Proving Laplacian in Cylindrical Coordinates

In summary, it seems that the laplacian in cylindrical coordinates is just (1/r²)d²f/dO² + d^2f/dz^2.
  • #1
quasar987
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I'm supposed to prove the laplacian in cylindrical coord. is what it is. I tried tackling the problem in two ways and none work! I have no idea what's the matter. The first way is to calculate d²f/dr² , d²f/dO² and d²f/dz² and isolate d²f/dx² , d²f/dy² and d²f/dz². In cylindrical coord.,

x=rcosO
y=rsinO
z=z

(I'm not writting partial derivatives because in latex it's a pain in the @)

[tex]df/dr = df/dx dx/dr + df/dy dy/dr[/tex]
[tex]\Rightarrow d^2f/dr^2 = d^2f/dx^2 dx/dr + df/dx d^2x/dr^2 + d^2f/dy^2 dy/dr + df/dy d^2y/dr^2[/tex]
[tex]d^2f/dr^2 = d^2f/dx^2 cos^2\theta +d^2f/dy^2 sin^2\theta[/tex]

In the same way,

[tex]d^2f/d\theta^2 = r^2 (d^2f/dx^2 d^2f/dx^2)[/tex]

Hence, it would seem that the laplacian in cylindrical is just (1/r²)d²f/dO² + d^2f/dz^2. :frown:
 
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  • #2
Have you tried

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

etc.?
 
  • #3
You are doing it backwards!
You know that the Laplacian is
[tex]\frac{\partial^2U}{/partial x^2}+ \frac{\partial^2U}{\partial y^2}+ \frac{\partial^2U}{\partial z^2}[/tex]
in Cartesian coordinates so you should be writing
[tex]\frac{\partialU}{\partial z}=\frac{\partial U}{\partial r}\frac{\partial r}{\partial x}+ \frac{\partial U}{\partial \theta}\frac{\partial \theta}{\partial x}+ \frac{\partial U}{\partial z}\frac{\partial z}{\partial x}[/tex]
etc. so that you can substitute into that.
 
  • #4
Yeah that's the "second way" I was referring to! Why do you call "backward" the first way? It's just as good imo. It's not guarenteed that you'll be able to isolate the cartesian laplancian, but if you can, shouldn't it give the right answer too ?!
 
  • #5
Ok, after working out the algebra completely and correcting a few mistake I had made in the second way (the one advetised by HallsofIvy), it works. But I'm still very interested to know why the "backwards" way doesn't work.
 

1. What is the Laplacian operator in cylindrical coordinates?

The Laplacian operator in cylindrical coordinates is a mathematical operator that represents the sum of the second partial derivatives of a function with respect to its three variables - radius (r), azimuthal angle (θ), and height (z).

2. Why is it important to prove the Laplacian in cylindrical coordinates?

Proving the Laplacian in cylindrical coordinates is important because it allows us to solve differential equations involving cylindrical geometries, which are commonly found in engineering and physics problems. It also allows us to understand the behavior of functions in cylindrical coordinate systems.

3. How is the Laplacian in cylindrical coordinates derived?

The Laplacian in cylindrical coordinates is derived by applying the chain rule to the Laplacian operator in Cartesian coordinates. This involves converting the Cartesian partial derivatives to their cylindrical equivalents and then simplifying the resulting expression.

4. What are the steps involved in proving the Laplacian in cylindrical coordinates?

The steps involved in proving the Laplacian in cylindrical coordinates include converting the Cartesian partial derivatives to cylindrical equivalents, applying the chain rule, simplifying the resulting expression, and then combining like terms to obtain the final expression for the Laplacian in cylindrical coordinates.

5. Can the Laplacian in cylindrical coordinates be extended to other coordinate systems?

Yes, the Laplacian operator can be extended to other coordinate systems such as spherical and polar coordinates. The process of proving the Laplacian in these coordinate systems follows a similar approach as in cylindrical coordinates, by converting the Cartesian partial derivatives to their respective equivalents and then applying the chain rule.

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