- #1

Mayhem

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- 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.