5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The title of the question makes it sound like you think the curl of the current density is always zero, and you want to know why. It's not always zero. It's just zero in this example (because the velocity is assumed constant). $\endgroup$ – Ben Crowell Mar 15 at 15:31
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I love your first sentence. However, you should be very cautious about saying wether $v$ depends on the coordinates or not. In a fluid, for example, it does, depending on how you approach the problem. $\endgroup$ – FGSUZ Mar 15 at 11:59
7
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ @GDumphart: $\hat r$ is the unit vector, some authors choose to bold this, $\hat{\mathbf r}$, but I often neglect that as the hat signifies the unit vector. Since $\hat r=\mathbf r/\vert\vert\mathbf r\vert\vert$, then $\hat r/r^2\equiv\mathbf r/\vert\mathbf r\vert^3$. $\endgroup$ – Kyle Kanos Oct 14 '14 at 15:16
  • 2
    $\begingroup$ I am still suspicious about your initial formula being a valid Biot-Savart without using $(\boldsymbol r-\boldsymbol{r}') / \|\boldsymbol r-\boldsymbol{r}'\|^3$ in the integrand. Your way, one could even pull the $\hat{r}/\boldsymbol r^2$ out of the integral, rendering the distance of a current $\boldsymbol J( \boldsymbol r')$ to the magnetic field at $\boldsymbol r$ irrelevant. That most likely doesn't affect the correctness of all the other steps though. $\endgroup$ – GDumphart Oct 14 '14 at 15:31
  • $\begingroup$ @GDumphart: The wikipedia link (it was there originally, but not active somehow until I just fixed it) shows this is a valid Biot-Savart law, as do all 3 of my E&M texts. You can take $\hat r/r^2$ out of the integral (taking care of minus signs), there is nothing wrong with doing that. $\endgroup$ – Kyle Kanos Oct 14 '14 at 15:38
  • $\begingroup$ This is either a question of either notation or assumption vs. generality. @GDumphart's version is more correct in general. The introductory texts that introduce the Biot-Savart law probably write it that way because they consider it implicit that $\vec{r}$ is actually $\vec{r}-\vec{r}'$, where $r$ is the point at which $\vec{B}(\vec{r})$ is sampled, and $\vec{r}'$ is a point in the integral being integrated over. In my opinion, this inconsistent use of $\vec{r}$ and $\vec{r}'$ muddies their notation and causes confusion. Check out Jackson's E&M. Wikipedia fixed it since your post. $\endgroup$ – jvriesem Oct 29 '17 at 19:10
  • $\begingroup$ It is NOT true to say that $\hat{r}/r^2$ may be taken out of the integral in general. The reason for this is that the actual quantity is $(\vec{r}-\vec{r}')/\|\vec{r}-\vec{r}'\|$, which is most certainly a function of $\vec{r}'$. The case in which you could take it out of the integral is if you were integrating in a circular loop (the typical example application). In this case, you could treat the vector integral as a scalar integral, because $\|\vec{r}-\vec{r}'\|$ and the cross product are fixed. Wikipedia fixed its eqn. since your post. Jackson's E&M has it right and provides discussion. $\endgroup$ – jvriesem Oct 29 '17 at 19:19
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Hi Jaseon, note that by accepting @IamAStudent's edit, you've deleted some of your commentary and changed some of the notation used. $\endgroup$ – Kyle Kanos Mar 15 at 1:42
  • $\begingroup$ Just to clarify, I only supplied the math notation edits. Everything else is by someone else. $\endgroup$ – wcc Mar 15 at 1:43
  • $\begingroup$ @IamAStudent why did you change "$\LaTeX$ed $r$-prime" to "bold text $r$-backtick"? Backtick is not even similar to a prime, and looks ugly in its place. $\endgroup$ – Ruslan Mar 15 at 5:42
  • $\begingroup$ @Ruslan, that's also by OP. Hopefully OP has learned how to typeset mathematical expressions by now! $\endgroup$ – wcc Mar 15 at 6:36
  • 1
    $\begingroup$ @IamAStudent please look at the revision history more closely. The OP changed **r`** to $r'$ at 1:07:41Z, while your edit has been approved at 1:31:42Z and reverts this particular piece. Might have been a collision in the times of editing vs approval though. I think I'll rollback your edit, since with current history it looks as if being destructive. $\endgroup$ – Ruslan Mar 15 at 9:54
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.