, we simply treat $x,y$ (and hence $r,\theta$) as constants, and this never mixes with the computation in the $r,\theta$ directions ($r_z=\theta_z=0$ and $z_x=z_y=0$). Hence, in the three-dimensional Cartesian Laplacian

$$ \nabla^2u=u_{xx}+u_{yy}+u_{zz} $$

the part $u_{xx}+u_{yy}$ is replaced as it stands by an expression in $r,\theta$ through Formula 44.1 (Eq. \eqref{eq:44-laplacian-polar}), while the part $u_{zz}$ remains entirely unchanged.

Derivation (direct method): The nine first-order partial derivatives $r_x,\theta_x,\varphi_x$, etc.

We compute the partial derivatives of the inverse transformation in Eq. \eqref{eq:44-sph-def}, $r=\sqrt{x^2+y^2+z^2}$, $\theta=\tan^{-1}\bigl(\sqrt{x^2+y^2}/z\bigr)$, $\varphi=\tan^{-1}(y/x)$, in exactly the same way as in Section 44.2 (fix one of the variables and differentiate partially with respect to the remaining ones). First, differentiating $r=\sqrt{x^2+y^2+z^2}$ partially with respect to $x$ gives $r_x$:

\begin{equation} r_x=\frac{x}{\sqrt{x^2+y^2+z^2}}=\frac xr=\frac{r\sin\theta\cos\varphi}{r}=\sin\theta\cos\varphi \label{eq:44-sph-rx} \end{equation}

(We substituted $x=r\sin\theta\cos\varphi$.) To find $\theta_x$, we differentiate $\theta=\tan^{-1}(\sqrt{x^2+y^2}/z)$ partially with respect to $x$. Letting $f=\sqrt{x^2+y^2}/z$, we have $f_x=\dfrac{x}{z\sqrt{x^2+y^2}}$; applying the derivative formula for the arctangent, $(\tan^{-1}f)_x=f_x/(1+f^2)$, together with $1+f^2=1+\dfrac{x^2+y^2}{z^2}=\dfrac{r^2}{z^2}$ (where we used $x^2+y^2+z^2=r^2$), we get

\begin{equation} \theta_x=\frac{x}{z\sqrt{x^2+y^2}}\cdot\frac{z^2}{r^2}=\frac{xz}{r^2\sqrt{x^2+y^2}} =\frac{(r\sin\theta\cos\varphi)(r\cos\theta)}{r^2\cdot r\sin\theta}=\frac{\cos\theta\cos\varphi}{r} \label{eq:44-sph-thetax} \end{equation}

(We used $\sqrt{x^2+y^2}=\sqrt{r^2\sin^2\theta}=r\sin\theta$ (since for $0\le\theta\le\pi$ we have $\sin\theta\ge0$, the absolute value can be dropped).) $\varphi_x$ is the $x$-derivative of $\varphi=\tan^{-1}(y/x)$, so the computation is exactly the same as that for $\theta_x$ in Section 44.2 (Eq. \eqref{eq:44-thetax}), except that in place of “the $r$ of polar coordinates” the radial coordinate in the $xy$-plane, $\rho$ (rho) $=\sqrt{x^2+y^2}=r\sin\theta$, appears:

\begin{equation} \varphi_x=-\frac{y}{x^2+y^2}=-\frac{y}{\rho^2}=-\frac{r\sin\theta\sin\varphi}{r^2\sin^2\theta}=-\frac{\sin\varphi}{r\sin\theta} \label{eq:44-sph-phix} \end{equation}

In exactly the same way (with the roles $x\leftrightarrow y$ interchanged), the partial derivatives with respect to $y$ can also be computed:

\begin{equation} r_y=\sin\theta\sin\varphi,\qquad \theta_y=\frac{\cos\theta\sin\varphi}{r},\qquad \varphi_y=\frac{\cos\varphi}{r\sin\theta} \label{eq:44-sph-y} \end{equation}

Finally, for the partial derivatives with respect to $z$: from $r=\sqrt{x^2+y^2+z^2}$ we get $r_z=z/r=\cos\theta$. Since $\varphi=\tan^{-1}(y/x)$ does not contain $z$, $\varphi_z=0$. For $\theta_z$, differentiating $\theta=\tan^{-1}(\sqrt{x^2+y^2}/z)$ partially with respect to $z$ (with $f=\sqrt{x^2+y^2}/z$, $f_z=-\sqrt{x^2+y^2}/z^2$) gives

\begin{equation} \theta_z=\frac{-\sqrt{x^2+y^2}/z^2}{1+(x^2+y^2)/z^2}=-\frac{\sqrt{x^2+y^2}}{r^2}=-\frac{r\sin\theta}{r^2}=-\frac{\sin\theta}{r} \label{eq:44-sph-z} \end{equation}

and the expression collapses to this simple form. With this, all nine first-order partial derivatives (also verified with SymPy) are now in hand:

\begin{equation} \begin{array}{lll} r_x=\sin\theta\cos\varphi, & \theta_x=\dfrac{\cos\theta\cos\varphi}{r}, & \varphi_x=-\dfrac{\sin\varphi}{r\sin\theta},\\[6pt] r_y=\sin\theta\sin\varphi, & \theta_y=\dfrac{\cos\theta\sin\varphi}{r}, & \varphi_y=\dfrac{\cos\varphi}{r\sin\theta},\\[6pt] r_z=\cos\theta, & \theta_z=-\dfrac{\sin\theta}{r}, & \varphi_z=0. \end{array} \label{eq:44-sph-all9} \end{equation}

Here we notice something interesting. The forms of $\varphi_x,\varphi_y$ (Eqs. \eqref{eq:44-sph-phix} and \eqref{eq:44-sph-y}) are exactly the same as those of $\theta_x,\theta_y$ in Section 44.2 (Eqs. \eqref{eq:44-thetax}, \eqref{eq:44-thetay}), with only $r$ replaced by $\rho=r\sin\theta$. This is no coincidence. If we keep $\theta$ (and hence $z,\rho$) fixed and vary only $\varphi$, the point moves on a circle of radius $\rho=r\sin\theta$ parallel to the $xy$-plane—that is, the computation that measures $\varphi$ is exactly the same problem as the computation that measures $\theta$ in plane polar coordinates of radius $\rho$. This observation foreshadows the shortcut introduced next.

From here on, if we carry out the same procedure as in Section 44.2 (the computations corresponding to Eqs. \eqref{eq:44-uxx-setup} through \eqref{eq:44-uxx-final}) for each of the three directions $x,y,z$, we can in principle arrive at the Laplacian in spherical coordinates. However, since the number of variables increases from two to three, the number of terms in the second derivatives also grows multiplicatively (six kinds of terms are involved: $u_{rr},u_{\theta\theta},u_{\varphi\varphi},u_{r\theta},u_{r\varphi},u_{\theta\varphi}$), and the hand computation becomes very long. We leave the task of completing this direct route to Exercise 44.4 at the end of the chapter, and here introduce another route that reaches the same conclusion more quickly.

Derivation (shortcut): Reusing the results of Section 44.2 twice

The radial coordinate $r$ and the zenith angle $\theta$ of spherical coordinates are related to the radial coordinate $\rho$ of cylindrical coordinates (the distance from the $z$-axis) and the height $z$ as follows,

\begin{equation} \rho=r\sin\theta,\qquad z=r\cos\theta \label{eq:44-sph-cyl-relation} \end{equation}

(from the expressions for $x,y$ in Definition 44.3 we have simply picked out $\sqrt{x^2+y^2}=r\sin\theta=\rho$ and $z=r\cos\theta$). Now look closely at Eq. \eqref{eq:44-sph-cyl-relation}. It says “$z$ is $r\cos\theta$ and $\rho$ is $r\sin\theta$,” which is a relation of exactly the same form as $x=r\cos\theta,\ y=r\sin\theta$ in Section 44.2 (Eq. \eqref{eq:44-polar-def}); the only difference is the names of the variables ($x\to z$, $y\to\rho$). The chain-rule computation is a purely formal algebraic procedure that has nothing to do with what we name the variables, so the results obtained in Section 44.2 can be reused as they are simply by reading $x\to z,\ y\to\rho$. Specifically, reading Eqs. \eqref{eq:44-ry}, \eqref{eq:44-thetay} ($r_y=\sin\theta,\ \theta_y=\cos\theta/r$) in this way, we obtain

\begin{equation} r_\rho=\sin\theta,\qquad \theta_\rho=\frac{\cos\theta}{r} \label{eq:44-r-rho} \end{equation}

(this is a consistent correspondence, which holds for the same reason that the form of $\varphi_x,\varphi_y$ in Eq. \eqref{eq:44-sph-all9} was the same as that of $\theta_x,\theta_y$ in Section 44.2). Similarly, reading Formula 44.1 (Eq. \eqref{eq:44-laplacian-polar}) with $x\to z,\ y\to\rho$, “the sum of the second partial derivatives in the $z,\rho$-plane (the meridian plane)” can be rewritten as an expression in $r,\theta$:

\begin{equation} u_{zz}+u_{\rho\rho}=u_{rr}+\frac1r u_r+\frac{1}{r^2}u_{\theta\theta} \label{eq:44-meridian-flat} \end{equation}

(We verified with SymPy, by substituting many specific functions, that this reinterpretation is correct.)

Now, rewriting Formula 44.2 (the Laplacian in cylindrical coordinates, Eq. \eqref{eq:44-laplacian-cyl}) with the radial coordinate written as $\rho$, we have

\begin{equation} \nabla^2u=\underbrace{u_{\rho\rho}+u_{zz}}_{\text{already replaced by Eq. \eqref{eq:44-meridian-flat}}}+\ \frac1\rho u_\rho\ +\ \frac{1}{\rho^2}u_{\varphi\varphi} \label{eq:44-cyl-with-rho} \end{equation}

The first two terms on the right-hand side have been taken care of by Eq. \eqref{eq:44-meridian-flat}, so only the two terms $\dfrac1\rho u_\rho$ and $\dfrac{1}{\rho^2}u_{\varphi\varphi}$ remain. As for the $u_{\varphi\varphi}$ term, since $\varphi$ is an independent variable unrelated to both $\rho,z$ and $r,\theta$ (if we regard $\varphi$ as fixed, the correspondence $(\rho,z)\leftrightarrow(r,\theta)$ is nothing but the ordinary plane polar coordinate relation), simply substituting $\rho=r\sin\theta$ lets us rewrite it as

\begin{equation} \frac{1}{\rho^2}u_{\varphi\varphi}=\frac{1}{r^2\sin^2\theta}u_{\varphi\varphi} \label{eq:44-phiphi-term} \end{equation}

As for the remaining term $\dfrac1\rho u_\rho$, noting that $u_\rho=u_r r_\rho+u_\theta\theta_\rho$ (the chain rule, using Eq. \eqref{eq:44-r-rho}), we have

\begin{equation} \frac1\rho u_\rho=\frac{1}{r\sin\theta}\left(u_r\sin\theta+u_\theta\frac{\cos\theta}{r}\right)=\frac1r u_r+\frac{\cos\theta}{r^2\sin\theta}u_\theta \label{eq:44-leftover-term} \end{equation}

Collecting all of Eqs. \eqref{eq:44-meridian-flat}, \eqref{eq:44-phiphi-term}, and \eqref{eq:44-leftover-term} into Eq. \eqref{eq:44-cyl-with-rho}, we obtain

$$ \nabla^2u=u_{rr}+\frac1r u_r+\frac{1}{r^2}u_{\theta\theta}+\frac1r u_r+\frac{\cos\theta}{r^2\sin\theta}u_\theta+\frac{1}{r^2\sin^2\theta}u_{\varphi\varphi} $$

Since the two $u_r$ terms have the same coefficient $1/r$, they can be combined ($\dfrac1r+\dfrac1r=\dfrac2r$), and we get

\begin{equation} \nabla^2u=u_{rr}+\frac2r u_r+\frac{1}{r^2}u_{\theta\theta}+\frac{\cos\theta}{r^2\sin\theta}u_\theta+\frac{1}{r^2\sin^2\theta}u_{\varphi\varphi} \label{eq:44-laplacian-sph-derived} \end{equation}

This is the Laplacian in spherical coordinates. (Verified with SymPy that it agrees with the direct computation from Cartesian coordinates for several test functions.)