Problem about the usage of Gauss' law involving the curl of a B field

[georg gill]

I am trying to derive that

$$\nabla \times B=\mu_0 J$$

First the derivation starts with the electric field

$$dS=rsin\varphi d\theta r d\varphi $$

$$ \iint\limits_S E \cdot dS = \frac{q}{4 \pi \varepsilon_0} \iint\limits_S \frac{r}{|r|^3} \cdot dS $$

$$\textbf{r}=x\textbf{i}+y\textbf{j}+z\textbf{k}=\textbf{r}=rcos\theta sin \varphi\textbf{i}+rsin\theta sin \varphi\textbf{j}+rcos\varphi\textbf{k}$$$$\frac{\frac{\partial \textbf{r}}{\partial r}} {|\frac{\partial \textbf{r}}{\partial r}|}=cos\theta sin \varphi\textbf{i}+sin\theta sin \varphi\textbf{j}+cos\varphi\textbf{k}$$$$ \iint\limits_S E \cdot dS = \frac{q}{4 \pi \varepsilon_0 |r|^3} \iint\limits_S rcos\theta sin \varphi\textbf{i}+rsin\theta sin \varphi\textbf{j}+rcos\varphi\textbf{k} \cdot dS$$
$$ =\frac{q}{4 \pi \varepsilon_0 |r|^2}\iint\limits_S cos\theta sin \varphi\textbf{i}+sin\theta sin \varphi\textbf{j}+cos\varphi\textbf{k} \cdot dS $$$$= \frac{q}{4 \pi \varepsilon_0 |r|^2}\iint\limits_S (cos\theta sin \varphi\textbf{i}+sin\theta sin \varphi\textbf{j}+cos\varphi\textbf{k}) \cdot (cos\theta sin \varphi\textbf{i}+sin\theta sin \varphi\textbf{j}+cos\varphi\textbf{k}) rsin\varphi d\theta d\varphi $$

$$ \iint\limits_S E \cdot dS = \frac{q}{4 \pi \varepsilon_0 }\iint\limits_S (cos^2\theta sin^2 \varphi+sin^2\theta sin^2 \varphi+cos^2\varphi) sin\varphi d\theta d\varphi $$
$$ \iint\limits_S E \cdot dS = \frac{q}{4 \pi \varepsilon_0 }\int_0^\pi \int_0^{2\pi} sin\varphi d\theta r d\varphi=\frac{q}{\varepsilon_0} $$

$$\iiint_V \nabla \cdot E dxdydz=\iint\limits_S E \cdot dS = \frac{q}{\varepsilon_0}$$

As for divergence of E:

$$E=\frac{q}{4 \pi \varepsilon_0 r^2 }[\frac{x}{r}\textbf{i}+\frac{y}{r}\textbf{j}+\frac{z}{r}\textbf{k}]$$
after some calculations:
$$\nabla \cdot E=\frac{q}{4 \pi \varepsilon_0}\frac{3r^2-3r^2}{r^5}$$

Which is 0 except in origo

From problem 2.7 and problem 2.8 in Griffiths introduction to electrodynamics we obtain that the E field inside a charged sphere is pointing radially outwards and is given as

$$E=\frac{Q\textbf{r}}{4\pi\varepsilon_0 R^3}$$

by using unit vector and obtaining the divergence inside the charged sphere one obtains:

$$\nabla \cdot E=\nabla \cdot \frac{Qr}{4\pi\varepsilon_0 R^3}[\frac{x}{r}\textbf{i}+\frac{y}{r}\textbf{j}+\frac{z}{r}\textbf{k}]=3\frac{Q}{4\pi\varepsilon_0 R^3}=\frac{\rho}{\varepsilon_0}$$

Introducing the electric potential

$$E=\frac{q}{4 \pi \varepsilon_0 r^2 }[\frac{x}{r}\textbf{i}+\frac{y}{r}\textbf{j}+\frac{z}{r}\textbf{k}]=-\frac{q}{4 \pi \varepsilon_0} \nabla \frac{1}{r}=-\nabla V$$

It can easily be obtained that

$$\frac{\textbf{r}}{r^3}=-\nabla \frac{1}{r}$$

From that we obtain

$$\nabla \cdot E=-\frac{q}{4 \pi \varepsilon_0} \nabla^2 \frac{1}{r} $$

From results above we obtain that for a point charge in origo we have that

$$\nabla^2 \frac{1}{r}=-4 \pi $$

And inside our charged sphere

$$\nabla^2 \frac{1}{r}=-\frac{4 \pi}{V} $$

Above V is the volume of the charged sphere.
from example 5.11 of Griffiths introduction to electrodynamics 4th edition. And problem 5.29 of Griffiths introduction to electrodynamics third edition one obtains the B field inside a rotating charged sphere. By taking the curl of that one obtains that the curl of the B field inside a rotating charged sphere is $\nabla \times B=\mu_0 J$

If we instead view the charge as a point charge

The magnetic vector potential gives that B field is the curl of the magnetic vector potential so that:

$$B(r)=\frac{\mu _0}{4\pi}\nabla \times \int \frac{ J(r')}{|r-r'|} dV'$$

$$\nabla \times B(r)=\frac{\mu _0}{4\pi}\nabla \times \nabla \times \int \frac{ J(r')}{|r-r'|} dV'$$

We use the identity:

$$\nabla \times (\nabla \times B)=\nabla(\nabla \cdot B)- \nabla^2 B$$

$$\nabla \times B(r)=\frac{\mu _0}{4\pi}\nabla \nabla \cdot \int \frac{ J(r')}{|r-r'|} dV' - \frac{\mu _0}{4\pi} \nabla^2 \int \frac{ J(r')}{|r-r'|} dV'$$

Since it is the current that is the vector we can rewrite the first part when we also use Leibniz integral rule to get the divergence inside the integral:

$$\nabla \times B(r)=\frac{\mu _0}{4\pi}\nabla \int J(r') \cdot \nabla \frac{ 1}{|r-r'|} dV' - \frac{\mu _0}{4\pi} \nabla^2 \int \frac{ J(r')}{|r-r'|} dV'$$
Since we assume that the current density is constant

$$\nabla \times B(r)=-\frac{\mu _0}{4\pi}\nabla \int \nabla' \cdot (J(r') \frac{ 1}{|r-r'|}) dV' - \frac{\mu _0}{4\pi} \nabla^2 \int \frac{ J(r')}{|r-r'|} dV'$$

Divergence theorem gives that the first part is 0 since we are integrating over the current density volume and it is steady so that no current density goes out of the volume:

$$\nabla \times B(r)= - \frac{\mu _0}{4\pi} \nabla^2 \int \frac{ J(r')}{|r-r'|} dV'$$

I have added all these derivations so that if someone would want to answer they could use the derivation that made me approach my problem which is:

Now they use that in the point charge:

$$\nabla^2 \frac{1}{|r-r'|}=-4 \pi $$

$$\nabla \times B(r)= \mu _0 \int J(r') dV'$$

How can they do that since $\nabla^2 \frac{1}{|r-r'|}=-4 \pi$ is taken from electric forces from the E field and we are looking at the magnetic field?

 

[robphy]

Note that what you call $$\nabla^2 \frac{1}{|r-r'|}=-4 \pi $$
comes from a mathematical calculation...
it's not the result of a law of physics... but can be applied to (say) the electric field of a point charge.

Look at Griffiths, 4th edition, p. 50, Eq.(1.99)--(1.102), reproduced here:

We are now in a position to resolve the paradox introduced in Sect. 1.5.1.
As you will recall, we found that the divergence of \hat r/r^2 is zero everywhere except at the origin, and yet its integral over any volume containing the origin is a
constant (to wit: 4\pi). These are precisely the defining conditions for the Dirac delta function; evidently
$$\vec\nabla \cdot \left( \frac{\hat r}{r^2}\right) = 4\pi \delta^3(\vec r)\qquad(1.99)$$
More generally,
\def\rvar{\large\mathscr{ r}}<br /> \def\vrvar{\ \ \vec{\!\!\!\!\large\mathscr{ r}}} <br /> \def\hrvar{\ \ \hat{\!\!\!\!\large\mathscr{ r}}}
$$\vec\nabla \cdot \left( \frac{\hrvar}{\rvar^2}\right) = 4\pi \delta^3(\vrvar)\qquad(1.100)$$
where, as always, \vrvar is the separation vector:\ \vrvar \equiv \vec r -\vec r&#039;. Note that differentiation here is with respect to r, while r&#039; is held constant. Incidentally, since
$$\vec\nabla \left( \frac{1}{\rvar}\right) =-\frac{\hrvar}{\rvar^2}\qquad (1.101)$$
(Prob. 1.13b), it follows that
$$\nabla^2 \left( \frac{1}{\rvar}\right) =- 4\pi \delta^3 (\vrvar) \qquad (1.102)$$
...all without using the Electric Field.

This (Eq. (1.100)) is applied
on p. 71 to compute the divergence of E (see Eq.(2.16)),
and on p.232 to compute the curl of B (see Eq.(5.53)).

 

[vanhees71]

That equation is obviously wrong (already dimensionally). Correct is
$$\Delta \frac{1}{|\vec{x}-\vec{x}'|}=-\vec{\nabla} \cdot \frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3} = -4 \pi \delta^{(3)}(\vec{x}-\vec{x}'),$$
where $\delta^{(3)}$ is the Dirac-$\delta$ distribution.

It's also clear that you cannot derive Ampere's circuital law of magnetostatics from electrostatics. It's one of the fundamental Maxwell equations for the static case. What you can however derive from the above Green's function of the Laplace operator is Biot-Savart's Law, using Ampere's law.

The most simple way is to introduce the vector potential, which must exist because beside Ampere's Law also Gauss's Law $\vec{\nabla} \cdot \vec{B}=0$ must hold:
$$\vec{B}=\vec{\nabla} \times \vec{A}.$$
Since $\vec{A}$ is defined only up to a gradient field, you can impose the Coulomb-gauge condition, $\vec{\nabla} \cdot \vec{A}=0$ and then use
$$\vec{\nabla} \times \vec{B} = \vec{\nabla} \times (\vec{\nabla} \times \vec{A}) = \vec{\nabla} (\vec{\nabla} \cdot \vec{A}) - \Delta \vec{A}=-\Delta \vec{A}=\mu_0 \vec{J}.$$
Using the above give Green's function of the Laplacian then indeed yields
$$\vec{A}(\vec{x})=\frac{\mu_0}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d^3} x' \frac{\vec{J}(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
For $\vec{B}$ you get
$$\vec{B}(\vec{x}) = \vec{\nabla} \times \vec{A}(\vec{x}) = -\frac{\mu_0}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d^3} x' \vec{J}(\vec{x}') \times \vec{\nabla}\frac{1}{|\vec{x}-\vec{x}'|}$$
and thus finally
$$\vec{B}(\vec{x})=+\frac{\mu_0}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{J}(\vec{x}') \times (\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|^3}.$$

 

[georg gill]

Since $\vec{A}$ is defined only up to a gradient field, you can impose the Coulomb-gauge condition, $\vec{\nabla} \cdot \vec{A}=0$

I have tried to go through a proof for this:

$$\nabla \cdot \vec{A}=\frac{\mu _0}{4\pi}\nabla \cdot \int \frac{ J(r')}{|r-r'|} dV'$$

again we use

$$ \nabla \cdot \frac{ 1}{|r-r'|}=-\nabla' \cdot \frac{ 1}{|r-r'|}$$

And by similar reasoning as in my first post in the 5th last equation

$$\nabla \cdot \vec{A}=-\frac{\mu _0}{4\pi} \int J(r') \cdot \nabla' \frac{ 1}{|r-r'|} dV'$$

Then they look at the integration as from integration by parts

$$\int_{-\infty}^\infty g \frac{\partial f}{\partial x} dx = [gf]_{-\infty}^\infty - \int_{-\infty}^\infty f \frac{\partial g}{\partial x} dx$$

initially the integral is over the current which does not go from $-\infty$ to $\infty$. I could perhaps reason and say that integrating from $-\infty$ to $\infty$ would get the same result. But if we extend the reasoning I did get to

For the $-\infty$ to $\infty$ we did get:

$$\nabla \cdot \vec{A}=-\frac{\mu _0}{4\pi} \int_{-\infty}^\infty J(r') \cdot \nabla' \frac{1 }{|r-r'|} dV'$$$$ J(r') \cdot \nabla' \frac{1 }{|r-r'|}=J(r') \cdot[\frac{\partial}{\partial x'}\textbf{i}+\frac{\partial}{\partial y'}\textbf{j}+\frac{\partial}{\partial z'}\textbf{k}] \frac{1}{[(x-x')^2+(y-y')^2+(z-z')^2]^{0.5}}$$

for the first component:
$$ g=J_x(r')$$ $$f=\frac{1}{[(x-x')^2+(y-y')^2+(z-z')^2]^{0.5}}$$

If we would have kept the integration limits from dV' it would be apparent that
$$ [gf] \neq0$$

If we increase to from $-\infty$ to $\infty$ it would be apparent that

$$\int_{-\infty}^\infty g \frac{\partial f}{\partial x} dx $$

and
$$- \int_{-\infty}^\infty f \frac{\partial g}{\partial x} dx$$
would not change after the increase in integration limits to $-\infty$ to $\infty$

but with from $-\infty$ to $\infty$ we would get

$$ [gf]_{-\infty}^\infty =0$$

which is a change? How is this possible?

 

[vanhees71]

You can of course also use
$$\vec{\nabla}' \left (\frac{\vec{J}(\vec{x}')}{|\vec{x}-\vec{x}'|} \right) = \frac{1}{|\vec{x}-\vec{x}'|} \vec{\nabla}' \cdot \vec{J}(\vec{x}') + \vec{J}(\vec{x}') \vec{\nabla}' \frac{1}{|\vec{x}-\vec{x}'|}.$$
Since now necessarily
$$\vec{\nabla} \cdot \vec{J}=0$$
you have
$$\vec{\nabla}' \left (\frac{\vec{J}(\vec{x}')}{|\vec{x}-\vec{x}'|} \right) = \vec{J}(\vec{x}') \vec{\nabla}' \frac{1}{|\vec{x}-\vec{x}'|}.$$
and you can use Gauss's theorem to get the final result that $\vec{\nabla} \cdot \vec{A}=0$, i.e., the integral indeed gives the solution in the Coulomb gauge.

 

[georg gill]

Sorry I have updated my last post it was not what I wanted to ask that was in the initial post. Please look at my last post above if interested.

and you can use Gauss's theorem to get the final result that $\vec{\nabla} \cdot \vec{A}=0$, i.e., the integral indeed gives the solution in the Coulomb gauge.

The magnetic vector potential is 0 when $r'=r$ in the denumerator. Is not that the same issue as in:

$$ \frac{\mu _0}{4\pi} \nabla^2 \int \frac{ J(r')}{|r-r'|} dV'$$

So how can you use divergence theorem on the first part

$$\frac{\mu _0}{4\pi}\nabla \int \nabla' \cdot (J(r') \frac{ 1}{|r-r'|}) dV' $$

when it is not defined when $r'=r$?

 

[vanhees71]

Of course there's a singularity at $\vec{r}=\vec{r}'$, and this singularity must be there, because it's the Green's function of the Laplacian in the sense that
$$-\Delta \frac{1}{|\vec{r}-\vec{r}'|}=-\Delta' \frac{1}{|\vec{r}-\vec{r}'|} = 4 \pi \delta^{(3)}(\vec{r}-\vec{r}').$$
To get the "singularity" of the $\delta$ distribution there must be a singularity in the Green's function too.