Partial derivative in Spherical Coordinates

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  • #1
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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
 

Answers and Replies

  • #2
STEMucator
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I'm slightly confused about the question, but if you're just changing co-ordinates and taking partials it looks fine.
 
  • #3
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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
 
  • #4
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You are just differentiating wrt different functions, any calculations done by either method should give you same end results.
 
  • #5
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Thanks
 
  • #6
vela
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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.
 
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  • #7
vanhees71
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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]
 
  • #8
vela
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D'oh!
 
  • #9
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Yes, I know Divergence, Gradient and Laplace/Poisson equation in Spherical. Just the simple partial derivative.
 

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