Laplacian over a radial function for charge density

In summary, the conversation discusses the determination of charge density using the Laplacian over the potential (phi) and the confusion surrounding the presence of a factor of 4pi in the denominator. The reasoning for including the factor of 4pi is explained as it accounts for spherical symmetry in the function. The conversation also mentions the absence of the permittivity constant (epsilon naught) and the potential being for a hydrogen atom. It is clarified that the 4pi comes from the potential itself and is constant, similarly to the epsilon_0. It is also noted that the answer is incorrect and should include a delta function.
  • #1
mateomy
307
0
ScreenShot2012-04-26at112622AM.png



As you probably can see from the above shot, I'm determining charge density via the Laplacian over the potential (phi). I understand the mathematical steps, just confused on the factor of 4pi that pops up in the denominator. I think I understand why you would do that and here's my reasoning...

You're taking a function that is radially outward and operating on it. Since we are assuming this is a spherically symmetric area we can neglect the aspect of the spherical coordinate system that would have symmetry and only focus on the radial change. Since again, you assume spherical symmetry the 4pi covers it. Would this be (semi) sound reasoning? Also, where did the permittivity constant go (epsilon naught)?

Thanks.
 
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  • #2
This is the (time-averaged) potential of a hydrogen atom. The 4pi comes from the potential itself, and then pops out of all the differentials because it is constant. Same as where the epsilon_0 comes from. By the way, the answer is actually wrong -- there should be a delta function in there too.
 

1. What is the Laplacian operator?

The Laplacian operator is a mathematical operator that is used to describe the second-order spatial variation of a function. It is often denoted as ∇² and is used in various fields of science, such as physics, engineering, and mathematics.

2. What is a radial function?

A radial function is a mathematical function that depends only on the distance from the origin, rather than on the direction in which the distance is measured. It is commonly used to describe the radial distribution of a physical quantity, such as charge density or probability density.

3. How is the Laplacian operator applied to a radial function?

The Laplacian operator is applied to a radial function by first converting the function into polar coordinates. The Laplacian operator is then applied to the radial part of the function, while keeping the angular part constant. This results in a differential equation that can be solved to determine the behavior of the function.

4. What is the significance of using the Laplacian over a radial function for charge density?

The use of the Laplacian over a radial function for charge density allows for a more accurate description of the spatial variation of the charge distribution. This is because the Laplacian takes into account the second-order derivatives of the function, providing a more detailed understanding of the distribution of charge.

5. Are there any practical applications of the Laplacian over a radial function for charge density?

Yes, the Laplacian over a radial function for charge density has many practical applications in various fields, such as electrostatics, quantum mechanics, and molecular dynamics. It is used to calculate the electric potential and energy of charged particles, as well as to model the behavior of atoms and molecules in different environments.

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