- #1
physicss
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- Homework Statement
- Hello, I had to calculate ∆r and ∆Theta,phi. Is the answer on the second picture correct?
- Relevant Equations
- ∆r, ∆Theta,phi
Laplace operator in spherical coordinates
- Thread starter physicss
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- Laplace
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- #2
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It's not clear which is the first and which is the second picture. You can find all you need here
https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates
Please don't post pictures sideways. They are hard to read. Thanks.
https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates
Please don't post pictures sideways. They are hard to read. Thanks.
FAQ: Laplace operator in spherical coordinates
What is the Laplace operator in spherical coordinates?
The Laplace operator, also known as the Laplacian, in spherical coordinates (r, θ, φ) is given by the formula: Δf = (1/r^2) ∂/∂r (r^2 ∂f/∂r) + (1/r^2 sinθ) ∂/∂θ (sinθ ∂f/∂θ) + (1/r^2 sin^2θ) ∂^2f/∂φ^2. This expression accounts for the geometry of spherical coordinates and is used to solve partial differential equations in fields like quantum mechanics, electromagnetism, and fluid dynamics.
How do you derive the Laplace operator in spherical coordinates?
The derivation of the Laplace operator in spherical coordinates involves transforming the Cartesian coordinate expression of the Laplacian into spherical coordinates. This process includes converting the Cartesian partial derivatives to spherical partial derivatives using the chain rule and the relationships between Cartesian and spherical coordinates (x = r sinθ cosφ, y = r sinθ sinφ, z = r cosθ). The detailed steps involve considerable algebraic manipulation and the use of trigonometric identities.
Why is the Laplace operator important in spherical coordinates?
The Laplace operator in spherical coordinates is crucial because many physical problems exhibit spherical symmetry, such as the gravitational potential around a planet, the electric potential around a charged sphere, and the wave functions of atoms. Using spherical coordinates simplifies the mathematical formulation and solution of these problems, making the Laplacian a vital tool in theoretical and applied physics.
What are the boundary conditions for solving the Laplace equation in spherical coordinates?
Boundary conditions for solving the Laplace equation in spherical coordinates depend on the specific physical problem. Common types include Dirichlet boundary conditions, where the function value is specified on the boundary; Neumann boundary conditions, where the derivative of the function is specified on the boundary; and mixed boundary conditions, which combine both. Properly defining these conditions is essential for obtaining meaningful and accurate solutions.
Can the Laplace operator in spherical coordinates be separated into simpler equations?
Yes, the Laplace operator in spherical coordinates can often be separated into simpler equations using the method of separation of variables. This technique assumes that the solution can be written as a product of functions, each depending on a single coordinate (r, θ, φ). By substituting this product into the Laplace equation and separating the variables, one can obtain ordinary differential equations for each coordinate, which are typically easier to solve.