Radiating quadrupole [closed]

Question

I'm trying to solve the following problem.

Two charges $+q$ are located along the $z$ axis at $z=\pm a \sin \omega t$. Determine the lowest non-vanishing multipole moments, the vector potential and the angular distribution of radiation. Assume that $ka \ll 1$.

It's easy to show that both the electric and magnetic dipole moments vanish. I was also able to derive the quadrupole tensor for this system:

$$ Q_{ij} (t) = \int d^3 x\left(3x_{i}x_{j}-r^{2}\delta_{ij}\right)\rho\left(\mathbf{x} , t\right) = \left(3\delta_{i3}\delta_{j3}-\delta_{ij}\right)qa^{2}\left(1-\cos2\omega t\right) $$

where $\rho\left(\mathbf{x} , t\right)=q\delta\left(\mathbf{x}-a\sin\omega t\hat{\mathbf{z}}\right)+q\delta\left(\mathbf{x}+a\sin\omega t\hat{\mathbf{z}}\right)$.

Clearly the quadrupole moment consists of a static part and an oscillating (harmonic) part

$$ Q_{ij}^{\prime} (t) = qa^{2}\begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -2 \end{pmatrix} \Re\left(e^{-2i\omega t}\right) $$

only the latter contributes to the radiation.

Now according to Eq. (9.38) in "Classical Electrodynamics" by Jackson, the vector potential is proportional to

$$ \int d^3 {x^{\prime}}\mathbf{x}^{\prime}\left(\mathbf{n}\cdot\mathbf{x}^{\prime}\right)\rho\left(\mathbf{x}^{\prime}\right) \tag{1} $$

where $\mathbf{n}=\left\langle \sin\theta\cos\varphi,\sin\theta\sin\varphi,\cos\theta\right\rangle$ is the direction of observation. Jackson also claims that

$$ \int d^3 x^{\prime } \mathbf{x}^{\prime}\left(\mathbf{n}\cdot\mathbf{x}^{\prime}\right)\rho\left(\mathbf{x}^{\prime}\right)=\frac{1}{3}\mathbf{Q}\left(\mathbf{n}\right) \tag{2} $$

where $Q_\alpha = \sum_\beta Q_{\alpha \beta} n_{\beta}$.

I tried calculating the integral both ways (and with/without the time dependence).

Direct calculation: $$ \begin{align*} \int d^{3}x^{\prime}\mathbf{x}^{\prime}\left(\mathbf{n}\cdot\mathbf{x}^{\prime}\right)\rho\left(\mathbf{x}^{\prime}\right) & =\int d^{3}x\begin{pmatrix}x\\ y\\ z \end{pmatrix}\left(x\sin\theta\cos\varphi+y\sin\theta\sin\varphi+z\cos\theta\right)\rho\left(\mathbf{x}\right)\\ & =qa^{2}\left(1-\cos2\omega t\right)\cos\theta\hat{\mathbf{z}} \end{align*} $$

Calculation using $(2)$:

$$ \frac{1}{3}\mathbf{Q}\left(\mathbf{n}\right)=\frac{1}{3} qa^{2}\left(1-\cos2\omega t\right)\begin{pmatrix}-1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & 2 \end{pmatrix}\begin{pmatrix}\sin\theta\cos\varphi\\ \sin\theta\sin\varphi\\ \cos\theta \end{pmatrix}=\frac{1}{3} qa^{2}\left(1-\cos2\omega t\right)\begin{pmatrix}-\sin\theta\cos\varphi\\ -\sin\theta\sin\varphi\\ 2\cos\theta \end{pmatrix} $$

The results are very different - in the first case it seems that the first two diagonal elements vanish whereas in the second case they don't. Where's my mistake?

Answer 1

To sumarize my comments, in general we have \begin{eqnarray} \int \mathbf{x'} (\mathbf{n} \cdot \mathbf{x'}) \rho(\mathbf{x'}) d^3 x' &=& \mathbf{\hat{e}}_\alpha \int x'_\alpha n_\beta x'_\beta \rho(\mathbf{x'}) d^3 x' \\ &=& \frac{1}{3} \mathbf{\hat{e}}_\alpha n_\beta \int 3 x'_\alpha x'_\beta \rho(\mathbf{x'}) d^3 x' \\ &=& \frac{1}{3} \mathbf{\hat{e}}_\alpha n_\beta \int \left[ 3 x'_\alpha x'_\beta -r'^2 \delta_{\alpha \beta} + r'^2 \delta_{\alpha \beta}\right] \rho(\mathbf{x'}) d^3 x' \\ &=& \frac{1}{3} \mathbf{\hat{e}}_\alpha n_\beta Q_{\alpha \beta} + \frac{1}{3} \mathbf{\hat{e}}_\alpha n_\beta \int r'^2 \delta_{\alpha \beta} \rho(\mathbf{x'}) d^3 x' \\ &=& \frac{1}{3} \mathbf{Q}(\mathbf{n}) + \frac{1}{3} \mathbf{n} \int r'^2 \rho(\mathbf{x'}) d^3 x' \, . \end{eqnarray} Hence, both results match up to a component along $\mathbf{n}$. Notice that I didn't assume anything about the charge distribution itself. This second contribution can, in principle, vanish for some charge distributions, but that is not the case for your particular example.