Deriving the Curl of the Magnetic Field, Role of the Nabla Operator

[deuteron]

TL;DR Summary

.

We know that the magnetic field can be written in the following way:

$$\nabla_{\vec r}\times \vec B(\vec r) =\frac 1 c \nabla_{\vec r} \times\int d^3\vec r_q\ \vec j(\vec r_q) \times \frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3}$$

and, using the $BAC-CAB$ identity, the curl of this becomes:
$$=\frac 1 c \int d^3\vec r_q\ \vec j(\vec r_q) (\nabla_{\vec r}\cdot \frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3} ) \ -\ \frac 1c \int d^3r_q\ \frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3} ( \nabla_{\vec r} \cdot \vec j(\vec r_q) ) $$

With the property, $\nabla_{\vec r} \cdot \frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3}= 4\pi\delta^3(\vec r-\vec r_q)$, we can write this as:

$$= \frac 1 c \int d^3\vec r_q\ \vec j(\vec r_q)\ 4\pi\ \delta^3(\vec r-\vec r_q) \ -\ \frac 1c \int d^3\vec r_q\ \frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3} (\nabla_{\vec r}\cdot \vec j(\vec r_q))$$

from here, what is usually done is to obsevre the second integral, and rewrite the integrand as:

$$(\nabla_{\vec r}\cdot \vec j(\vec r_q))\frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3}= (\vec j(\vec r_q) \cdot \nabla_{\vec r}) \frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3} = -(\vec j(\vec r_q) \cdot \nabla_{\vec r_q} ) \frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3}$$

and from here, using partial integration and continuity equation for magnetostatics, one get $0$ for the integrand, and thus the integral

What I would have done is to say, since $\nabla_{\vec r}$ is a differentiation with respect to $\vec r$, applying it to $\vec j(\vec r_q)$ would give $0$ since $\vec j$ is not dependent on $\vec r$.

I understand that if this were true, then people wouldn't get in the trouble to come to $0$ with the above steps, but what I don't understand is the reason why my explanation is false. How can we say that $\nabla_{\vec r}$ acts on $\frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3}$, when it can also be interpreted as acting on $\vec j(\vec r_q)$?

 

[TSny]

using the $BAC-CAB$ identity, . . .

The "BAC - CAB" identity is not valid if A is $\nabla$. See the second identity here.

 

[kuruman]

The "BAC - CAB" identity is not valid if A is $\nabla$. See the second identity here.

One can use BAC-CAB if one realizes that for each of the two terms in it there are two more because of the product rule of differentiation. To keep the terms sorted out, one would first split the del operator into two, one operating on A alone and one on B alone. Then use BAC-CAB treating everything as vectors but making sure that the del operator operates on the right item on its right. Here is what I mean.

$\vec{\nabla}\times(\vec A\times \vec B)= \vec{\nabla_A}\times(\vec A\times \vec B)+\vec{\nabla_B}\times(\vec A\times \vec B)$
Work on the first term and apply BAC-CAB
$\vec{\nabla_A}\times(\vec A\times \vec B)\rightarrow \vec A(\vec{\nabla_A}\cdot \vec B)-\vec B(\vec{\nabla_A}\cdot \vec A)$
The second term above is fine but the first term is not. To fix it, move $\vec A$ past the scalar to the other side of the parentheses and then flip the order in the dot product of the two vectors ($\vec{\nabla_A}$ commutes with $\vec B$) to get
$\vec{\nabla_A}\times(\vec A\times \vec B)\rightarrow (\vec B\cdot \vec{\nabla_A} )\vec A -\vec B(\vec{\nabla_A}\cdot \vec A).$

Similarly,
$\vec{\nabla_B}\times(\vec A\times \vec B)\rightarrow \vec A(\vec{\nabla_B}\cdot \vec B)-\vec B(\vec{\nabla_B}\cdot \vec A)\rightarrow\vec A(\vec{\nabla_B}\cdot \vec B)-(\vec A\cdot \vec{\nabla_B})\vec B.$

Now we can drop the subscripts and put the two expressions together to get
$\vec{\nabla}\times(\vec A\times \vec B)=(\vec B\cdot \vec{\nabla} )\vec A -\vec B(\vec{\nabla}\cdot \vec A)+\vec A(\vec{\nabla}\cdot \vec B)-(\vec A\cdot \vec{\nabla})\vec B.$

Of course, this is not a proof of the identity, just a way to reconstruct it if one doesn't have internet access.

 

[Meir Achuz]

TL;DR Summary: .

We know that the magnetic field can be written in the following way:

$$\nabla_{\vec r}\times \vec B(\vec r) =\frac 1 c \nabla_{\vec r} \times\int d^3\vec r_q\ \vec j(\vec r_q) \times \frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3}$$

and, using the $BAC-CAB$ identity, the curl of this becomes:
$$=\frac 1 c \int d^3\vec r_q\ \vec j(\vec r_q) (\nabla_{\vec r}\cdot \frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3} ) \ -\ \frac 1c \int d^3r_q\ \frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3} ( \nabla_{\vec r} \cdot \vec j(\vec r_q) ) $$

With the property, $\nabla_{\vec r} \cdot \frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3}= 4\pi\delta^3(\vec r-\vec r_q)$, we can write this as:

$$= \frac 1 c \int d^3\vec r_q\ \vec j(\vec r_q)\ 4\pi\ \delta^3(\vec r-\vec r_q) \ -\ \frac 1c \int d^3\vec r_q\ \frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3} (\nabla_{\vec r}\cdot \vec j(\vec r_q))$$

from here, what is usually done is to obsevre the second integral, and rewrite the integrand as:

$$(\nabla_{\vec r}\cdot \vec j(\vec r_q))\frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3}= (\vec j(\vec r_q) \cdot \nabla_{\vec r}) \frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3} = -(\vec j(\vec r_q) \cdot \nabla_{\vec r_q} ) \frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3}$$

and from here, using partial integration and continuity equation for magnetostatics, one get $0$ for the integrand, and thus the integral

What I would have done is to say, since $\nabla_{\vec r}$ is a differentiation with respect to $\vec r$, applying it to $\vec j(\vec r_q)$ would give $0$ since $\vec j$ is not dependent on $\vec r$.

I understand that if this were true, then people wouldn't get in the trouble to come to $0$ with the above steps, but what I don't understand is the reason why my explanation is false. How can we say that $\nabla_{\vec r}$ acts on $\frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3}$, when it can also be interpreted as acting on $\vec j(\vec r_q)$?

$\nabla_{\vec r}$ acts ONLY on $\vec r$.

 

[deuteron]

One can use BAC-CAB if one realizes that for each of the two terms in it there are two more because of the product rule of differentiation. To keep the terms sorted out, one would first split the del operator into two, one operating on A alone and one on B alone. Then use BAC-CAB treating everything as vectors but making sure that the del operator operates on the right item on its right. Here is what I mean.

$\vec{\nabla}\times(\vec A\times \vec B)= \vec{\nabla_A}\times(\vec A\times \vec B)+\vec{\nabla_B}\times(\vec A\times \vec B)$
Work on the first term and apply BAC-CAB
$\vec{\nabla_A}\times(\vec A\times \vec B)\rightarrow \vec A(\vec{\nabla_A}\cdot \vec B)-\vec B(\vec{\nabla_A}\cdot \vec A)$
The second term above is fine but the first term is not. To fix it, move $\vec A$ past the scalar to the other side of the parentheses and then flip the order in the dot product of the two vectors ($\vec{\nabla_A}$ commutes with $\vec B$) to get
$\vec{\nabla_A}\times(\vec A\times \vec B)\rightarrow (\vec B\cdot \vec{\nabla_A} )\vec A -\vec B(\vec{\nabla_A}\cdot \vec A).$

Similarly,
$\vec{\nabla_B}\times(\vec A\times \vec B)\rightarrow \vec A(\vec{\nabla_B}\cdot \vec B)-\vec B(\vec{\nabla_B}\cdot \vec A)\rightarrow\vec A(\vec{\nabla_B}\cdot \vec B)-(\vec A\cdot \vec{\nabla_B})\vec B.$

Now we can drop the subscripts and put the two expressions together to get
$\vec{\nabla}\times(\vec A\times \vec B)=(\vec B\cdot \vec{\nabla} )\vec A -\vec B(\vec{\nabla}\cdot \vec A)+\vec A(\vec{\nabla}\cdot \vec B)-(\vec A\cdot \vec{\nabla})\vec B.$

Of course, this is not a proof of the identity, just a way to reconstruct it if one doesn't have internet access.

Thank you very much! That was very helpful! But I still didn't understand these:
Why do we say that $(\vec B\cdot \vec{\nabla_A} )\vec A$ this acts on $\vec A$, when the dot product is supposed to be executed first?
And why are we now allowed to say here $\vec A(\vec{\nabla_A}\cdot \vec B)$ that $\vec{\nabla_A}\cdot \vec B$ is zero, since we say it, e.g. for differentiating $\rho(\vec r)$ with respect to $t$, since $\rho$ is not dependent on $t$? What is the difference between these two situations?

 

[kuruman]

Why do we say that $(\vec B\cdot \vec{\nabla_A} )\vec A$ this acts on $\vec A$, when the dot product is supposed to be executed first?

There is no conflict here. By "acts on $\vec A$" I mean
$(\vec B\cdot \vec{\nabla_A} )\vec A=(B_x\frac{\partial}{\partial x}+B_y\frac{\partial}{\partial y}+B_z\frac{\partial}{\partial z})\vec A$
$(\vec B\cdot \vec{\nabla_A} )\vec B=0.$
In other words, as far as $\vec{\nabla}_A$ is concerned, $\vec B$ is a constant. That's how the product rule of differentiation is sorted out.

What is the difference between these two situations?

No difference. It's just a way to say the same thing. I wanted to show how one can get the identity that @TSny suggested that you use in order to answer this question using the BAC-CAB rule.

 

[deuteron]

There is no conflict here. By "acts on $\vec A$" I mean
$(\vec B\cdot \vec{\nabla_A} )\vec A=(B_x\frac{\partial}{\partial x}+B_y\frac{\partial}{\partial y}+B_z\frac{\partial}{\partial z})\vec A$
$(\vec B\cdot \vec{\nabla_A} )\vec B=0.$
In other words, as far as $\vec{\nabla}_A$ is concerned, $\vec B$ is a constant. That's how the product rule of differentiation is sorted out.

No difference. It's just a way to say the same thing. I wanted to show how one can get the identity that @TSny suggested that you use in order to answer this question using the BAC-CAB rule.

But why do we not say $\vec A (\nabla_A\cdot\vec B)=0$, since here $\nabla_A$ acts on $\vec B$, and sees it as a constant, what motivates us to do all the other algebra to reform the equation such that $\nabla_A$ acts on $\vec A$?

 

[kuruman]

What motivates us is this. You have essentially
$$\nabla_{\vec r}\times \left[ \vec j(\vec r_q) \times \frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3}\right]$$and you want to apply the BAC-CAB rule correctly. If you do that strictly by the rule, you get $$\vec j(\vec r_q)\left(\nabla_{\vec r}\cdot \frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3}\right)-\frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3}\left[\nabla_{\vec r}\cdot\vec j(\vec r_q)\right]$$which happens to give you the answer that you want in this case. In general, the curl of the triple cross product has four terms which reduce to two if one of the vectors is constant. If you want to play it safe, you use the identity to write out the four terms and then eliminate the two that are zero. The motivation is avoidance of possible error.

 

[TSny]

The correct identity is $$\vec{\nabla}_{\vec r}\times(\vec A\times \vec B)=(\vec B\cdot \vec{\nabla}_{\vec r} )\vec A -\vec B(\vec{\nabla}_{\vec r}\cdot \vec A)+\vec A(\vec{\nabla}_{\vec r}\cdot \vec B)-(\vec A\cdot \vec{\nabla}_{\vec r})\vec B.$$

For our situation, $\vec A = \vec j(\vec r_q)$ is a function of $\vec r_q$ while $\vec B = \frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3}$ is a function of $\vec r$ and $\vec r_q$. So the first two terms are zero in the identity above. Thus, we get for the curl of the magnetic field:
$$\frac 1 c \int d^3\vec r_q\ \vec j(\vec r_q) (\nabla_{\vec r}\cdot \frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3} ) \ -\ \frac 1c \int d^3r_q\ \left(\vec j(\vec r_q)\cdot \nabla_{\vec r} \right) \frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3}$$
The last integral equals zero using the steps that you outlined in the first post where $\nabla_{\vec r}$ acting on $\frac {\vec r-\vec r_q}{|\vec r-\vec r_q|^3}$ can be replaced by $-\nabla_{\vec r_q} \,$, then integration by parts, and finally use of the continuity equation.