$\nabla \cdot (\mathbf{B}\mathbf{B} - \frac{1}{2}B^2 \tilde{1})=(\nabla \cdot \mathbf{B})\mathbf{B} - \mathbf{B} \times (\nabla \times \mathbf{B})$

Question

Does someone know how to show this identity?

$\nabla \cdot (\mathbf{B}\mathbf{B} - \frac{1}{2}B^2 \tilde{1})=(\nabla \cdot \mathbf{B})\mathbf{B} - \mathbf{B} \times (\nabla \times \mathbf{B})$

Answer 1

I will work in index notation and with the Einstein summation convention.

First, let's write the LHS in index notation and simplify: \begin{align} \partial_i \left(B_i B_j -\frac{1}{2} B_k B_k \delta_{i j}\right) =(\partial_i B_i)B_j +B_i (\partial_i B_j)-B_k( \partial_j B_k). \end{align} Here we have applied the product rule and summed over the delta function. Note that the first term is the $j$th component of $(\nabla\cdot \mathbf{B})\mathbf{B}$. That leaves the $-\mathbf{B}\times \nabla\times \mathbf{B}$ term, which in index notation may be written as

\begin{align} -\epsilon_{jlm}\epsilon_{mkn}B_l \, \partial_k B_{n} &= -(\delta_{jk}\delta_{ln}-\delta_{jn}\delta_{lk})B_l \, \partial_k B_{n} = -B_n(\partial_j B_n) +B_k (\partial_k B_j) \end{align} where we have again eliminated the delta function. Accounting for dummy indices, this is exactly the two remaining terms. Hence both sides agree and the identity is valid.

Answer 2

As an alternative to Semiclassical, the problem can be tackled entirely using the machinery of clifford algebra. Let $a$ be an arbitrary vector, and let $\underline B(a) = \overline B(a) = (B \cdot a) B$. Then we have

$$\nabla \cdot [\underline B(a) - \frac{1}{2} B^2 a] = a \cdot [\overline B(\nabla) - \frac{1}{2} \nabla B^2] = a \cdot [B (\nabla \cdot B) + (B \cdot \nabla) B - \frac{1}{2} \nabla B^2]$$

Expansion of the $\nabla B^2$ term requires a common identity:

$$\nabla (A \cdot B) = (\nabla \wedge B) \cdot A + (\nabla \wedge A) \cdot B + (A \cdot \nabla) B + (B \cdot \nabla) A$$

You can see that, once this is proved, for $A = B$, the required expansion falls out immediately. The hard part, then, is merely the proof of this identity:

This is where clifford algebra becomes useful. We can write

$$\nabla (A \cdot B) = \nabla (AB) - \nabla \cdot (A \wedge B)$$

where $AB$ is understood as using the geometric product. The resulting expansion gives

$$\nabla (AB) = (\nabla \cdot A) B + (\nabla \wedge A) \cdot B + (A \cdot \nabla) B - (A \wedge \nabla) \cdot B$$

and

$$\nabla \cdot (A \wedge B) = (A \cdot \nabla) B + (\nabla \cdot A) B - (B \cdot \nabla) A - (\nabla \cdot B) A$$

We can see some terms canceling already. We need a slightly better expansion of terms like $(A \wedge \nabla) \cdot B$, though. See that

$$\langle A (\nabla B )\rangle_1 = (A \cdot \nabla) B + (A \wedge \nabla) \cdot B =A (\nabla \cdot B) + A \cdot (\nabla \wedge B) $$

Putting this into the $\nabla(AB)$ expansion gives

$$\nabla(AB) = (\nabla \cdot A) B + (\nabla \wedge A) \cdot B + 2 (A \cdot \nabla) B - A (\nabla \cdot B) - A \cdot (\nabla \wedge B)$$

This, combined with the previous expansion for $\nabla \cdot (A \wedge B)$, immediately gives the desired result. In case it's hard to keep track of all the terms, though, the following should be better formatted to convince you of the result:

$$\begin{alignat*}{6} \nabla(AB) &= (\nabla \cdot A) B &- (\nabla \cdot B) A &+ 2(A \cdot \nabla) B &+ 0 &+ (\nabla \wedge A) \cdot B &+ (\nabla \wedge B) \cdot A \\ \nabla \cdot (A \wedge B) &= (\nabla \cdot A) B &- (\nabla \cdot B) A &+ (A \cdot \nabla) B &- (B \cdot \nabla) A & \end{alignat*}$$

The resulting identify for $\nabla (A \cdot B)$ is then proven, and the special case needed for $A = B$ can be taken to prove the result. All of this was proven, without resorting to index notation, thanks to the power of clifford algebra and its associated calculus.