I am looking at an example problem from Greiner's Classical Electrodynamics (chapter 21 , page 441) about the Hertzian Dipole where the radiation will require a cross product ($\vec{d} \times \hat{n}$, where) calculation as is shown below:

enter image description here

So, if this cross product was done in Cartesian coordinates, then we would need the component information of the $\hat{n}$ vector, $(n_x, n_y, n_z)$. Now, both vectors in the cross product, $\vec{d}$ and $\hat{n}$, are on equal footing and we would need to replace each cartesian unit vector with its corresponding linear combination of spherical unit vectors.

My questions:

  1. Why does the text solution only give the $\hat{n}$ vector only a component for $\hat{r}$ and nothing in $\hat{\theta}$ and $\hat{\phi}$?Is it because it assumes those coordinates are zero and automatically aligns the spherical coordinate unit vectors around the $\hat{n}$ vector so that $\theta=\phi=0$?

  2. If I were to do the reverse and define the $\vec{d}$ vector as a vector with only a $\vec{r}$ component and the $\vec{n}$ vector having all three components $(\hat{r}, \hat{\theta}, \hat{\phi})$, would this allowing me to calculate the $\vec{H}$ vector for any orientation of $\hat{n}$?

I really need some elucidation on this to fully understand whats going on! Thanks in advance to anyone willing to help!

  • $\begingroup$ Jeez, who read the proof for that book? The equations for $\hat e_x$ and $\hat e_y$ are wrong, and equation $(21.33)$ has the wrong sign. The formula only needs $\hat e_r$ because that is the location of the field point. $\endgroup$ Commented Jul 17, 2018 at 1:59
  • $\begingroup$ @user5713492 Really? Because (plaza.obu.edu/corneliusk/mp/suv.pdf) concurs. $\endgroup$ Commented Jul 17, 2018 at 4:13
  • $\begingroup$ A first note : usually we define the azimuthal angle $\:\phi\:$ with respect to the $\:x-$axis. But Greiner defines the azimuthal angle $\:\phi\:$ with respect to the $\:y-$axis as shown in the Figure 21.9. In the link you gave we have the former definition instead of the latter (Greiner's). That's the reason to disagree with @user5713492 about signs. $\endgroup$
    – Frobenius
    Commented Jul 17, 2018 at 5:21
  • $\begingroup$ @Frobenius No, careful analysis shows that Greiner's Figure $21.9$ is in error. I'll have to write out an answer to change his diapers. $\endgroup$ Commented Jul 17, 2018 at 7:43
  • $\begingroup$ @user5713492 : Ok. I post my note after a first glance. Please make your careful analysis to correct the error and help the OP to realize what is going wrong with Greiner's. Oh, now I saw your answer. $\endgroup$
    – Frobenius
    Commented Jul 17, 2018 at 8:33

1 Answer 1


First off, Greiner's Figure $21.9$ is simply wrong. It implies that $$\vec r=\langle-r\sin\theta\sin\phi,r\sin\theta\cos\phi,r\cos\theta\rangle$$ So $$\begin{align}d\vec r&=\langle-\sin\theta\sin\phi,\sin\theta\cos\phi,\cos\theta\rangle dr\\ &\quad+r\langle-\cos\theta\sin\phi,\cos\theta\cos\phi,-\sin\theta\rangle d\theta\\ &\quad+r\sin\theta\langle-\cos\phi,-\sin\phi,0\rangle d\phi\end{align}$$ From which we may read off $$\begin{align}\hat e_r&=\langle-\sin\theta\sin\phi,\sin\theta\cos\phi,\cos\theta\rangle\\ \hat e_{\theta}&=\langle-\cos\theta\sin\phi,\cos\theta\cos\phi,-\sin\theta\rangle\\ \hat e_{\phi}&=\langle-\cos\phi,-\sin\phi,0\rangle\end{align}$$ And then solve to obtain $$\begin{align}\hat e_x&=-\sin\theta\sin\phi\hat e_r-\cos\theta\sin\phi\hat e_{\theta}-\cos\phi\hat e_{\phi}\\ \hat e_y&=\sin\theta\cos\phi\hat e_r+\cos\theta\cos\phi\hat e_{\theta}-\sin\phi\hat e_{\phi}\\ \hat e_z&=\cos\theta\hat e_r-\sin\theta\hat e_{\theta}\end{align}$$ And that's miles away from what Greiner has. The figure Greiner is actually working from is as you say http://plaza.obu.edu/corneliusk/mp/suv.pdf . From there $$\vec r=\langle r\sin\theta\cos\phi,r\sin\theta\sin\phi,r\cos\theta\rangle$$ And now we get $$\begin{align}d\vec r&=\langle\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta\rangle dr\\ &\quad+r\langle\cos\theta\cos\phi,\cos\theta\sin\phi,-\sin\theta\rangle d\theta\\ &\quad+r\sin\theta\langle-\sin\phi,\cos\phi,0\rangle d\phi\end{align}$$ And then $$\begin{align}\hat e_r&=\langle\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta\rangle\\ \hat e_{\theta}&=\langle\cos\theta\cos\phi,\cos\theta\sin\phi,-\sin\theta\rangle\\ \hat e_{\phi}&=\langle-\sin\phi,\cos\phi,0\rangle\end{align}$$ And solve to get $$\begin{align}\hat e_x&=\sin\theta\cos\phi\hat e_r+\cos\theta\cos\phi\hat e_{\theta}-\sin\phi\hat e_{\phi}\\ \hat e_y&=\sin\theta\sin\phi\hat e_r+\cos\theta\sin\phi\hat e_{\theta}+\cos\phi\hat e_{\phi}\\ \hat e_z&=\cos\theta\hat e_r-\sin\theta\hat e_{\theta}\end{align}$$ You see the difference now, right? Greiner has the trig functions of the wrong argument multiplying $\hat e_{\phi}$, while the *.pdf you linked to has the right stuff. Also we can work out that $$\begin{align}\hat e_r\times\hat e_r&=\vec0\\ \hat e_{\theta}\times\hat e_r&=-\hat e_{\phi}\\ \hat e_{\phi}\times\hat e_r&=\hat e_{\theta}\end{align}$$ And this would lead us to conclude that the overall sign of $\vec H$ in Equation $21.33$ is inconsistent with the previous equation. I think that may be due to forgetting a convention that the charge of the electron is $-e$ earlier on, but then copying the formula into the final result. A very sloppy derivation.

Again, $\vec r=r\hat e_r$ is the vector from the source point to the field point so $\hat e_r$ is the one we need. Where would $\hat e_{\theta}$ or $\hat e_{\phi}$ come in?


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