Line 8: Line 8:
 
</math>
 
</math>
  
The basis for this conversion is, in the opinion of the authors, quite clear. The dot product of the vectors: <math> \nabla f = \Bigg[\frac{\partial f}{\partial x_{1}},\cdots,\frac{\partial f}{\partial x_{n}}\Bigg] </math>
+
The basis for this conversion is, in the opinion of the authors, quite clear. The dot product of the vectors: <math> \nabla f = \Bigg[\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\Bigg] </math> and <math> \left[\begin{array}{l}
 +
\frac{\partial}{\partial x} \\
 +
\frac{\partial}{\partial y} \\
 +
\frac{\partial}{\partial z}
 +
\end{array}\right]

Revision as of 15:13, 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: $ \nabla f = \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] $

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett