# Partial derivative in Spherical Coordinates

1. Sep 1, 2013

### 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

2. Sep 1, 2013

### Zondrina

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

3. Sep 1, 2013

### yungman

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

4. Sep 1, 2013

### Enigman

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

5. Sep 1, 2013

Thanks

6. Sep 1, 2013

### vela

Staff Emeritus
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.

No.

Last edited: Sep 1, 2013
7. Sep 1, 2013

### vanhees71

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

8. Sep 1, 2013

### vela

Staff Emeritus
D'oh!

9. Sep 1, 2013

### yungman

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