Applying the divergence to Eq. $47$, we obtain

$$ \mathbf{\nabla} \cdot \mathbf{B} = \frac{\mu_{0}}{4\pi} \int \nabla \cdot \left( \mathbf{J} \times \ \frac{\hat{\mathbf{r}}}{r^2}\right) d\tau^{'}. \tag{48}$$

Invoking the product rule number $6$,

$$\nabla \cdot \left(\mathbf{J} \times \ \frac{\hat{\mathbf{r}}}{r^2}\right)=\frac{\hat{\mathbf{r}}}{r^2}\cdot(\nabla \times \mathbf{J})-\mathbf{J}\cdot\left(\mathbf{\nabla} \times \frac{\hat{\mathbf{r}}}{r^2}\right)\; .\tag{49} $$

But $\nabla \times \mathbf{J}=0$, because $\mathbf{J}$ doesn't depend on the unprimed variables, while $\mathbf{\nabla} \times \frac{\hat {\mathbf{r}}}{r^2}=0$, so

$$\mathbf{\nabla} \cdot \mathbf{B}=0\;.\tag{50}$$

I have the Fourth Edition of Introduction to Electrodynamics by David J. Griffiths, published by Pearson.

I have found this erratum in Chapter 5 Magnetostatics, page 234 .

It's just a normal triple product manipulation, i.e. $$ \vec a \cdot \left(\vec b \times \vec c \right) = \vec b \cdot \left( \vec a \times \vec c \right) = \vec c \left(\vec b \times \vec a\right)$$ Therefore, I think that an "equal to" sign is missing between the expressions on the RHS of Eq. $(49)$.

I searched for errata but couldn't find it.

  • $\begingroup$ This is just the product rule of the derivative. It's not an erratum $\endgroup$ Commented Feb 8, 2020 at 17:08
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    $\begingroup$ Looks fine to me: $\nabla \cdot \left( A \times B \right) = (\nabla \times A) \cdot B - A \cdot ( \nabla \times B)$. $\endgroup$ Commented Feb 8, 2020 at 17:19
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    $\begingroup$ You should revisit chapter 1 where it talks about $\nabla$ not being an actual vector. $\endgroup$ Commented Feb 8, 2020 at 17:25
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    $\begingroup$ This says product rules are in section 1.2.6. $\endgroup$
    – G. Smith
    Commented Feb 8, 2020 at 18:04
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    $\begingroup$ This is not a site for spreading awareness of textbook errata. It is a site for asking questions about physics. If you want to mention the new wrong reference you found, I suggest editing the question where you reported the other wrong reference. Please do not post a third question. $\endgroup$
    – G. Smith
    Commented Feb 9, 2020 at 7:24

1 Answer 1


As others have pointed out, this is not an erratum.

It can sometimes help to skip the fancy products and use a more sophisticated notation. So that is what this answer will do.

Abstract Index Notation

In one, called abstract index notation, we denote vectors with raised indexes, and covectors (linear mappings from vectors to scalars) with lowered indexes. We only add two vectors when the indexes are the same.

The dot product takes two vectors and produces a third, so we write it $g_{ab}$ with two lowered indexes. It is symmetric, so $g_{ab}=g_{ba}.$ Not all tensors are. In particular the cross product as you have noted takes three vectors and makes a scalar out of them, so we write that with three indexes, $$ \epsilon_{abc} u^av^bw^c = \vec u\cdot(\vec v\times\vec w) $$ And it is as you have noticed totally antisymmetric, so it satisfies $$ \epsilon_{abc}=-\epsilon_{bac}=\epsilon_{bca}. $$ When a lowered index is labeled with the same letter as an upper index, that means we are supposed to contract the two indices, where we apply the tensor to the vector to get a smaller tensor. So the indices help us “wire together” the tensors in an expression.

To complete this picture there is also an index relabeling isomorphism, so if you wrote $v^a$ but you meant $v^b$ or so, that we write $v^b = \delta^b_a v^a$, and the dot product also has an inverse which we call $g^{ab}$ satisfying $g^{ab}g_{bc} =\delta^a_c$ when contracted with the dot product. One can use this to “raise” an arbitrary index and the dot product to “lower” one, in any expression.

So for example I gave you the cross product tensor in it's “all-down” or “triple product” form, but the normal cross product that returns a vector, if you prefer that, is $\epsilon^a_{~~bc}=g^{as}\epsilon_{sbc}.$ It maps two vectors wired into its $b,c$ inputs into one vector with an $a$ index.

For one example of how this formalism becomes easier, consider the BAC-CAB identity that $$ \vec A \times (\vec B \times \vec C) = \vec B~(\vec A\cdot\vec C)-\vec C~(\vec A\cdot\vec B). $$ Written in this notation at first it looks painful, that $$ \epsilon^a_{~~bc} ~A^b ~\epsilon^c_{~~de} ~B^d ~C^e = \delta^a_d~B^d~g_{be}~ A^b~ C^e - \delta^a_e ~C^e~g_{bd}~ A^b~ B^d $$ but we can remove these “test vectors” from it and just write, $$ \epsilon^a_{~~bc} ~\epsilon^c_{~~de} = \delta^a_d~g_{be}- \delta^a_e ~g_{bd} $$ or if you want to manipulate even a little further you can massage it down to$$\epsilon^{abc}\epsilon_{cde} =\delta^a_d\delta^b_e-\delta^a_e\delta^b_d.$$

Adding Del into the mix.

For various reasons del is added into this algebra as a covector operator $\nabla_\bullet$. (Basically the reason is that the geometrically minded of us find it very easy to think about the directional derivatives $\vec v\cdot\nabla$ and it saves us some hoops to jump through if we just define that those are what vector fields are and then this operator just tells a vector field to operate on a scalar field.) It derives potentially every term in an expression by the product rule, even if that term has a different index. So $\nabla_a (u^a v^b) = v^b (\nabla_a u^a) + u^a(\nabla_a v^b).$

In vector notation you would have to write for this,$$\nabla\cdot(\vec u \otimes\vec v)=(\nabla\cdot\vec u) \vec v + (\vec u\cdot\nabla) \vec v$$ and it looks okay but a little confusing. As you can see, $\nabla_a v^a$ is the divergence of $\vec v$ while you might imagine that its curl is indeed written $\epsilon^{ab}_{~~~~c}\nabla_b v^c.$

The tensors you have seen $g_{\bullet\bullet},\epsilon_{\bullet\bullet\bullet}, \delta^\bullet_\bullet,$ and $ g^{\bullet\bullet}$ are all to be regarded as being constant over space, so their spatial derivative with $\nabla_\bullet$ is 0.

The advantage is that we can now actually apply our BAC-CAB law to the curl of a curl without losing our heads because we have separated the structure of the cross product from the action of the derivative. So we write out$$ \epsilon^{ab}_{~~~~c}\nabla_b \epsilon^{cd}_{~~~~e}\nabla_d v^e, $$we mess around with raising and lowering some indices,$$ \begin{align} \epsilon^{ab}_{~~~~c}\nabla_b \epsilon^{cd}_{~~~~e}\nabla_d v^e&= \epsilon^{abc}\nabla_b g^{kd} \epsilon_{cke}\nabla_d v^e\\ &=\delta^a_k\delta^b_e g^{kd} \nabla_b \nabla_d v^e - \delta^a_e\delta^b_k g^{kd}\nabla_b\nabla_d v^e\\ &= g^{ad}\nabla_d \nabla_e v^e - g^{bd}\nabla_b\nabla_d v^a,\end{align}$$ or as we would write with vectors,$$\nabla\times(\nabla\times \vec v) = \nabla(\nabla\cdot v)-\nabla^2 \vec v.$$

How this answers your question

Written in this notation we are trying to compute $$\nabla\cdot(\vec J\times\vec R)= \nabla_n g^{an} (\epsilon_{abc} J^b R^c).$$ By the product rule this splits into two terms, $$g^{an} \nabla_n (\epsilon_{abc} J^b R^c) = g^{an} \epsilon_{abc} (\nabla_n J^b) R^c + g^{an} \epsilon_{abc} J^b (\nabla_n R^c) $$ We can rearrange these two terms to look like curls by using antisymmetry to permute the indexes, $$g^{an} \nabla_n (\epsilon_{abc} J^b R^c) = g^{an} R^c \epsilon_{cab} (\nabla_n J^b) - g^{an} J^b \epsilon_{bac} (\nabla_n R^c) $$ And replacing $R^c=R^m \delta^c_m=R^m g^{c\ell}g_{\ell m}$ in the first term and similarly with $J^b$ in the second this becomes $$g^{an} \nabla_n (\epsilon_{abc} J^b R^c) = R^m g_{m\ell} \epsilon^{\ell n}_{~~~~b} \nabla_n J^b - J^m g_{m\ell} \epsilon^{\ell n}_{~~~~c} \nabla_n R^c $$or as we would write with vectors, $$\nabla\cdot(\vec J\times\vec R)= \vec R\cdot (\nabla\times\vec J) - \vec J \cdot(\nabla\times \vec R).$$ As you can see, in some sense this is just the product rule applied to a particular circumstance. Both sides have a dot product and a cross product, so none of those details have changed, it is just that in distributing the derivative over the two terms, each term picked up some of the cross product tensor to become a curl itself.


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