Why is curl of current density $\nabla \times \vec{J}$ equal zero?

Question

I am revisiting the derivation for $\nabla \cdot \vec{B} = 0$ in magnetostatics for the field $\vec{B}(\vec{r})$ of a charge $q$ at position $\vec{0}$ with velocity $\vec{v}$. It proceeds like

\begin{align} \nabla \cdot \vec{B} &= \nabla \cdot \frac{\mu_0 q}{4\pi} \frac{\vec{v} \times \vec{r}}{r^3} \propto \nabla \cdot \frac{\vec{v} \times \vec{r}}{r^3} = \frac{\vec{r}}{r^3} \cdot \underbrace{(\nabla \times \vec{v})}_{=0} - \vec{v} \cdot \left( \nabla \times \frac{\vec{r}}{r^3}\right) \\ &= \vec{v} \cdot \left( \nabla \times \left(-\frac{\vec{r}}{r^3}\right) \right) = \vec{v} \cdot \left( \nabla \times \nabla \frac{1}{r} \right) = \vec{v} \cdot \vec{0} = 0 \end{align}

So far so good. The problem I have is with the step $\nabla \times \vec{v} = \vec{0}$, i.e. $\nabla \times \vec{J} = \vec{0}$. My main text discards the respective term without any comment and another derivation I looked up says this is obvious. Why does this hold? And is it really obvious? After all there is a phenomenon called circular eddy currents.

Answer 1

I think the better way to derive this is to first observe the Biot-Savart law, $$ \mathbf B(\mathbf r)=\frac{\mu_0}{4\pi}\int\mathbf J(\mathbf r')\times\frac{\hat{r}}{r^2}\,\mathrm dV'\tag{1} $$ Since $$ \frac{\hat r}{r^2}=-\nabla_r\left(\frac1r\right) $$ (your text may derive this, if not you can prove it by starting with the RHS), we can write (1) as $$ \mathbf B(\mathbf r)=-\frac{\mu_0}{4\pi}\int\mathbf J(\mathbf r')\times\nabla_r\left(\frac{1}{r}\right)\,\mathrm dV'\tag{2} $$ Since $\mathbf J$ is a function or $r'$ and not $r$, we can put it inside the parenthesis and swap the order of the cross product (i.e., $\mathbf J\times\nabla=-\nabla\times\mathbf J$), $$ \mathbf B(\mathbf r)=\frac{\mu_0}{4\pi}\int\nabla_r\times\frac{\mathbf J(r')}{r}\,\mathrm dV'\tag{3}=\nabla_r\times\frac{\mu_0}{4\pi}\int\frac{\mathbf J(r')}{r}\,\mathrm dV' $$ Then we can define the vector potential as $$ \mathbf A(\mathbf r)=\frac{\mu_0}{4\pi}\int\frac{\mathbf J(r')}{r}\,\mathrm dV' $$ To get $$ \mathbf B(\mathbf r)=\nabla\times\mathbf A(\mathbf r)\tag{4} $$ where we drop the subscript $r$ because it's implied that it's over $\mathbf r$.

That proof over, we can take the divergence of (4): $$ \nabla\cdot\mathbf B=\nabla\cdot\nabla\times\mathbf A\equiv0 $$ by the fact that the divergence of every curl is identically zero (worth the effort to prove this).

Answer 2

I also dislike when authors claim things to be obvious. If it's so simple, then why not just write it out.

Anyhow, regarding this specific case. If you go to the definition of the curl you will see that this is a collection of partial derivatives with respect to position.

So to claim that the curl is zero is to claim that the velocity is independent of the particles position, ie. it is assumed that there are no other fields present, be it gravitaional or electrical.

Answer 3

It is the divergence of the B-field and not the actual source. He should have written $\boldsymbol u'$ for the velocity vector.

$\boldsymbol J$ can be defined as curl-free, but in reality there are no such thing as a curl-free current density.

Even on the inside of a current you will find that the current tend to spiral around the axis of the current. Plasma physics is very complex.

Answer 4

Kanos' answer is good. To better understand it, notice the B-S law he mentioned at first $$ \mathbf B(\mathbf r)=\frac{\mu_0}{4\pi}\int\mathbf J(\mathbf r')\times\frac{\hat{r}}{r^2}\,\mathrm dV'\tag{1} $$ The r of B at left is the radius vector from origin, your observation point. But the r in the integral formula should be reckoned as the distance from the source ($r'$) to the position you presently consider. Thus, (according to my teacher's habit) I'd write $$ \mathbf B(\mathbf x)=\frac{\mu_0}{4\pi}\iiint\mathbf J(\mathbf r')\times\frac{\mathbf {r}}{r^3}\,\mathrm dV'\tag{1} $$ where $$ \frac{\mathbf {r}}{r^3} = \frac{\hat {r}}{r^2} = \nabla \left(-\frac {1}{r}\right) $$ Besides, His answer exactly shows the logic when you introduce the vector potential.

Answer 5

This strikes me as, "Right answer, wrong reason." Consider the classic magnetostatics problem that you can solve using Ampere's law, the infinitely long current carrying wire with uniform current density $J$ and radius $R$. Using Ampere's law you'll find that \begin{align} \mathbf{B}(r) &= \left\{\begin{array}{ll} \mu_0\frac{J\pi R^2}{2\pi r} \hat{\phi} & r \ge R \\ \mu_0\frac{J\pi r^2}{2\pi r}\hat{\phi} & r < R. \end{array}\right. \end{align}

Inside the wire the curl of $\mathbf{J}$ is zero, same for outside the wire. At the surface of the wire, though, the curl of $\mathbf{J}$ has a spike (Dirac delta function) in it that you can verify using Stoke's theorem.

The correct answer is that the derivation the book gave is ambiguous. Kanos's answer provides one alternative, but the vector potential is not needed. What you really need is to express the relationship using notation that is a little more detailed, but unambiguous. We split the $r$ from the Biot-Savart law to get two independent variables - one that is an integration variable, the other that is not. $$\mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int \mathbf{J}(\mathbf{r}')\times \frac{\mathbf{r}-\mathbf{r}'}{\left|\mathbf{r}-\mathbf{r}'\right|^3} \operatorname{d}^3 r'.$$ Notice: $\mathbf{J}(\mathbf{r}')$ is not a function of $\mathbf{r}$, so when you try to take any derivatives with respect to any $\mathbf{r}$ coordinate you'll get zero.

So, we get: \begin{align} \nabla\cdot \mathbf{B}(\mathbf{r}) & = \frac{\mu_0}{4\pi} \int \nabla\cdot \left[\mathbf{J}(\mathbf{r}')\times \frac{\mathbf{r}-\mathbf{r}'}{\left|\mathbf{r}-\mathbf{r}'\right|^3} \right] \operatorname{d}^3 r'. \end{align} Now, apply the vector calculus cross product identity, $\nabla\cdot(\mathbf{A}\times\mathbf{B}) = (\nabla\times \mathbf{A})\cdot \mathbf{B} - \mathbf{A}\cdot(\nabla\times\mathbf{B})$ with $\mathbf{A}=\mathbf{J}$ and $\mathbf{B}=\frac{\mathbf{r}-\mathbf{r}'}{\left|\mathbf{r}-\mathbf{r}'\right|^3}$ to get \begin{align} \nabla\cdot \mathbf{B}(\mathbf{r}) & = \frac{\mu_0}{4\pi} \int \left[(\nabla\times\mathbf{J}(\mathbf{r}'))\cdot \frac{\mathbf{r}-\mathbf{r}'}{\left|\mathbf{r}-\mathbf{r}'\right|^3} - \mathbf{J}(\mathbf{r}')\cdot \left(\nabla\times\frac{\mathbf{r}-\mathbf{r}'}{\left|\mathbf{r}-\mathbf{r}'\right|^3} \right)\right] \operatorname{d}^3 r'. \end{align} The first cross product vanishes because $\mathbf{J}(\mathbf{r}')$ is not a function of $\mathbf{r}$ and $\nabla$ is a derivative in the $\mathbf{r}$ coordinates. That the second cross product vanishes can be shown using a little algebra, or some tricks discussed in other answers.

Answer 6

The curl of a vector field $\vec v$ $$\nabla \times \vec v$$ measures the rotational motion of the vector field.

Take your hand extend your thumb and curl your fingers.

If the thumb is the model for the flow of the vector field, then $$\nabla \times \vec v =0.$$

If the curling of your fingers is the model for the flow of the vector field then $$\nabla \times \vec v \neq 0$$

and the measures the rotational motion of the vector field.

Hence the name "curl".