Laplacian operator in different coordinates

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Homework Help Overview

The discussion revolves around the formulation of the Laplacian operator in different coordinate systems, specifically spherical and cylindrical coordinates, starting from a Cartesian basis. Participants are exploring the mathematical expressions and transformations involved in this process.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Some participants attempt to derive the Laplacian in cylindrical coordinates using substitutions for radius and angle, while others provide the expression for spherical coordinates. There are also references to general principles learned in vector analysis.

Discussion Status

The discussion includes various attempts to express the Laplacian operator, with some participants questioning the validity of certain transformations and expressions. There is an ongoing exploration of different formulations, but no consensus has been reached on the correctness of the approaches presented.

Contextual Notes

Some participants note potential confusion regarding the derivations and the assumptions made in the transformations, indicating a need for further clarification on the underlying principles.

captain
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how do you write the laplacian operator in spherical coordinates and cylindrical coordinates from a cartesian basis?
 
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captain said:
how do you write the laplacian operator in spherical coordinates and cylindrical coordinates from a cartesian basis?

Cylindrical: Use the substitution [tex]r=\sqrt{x^2+y^2}[/tex] and [tex]\theta = \tan^{-1} \frac{y}{x}[/tex] assuming this is valid on this region.

This leads to,
[tex]\nabla^2 u = \frac{\partial ^2 u}{\partial r^2} +\frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2<br /> }\frac{\partial ^2 u}{\partial \theta ^2} + \frac{\partial ^2 u}{\partial z^2}=0[/tex]
For the most part [tex]z[/tex] coordinate is not taken and that term vanished.

Spherical: Using Spherical Coordinate substitutions:
[tex]\nabla^2 u = \frac{1}{r^2} \left\{ \frac{\partial (r^2u_r)}{\partial r}+\csc^2 \theta \frac{\partial ^2 y}{\partial \theta^2}+\csc \theta \frac{\partial (\sin \phi u_{\phi})}{\partial \phi} \right\} = 0[/tex]
 
in general (which is something you learn in vector analysis for physicists):
[tex]\nabla^2=(\frac{h_3h_2}{h_1}\frac{\partial}{\partial u_1}\frac{\partial h_1}{\partial u_1},\frac{h_3h_1}{h_2}\frac{\partial}{\partial u_2}\frac{\partial h_2}{\partial u_2},\frac{h_1h_2}{h_3}\frac{\partial}{\partial u_3}\frac{\partial h_3}{\partial u_3})[/tex] or something like this.
where:
r=xi+yj+zk
and h_i=|dr/du_i|
i.e you take the norm of the vector.
 
Last edited:
Cheat sheet (well not really cheating, unless you like deriving these things)
 
Simply use the rules of change of variables in partial differentials. For example

[tex]\frac{\partial}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial}{\partial r} +...[/tex]

and then sub everyting in terms of the spherical coordinates. Then compute the 2-nd partial wrt to x and the same for y and z.
 
MathematicalPhysicist said:
in general (which is something you learn in vector analysis for physicists):
[tex]\nabla^2=(\frac{h_3h_2}{h_1}\frac{\partial}{\partial u_1}\frac{\partial h_1}{\partial u_1},\frac{h_3h_1}{h_2}\frac{\partial}{\partial u_2}\frac{\partial h_2}{\partial u_2},\frac{h_1h_2}{h_3}\frac{\partial}{\partial u_3}\frac{\partial h_3}{\partial u_3})[/tex] or something like this.
where:
r=xi+yj+zk
and h_i=|dr/du_i|
i.e you take the norm of the vector.

Either I am confused at the moment, or it is not right. In general case it should be

[tex]\nabla^2 \Phi= \frac{1}{h_1h_2h_3} \left[ \frac{\partial}{\partial u_1} \left( \frac{h_2h_3}{h_1} \frac{\partial \Phi}{\partial u_1} \right) +<br /> <br /> \frac{\partial}{\partial u_2} \left( \frac{h_3h_1}{h_2}\frac{\partial \Phi}{\partial u_2} \right)<br /> <br /> +\frac{\partial}{\partial u_3} \left(\frac{h_1h_2}{h_3}\frac{\partial \Phi}{\partial u_3}\right)<br /> <br /> \right][/tex]

Source: Hobson, Mathematical methods for Physics and Engineering, pg. 374.
 

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