# Partial derivative in Spherical Coordinates

• yungman
In summary, the conversation discusses the concept of partial derivatives in Cartesian and Spherical coordinates. The main question is whether the partial derivative of a function in Spherical coordinates is equal to the sum of its partial derivatives in each coordinate direction. The conversation concludes that this is not the case, as the expression given is for the divergence directional derivative in the (1,1,1) direction.

#### yungman

Is partial derivative of ##u(x,y,z)## equals to
$$\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}$$
Is partial derivative of ##u(r,\theta,\phi)## in Spherical Coordinates equals to
$$\frac{\partial u}{\partial r}+\frac{\partial u}{\partial \theta}+\frac{\partial u}{\partial \phi}$$

Thanks

I'm slightly confused about the question, but if you're just changing co-ordinates and taking partials it looks fine.

Zondrina said:
I'm slightly confused about the question, but if you're just changing co-ordinates and taking partials it looks fine.

Thanks for the reply. I just want to verify partial derivative in TRUE Spherical Coordinates system is $$\frac{\partial u}{\partial r}+\frac{\partial u}{\partial \theta}+\frac{\partial u}{\partial \phi}$$

Not the spherical amplitude representation of (x,y,z) where
$$\vec r=\hat x x +\hat y y+\hat z z\;\hbox { to which}\; x=r\cos\phi\sin\theta,\;y=r\sin\phi\sin\theta, \;z=r\cos\theta$$
In the ordinary Calculus III class.

Thanks

You are just differentiating wrt different functions, any calculations done by either method should give you same end results.

Thanks

yungman said:
Is partial derivative of ##u(x,y,z)## equals to
$$\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}$$
You need to get your terminology straight if you want people to understand you. The expression you wrote above is for the [strike]divergence[/strike] directional derivative of u in the (1,1,1) direction times the square root of 3. Each term is a partial derivative. So the answer to your question is no.

Is partial derivative of ##u(r,\theta,\phi)## in Spherical Coordinates equals to
$$\frac{\partial u}{\partial r}+\frac{\partial u}{\partial \theta}+\frac{\partial u}{\partial \phi}$$
No.

Last edited:
This expression is not a divergence. The divergence is defined for a vector field. In Cartesian coordinates it reads
$$\vec{\nabla} \cdot \vec{V}=\frac{\partial V_x}{\partial x}+\frac{\partial V_y}{\partial y}+\frac{\partial V_z}{\partial z}.$$

D'oh!

Yes, I know Divergence, Gradient and Laplace/Poisson equation in Spherical. Just the simple partial derivative.

## 1. What is a partial derivative in spherical coordinates?

A partial derivative in spherical coordinates is a mathematical tool used to measure the rate of change of a function with respect to one variable while holding all other variables constant. In spherical coordinates, the function is defined in terms of the radial distance, the polar angle, and the azimuthal angle.

## 2. How is a partial derivative calculated in spherical coordinates?

In spherical coordinates, a partial derivative is calculated using the chain rule, which involves taking the derivative of each variable with respect to its corresponding coordinate. The resulting expression is then multiplied by the partial derivative of the function with respect to that coordinate.

## 3. What is the physical significance of a partial derivative in spherical coordinates?

Partial derivatives in spherical coordinates are used to describe the behavior of physical phenomena in three-dimensional space. For example, in physics, they are commonly used to describe the rate of change of temperature or pressure at a specific point in space.

## 4. Can a partial derivative in spherical coordinates be negative?

Yes, a partial derivative in spherical coordinates can be negative. This indicates that the function is decreasing in the direction of that particular coordinate.

## 5. How is a partial derivative in spherical coordinates related to the gradient?

In spherical coordinates, the partial derivatives in each direction make up the components of the gradient vector. The gradient represents the direction of steepest increase of a function, and the magnitude of the gradient is equal to the rate of change in that direction.