- #1
Krikri
- 9
- 0
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
[tex] \textbf{A}_{\omega}(x) = -ik \frac{e^{ikr}}{r} \textbf{P}_{\omega} [/tex]
with [tex]\textbf{P}_{\omega} = \int{d^{3}x^{\prime} \textbf{x}^{\prime} \rho_{\omega} (\textbf{x}^{\prime})} [/tex] the electric dipole moment. The fields are the calculated from
[tex]\textbf{B}_{\omega}(x) = \nabla\times\textbf{A}_{\omega}(x) [/tex] and [tex]\textbf{E}_{\omega}(x)= \frac{i}{k}\nabla\times\textbf{B}_{\omega}(x)[/tex]
So first I tried to computed the B fiels and I get to an expression like
[tex]\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][/tex]
and it seems that [tex]\nabla\times\textbf{P}_{\omega} =0 [/tex] 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 [tex]\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][/tex]
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
[tex] \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} [/tex]
Is this thing zero and why?
[tex] \textbf{A}_{\omega}(x) = -ik \frac{e^{ikr}}{r} \textbf{P}_{\omega} [/tex]
with [tex]\textbf{P}_{\omega} = \int{d^{3}x^{\prime} \textbf{x}^{\prime} \rho_{\omega} (\textbf{x}^{\prime})} [/tex] the electric dipole moment. The fields are the calculated from
[tex]\textbf{B}_{\omega}(x) = \nabla\times\textbf{A}_{\omega}(x) [/tex] and [tex]\textbf{E}_{\omega}(x)= \frac{i}{k}\nabla\times\textbf{B}_{\omega}(x)[/tex]
So first I tried to computed the B fiels and I get to an expression like
[tex]\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][/tex]
and it seems that [tex]\nabla\times\textbf{P}_{\omega} =0 [/tex] 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 [tex]\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][/tex]
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
[tex] \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} [/tex]
Is this thing zero and why?