Divergence of ##\vec{x}/\vert\vec{x}\vert^3##

  • #1
PhysicsRock
117
18
Homework Statement
Calculate ##\displaystyle \frac{\partial}{\partial x_i} \frac{x_i}{\vert\vec{x}\vert^3}##
Relevant Equations
##\nabla \cdot \vec{F} = \sum_i \frac{\partial f_i}{\partial x_i}##
As you can see in the homework statement, I am asked to calculate what's effectively the divergence of the vector field ##\vec{F} = \vec{x}/\vert\vec{x}\vert^3## over ##\mathbb{R}^3##. I have done that, the calculation itself isn't that difficult after all. However, I can't make sense of the result, which makes me wonder whether I've made a mistake. What I find is that ##\nabla \cdot \vec{F} = 0##. I used GeoGebra to plot the field, and what I see is vectors "coming out" of the origin. Is it as simple as to say that, since you can't divide by 0, the origin isn't part of the field, and thus, there is no particular source? Or have I made a mistake in my calculations and the result isn't 0?
 
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  • #2
See this thread
Lambda96 said:
Homework Statement: Calculate the following ##\vec{\nabla}\cdot \vec{E}(\vec{r})##
Relevant Equations: none

Hi,

unfortunately, I am not sure if I have calculated the task correctly

View attachment 325180
The electric field of a point charge looks like this ##\vec{E}(\vec{r})=\frac{Q}{4 \pi \epsilon_0}\frac{\vec{r}}{|\vec{r}|^3}## I have now simply divided the electric field into its components i.e. #E_x , E-y, E_z#.

$$\vec{E}(\vec{r})=\frac{Q}{4 \pi \epsilon_0}\left( \begin{array}{rrr}
\frac{x}{(x^2+y^2+z^3)^{\frac{3}{2}}} \\
\frac{y}{(x^2+y^2+z^3)^{\frac{3}{2}}} \\
\frac{z}{(x^2+y^2+z^3)^{\frac{3}{2}}} \\
\end{array}\right)$$

Then I calculated the divergence

$$\vec{\nabla}\cdot \vec{E}(\vec{r})=\frac{Q}{4 \pi \epsilon_0} \Bigl( \frac{\partial}{\partial x}\frac{x}{(x^2+y^2+z^2)^{\frac{3}{2} }} +\frac{\partial}{\partial y}\frac{y}{(x^2+y^2+z^2)^{\frac{3}{2} }} +\frac{\partial}{\partial z}\frac{z}{(x^2+y^2+z^2)^{\frac{3}{2} }} \Bigr)=\Bigl( \frac{-2x^2+y^2+z^2}{(x^2+y^2+z^2)^{\frac{5}{2} }} +\frac{x^2-2y^2+z^2}{(x^2+y^2+z^2)^{\frac{5}{2} }} +\frac{x^2+y^2-2z^2}{(x^2+y^2+z^2)^{\frac{5}{2} }} \Bigr)=0$$

With the result of 0 I am a bit confused, in the task is not mentioned the sign of the charge, but with a positive charge I would expect as a divergence a source, so the divergence would have to be positive and with a negative charge, a sink, so a negative divergence.

Have I somehow miscalculated, or do I have a thinking error in the physical interpretation of the result?
 
  • #3
The vector field is indeed solenoidal, without divergence. All vector fields of type ##\dfrac{\vec{c}\times \vec{r}}{\|\vec{r}\|^3}## are.

PhysicsRock said:
Is it as simple as to say that, since you can't divide by 0, the origin isn't part of the field, and thus, there is no particular source? Or have I made a mistake in my calculations and the result isn't 0?

We always have a ball of some positive radius and vectors pointing outward of equal length. You could as well say that there is no way to lose or gain energy on a closed path through that field.
 
  • #4
  • #5
PhysicsRock said:
I am asked to calculate what's effectively the divergence of the vector field ##\vec{F} = \vec{x}/\vert\vec{x}\vert^3## over ##\mathbb{R}^3##.
I think that the origin should be omitted.
PhysicsRock said:
Is it as simple as to say that, since you can't divide by 0, the origin isn't part of the field, and thus, there is no particular source?
IMO, you are correct. The initial statement of the problem seems a little careless by including the origin.
I would like to hear the opinion on this of others who have more expertise.
 
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FAQ: Divergence of ##\vec{x}/\vert\vec{x}\vert^3##

What is the divergence of ##\vec{x}/\vert\vec{x}\vert^3##?

The divergence of ##\vec{x}/\vert\vec{x}\vert^3## is zero everywhere except at the origin. Mathematically, it is represented as ##\nabla \cdot \left( \frac{\vec{x}}{\vert \vec{x} \vert^3} \right) = 0## for ##\vec{x} \neq \vec{0}##.

Why is the divergence of ##\vec{x}/\vert\vec{x}\vert^3## zero outside the origin?

The vector field ##\vec{x}/\vert\vec{x}\vert^3## represents a radial field that decays as the inverse square of the distance from the origin. When calculating the divergence in spherical coordinates, the radial and angular components cancel out, resulting in a zero divergence everywhere except at the origin.

What happens to the divergence of ##\vec{x}/\vert\vec{x}\vert^3## at the origin?

At the origin, the divergence of ##\vec{x}/\vert\vec{x}\vert^3## is not well-defined because the field ##\vec{x}/\vert\vec{x}\vert^3## becomes singular. The origin is a point of discontinuity where the field's magnitude approaches infinity.

How is the divergence of ##\vec{x}/\vert\vec{x}\vert^3## related to the Dirac delta function?

The divergence of ##\vec{x}/\vert\vec{x}\vert^3## can be interpreted in the sense of distributions. It is given by ##\nabla \cdot \left( \frac{\vec{x}}{\vert \vec{x} \vert^3} \right) = 4\pi \delta^3(\vec{x})##, where ##\delta^3(\vec{x})## is the three-dimensional Dirac delta function. This indicates that the field has a singularity at the origin, contributing to the divergence as a point source.

What is the physical significance of the divergence of ##\vec{x}/\vert\vec{x}\vert^3##?

The physical significance of the divergence of ##\vec{x}/\vert\vec{x}\vert^3## is often encountered in electrostatics and gravitational fields. For example, it represents the electric field due to a point charge or the gravitational field due to a point mass. The divergence theorem relates the flux of this field through a closed surface to the charge or mass enclosed, with the Dirac delta function representing the source at the origin.

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