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\end{array}\right] </math> will of course be <math> \frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}+\frac{\partial^{2} f}{\partial z^{2}}\ </math>. | \end{array}\right] </math> will of course be <math> \frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}+\frac{\partial^{2} f}{\partial z^{2}}\ </math>. | ||
| − | It became clear to the authors that the derivations for cylindrical and spherical coordinates are outside of their field of ability. However, their derivations are below, and if so interested, the reader can refer to the link titled "Laplacian Derivations in Cylindrical and Spherical Coordinates". | + | It became clear to the authors that the derivations for cylindrical and spherical coordinates are outside of their field of ability. However, their derivations are below, and if so interested, the reader can refer to the link titled "Laplacian Derivations in Cylindrical and Spherical Coordinates" in the section "Links for Further Reading". |
In cylindrical coordinates, the Laplace Operator is: | In cylindrical coordinates, the Laplace Operator is: | ||
Revision as of 14:51, 6 December 2020
Coordinate Conversions for the Laplace Operator
It is most common to use the Laplace Operator $ \Delta $ in three-dimensions, as that is the dimensionality of our physical universe. Thus, the Laplace Operator is often used in 3-D Cartesian coordinates, cylindrical coordinates, and spherical coordinates.
$ \Delta f=\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}+\frac{\partial^{2} f}{\partial z^{2}}\ $
The basis for this conversion is, in the opinion of the authors, quite clear. The dot product of the vectors: $ \Bigg[\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\Bigg] $ and $ \left[\begin{array}{l} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{array}\right] $ will of course be $ \frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}+\frac{\partial^{2} f}{\partial z^{2}}\ $.
It became clear to the authors that the derivations for cylindrical and spherical coordinates are outside of their field of ability. However, their derivations are below, and if so interested, the reader can refer to the link titled "Laplacian Derivations in Cylindrical and Spherical Coordinates" in the section "Links for Further Reading".
In cylindrical coordinates, the Laplace Operator is:
$ \Delta f=\frac{1}{\rho} \frac{\partial}{\partial \rho}\left(\rho \frac{\partial f}{\partial \rho}\right)+\frac{1}{\rho^{2}} \frac{\partial^{2} f}{\partial \varphi^{2}}+\frac{\partial^{2} f}{\partial z^{2}} $
In spherical coordinates, the Laplace Operator is:
$ \Delta f=\frac{1}{r} \frac{\partial^{2}}{\partial r^{2}}(r f)+\frac{1}{r^{2} \sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial f}{\partial \theta}\right)+\frac{1}{r^{2} \sin ^{2} \theta} \frac{\partial^{2} f}{\partial \varphi^{2}} $
