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Explicitly use partial derivatives of the gradient operator for divergence?

Daniel Shores
Posted 10 years ago

I would like to explicitly calculate the divergence in spherical coordinates in symbolic form. This involves taking the dot product of the gradient operator and a vector. In order to define the gradient operator explicitly in spherical coordinates there must be a way to define a partial derivative without an argument. This is where I am stuck. How do you define a partial derivative without an argument within a vector that will later need to evaluate the partial derivative once the dot product has been applied? Here is an example:

Equation

How can this be evaluated in Mathematica?
POSTED BY: Daniel Shores
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I still don't know what kind ot objects ur and er are going to be in your calculations. The standard Dot product has built-in meanings, so I don't know if you can adapt it to your needs, but you can use another similar symbol, "[CenterDot]", and assign a rule for expressions like this:

Subscript[Del, r] \[CenterDot] ur := your definition

Without some understanding of your specific problem I can't help you further.
POSTED BY: Gianluca Gorni
Daniel Shores
Daniel Shores
Posted 10 years ago

Thank you for your response. I am aware of the built-in Div operator, but what I am trying to do is solve the divergence step by step. Maybe a better way to explain what I want to do is create a partial derivative through a dot product ( such as the partial of "ur" w.r.t. "r" and the partial of "er" w.r.t. "r" in the equation I posted).
POSTED BY: Daniel Shores

I don't quite understand what you mean, but have you tried the built-in Div operator, which accepts spherical coordinates?

Div[{f[r, \[Theta], \[CurlyPhi]], g[r, \[Theta], \[CurlyPhi]], 
  h[r, \[Theta], \[CurlyPhi]]}, {r, \[Theta], \[CurlyPhi]}, 
"Spherical"]

Look up Div in the documentation.
POSTED BY: Gianluca Gorni
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