Gradient in spherical coordinates problem

In summary, the conversation revolves around understanding the gradient in spherical coordinates. The individual has a scalar field in spherical coordinates and is trying to find the gradient using the chain rule. They are confused about the terms involving \frac{1}{r} and \frac{1}{r \sin(\theta)} and are seeking help to better understand them. The conversation concludes with a suggestion to define the gradient operator and use the known spherical coordinates to find the desired result.
  • #1
Theoretiker
22
0
Hello,

I need help. The topic is a gradient in spherical coordinates. In cartesian it is clear but in spherical coordinates I have two terms which I don't understand from where they come.

Okay, I have a scalar field in spherical coordinates:

[tex] \Phi = \Phi(r, \theta, \phi) [/tex]

I thought that this is the gradient but it is wrong and I don't know why :(

[tex] grad \Phi = \frac{\partial \phi}{\partial r} \vec{e}_{r} + \frac{\partial \phi}{\partial \theta} \vec{e}_{\theta} + \frac{\partial \phi}{\partial \phi} \vec{e}_{\phi}[/tex]

My mathbook tells me that this is the gradient in spherical coordinates but I don't understand the terms [tex] \frac{1}{r} [/tex] and [tex] \frac{1}{r \sin(\theta)} [/tex]

[tex] grad \Phi = \frac{\partial \phi}{\partial r} \vec{e}_{r} + \frac{1}{r} ~ \frac{\partial \phi}{\partial \theta} \vec{e}_{\theta} + \frac{1}{r \sin(\theta)} ~ \frac{\partial \phi}{\partial \phi} \vec{e}_{\phi}[/tex]


I would be thank you for helping :)

greetings

P.S.
Sorry for my bad english. I will practice and learn grammar for better english in the future ;)
 
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  • #2
In Cartesian coordinates
[tex]\grad \phi= \frac{\partial \phi}{\partial x}\vec{i}+ \frac{\partial \phi}{\partial y}\vec{j}+ \frac{\partial \phi}{\partial z}\vec{k}[/tex]
Now you have to use the chain rule to convert those derivatives to spherical coordinates:
[tex]\frac{\partial\Phi}{\partial x}= \frac{\partial \Phi}{\partial \rho}\frac{\partial \rho}{\partial x}+ \frac{\partial \Phi}{\partial \theta}\frac{\partial \theta}{\partial x}+ \frac{\partial \Phi}{\partial \phi}\frac{\partial \phi}{\partial x}[/tex]
It's tedious but doable.
 
  • #3
HallsofIvy said:
Now you have to use the chain rule to convert those derivatives to spherical coordinates:

This is what I don't saw. Thank you :)
 
  • #4
you can define the gradient operator such that :

[tex] d \Phi = \left< grad \Phi , d\vec{r}\right > [/tex]

knowing that in spherical coordinates :

[tex] d\vec{r}\right = \vec{e}_{r} dr + \vec{e}_{\theta} \cdot r d\theta + \vec{e}_{\phi} \cdot r \cdot sin(\theta) d\phi[/tex]

then you should find what you want.
 

1. What is a gradient in spherical coordinates?

A gradient in spherical coordinates is a vector operator that represents the rate of change of a scalar field in three-dimensional space. It is used to determine the direction and magnitude of the steepest slope of a function at a given point.

2. How is the gradient in spherical coordinates different from the gradient in Cartesian coordinates?

The gradient in spherical coordinates takes into account the curvature of the coordinate system, while the gradient in Cartesian coordinates assumes a flat plane. This means that the gradient in spherical coordinates is dependent on the choice of coordinate system, while the gradient in Cartesian coordinates is not.

3. How do you calculate the gradient in spherical coordinates?

The gradient in spherical coordinates can be calculated using the formula:

grad f = (1/r) ∂f/∂r + (1/rsinθ) ∂f/∂θ + (1/rsinθ) ∂f/∂φ

where r, θ, and φ represent the radial, polar, and azimuthal coordinates, respectively.

4. What is the physical significance of the gradient in spherical coordinates?

The gradient in spherical coordinates has physical significance in fields such as fluid mechanics, electromagnetism, and quantum mechanics. It is used to determine the direction and magnitude of forces, electric or magnetic fields, and wave functions, respectively.

5. Are there any limitations to using the gradient in spherical coordinates?

One limitation of using the gradient in spherical coordinates is that it only applies to functions that are continuous and differentiable in three-dimensional space. Additionally, it may become more complicated to solve problems involving the gradient in spherical coordinates compared to using Cartesian coordinates.

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