Laplacian operator in different coordinates

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The discussion focuses on expressing the Laplacian operator in cylindrical and spherical coordinates based on Cartesian coordinates. In cylindrical coordinates, the operator is defined as ∇²u = ∂²u/∂r² + (1/r)∂u/∂r + (1/r²)∂²u/∂θ² + ∂²u/∂z², with the z-term often omitted. For spherical coordinates, the Laplacian is given by ∇²u = (1/r²)(∂(r²u_r)/∂r + csc²θ∂²y/∂θ² + cscθ∂(sinφu_φ)/∂φ). The discussion emphasizes the importance of using change of variables in partial derivatives to derive these forms. Overall, the conversation highlights the mathematical transformations necessary to adapt the Laplacian operator across different coordinate systems.
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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 r=\sqrt{x^2+y^2} and \theta = \tan^{-1} \frac{y}{x} assuming this is valid on this region.

This leads to,
\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
For the most part z coordinate is not taken and that term vanished.

Spherical: Using Spherical Coordinate substitutions:
\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
 
in general (which is something you learn in vector analysis for physicists):
\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}) 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

\frac{\partial}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial}{\partial r} +...

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):
\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}) 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

\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]<br />

Source: Hobson, Mathematical methods for Physics and Engineering, pg. 374.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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