Partial derivative in Spherical Coordinates

[yungman]

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

Thanks

 

[STEMucator]

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

 

[yungman]

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 [tex]\frac{\partial u}{\partial r}+\frac{\partial u}{\partial \theta}+\frac{\partial u}{\partial \phi}[/tex]

Not the spherical amplitude representation of (x,y,z) where
[tex]\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[/tex]
In the ordinary Calculus III class.

Thanks

 

[Enigman]

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

 

[yungman]

Thanks

 

[vela]

Is partial derivative of ##u(x,y,z)## equals to
[tex]\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}[/tex]

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
[tex]\frac{\partial u}{\partial r}+\frac{\partial u}{\partial \theta}+\frac{\partial u}{\partial \phi}[/tex]

No.

 

[vanhees71]

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

 

[vela]

D'oh!

 

[yungman]

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