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.