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For a z-directed Hertizan dipole at the origin, the field equations in spherical coordinates are $$E_r\left(r\right)=\frac{Il}{2\pi}e^{-jkr}\left(\frac{\eta}{r^2}+\frac{1}{j\omega \epsilon r^3}\right)\cos\theta$$

$$E_\theta\left(r\right)=\frac{Il}{4\pi}e^{-jkr}\left(\frac{j\omega\mu}{r}+\frac{\eta}{r^2}+\frac{1}{j\omega{\varepsilon r}^3}\right)\sin\theta$$

$$H_\varphi\left(r\right)=\frac{Il}{4\pi}e^{-jkr}\left(\frac{jk}{r}+\frac{1}{r^3}\right)\sin\theta$$

Let's say that we displace this Hertzian dipole from the origin and at the same rotate it pointing in an arbitrary direction. How then these equations will be transformed with respect to (r,θ,φ)? Can you recommend a book on these kind of transformations?

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  • $\begingroup$ You can do the rotation in Cartesian coordinates with a standard rotation matrix, and then convert back to spherical. $\endgroup$
    – CWPP
    Aug 6, 2021 at 14:39
  • $\begingroup$ Ok but the equations are in spherical coordinates so i can't use a cartesian rotation matrix on them. $\endgroup$ Aug 6, 2021 at 15:42

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You could project the spherical unit vectors onto the Cartesian axes to do this. But I think maybe that's the wrong way to look at it. The whole power of vector notation is that it is independent of the co-ordinate system, so we should express these fields as vectors: $\mathbf{E}=E_r\mathbf{\hat r}+E_\theta\mathbf{\hat\theta}$ and $\mathbf{H}=H_\phi\hat\phi$. This dipole as a vector is $\mathbf{l}=l\mathbf{\hat z}$. Then $l\sin\theta\hat\phi$ is $\mathbf{l}\times\mathbf{\hat r}$, $l\cos\theta\mathbf{\hat r}=\mathbf{l}\cdot\mathbf{\hat r}\mathbf{\hat r}$ and $l\sin\theta\hat\theta=\mathbf{l}\cdot\mathbf{\hat r}\mathbf{\hat r}-\mathbf{l}$ (you should check these expressions!) Then these are correct for any direction of $\mathbf l$.

The more difficult part is the translation. Consider the following thought experiment: we represent a very slightly displaced dipole as the original dipole at the origin plus a quadrupole consisting of the displaced dipole and the dipole at the origin subtracted. Obviously this sightly displaced dipole will generate both dipole and quadrupole fields. If we iterate this approach to make up a finite displacement we can see that we will generate all possible multipole fields.

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