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Note on finding the Laplacian in Polar, Cylinderical and Spherical coordinates using Tensor calculus

Nasser M. Abbasi

April 10, 2019   Compiled on April 10, 2019 at 9:20pm

Contents

1 Introduction
2 2D Polar
3 3D Spherical
4 3D Cylindrical
 4.1 References

1 Introduction

I wrote this note to help me learn tensors. The goal is to derive the Laplacian \(\nabla ^{2}\) using tensor calculus for 2D Polar, 3D Cylindrical and in 3D Spherical coordinates.

The Laplacian in Cartesian coordinates is given by \(\nabla ^{2}=\frac{\partial ^{2}}{\partial x^{2}}+\frac{\partial ^{2}}{\partial y^{2}}+\frac{\partial ^{2}}{\partial z^{2}}\). The following diagram shows \(\nabla ^{2}u\) in Polar, Cylindrical and Spherical coordinates

pict

Figure 1: Laplacian in different coordinate systems

The Laplacian operator in any orthogonal coordinate system is given by\begin{equation} \nabla ^{2}=\frac{1}{\sqrt{\det \left ( g\right ) }}\frac{\partial }{\partial x_{i}}\left ( \frac{\sqrt{\det \left ( g\right ) }}{g_{ii}}\frac{\partial }{\partial x^{i}}\right ) \tag{1} \end{equation}

Where \(g\) is the metric tensor and \(\left \vert g\right \vert \) is the determinant of \(g\).  The derivation of the above is not shown here. References contains the derivation of the above. Only the use of (1) is shown here.

2 2D Polar

The coordinates in the Cartesian system are \(\zeta ^{1}=x,\zeta ^{2}=y\) and the coordinates in the other system (Polar) are \(x^{1}=r,x^{2}=\theta \). The relation between these must be known and invertible also, meaning \(\zeta \equiv \zeta \left ( x\right ) \) and \(x\equiv x\left ( \zeta \right ) \). This relation can be found from geometry as\begin{align*} \zeta ^{1} & =r\cos \theta \\ \zeta ^{2} & =r\sin \theta \end{align*}

The first step is to determine the metric tensor \(g\) for the Polar coordinates. This is given by\[ g_{kl}=\delta _{ij}\frac{\partial \zeta ^{i}}{\partial x^{k}}\frac{\partial \zeta ^{j}}{\partial x^{l}}\] The  above using Einstein summation notation. Since the coordinate system is orthogonal, \(g_{kl}\) will be diagonal, hence only \(g_{11},g_{22}\) are non zero. This is not the case for all coordinates systems. For general curvilinear coordinates system, \(g\) can contain all components. But for the coordinates systems used here, \(g\) will always be diagonal.\begin{align*} g_{11} & =\frac{\partial \zeta ^{1}}{\partial x^{1}}\frac{\partial \zeta ^{1}}{\partial x^{1}}+\frac{\partial \zeta ^{2}}{\partial x^{1}}\frac{\partial \zeta ^{2}}{\partial x^{1}}\\ & =\frac{\partial \zeta ^{1}}{\partial r}\frac{\partial \zeta ^{1}}{\partial r}+\frac{\partial \zeta ^{2}}{\partial r}\frac{\partial \zeta ^{2}}{\partial r}\\ & =\left ( \frac{\partial \zeta ^{1}}{\partial r}\right ) ^{2}+\left ( \frac{\partial \zeta ^{2}}{\partial r}\right ) ^{2}\\ & =\cos ^{2}\theta +\sin ^{2}\theta \\ & =1 \end{align*}

And\begin{align*} g_{22} & =\frac{\partial \zeta ^{1}}{\partial x^{2}}\frac{\partial \zeta ^{1}}{\partial x^{2}}+\frac{\partial \zeta ^{2}}{\partial x^{2}}\frac{\partial \zeta ^{2}}{\partial x^{2}}\\ & =\frac{\partial \zeta ^{1}}{\partial \theta }\frac{\partial \zeta ^{1}}{\partial \theta }+\frac{\partial \zeta ^{2}}{\partial \theta }\frac{\partial \zeta ^{2}}{\partial \theta }\\ & =\left ( \frac{\partial \zeta ^{1}}{\partial \theta }\right ) ^{2}+\left ( \frac{\partial \zeta ^{2}}{\partial \theta }\right ) ^{2}\\ & =\left ( -r\sin \theta \right ) ^{2}+\left ( r\cos \theta \right ) ^{2}\\ & =r^{2} \end{align*}

Hence \(ds^{2}\) in polar coordinates is\begin{align*} ds^{2} & =g_{kl}dx^{k}dx^{l}\\ & =g_{11}\left ( dx^{1}\right ) ^{2}+g_{22}\left ( dx^{2}\right ) ^{2}\\ & =g_{11}\left ( dr\right ) ^{2}+g_{22}\left ( d\theta \right ) ^{2}\\ & =\left ( dr\right ) ^{2}+r^{2}\left ( d\theta \right ) ^{2} \end{align*}

From the above we see that\[ g=\begin{pmatrix} g_{11} & g_{12}\\ g_{21} & g_{22}\end{pmatrix} =\begin{pmatrix} 1 & 0\\ 0 & r^{2}\end{pmatrix} \] Hence \(\det \left ( g\right ) =r^{2}\). We are now ready to apply (1)\begin{align*} \nabla ^{2} & =\frac{1}{\sqrt{\det \left ( g\right ) }}\frac{\partial }{\partial x_{i}}\left ( \frac{\sqrt{\det \left ( g\right ) }}{g_{ii}}\frac{\partial }{\partial x^{i}}\right ) \\ & =\frac{1}{\sqrt{r^{2}}}\frac{\partial }{\partial x_{1}}\left ( \frac{\sqrt{r^{2}}}{g_{11}}\frac{\partial }{\partial x^{1}}\right ) +\frac{1}{\sqrt{r^{2}}}\frac{\partial }{\partial x_{2}}\left ( \frac{\sqrt{r^{2}}}{g_{22}}\frac{\partial }{\partial x^{2}}\right ) \\ & =\frac{1}{r}\frac{\partial }{\partial r}\left ( r\frac{\partial }{\partial r}\right ) +\frac{1}{r}\frac{\partial }{\partial \theta }\left ( \frac{r}{r^{2}}\frac{\partial }{\partial \theta }\right ) \\ & =\frac{1}{r}\frac{\partial }{\partial r}\left ( r\frac{\partial }{\partial r}\right ) +\frac{1}{r}\frac{\partial }{\partial \theta }\left ( \frac{1}{r}\frac{\partial }{\partial \theta }\right ) \\ & =\frac{1}{r}\left ( \frac{\partial }{\partial r}+r\frac{\partial ^{2}}{\partial r^{2}}\right ) +\frac{1}{r^{2}}\frac{\partial ^{2}}{\partial \theta ^{2}}\\ & =\frac{\partial ^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial }{\partial r}+\frac{1}{r^{2}}\frac{\partial ^{2}}{\partial \theta ^{2}} \end{align*}

Therefore\begin{align*} \nabla ^{2}u & =\frac{\partial ^{2}u}{\partial r^{2}}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^{2}}\frac{\partial ^{2}u}{\partial \theta ^{2}}\\ & =u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta \theta } \end{align*}

3 3D Spherical

The coordinates in the Cartesian system are \(\zeta ^{1}=x,\zeta ^{2}=y,\zeta ^{3}=z\). And the coordinates in the Spherical system are \(x^{1}=\phi ,x^{2}=r,x^{3}=\theta \). The relation between these is known as (Note that the following depends on convention used for which is \(\theta \) and which is \(\phi \). Physics convention as shown in the diagram above is used here).\begin{align*} \zeta ^{1} & =r\sin \theta \cos \phi \\ \zeta ^{2} & =r\sin \theta \sin \phi \\ \zeta ^{3} & =r\cos \theta \end{align*}

The first step is to determine the metric tensor \(g\) for the Spherical coordinates. This is given by\[ g_{kl}=\delta _{ij}\frac{\partial \zeta ^{i}}{\partial x^{k}}\frac{\partial \zeta ^{j}}{\partial x^{l}}\] Since the coordinate system are orthogonal, \(g_{kl}\) will be diagonal. Hence only \(g_{11},g_{22},g_{33}\) are non zero.\begin{align*} g_{11} & =\frac{\partial \zeta ^{1}}{\partial x^{1}}\frac{\partial \zeta ^{1}}{\partial x^{1}}+\frac{\partial \zeta ^{2}}{\partial x^{1}}\frac{\partial \zeta ^{2}}{\partial x^{1}}+\frac{\partial \zeta ^{3}}{\partial x^{1}}\frac{\partial \zeta ^{3}}{\partial x^{1}}\\ & =\frac{\partial \zeta ^{1}}{\partial \phi }\frac{\partial \zeta ^{1}}{\partial \phi }+\frac{\partial \zeta ^{2}}{\partial \phi }\frac{\partial \zeta ^{2}}{\partial \phi }+\frac{\partial \zeta ^{3}}{\partial \phi }\frac{\partial \zeta ^{3}}{\partial \phi }\\ & =\left ( \frac{\partial \zeta ^{1}}{\partial \phi }\right ) ^{2}+\left ( \frac{\partial \zeta ^{2}}{\partial \phi }\right ) ^{2}+\left ( \frac{\partial \zeta ^{3}}{\partial \phi }\right ) ^{2}\\ & =\left ( -r\sin \theta \sin \phi \right ) ^{2}+\left ( r\sin \theta \cos \phi \right ) ^{2}+\left ( 0\right ) ^{2}\\ & =r^{2}\sin ^{2}\theta \sin ^{2}\phi +r^{2}\sin ^{2}\theta \cos ^{2}\phi \\ & =r^{2}\sin ^{2}\theta \left ( \sin ^{2}\phi +\cos ^{2}\phi \right ) \\ & =r^{2}\sin ^{2}\theta \end{align*}

And\begin{align*} g_{22} & =\frac{\partial \zeta ^{1}}{\partial x^{2}}\frac{\partial \zeta ^{1}}{\partial x^{2}}+\frac{\partial \zeta ^{2}}{\partial x^{2}}\frac{\partial \zeta ^{2}}{\partial x^{2}}+\frac{\partial \zeta ^{3}}{\partial x^{2}}\frac{\partial \zeta ^{3}}{\partial x^{2}}\\ & =\frac{\partial \zeta ^{1}}{\partial r}\frac{\partial \zeta ^{1}}{\partial r}+\frac{\partial \zeta ^{2}}{\partial r}\frac{\partial \zeta ^{2}}{\partial r}+\frac{\partial \zeta ^{3}}{\partial r}\frac{\partial \zeta ^{3}}{\partial r}\\ & =\left ( \frac{\partial \zeta ^{1}}{\partial r}\right ) ^{2}+\left ( \frac{\partial \zeta ^{2}}{\partial r}\right ) ^{2}+\left ( \frac{\partial \zeta ^{3}}{\partial r}\right ) ^{2}\\ & =\left ( \sin \theta \cos \phi \right ) ^{2}+\left ( \sin \theta \sin \phi \right ) ^{2}+\left ( \cos \theta \right ) ^{2}\\ & =\sin ^{2}\theta \cos ^{2}\phi +\sin ^{2}\theta \sin ^{2}\phi +\cos ^{2}\theta \\ & =\sin ^{2}\theta \left ( \cos ^{2}\phi +\sin ^{2}\phi \right ) +\cos ^{2}\theta \\ & =\sin ^{2}\theta +\cos ^{2}\theta \\ & =1 \end{align*}

And\begin{align*} g_{33} & =\frac{\partial \zeta ^{1}}{\partial x^{3}}\frac{\partial \zeta ^{1}}{\partial x^{3}}+\frac{\partial \zeta ^{2}}{\partial x^{3}}\frac{\partial \zeta ^{2}}{\partial x^{3}}+\frac{\partial \zeta ^{3}}{\partial x^{3}}\frac{\partial \zeta ^{3}}{\partial x^{3}}\\ & =\frac{\partial \zeta ^{1}}{\partial \theta }\frac{\partial \zeta ^{1}}{\partial \theta }+\frac{\partial \zeta ^{2}}{\partial \theta }\frac{\partial \zeta ^{2}}{\partial \theta }+\frac{\partial \zeta ^{3}}{\partial \theta }\frac{\partial \zeta ^{3}}{\partial \theta }\\ & =\left ( \frac{\partial \zeta ^{1}}{\partial \theta }\right ) ^{2}+\left ( \frac{\partial \zeta ^{2}}{\partial \theta }\right ) ^{2}+\left ( \frac{\partial \zeta ^{3}}{\partial \theta }\right ) ^{2}\\ & =\left ( r\cos \theta \cos \phi \right ) ^{2}+\left ( r\cos \theta \sin \phi \right ) ^{2}+\left ( -r\sin \theta \right ) ^{2}\\ & =r^{2}\cos ^{2}\theta \left ( \cos ^{2}\phi +\sin ^{2}\phi \right ) +r^{2}\sin ^{2}\theta \\ & =r^{2}\cos ^{2}\theta +r^{2}\sin ^{2}\theta \\ & =r^{2} \end{align*}

Hence \(ds^{2}\) in Spherical coordinates is\begin{align*} ds^{2} & =g_{kl}dx^{k}dx^{l}\\ & =g_{11}\left ( dx^{1}\right ) ^{2}+g_{22}\left ( dx^{2}\right ) ^{2}+g_{33}\left ( dx^{3}\right ) ^{2}\\ & =g_{11}\left ( d\phi \right ) ^{2}+g_{22}\left ( dr\right ) ^{2}+g_{33}\left ( d\theta \right ) ^{2}\\ & =r^{2}\sin ^{2}\theta \left ( d\phi \right ) ^{2}+\left ( dr\right ) ^{2}+r^{2}\left ( d\theta \right ) ^{2} \end{align*}

From the above we see that\[ g=\begin{pmatrix} g_{11} & g_{12} & g_{13}\\ g_{21} & g_{22} & g_{23}\\ g_{31} & g_{32} & g_{33}\end{pmatrix} =\begin{pmatrix} r^{2}\sin ^{2}\theta & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & r^{2}\end{pmatrix} \] Hence \(\det \left ( g\right ) =r^{4}\sin ^{2}\theta \). We are now ready to apply (1)\begin{align*} \nabla ^{2} & =\frac{1}{\sqrt{\det \left ( g\right ) }}\frac{\partial }{\partial x_{i}}\left ( \frac{\sqrt{\det \left ( g\right ) }}{g_{ii}}\frac{\partial }{\partial x^{i}}\right ) \\ & =\frac{1}{\sqrt{r^{4}\sin ^{2}\theta }}\frac{\partial }{\partial x_{1}}\left ( \frac{\sqrt{r^{4}\sin ^{2}\theta }}{g_{11}}\frac{\partial }{\partial x^{1}}\right ) +\frac{1}{\sqrt{r^{4}\sin ^{2}\theta }}\frac{\partial }{\partial x_{2}}\left ( \frac{\sqrt{r^{4}\sin ^{2}\theta }}{g_{22}}\frac{\partial }{\partial x^{2}}\right ) +\frac{1}{\sqrt{r^{4}\sin ^{2}\theta }}\frac{\partial }{\partial x_{3}}\left ( \frac{\sqrt{r^{4}\sin ^{2}\theta }}{g_{33}}\frac{\partial }{\partial x^{3}}\right ) \\ & =\frac{1}{r^{2}\sin \theta }\frac{\partial }{\partial \phi }\left ( \frac{r^{2}\sin \theta }{r^{2}\sin ^{2}\theta }\frac{\partial }{\partial \phi }\right ) +\frac{1}{r^{2}\sin \theta }\frac{\partial }{\partial r}\left ( \frac{r^{2}\sin \theta }{1}\frac{\partial }{\partial r}\right ) +\frac{1}{r^{2}\sin \theta }\frac{\partial }{\partial \theta }\left ( \frac{r^{2}\sin \theta }{r^{2}}\frac{\partial }{\partial \theta }\right ) \\ & =\frac{1}{r^{2}\sin \theta }\frac{\partial }{\partial \phi }\left ( \frac{1}{\sin \theta }\frac{\partial }{\partial \phi }\right ) +\frac{1}{r^{2}\sin \theta }\frac{\partial }{\partial r}\left ( r^{2}\sin \theta \frac{\partial }{\partial r}\right ) +\frac{1}{r^{2}\sin \theta }\frac{\partial }{\partial \theta }\left ( \sin \theta \frac{\partial }{\partial \theta }\right ) \\ & =\frac{1}{r^{2}\sin ^{2}\theta }\frac{\partial ^{2}}{\partial \phi ^{2}}+\frac{1}{r^{2}}\left ( 2r\frac{\partial }{\partial r}+r^{2}\frac{\partial ^{2}}{\partial r^{2}}\right ) +\frac{1}{r^{2}\sin \theta }\left ( \cos \theta \frac{\partial }{\partial \theta }+\sin \theta \frac{\partial ^{2}}{\partial \theta ^{2}}\right ) \\ & =\frac{1}{r^{2}\sin ^{2}\theta }\frac{\partial ^{2}}{\partial \phi ^{2}}+\frac{2}{r}\frac{\partial }{\partial r}+\frac{\partial ^{2}}{\partial r^{2}}+\frac{\cos \theta }{r^{2}\sin \theta }\frac{\partial }{\partial \theta }+\frac{1}{r^{2}}\frac{\partial ^{2}}{\partial \theta ^{2}}\\ & =\frac{\partial ^{2}}{\partial r^{2}}+\frac{2}{r}\frac{\partial }{\partial r}+\frac{1}{r^{2}}\left ( \frac{\cos \theta }{\sin \theta }\frac{\partial }{\partial \theta }+\frac{\partial ^{2}}{\partial \theta ^{2}}\right ) +\frac{1}{r^{2}\sin ^{2}\theta }\frac{\partial ^{2}}{\partial \phi ^{2}} \end{align*}

Therefore\begin{align*} \nabla ^{2}u & =\frac{\partial ^{2}u}{\partial r^{2}}+\frac{2}{r}\frac{\partial u}{\partial r}+\frac{1}{r^{2}}\left ( \frac{\cos \theta }{\sin \theta }\frac{\partial u}{\partial \theta }+\frac{\partial ^{2}u}{\partial \theta ^{2}}\right ) +\frac{1}{r^{2}\sin ^{2}\theta }\frac{\partial ^{2}u}{\partial \phi ^{2}}\\ & =u_{rr}+\frac{2}{r}u_{r}+\frac{1}{r^{2}}\left ( \frac{\cos \theta }{\sin \theta }u_{\theta }+u_{\theta \theta }\right ) +\frac{1}{r^{2}\sin ^{2}\theta }u_{\phi \phi } \end{align*}

4 3D Cylindrical

The coordinates in the Cartesian system are \(\zeta ^{1}=x,\zeta ^{2}=y,\zeta ^{3}=z\). And the coordinates in the Cylindrical system are \(x^{1}=\phi ,x^{2}=r,x^{3}=z\). The relation between these is known as \begin{align*} \zeta ^{1} & =r\cos \phi \\ \zeta ^{2} & =r\sin \phi \\ \zeta ^{3} & =z \end{align*}

The first step is to determine the metric tensor \(g\) for the Spherical coordinates. This is given by\[ g_{kl}=\delta _{ij}\frac{\partial \zeta ^{i}}{\partial x^{k}}\frac{\partial \zeta ^{j}}{\partial x^{l}}\] Since the coordinate system are orthogonal, \(g_{kl}\) will be diagonal. Hence only \(g_{11},g_{22},g_{33}\) are non zero.\begin{align*} g_{11} & =\frac{\partial \zeta ^{1}}{\partial x^{1}}\frac{\partial \zeta ^{1}}{\partial x^{1}}+\frac{\partial \zeta ^{2}}{\partial x^{1}}\frac{\partial \zeta ^{2}}{\partial x^{1}}+\frac{\partial \zeta ^{3}}{\partial x^{1}}\frac{\partial \zeta ^{3}}{\partial x^{1}}\\ & =\frac{\partial \zeta ^{1}}{\partial \phi }\frac{\partial \zeta ^{1}}{\partial \phi }+\frac{\partial \zeta ^{2}}{\partial \phi }\frac{\partial \zeta ^{2}}{\partial \phi }+\frac{\partial \zeta ^{3}}{\partial \phi }\frac{\partial \zeta ^{3}}{\partial \phi }\\ & =\left ( \frac{\partial \zeta ^{1}}{\partial \phi }\right ) ^{2}+\left ( \frac{\partial \zeta ^{2}}{\partial \phi }\right ) ^{2}+\left ( \frac{\partial \zeta ^{3}}{\partial \phi }\right ) ^{2}\\ & =\left ( -r\sin \phi \right ) ^{2}+\left ( r\cos \phi \right ) ^{2}+\left ( 0\right ) ^{2}\\ & =r^{2} \end{align*}

And\begin{align*} g_{22} & =\frac{\partial \zeta ^{1}}{\partial x^{2}}\frac{\partial \zeta ^{1}}{\partial x^{2}}+\frac{\partial \zeta ^{2}}{\partial x^{2}}\frac{\partial \zeta ^{2}}{\partial x^{2}}+\frac{\partial \zeta ^{3}}{\partial x^{2}}\frac{\partial \zeta ^{3}}{\partial x^{2}}\\ & =\frac{\partial \zeta ^{1}}{\partial r}\frac{\partial \zeta ^{1}}{\partial r}+\frac{\partial \zeta ^{2}}{\partial r}\frac{\partial \zeta ^{2}}{\partial r}+\frac{\partial \zeta ^{3}}{\partial r}\frac{\partial \zeta ^{3}}{\partial r}\\ & =\left ( \frac{\partial \zeta ^{1}}{\partial r}\right ) ^{2}+\left ( \frac{\partial \zeta ^{2}}{\partial r}\right ) ^{2}+\left ( \frac{\partial \zeta ^{3}}{\partial r}\right ) ^{2}\\ & =\left ( \cos \phi \right ) ^{2}+\left ( \sin \phi \right ) ^{2}+\left ( 1\right ) ^{2}\\ & =1 \end{align*}

And\begin{align*} g_{33} & =\frac{\partial \zeta ^{1}}{\partial x^{3}}\frac{\partial \zeta ^{1}}{\partial x^{3}}+\frac{\partial \zeta ^{2}}{\partial x^{3}}\frac{\partial \zeta ^{2}}{\partial x^{3}}+\frac{\partial \zeta ^{3}}{\partial x^{3}}\frac{\partial \zeta ^{3}}{\partial x^{3}}\\ & =\frac{\partial \zeta ^{1}}{\partial z}\frac{\partial \zeta ^{1}}{\partial z}+\frac{\partial \zeta ^{2}}{\partial z}\frac{\partial \zeta ^{2}}{\partial z}+\frac{\partial \zeta ^{3}}{\partial z}\frac{\partial \zeta ^{3}}{\partial z}\\ & =\left ( \frac{\partial \zeta ^{1}}{\partial z}\right ) ^{2}+\left ( \frac{\partial \zeta ^{2}}{\partial z}\right ) ^{2}+\left ( \frac{\partial \zeta ^{3}}{\partial z}\right ) ^{2}\\ & =\left ( 0\right ) ^{2}+\left ( 0\right ) ^{2}+\left ( 1\right ) ^{2}\\ & =1 \end{align*}

Hence \(ds^{2}\) in Cylindrical coordinates is\begin{align*} ds^{2} & =g_{kl}dx^{k}dx^{l}\\ & =g_{11}\left ( dx^{1}\right ) ^{2}+g_{22}\left ( dx^{2}\right ) ^{2}+g_{33}\left ( dx^{3}\right ) ^{2}\\ & =g_{11}\left ( d\phi \right ) ^{2}+g_{22}\left ( dr\right ) ^{2}+g_{33}\left ( dz\right ) ^{2}\\ & =r^{2}\left ( d\phi \right ) ^{2}+\left ( dr\right ) ^{2}+\left ( dz\right ) ^{2} \end{align*}

From the above we see that\[ g=\begin{pmatrix} g_{11} & g_{12} & g_{13}\\ g_{21} & g_{22} & g_{23}\\ g_{31} & g_{32} & g_{33}\end{pmatrix} =\begin{pmatrix} r^{2} & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix} \] Hence \(\det \left ( g\right ) =r^{2}\). We are now ready to apply (1)\begin{align*} \nabla ^{2} & =\frac{1}{\sqrt{\det \left ( g\right ) }}\frac{\partial }{\partial x_{i}}\left ( \frac{\sqrt{\det \left ( g\right ) }}{g_{ii}}\frac{\partial }{\partial x^{i}}\right ) \\ & =\frac{1}{\sqrt{r^{2}}}\frac{\partial }{\partial x_{1}}\left ( \frac{\sqrt{r^{2}}}{g_{11}}\frac{\partial }{\partial x^{1}}\right ) +\frac{1}{\sqrt{r^{2}}}\frac{\partial }{\partial x_{2}}\left ( \frac{\sqrt{r^{2}}}{g_{22}}\frac{\partial }{\partial x^{2}}\right ) +\frac{1}{\sqrt{r^{2}}}\frac{\partial }{\partial x_{3}}\left ( \frac{\sqrt{r^{2}}}{g_{33}}\frac{\partial }{\partial x^{3}}\right ) \\ & =\frac{1}{r}\frac{\partial }{\partial \phi }\left ( \frac{1}{r}\frac{\partial }{\partial \phi }\right ) +\frac{1}{r}\frac{\partial }{\partial r}\left ( r\frac{\partial }{\partial r}\right ) +\frac{1}{r}\frac{\partial }{\partial z}\left ( r\frac{\partial }{\partial z}\right ) \\ & =\frac{1}{r^{2}}\frac{\partial ^{2}}{\partial \phi ^{2}}+\frac{1}{r}\frac{\partial }{\partial r}\left ( r\frac{\partial }{\partial r}\right ) +\frac{\partial ^{2}}{\partial z^{2}}\\ & =\frac{1}{r^{2}}\frac{\partial ^{2}}{\partial \phi ^{2}}+\frac{1}{r}\frac{\partial }{\partial r}+\frac{\partial ^{2}}{\partial r^{2}}+\frac{\partial ^{2}}{\partial z^{2}} \end{align*}

Therefore\begin{align*} \nabla ^{2}u & =\frac{\partial ^{2}u}{\partial r^{2}}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^{2}}\frac{\partial ^{2}u}{\partial \phi ^{2}}+\frac{\partial ^{2}u}{\partial z^{2}}\\ & =u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\phi \phi }+u_{zz} \end{align*}

4.1 References

  1. Lecture notes, Physics 5041. UMN Spring 2019 by Professor Kapusta
  2. Appendix A, Einstein’s Theory, A rigorous introduction. By Gron and Naess. Springer publisher.