Divergence of the Electric field of a point charge

[Lambda96]

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

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?

 

[TSny]

Your result is correct.

For a point charge, the divergence of $\vec E$ is equal to zero at all points of space except at the point of location of the charge ($r = 0$). At that point, both $\vec E$ and $\vec{\nabla}\cdot \vec{E}$ are singular; i.e., they do not have values at that point.

 

[jtbell]

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.

True, but only at the location of the charge (the origin, in this example). At any location where there is no charge, the divergence of $\vec E$ must be zero.

Have you encountered the differential version of Gauss's Law yet? $$\nabla \cdot \vec E = \frac {\rho (\vec r)} {\varepsilon_0}$$ where $\rho$ is the charge density. For a point charge at the origin, $\rho = 0$ everywhere except at the origin.

 

[Lambda96]

Thanks TSny and jtbell for your explanations

In the book of David J. Griffiths Introduction to Electrodynamics in chapter 1.5.1 the special feature of $\frac{\vec{r}}{r^3}$ at the point $\vec{r}=(0,0,0)$ is discussed and if I have understood everything correctly now, this behavior of the divergence occurs only at this point. If I place the charge now at another place, the divergence would not be zero.

 

[TSny]

If I place the charge now at another place, the divergence would not be zero.

The divergence would not be zero at what point(s)?

 

[PhDeezNutz]

The divergence blows up to 4pi wherever you place the charge and is zero everywhere else. Regardless if you place the charge at the origin or not. This has to be true by the divergence theorem and gauss’s law.

 

[vanhees71]

That's not entirely right: In the sense of usual functions, the divergence is 0 everywhere and undefined at $\vec{r}=0$, where the field has a singularity.

On the other hand, the surface integral over an arbitrary surface including the origin in its interior is $4 \pi$. Thus, if you calculate the divergence using its invariant definition through surface integrals over the boundary of volumes shrinking to the origin, you get, in the sense of distributions
$$\vec{\nabla} \cdot \frac{\vec{r}}{|\vec{r}|^3}=4 \pi \delta^{(3)}(\vec{r}).$$

 

[Lambda96]

Thanks TSny, PhDeezNutz and vanhees71 for your help

Sorry, but I still don't fully understand the reason. I think my big problem is that the following holds for Gauss' theorem.

$$\iint_S \textbf{E} \cdot d\textbf{A}=\iiint_V \nabla \cdot \textbf{E} dv$$

Now if using the theorem for Gauss in the integral form to calculate the flux of a point charge I get $\frac{q}{\epsilon_0}$ but the differential form I would get 0 since yes $\nabla \cdot E=0$. But the two sides should actually be the same.

That's why I thought that simply the origin, that is $r=(0,0,0)$ is a special point and if I put, for example, the point charge on $r=(1,2,3)$, the divergence would not be zero and if I now used theorem of Gauss in the differential form, get the same result as in the integral form.

 

[PhDeezNutz]

Let's forget where you put the point charge and just put it at the origin. You can generalize afterwards.

I suggest switching to spherical coordinates to address the question of $\nabla \cdot \left(\frac{1}{r^2} \right) \hat{r}$

The expression for $\nabla \cdot \vec{v}$ in spherical coordinates is the following for a purely radial vector field

$$ \nabla \cdot \vec{v} = \frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 v_r \right)$$

You plug in $v_r = \frac{1}{r^2}$you will get 0 for the divergence (This is valid everywhere but the origin and soon you'll see why)

BUT if you apply the divergence theorem over a sphere of nonzero radius (say $R$)

$$\int \nabla \cdot \left(\frac{1}{r^2}\right) \hat{r} \, d \tau = \oint \vec{v} \cdot d\vec{a} = \int \frac{1}{r^2} \hat{r} \cdot r^2 \sin \theta \, d\theta d\phi \hat{r}= 4 \pi$$

Notice the integrand is independent of $r$ because the $r^2$ in the numerator and denominator cancel out.

So what if we shrink this sphere as $R \rightarrow 0$? We'd still get $4 \pi$. (THIS IS THE MOST IMPORTANT THING)

In order for this "most important thing" to be true the $\nabla \cdot \left(\frac{1}{r^2} \hat{r} \right)$ must blow up to $4 \pi$ at the origin and be 0 everywhere else

Therefore we say

$\nabla \cdot \frac{1}{r^2} \hat{r} = 4 \pi \delta \left( \vec{r} \right)$

Edit: I think this is what @vanhees71 was getting at

 

[PhDeezNutz]

Edit: It must blow up at the origin so large that when you integrate over the tiny sphere we talked about earlier you still get 4pi

That is the purpose of the delta function.

 

[vanhees71]

Thanks TSny, PhDeezNutz and vanhees71 for your help

Sorry, but I still don't fully understand the reason. I think my big problem is that the following holds for Gauss' theorem.

$$\iint_S \textbf{E} \cdot d\textbf{A}=\iiint_V \nabla \cdot \textbf{E} dv$$

Now if using the theorem for Gauss in the integral form to calculate the flux of a point charge I get $\frac{q}{\epsilon_0}$ but the differential form I would get 0 since yes $\nabla \cdot E=0$. But the two sides should actually be the same.

That's why I thought that simply the origin, that is $r=(0,0,0)$ is a special point and if I put, for example, the point charge on $r=(1,2,3)$, the divergence would not be zero and if I now used theorem of Gauss in the differential form, get the same result as in the integral form.

The problem is the notion of "point particle". It's a stranger in field theories, and that's why you need to extent your mathematical toolbox to deal with it in a convenient way.

Let's only discuss the electrostatic case first. Then we can start backward, i.e., from the solution. A point charge, sitting at the origin of our Cartesian coordinate system, is described by the Coulomb field,
$$\vec{E}=\frac{q}{4 \pi \epsilon_0} \frac{\vec{r}}{r^3}.$$
As you see, there's a singularity at the origin (i.e., where the "point charge" sits), and that's indicating, why the notion of point charges make trouble, and it makes trouble, which cannot be fully resolved within classical physics.

We can, however, get a bit further in describing this overidealized idea. Within standard calculus the Coulomb field is defined only at $\mathbb{R}^3 \setminus \{\vec{0} \}$, and there it's easy to show that indeed
$$\mathrm{div} \vec{E}=\vec{\nabla} \cdot \vec{E}=0, \quad \vec{r} \neq \vec{0}.$$
On the other hand, there's the singularity, and the apparent paradox you indicate above with Gauss's theorem, is in fact not there, because you cannot integrate $\vec{\nabla} \cdot \vec{E}$ over any volume which contains the origin, because there neither $\vec{E}$ nor $\vec{\nabla} \cdot \vec{E}$ is defined within standard calculus, and thus Gauss's theorem doesn't hold in the usual sense.

You can, however integrate over $V$, containing the origin in its interior, when taking out a little sphere $K_{\epsilon}$ of radius $\epsilon$ around the origin contained completely in $V$. Then you can apply Gauss's theorem,
$$\int_{V \setminus K_{\epsilon}} \mathrm{d}^3 r \vec{\nabla} \cdot \vec{E}=\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E} - \int_{\partial K_{\epsilon}} \mathrm{d}^2 \vec{f} \cdot \vec{E}.$$
Here I used the usual orientation of the surface normal moment of the little spherical shell $\partial K_{\epsilon}$, pointing out of the sphere. That's why there's the minus sign on the righ-hand side. Now this expression is obviously 0, i.e.,
$$\int_{\partial V} \dd^2 \vec{f} \cdot \vec{E} =\int_{\partial K_{\epsilon}} \mathrm{d}^2 \vec{f} \cdot \vec{E}=\frac{q}{\epsilon_0},$$
where the latter result is easily found by direct integration of the Coulomb field.

Now remember the definition of $\mathrm{div} \vec{E}$ by an integral, i.e.,
$$\mathrm{div} \vec{E}(\vec{r}) = \lim_{V \rightarrow \{\vec{r} \}} \frac{1}{V} \int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E}.$$
If you apply this to the Coulomb field, you get of course infinity, because the surface integral stays constant at the value $q/\epsilon_0$ when making $V$ shrink to $\vec{r}=\vec{0}$, and thus the limit defining $\text{div} \vec{E}$ at $\vec{r}=0$ goes to infinity.

So the divergence in fact vanishes everywhere except at the origin, where it diverges. Now you can define a charge density in a generalized way, i.e., you define $\rho(\vec{r})$ to vanish everywhere except at the origin, where it diverges in such a way that the volume integral (!) over any volume containing the origin gives $q$. That defines (in a non-rigorous phycisists' way) the Dirac $\delta$-distribution. Thus in this case you have
$$\rho(\vec{r})=q \delta^{(3)}(\vec{r}),$$
and you thus define the divergence of the Coulomb field in this sense of "generalized functions" (or "distributions") as
$$\vec{\nabla} \cdot \vec{E} = \frac{1}{\epsilon_0} \rho=\frac{q}{\epsilon_0} \delta^{(3)}(\vec{r}).$$
With this mathematical extension from the notion of fields as functions to the notion of fields in the sense of distributions/generalized functions the apparent paradox with Gauss's integral theorem goes away.

Of course you have to be careful with how to handling the distributions. You can be pretty sloppy in many cases, but you have to be aware that you must be careful when problems occur. In electrostatics there are usually no severe problems. You can also use distributions to describe surface- or line-charge densities in a pretty straight-forward way, and you can extend the differential operators div, grad, and rot to such singular sources.

 

[Lambda96]

Thank you PhDeezNutz and vanhees71 for your help and thank you vanhees71 for your detailed explanation (PhDeezUse yours too, of course) which was even more detailed than my textbook

Now I finally got it

 

[PhDeezNutz]

My response was kind of trite to be honest, @vanhees71 's response was much better, detailed, and insightful.

In a sense the argument that Griffiths provides could be viewed as circular.

OP was right to ask for more clarification and @vanhees71 definitely delivered.

I feel compelled to apologize to OP for being so surface level.

 

[physwiz222]

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?

The electric field E is only defined for r>0 so that’s where the charge is 0 as the charge distribution is a delta function at the origin so 0 everywhere else where the electric field is defined so thats why the divergence is 0