homework and exercises - Angular momentum of the electric field of a point-like electric charge and the magnetic field of a monopole

I am currently reading "Magnetic Monopoles" of Ya. Shnir. My problem is I can not retrieve a result the author provides in the first chapter of the first part. In this chapter, he studies the non-relativistic scattering of an electric charge on a magnetic one.

The author writes [p.5, near eq. (1.13)]:

... the appearance of an additional term in the definition of the angular momentum $(1.11)$ originates from a non-trivial field contribution. Indeed, since a static monopole is placed at the origin, its magnetic field is given by $(1.1)$. Then the classical angular momentum of the electric field of a point-like electric charge, whose position is defined by its radius vector $\mathbf{r}$, and the magnetic field of a monopole is a volume integral involving the Poynting vector

$$\begin{align} \tilde{\mathbf{L}}_{eg} &= \dfrac{1}{4\pi}\int \mathbf{r'} \times \left [ \mathbf{E} \times \mathbf{B}\right] d^3r'\tag{L.1}\\& = - \dfrac{g}{4\pi} \int d^3r' \left( \mathbf{\nabla}'\cdot \mathbf{E}\right) \hat{\bf r}' \tag{L.2}\\ &= -eg\hat{\bf r} \tag{L.3} \end{align}$$

where we perform the integration by parts, take into account that the fields vanish asymptotically and invoke the Maxwell equation

$$\begin{equation}\left(\mathbf{\nabla}' . \mathbf{E} \right) = 4 \pi e \delta^{(3)}\left( \mathbf{r} - \mathbf{r}'\right)\end{equation}$$ ...

The magnetic field is

$\mathbf{B} = \dfrac{g}{r^3} \mathbf{r} \tag{1.1}$

The generalised angular momentum is

$\mathbf{L} = \mathbf{r} \times m\mathbf{v} - eg \hat{\bf r} \tag{1.11}$

The author gives how he got $(L.2)$ from $(L.1)$ but I do not know how to do? Have you any idea?

Answer

Answer expected by following author's hints.

$$\begin{align} \mathbf{L}_{eg} &= \dfrac{1}{4\pi}\int \mathbf{r'} \times \left [\mathbf{E} \times \mathbf{B} \right] d^3r'\\ & = \dfrac{1}{4\pi} \int \left [ \left (\mathbf{B.r'} \right)\mathbf{E} - \left (\mathbf{E . r'} \right) \mathbf{B} \right] d^3r'\\ & = \dfrac{1}{4\pi} \int \left [ \left (\dfrac{g}{r'^3}\mathbf{r'.r'} \right)\mathbf{E} - \left (\mathbf{E . r'} \right)\dfrac{g}{r'^3} \mathbf{r'} \right] d^3r'\\ & = \dfrac{g}{4\pi} \int \dfrac{1}{r'} \left [\mathbf{E} - \left (\mathbf{E . \hat{r}'} \right) \mathbf{\hat{r}'} \right] d^3r' \\ & = \dfrac{g}{4\pi} \int \left[ \mathbf{E.\nabla'}\right]\mathbf{\hat{r}'} d^3r'. \end{align}$$

Or, let $\mathbf{U}$ and $\mathbf{v}$ be arbitrary vectors:

$$\begin{equation} \left [ \mathbf{U.\nabla}\right]\mathbf{v} = \left[\mathbf{U.\nabla}v^i\right]\mathbf{e}_i, \end{equation}$$ where $(\mathbf{e}_i)_{1\leq i \leq 3}$ denotes the cartesian basis.

By integrating by parts we have :

$$\begin{align} \mathbf{L}_{eg} & = \dfrac{g}{4\pi} \int \left[ \mathbf{E.\nabla'}\right]\mathbf{\hat{r}'} d^3r'\\ & = \dfrac{g}{4\pi} \int \mathbf{E.\nabla'}(\hat{r}'^i) d^3r' \mathbf{e}_i \\ & = \dfrac{g}{4\pi} \int \left [\mathbf{\nabla' .} \left(\mathbf{E}\hat{r}'^i \right) \mathbf{e}_i - \left (\mathbf{\nabla' . E} \right)\mathbf{\hat{r}}'\right]d^3r'\\ & = \dfrac{g}{4\pi}\ \left[ \oint \mathbf{\hat{r}'} \left (\mathbf{E.da}\right) - \int \left ( \mathbf{\nabla' . E}\right) \mathbf{\hat{r}'} d^3r' \right]. \end{align}$$ But the field $\mathbf{E}$ vanishes at infinty so it comes : $$\begin{equation} \mathbf{L}_{eg} = -\dfrac{g}{4\pi} \int \left ( \mathbf{\nabla' . E}\right) \mathbf{\hat{r}'} d^3r'. \end{equation}$$ And finally, using the Maxwell equation :$\mathbf{\nabla'.E} = 4\pi e \delta^{(3)}(\mathbf{r} - \mathbf{r'})$, we get the result : $$\begin{equation} \mathbf{L}_{eg} = -eg\mathbf{\hat{r}}. \end{equation}$$