 #1
Mayhem
 274
 174
 Homework Statement:
 Derive the Laplacian in spherical coordinates using the Laplacian in rectangular coordinates
 Relevant Equations:

##\nabla = \left ( \frac{\partial }{\partial x}, \frac{\partial }{\partial y}, \frac{\partial }{\partial z}\right )##
##x=r \sin(\theta) \cos(\phi), y=r \sin(\theta) \sin(\phi), z = r \cos(\theta)##
As a part of my self study, I am trying to derive the Laplacian in spherical coordinates to gain a deeper understanding of the mathematics of quantum mechanics.
For reference, this the sphere I am using, where ##r## is constant and ##\theta = \theta (x,y, z), \phi = \phi(x,y)##.
Given the definitions in "Relevant Equations", we can define the partial derivatives of ##x, y## and ##z##$$\begin{align*}
\frac{\partial }{\partial x} &= r\left(\cos(\theta)\cos(\phi)\frac{\partial }{\partial \theta}  \sin(\theta)\sin(\phi)\frac{\partial }{\partial \phi}\right ) \\
\frac{\partial }{\partial y} &= r\left(\cos(\theta)\sin(\phi)\frac{\partial }{\partial \theta}  \sin(\theta)\cos(\phi)\frac{\partial }{\partial \phi}\right ) \\
\frac{\partial }{\partial z} &= r \sin(\theta) \frac{\partial }{\partial \theta}
\end{align*}$$
Thus the gradient in spherical coordinates becomes $$\nabla = \begin{pmatrix}
r\left(\cos(\theta)\cos(\phi)\frac{\partial }{\partial \theta}  \sin(\theta)\sin(\phi)\frac{\partial }{\partial \phi}\right )\\
r\left(\cos(\theta)\sin(\phi)\frac{\partial }{\partial \theta}  \sin(\theta)\cos(\phi)\frac{\partial }{\partial \phi}\right )
\\
r \sin(\theta) \frac{\partial }{\partial \theta}\end{pmatrix}
$$
The Laplacian is given as ##\Delta = \nabla \cdot \nabla = \nabla^2##.
My question is: once I take the dot product of the given gradient vector, will I be able to simplify into the Laplacian for spherical coordinates? I am asking, because it takes a lot of brute calculations to find out, and therefore I want to know if my initial math is correct.
For reference, this the sphere I am using, where ##r## is constant and ##\theta = \theta (x,y, z), \phi = \phi(x,y)##.
Given the definitions in "Relevant Equations", we can define the partial derivatives of ##x, y## and ##z##$$\begin{align*}
\frac{\partial }{\partial x} &= r\left(\cos(\theta)\cos(\phi)\frac{\partial }{\partial \theta}  \sin(\theta)\sin(\phi)\frac{\partial }{\partial \phi}\right ) \\
\frac{\partial }{\partial y} &= r\left(\cos(\theta)\sin(\phi)\frac{\partial }{\partial \theta}  \sin(\theta)\cos(\phi)\frac{\partial }{\partial \phi}\right ) \\
\frac{\partial }{\partial z} &= r \sin(\theta) \frac{\partial }{\partial \theta}
\end{align*}$$
Thus the gradient in spherical coordinates becomes $$\nabla = \begin{pmatrix}
r\left(\cos(\theta)\cos(\phi)\frac{\partial }{\partial \theta}  \sin(\theta)\sin(\phi)\frac{\partial }{\partial \phi}\right )\\
r\left(\cos(\theta)\sin(\phi)\frac{\partial }{\partial \theta}  \sin(\theta)\cos(\phi)\frac{\partial }{\partial \phi}\right )
\\
r \sin(\theta) \frac{\partial }{\partial \theta}\end{pmatrix}
$$
The Laplacian is given as ##\Delta = \nabla \cdot \nabla = \nabla^2##.
My question is: once I take the dot product of the given gradient vector, will I be able to simplify into the Laplacian for spherical coordinates? I am asking, because it takes a lot of brute calculations to find out, and therefore I want to know if my initial math is correct.