Laplacian of f equals zero and spherical harmonics equation

In summary: Legendre functions, which are only defined for integers).In summary, the equation \nabla^2 f=0 can be decomposed into two equations in spherical coordinates, one depending only on r and the other having the form of a spherical harmonics equation with an arbitrary constant C. However, in the literature, the case of C=l(l+1) is usually considered. This is because when separating variables, we get two ordinary differential equations, one in \theta and the other in \phi, which must each be equal to a constant. Since the function in \theta must be periodic with period 2\pi, C must be equal to l(l+1). While the
  • #1
mantysa
1
0
Lets consider the equation:
[tex]\nabla^2 f=0[/tex]
I know that in spherical coordinates this equation may be decomposed into two equations,
first which depends only on r, and the second one which has the form of spherical harmonics equation except that the [tex]l(l+1)[/tex] is an arbitrary constant, let's say C (and of course the same constant is present in the first equation).
I do not understand why we consider (in literature for example) only the case of
[tex]C=l(l+1)[/tex]

What if we have an equation of form:
[tex]\left(\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\sin\theta\frac{\partial}{\partial\theta}+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}+C\right)u=0[/tex]

is any way to decompose [tex]u[/tex] into spherical harmonics or to transform this equation into standard spherical harmonic equation?
 
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  • #2
The reason we only consider [itex]l(l+1)[/itex] is that, when we separate variables, we get two ordinary differential equations, one in [itex]\theta[/itex], the other in [itex]\phi[/itex]. Basically, we get a function in [itex]\theta[/itex] only equal to a function of [itex]\phi[/itex] only. Those must each be equal to a constant, C. And, since [itex]\theta[/itex] goes all the way from 0 to [itex]2\pi[/itex] the function in [itex]\theta[/itex] must be periodic with period [itex]2\pi[/itex]. That only happens when [itex]C= l(l+1)[/itex].
 
  • #3
HallsofIvy said:
The reason we only consider [itex]l(l+1)[/itex] is that, when we separate variables, we get two ordinary differential equations, one in [itex]\theta[/itex], the other in [itex]\phi[/itex]. Basically, we get a function in [itex]\theta[/itex] only equal to a function of [itex]\phi[/itex] only. Those must each be equal to a constant, C. And, since [itex]\theta[/itex] goes all the way from 0 to [itex]2\pi[/itex] the function in [itex]\theta[/itex] must be periodic with period [itex]2\pi[/itex]. That only happens when [itex]C= l(l+1)[/itex].

HallsofIvy said:
The reason we only consider [itex]l(l+1)[/itex] is that, when we separate variables, we get two ordinary differential equations, one in [itex]\theta[/itex], the other in [itex]\phi[/itex]. Basically, we get a function in [itex]\theta[/itex] only equal to a function of [itex]\phi[/itex] only. Those must each be equal to a constant, C. And, since [itex]\theta[/itex] goes all the way from 0 to [itex]2\pi[/itex] the function in [itex]\theta[/itex] must be periodic with period [itex]2\pi[/itex]. That only happens when [itex]C= l(l+1)[/itex].

The differential equations are:

[tex]\frac{1}{\Phi}\frac{d^2 \Phi}{d \phi^2}=-m^2
[/tex]

and

[tex]
(1-x^2)\frac{d^2\Theta}{dx^2}-2x\frac{d\Theta}{dx}+[l(l+1)-\frac{m^2}{1-x^2}]\Theta=0
[/tex]

where [tex]x=cos(\theta) [/tex].

The solution of the first equation requires that "m" is an integer for the solution [tex] e^{im\phi}[/tex] to be periodic.

However, I don't see why the second equation requires l to be an integer. If l is an integer then it's nice as you can find solutions that are polynomial (the equation is the associated Legendre equation). But does l have to be integral?
 
  • #4

Related to Laplacian of f equals zero and spherical harmonics equation

Q1: What is the Laplacian of a function?

The Laplacian of a function, denoted by ∇2f or Δf, is a mathematical operator that calculates the sum of the second-order partial derivatives of the function.

Q2: What does it mean for the Laplacian of a function to equal zero?

When the Laplacian of a function is equal to zero, it means that the function is harmonic, which indicates that the function is smooth and has no local maxima or minima.

Q3: What is the relationship between the Laplacian of a function and spherical harmonics?

The Laplacian of a function can be expressed in terms of spherical harmonics, which are a set of special functions that are used to describe solutions to Laplace's equation in spherical coordinates. The spherical harmonics equation is a specific form of Laplace's equation, where the function is separable into radial and angular components.

Q4: How are spherical harmonics used in physics and engineering?

Spherical harmonics are used to solve problems involving spherical symmetry, such as in heat transfer, electromagnetism, and fluid dynamics. They are also used in the analysis of satellite data and in computer graphics to generate realistic lighting and shading effects.

Q5: Can the Laplacian of a function and the spherical harmonics equation be applied to non-spherical systems?

Yes, the Laplacian of a function and the spherical harmonics equation can be extended to non-spherical systems by using different coordinate systems, such as cylindrical or Cartesian coordinates. However, the equations and solutions will be more complex and may not have the same physical interpretation as in spherical systems.

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