Why is the curl of the electric dipole moment equal to zero in the far field?

[Krikri]

Hello.Looking at Jackson's ch 9 on radiation, I am trying to calculate the fields E and B from the potentials in the far field but it is very confusing. Given now the approximation for he vector potential

\textbf{A}_{\omega}(x) = -ik \frac{e^{ikr}}{r} \textbf{P}_{\omega}

with \textbf{P}_{\omega} = \int{d^{3}x^{\prime} \textbf{x}^{\prime} \rho_{\omega} (\textbf{x}^{\prime})} the electric dipole moment. The fields are the calculated from

\textbf{B}_{\omega}(x) = \nabla\times\textbf{A}_{\omega}(x) and \textbf{E}_{\omega}(x)= \frac{i}{k}\nabla\times\textbf{B}_{\omega}(x)
So first I tried to computed the B fiels and I get to an expression like

\textbf{B}_{\omega}(x) = \Big[k^2 \frac{e^{ikr}}{r} + ik\frac{e^{ikr}}{r^2}\Big](\hat{n}\times\textbf{P}_{\omega} ) + \Big[-ik\frac{e^{ikr}}{r} (\nabla\times\textbf{P}_{\omega})\Big]

and it seems that \nabla\times\textbf{P}_{\omega} =0 but I am not sute why exactly. I think maybe that the dipole is calculated at the prime coordinates and so the curl is with respect to the not prime coordinates so maybe that's why. Also I am confused on how to proceed with the calculations for the E field \textbf{E}_{\omega}(x)= \frac{i}{k} \Big[\big(k^3\frac{e^{ikr}}{r} - k^2\frac{e^ikr}{r^2}\big)[\hat{n}\times(\hat{n}\times\textbf{P}_{\omega})] + k^2\frac{e^{ikr}}{r} \nabla\times(\hat{n}\times\textbf{P}_{\omega})\Big]

I believe the second term must vanish as i suspect from the result but don't why again. Also the dipole moment which direction has?

I tried to investigate the second term using the vector identity relevant here so

\nabla\times(\hat{n}\times\textbf{P}_{\omega}) =\hat{n}(\nabla\cdot\textbf{P}_{\omega}) - \textbf{P}_{\omega}(\nabla\cdot\hat{n}) + (\textbf{P}_{\omega}\cdot\nabla)\hat{n}- (\hat{n}\cdot\nabla)\textbf{P}_{\omega}

Is this thing zero and why?

 

[Charles Link]

All the vector operations on $P_{\omega}$ are zero because $P_{\omega}$ is just a constant. The vector operations on $\hat{n}$ I think need a little consideration, but I think $\hat{n}=1\hat{a_r}$ in spherical coordinates. You can google the operations of curl and divergence in spherical coordinates. I think the curl vanishes, but not the divergence. The $P_{\omega}\cdot \nabla \hat{n}$ term also needs to be carefully evaluated. I think this last term is zero because the "1" in front of $\hat{a_r}$ is a constant. $\\$ Just an additional comment: I think the first equation for $A_{\omega}$ comes from the polarization current formula $J_p=\dot{P}$ along with $A(x)=\int {\frac{J(x')}{c|x-x'|} } \, d^3x'$. (My formula for $A$ may be a steady state formula, so I may need to do some further reading on the subject). $\$ In the $x'$ integral, the sinusoidal dipole is assumed to be localized near the origin.