Derivation of Laplace in spherical co-ordinates

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SUMMARY

The discussion focuses on deriving the Laplace equation in spherical coordinates, specifically the equation d(phi)/dx = -sin(phi)/(r sin(theta)). The participants emphasize using the action principle, represented by the action A[φ] = ∫d³x[(∇φ)² + fφ], to derive the equation Δφ = f. The gradient in spherical coordinates is expressed as ∇φ = e_r ∂_r φ + (1/r)e_θ ∂_θ φ + (1/(r sin θ))e_φ ∂_φ φ, which is essential for the derivation. The volume element in spherical coordinates is also highlighted as d³r = dr dθ dφ r² sin θ.

PREREQUISITES
  • Understanding of spherical coordinates in calculus
  • Familiarity with the action principle in physics
  • Knowledge of partial differential equations (PDEs)
  • Basic concepts of vector calculus, particularly gradients
NEXT STEPS
  • Study the derivation of the Laplacian in spherical coordinates
  • Learn about the Euler-Lagrange equations in classical mechanics
  • Explore the applications of the action principle in field theory
  • Review partial differential equations and their transformations
USEFUL FOR

Students and researchers in physics, particularly those focusing on mathematical physics, differential equations, and field theory. This discussion is especially beneficial for beginners seeking to understand the derivation of the Laplace equation in spherical coordinates.

darwined
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I have been trying to derive the Laplace in spherical co ordinates.
I have attached a file which has basic equations.
I am trying to get the following equation.

d(phi)/dx= -sin(phi)/(r sin (theta)).

I have also attached the materials I am referring to.
Can someone please help me derive the equation.

Thank you.
 

Attachments

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  • The Laplacian Operator from Cartesian to Cylindrical to Spherical Coordinates.pdf
    The Laplacian Operator from Cartesian to Cylindrical to Spherical Coordinates.pdf
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This is a pretty cumbersome way. The most easy is to use the action principle. The action
A[\phi]=\int \mathrm{d}^3 \vec{x} [(\vec{\nabla} \phi)^2+f \phi]
leads to the equation
\Delta \phi=f.
Now you write the gradient in terms of spherical coordinates (which is easy to derive by your direct method)
\vec{\nabla} \phi=\vec{e}_r \partial_r \phi+\frac{1}{r} \vec{e}_{\theta} \partial_{\theta} \phi + \frac{1}{r \sin \theta} \vec{e}_{\varphi} \partial_{\varphi} \phi.
The volume element is
\mathrm{d}^3 {\vec{r}}=\mathrm{d} r \mathrm{d} \theta \mathrm{d} \varphi r^2 \sin \theta.
This you plug into the action integral and use the Euler-Lagrange equations to derive the field equation in terms of spherical coordinates. This leads to the Laplacian by identifying it with f[/tex].
 
darwined said:
I have been trying to derive the Laplace in spherical co ordinates.
I have attached a file which has basic equations.
I am trying to get the following equation.

d(phi)/dx= -sin(phi)/(r sin (theta)).

I have also attached the materials I am referring to.
Can someone please help me derive the equation.

Thank you.
Show us your attempt at deriving that equation.
 
I second Vela. Vanhees is focusing on ##\Delta##. If you want a hint for your ## {d\phi \over dx} ## question: ##\tan \phi = {y \over x} ## is a good starting point.
 
Thank you for your reply vanhees71. But I am not sure how you even got the equation.

A[ϕ]=∫d3x→[(∇→ϕ)2+fϕ]

I am a beginner trying to learn this derivation.

Thank you BvU, your idea helped.

Thank you all.
 
Your attached references are just a straightforward transformation of coordinates that you learn how to do in a course on partial differential equations. Have you had a course that covers partial differential equations yet?

Chet
 

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