# Deriving the Laplacian in spherical coordinates

• Mayhem
In summary, the Laplacian is a gradient vector that takes the form $$\Delta = \nabla \cdot \nabla = \nabla^2$$.
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.

Delta2
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.

docnet
PeroK said:
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?

Mayhem said:
Where exactly do I start then?
I just noticed that your derivatives are not right:

Mayhem said:
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:
$$\frac{\partial f }{\partial x} = \frac{\partial f}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial x} + \frac{\partial f}{\partial \phi}\frac{\partial \phi}{\partial x}$$

PeroK said:
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?

Mayhem said:
I am a little confused it seems. I cannot take ##r## to be constant?
Definitely not. ##r## is one of the three coordinates.

PeroK said:
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.

Mayhem said:
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.

PeroK said:
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.

## 1. What is the Laplacian operator in spherical coordinates?

The Laplacian operator in spherical coordinates is a mathematical operator used to describe the second-order derivatives of a function with respect to the spherical coordinates (r, θ, φ). It is denoted by ∇2 or ∆ and is defined as the sum of the second partial derivatives of the function with respect to each of the three coordinates.

## 2. How is the Laplacian operator derived in spherical coordinates?

The Laplacian operator in spherical coordinates can be derived using the chain rule and the product rule of differentiation. By applying these rules to the spherical coordinate system, the Laplacian can be expressed as a combination of the first and second derivatives with respect to r, θ, and φ.

## 3. What is the physical significance of the Laplacian in spherical coordinates?

The Laplacian operator in spherical coordinates is commonly used in physics and engineering to describe the rate of change of a physical quantity in a spherical system. It is particularly useful in solving problems involving heat transfer, fluid dynamics, and electrostatics in spherical geometries.

## 4. How is the Laplacian operator used in solving problems?

The Laplacian operator is used in solving problems by converting the given equations or functions into spherical coordinates and then applying the Laplacian operator. This allows for the simplification of complex equations and the solution of problems in spherical geometries.

## 5. Are there any alternative methods for deriving the Laplacian in spherical coordinates?

Yes, there are alternative methods for deriving the Laplacian in spherical coordinates, such as using vector calculus identities or using the divergence theorem. However, the chain rule and product rule method is the most commonly used and straightforward approach.

### Similar threads

• Advanced Physics Homework Help
Replies
1
Views
790
• Advanced Physics Homework Help
Replies
29
Views
85
• Advanced Physics Homework Help
Replies
3
Views
357
• Advanced Physics Homework Help
Replies
4
Views
344
• Advanced Physics Homework Help
Replies
2
Views
970
• Advanced Physics Homework Help
Replies
3
Views
678
• Advanced Physics Homework Help
Replies
8
Views
816
• Advanced Physics Homework Help
Replies
4
Views
916
• Advanced Physics Homework Help
Replies
1
Views
905
• Advanced Physics Homework Help
Replies
1
Views
381