How to Simplify the Laplace Equation in Spherical Coordinates?

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The discussion focuses on simplifying the Laplace equation in spherical coordinates using the function f(r,θ,φ)=Rl(r)Ylm(θ,φ). Participants emphasize the importance of understanding the Laplace operator's action on the function's radial and angular components. The concept of "separability" is highlighted as a key approach to solving the problem. Additionally, contributors request that the original poster demonstrate their attempts to solve the problem to facilitate more effective assistance. Overall, the conversation underscores the need for a clear problem statement and prior effort to engage in the discussion.
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Homework Statement
Hello, how can I simplify ∆f(r,θ,φ) by using f(r,θ,φ)=Rl(r)Ylm(θ,φ)?
Relevant Equations
f(r,θ,φ)=Rl(r)Ylm(θ,φ)
I know what the Laplace operator is and I also looked up how f(r,θ,φ)=Rl(r)Ylm(θ,φ) is defined but I still could not solve the problem.
 
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physicss said:
Homework Statement: Hello, how can I simplify ∆f(r,θ,φ) by using f(r,θ,φ)=Rl(r)Ylm(θ,φ)?
Relevant Equations: f(r,θ,φ)=Rl(r)Ylm(θ,φ)

I know what the Laplace operator is and I also looked up how f(r,θ,φ)=Rl(r)Ylm(θ,φ) is defined but I still could not solve the problem.
What does the Laplace operator look like in spherical polar coordinates? If you then have this operator act on the following
f(r,θ,φ)=Rl(r)Ylm(θ,φ)
what happens when the "r" part of the operator hits the "Y" part of the function? And similarly for the angle parts acting on Rl(r)?

The buzzword is "separability." You can probably get quite a lot of help by googling this.
 
physicss said:
I know what the Laplace operator is and I also looked up how f(r,θ,φ)=Rl(r)Ylm(θ,φ) is defined but I still could not solve the problem.
Can you state the problem that you still cannot solve?
According to our rules, to receive help, you need to show some credible effort towards answering the question. How about showing us what you tried and where you got stuck? We need something to work from.
 
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My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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