Deriving laplacian in spherical coordinates

In summary: =\vec{i}_{r}\frac{1}{r^2}\frac{\partial\nabla}{\partial{r^2}}+\vec{i}_{\phi}\frac{1}{r^3}\frac{\partial\nabla}{\partial{r^3}}+\vec{i}_{\theta}\frac{1}{r^4}\frac{\partial\nabla}{\partial{r^4}}=\vec{i}_{r}\frac{1}{4}(\vec{i}_{r}\frac{\partial}{\partial{r^2}}+\vec{i}_{\phi}\frac{1}{r^3
  • #1
mooshasta
31
0
Hey...

Could someone help me out with deriving the LaPlacian in spherical coordinates? I tried using the chain rule but it just isn't working out well.. any sort of hint would be appriciated. :)

[tex] \nabla^2 = \frac{1}{r^2} [ \frac{\partial}{\partial r} ( r^2 \frac{\partial}{\partial r} ) + \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} ( \sin \theta \frac{\partial}{\partial \theta\ ) + \frac{1}{\sin^2 \theta} \frac{\partial^2}{\partial \phi^2} ] [/tex]
 
Last edited:
Physics news on Phys.org
  • #2
I want to understand the derivation too. I understand the gradient in spherical coordinate, but the divergence formula bothers me. I got some info from some website saying that one can find a basis of vectors where its divergence is zero then take the divergence of the function based on those vectors...
 
  • #3
If anyone was curious about this, I found the solution online:

http://planetmath.org/encyclopedia/%3Chttp://planetmath.org/?method=l2h&from=collab&id=76&op=getobj
 
Last edited by a moderator:
  • #4
That's one way of doing it.
Here's another one:
[tex]\vec{i}_{r}=\sin\phi\cos\theta\vec{i}+\sin\phi\sin\theta\vec{j}+\cos\phi\vec{k}, \vec{i}_\phi}=\frac{\partial\vec{i}_{r}}{\partial\phi}=\cos\phi\cos\theta\vec{i}+\cos\phi\sin\theta\vec{j}-\sin\phi\vec{k},\vec{i}_{\theta}=\frac{1}{\sin\phi}\frac{\partial\vec{i}_{r}}{\partial\theta}=-\sin\theta\vec{i}+\cos\theta\vec{j}[/tex]
along with:
[tex]\frac{\partial\vec{i}_{r}}{\partial{r}}=\frac{\partial\vec{i}_{\phi}}{\partial{r}}=\frac{\partial\vec{i}_{\theta}}{\partial{r}}=\vec{0}[/tex]
[tex]\frac{\partial\vec{i}_{\theta}}{\partial\phi}=\vec{0},\frac{\partial\vec{i}_{\phi}}{\partial\phi}=-\vec{i}_{r}[/tex]
[tex]\frac{\partial\vec{i}_{\phi}}{\partial\theta}=\cos\phi\vec{i}_{\theta},\frac{\partial\vec{i}_{\theta}}{\partial\theta}=-\sin\phi\vec{i}_{r}-\cos\phi\vec{i}_{\phi}[/tex]

Now, given these relations, along with the expression for the gradient in spherical coordinates [tex]\nabla=\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\phi}\frac{1}{r}\frac{\partial}{\partial\phi}+\vec{i}_{\theta}\frac{1}{r\sin\phi}\frac{\partial}{\partial\theta}[/tex], we may easily derive the expression for the Laplacian by differentiating, and performing the dot products:
[tex]\nabla^{2}=\nabla\cdot\nabla=\vec{i}_{r}\cdot\frac{\partial\nabla}{\partial{r}}+\frac{1}{r}\vec{i}_{\phi}\cdot\frac{\partial\nabla}{\partial\phi}+\frac{1}{r\sin\phi}\vec{i}_{\theta}\cdot\frac{\partial\nabla}{\partial\theta}[/tex]
 
Last edited:

1. What is the purpose of deriving the Laplacian in spherical coordinates?

The Laplacian is a mathematical operator that is used to describe how a physical quantity changes over space and time. Deriving it in spherical coordinates allows us to analyze physical systems with spherical symmetry, such as planets or atoms, and understand how different variables (such as temperature or electric potential) change with distance from a central point.

2. How is the Laplacian expressed in spherical coordinates?

In spherical coordinates, the Laplacian is expressed as &nabla2 = (1/r2) ∂/∂r (r2 ∂/∂r) + (1/r2 sin θ ∂/∂θ (sin θ ∂/∂θ) + (1/r2 sin2 θ ∂2/∂φ2), where r is the distance from the origin, θ is the polar angle, and φ is the azimuthal angle.

3. What are the main steps in deriving the Laplacian in spherical coordinates?

The main steps in deriving the Laplacian in spherical coordinates involve converting the Cartesian coordinates (x, y, z) to spherical coordinates (r, θ, φ), expressing the Laplacian operator in terms of these coordinates, and then simplifying the resulting equations using trigonometric identities and the chain rule. This process leads to the expression mentioned in the answer to question 2.

4. Can the Laplacian be derived in other coordinate systems?

Yes, the Laplacian can be derived in any coordinate system. The specific form of the Laplacian will depend on the geometry of the coordinate system and the variables used to describe it. Some common coordinate systems in which the Laplacian is derived include Cartesian, cylindrical, and spherical coordinates.

5. How is the Laplacian used in real-world applications?

The Laplacian has many applications in physics, engineering, and other scientific fields. It is used to solve differential equations that describe physical phenomena, such as heat transfer, fluid flow, and electromagnetic fields. It is also used in image processing and computer graphics to enhance and analyze images. Additionally, the Laplacian is an important tool in quantum mechanics, where it is used to describe the behavior of particles in three-dimensional space.

Similar threads

  • Advanced Physics Homework Help
Replies
9
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
792
  • Advanced Physics Homework Help
Replies
29
Views
117
  • Advanced Physics Homework Help
Replies
30
Views
1K
  • Advanced Physics Homework Help
Replies
3
Views
374
  • Advanced Physics Homework Help
Replies
11
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
915
  • Advanced Physics Homework Help
Replies
10
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
984
Replies
6
Views
3K
Back
Top