Deriving the Laplacian in spherical coordinates

[Mayhem]

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.

 

[PeroK]

You've expressed the partial derivatives in terms of $\theta$ and $\phi$, but you've omitted the derivatives in terms of $r$. Also, the vector is still in terms of the Cartesian basis vectors, rather than $\hat r$ etc.

The Laplacian has second derivatives with respect to $x, y, z$, so you should expect second derivatives with respect to $r, \theta, \phi$ to appear.

So, you have some mistakes and misconceptions, I'm sorry to say.

 

[Mayhem]

You've expressed the partial derivatives in terms of $\theta$ and $\phi$, but you've omitted the derivatives in terms of $r$. Also, the vector is still in terms of the Cartesian basis vectors, rather than $\hat r$ etc.

The Laplacian has second derivatives with respect to $x, y, z$, so you should expect second derivatives with respect to $r, \theta, \phi$ to appear.

So, you have some mistakes and misconceptions, I'm sorry to say.

Where exactly do I start then?

 

[PeroK]

Where exactly do I start then?

I just noticed that your derivatives are not right:

Given the definitions in "Relevant Equations", we can define the partial derivatives of $x, y$ and $z$$$\begin{align*}
\frac{\partial }{\partial x} &= r\left(\cos(\theta)\cos(\phi)\frac{\partial }{\partial \theta} - \sin(\theta)\sin(\phi)\frac{\partial }{\partial \phi}\right ) \\
\frac{\partial }{\partial y} &= r\left(\cos(\theta)\sin(\phi)\frac{\partial }{\partial \theta} - \sin(\theta)\cos(\phi)\frac{\partial }{\partial \phi}\right ) \\
\frac{\partial }{\partial z} &= -r \sin(\theta) \frac{\partial }{\partial \theta}
\end{align*}$$

It's generally better to included test function to avoid mistakes:
$$\frac{\partial f }{\partial x} = \frac{\partial f}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial x} + \frac{\partial f}{\partial \phi}\frac{\partial \phi}{\partial x}$$

 

[Mayhem]

I just noticed that your derivatives are not right:


It's generally better to included test function to avoid mistakes:
$$\frac{\partial f }{\partial x} = \frac{\partial f}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial x} + \frac{\partial f}{\partial \phi}\frac{\partial \phi}{\partial x}$$

I am a little confused it seems. I cannot take $r$ to be constant?

 

[PeroK]

I am a little confused it seems. I cannot take $r$ to be constant?

Definitely not. $r$ is one of the three coordinates.

 

[Mayhem]

Definitely not. $r$ is one of the three coordinates.

I see. I found this and it seems what I am trying to accomplish isn't exactly a simple task (probably why my textbook omitted the derivation), but I will try to work through it.

 

[PeroK]

I see. I found this and it seems what I am trying to accomplish isn't exactly a simple task (probably why my textbook omitted the derivation), but I will try to work through it.

That's even more work than I recall. There are quicker ways to do it using results for general coordinate systems.

 

[Mayhem]

That's even more work than I recall. There are quicker ways to do it using results for general coordinate systems.

Hm.

I think I might have to backtrack a little and consider the same problem but in two dimensions, and then once I understand that, I can move on to the 3D case.

 

[PeroK]

Hm.

I think I might have to backtrack a little and consider the same problem but in two dimensions, and then once I understand that, I can move on to the 3D case.

Try this:

https://www.jfoadi.me.uk/documents/lecture_mathphys2_05.pdf