# The Laplacian of the potential q*exp(-r)/r

• Elder1994
In summary, the potential of a point charge is not differentiable along the polar axis, so you can't use the Laplacian operator.
Elder1994
Hello, I have a problem where I'm supposed to calculate the charge distribution ρ. I need to calculate it by applying the Laplacian operator to the potential Θ. The potential is the function: q*exp(-αr)/r
I found on the internet that for this type of potentials I cannot just apply the Laplacian operator because the function is not differentiable as r approaches to 0. So I'm not sure what I'm supposed to do with this.

Delta2
You have a potential that is the product of two functions, ##f = \exp(-\alpha r)## and ##g = \frac{1}{r}##, and you need to calculate ##\nabla^2 f g##. I would expand ##\nabla^2 f g##, then plug in your specific ##f## and ##g##.

Delta2 and vanhees71
Also carefully think about what ##\Delta(1/r)## is (of course, there's only a problem at the origin ##r=0##). Also note that to answer this question, you should be very careful and NOT use spherical but Cartesian coordinates in your calculations since spherical coordinates are singular along the polar axis including the origin ##r=0##!

jasonRF and Delta2
vanhees71,

When I solved this, I used knowledge of electrostatics to know what ##\nabla^2 \frac{1}{r}## is, but used spherical coordinates for the remaining terms and it certainly yielded what I was expecting (this is the potential of a point charge in a hot plasma so we know the equation it should satisfy). The other terms include
$$\left( \nabla \frac{1}{r}\right) \cdot \nabla e^{-\alpha r}$$
and
$$\nabla^2 e^{-\alpha r}$$
What is wrong with using spherical coordinates for those?

Last edited:
There's nothin a priori wrong. If you are lucky the coordinate singularities of spherical coordinates don't do any harm, but you can't express
$$\Delta \frac{1}{r}=-4 \pi \delta^{(3)}(\vec{x})$$
in terms of spherical coordinates, because there the coordinate singularities hit with maximal impact ;-)). I've seen this done wrong in otherwise respectable textbooks!

Well, the three dimensional delta function at the origin can certainly be expressed in Cartesian or spherical coordinates. Using your notation (assuming I understand it correctly), I get,
$$\delta^{(3)}(\overrightarrow{x}) = \delta(x)\delta(y)\delta(z) = \frac{\delta(r)}{4\pi r^2}$$

Perhaps we are getting off-top from the OP...

## 1. What is the Laplacian of the potential q*exp(-r)/r?

The Laplacian of the potential q*exp(-r)/r is a mathematical operator that is used to describe the rate of change of the potential q*exp(-r)/r at a given point in space. It is commonly used in electromagnetism and quantum mechanics to calculate the electric and magnetic fields, as well as the energy levels of particles.

## 2. How is the Laplacian of the potential q*exp(-r)/r calculated?

The Laplacian of the potential q*exp(-r)/r is calculated by taking the second partial derivatives of the potential function with respect to each of the spatial coordinates (x, y, z) and then adding them together. This can be represented by the equation: ∇²(q*exp(-r)/r) = (∂²/∂x² + ∂²/∂y² + ∂²/∂z²)(q*exp(-r)/r).

## 3. What is the physical significance of the Laplacian of the potential q*exp(-r)/r?

The Laplacian of the potential q*exp(-r)/r has several physical significances. It represents the curvature of the potential function at a given point, and can be used to determine the stability of a system. It also plays a crucial role in determining the electric and magnetic fields, as well as the energy levels of particles in a system.

## 4. How does the Laplacian of the potential q*exp(-r)/r relate to the Laplace equation?

The Laplacian of the potential q*exp(-r)/r is closely related to the Laplace equation, which is a partial differential equation that describes the behavior of scalar fields in three-dimensional space. The Laplacian operator is used in the Laplace equation to calculate the potential at a given point in space.

## 5. Can the Laplacian of the potential q*exp(-r)/r be used to solve real-world problems?

Yes, the Laplacian of the potential q*exp(-r)/r can be used to solve many real-world problems in physics and engineering. It is commonly used in electromagnetism to calculate the electric and magnetic fields, and in quantum mechanics to determine the energy levels of particles. It is also used in fluid dynamics to describe the flow of fluids in a system.

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