We are given the equation

\begin{equation*} \vec{E}=\hat{u}_\theta\ p_0 \sin{(\theta)}\frac{k^2}{4\pi\varepsilon_0r}\sin{(kr-\omega t)}~, \end{equation*}

that describes the far field radiated by linear electric dipole-antenna, and the figure

enter image description here

Just by looking at the equation, we have to explain why the wavefronts of the field are spherical.

The only thing I've come up with until now, is that the angle, $\varphi$, "doesn't exist" in the equation, meaning that the field would basically make a circular wavefront around the origin for every $r$ and $\theta$. The thing that messes a little with me is the term $\sin{(\theta)}$, where the electric field would be $\vec 0$ when $\theta=0$ and $\theta=\pi$. How can I take this into account when explaining why the wavefronts are spherical?

We also have to explain if its possible to to create a spherical wave that is at each point circularly polarized by using two orthogonal dipole antennas. Wouldn't the answer to this be something like that if both antennas create an electric field of equal strength, then by choosing their phase difference to be $\phi=\pm \pi/2$, the wave would be circularly polarized at each point?

Our lecturer doesn't really explain any of these things to us, so I would be really grateful if someone could explain this to me.


Just by looking at the equation, we have to explain why the wavefronts of the field are spherical.

It seems you are thinking too complicated.

A wavefront is defined to be a set of moving positions having all the same phase. See also Wikipedia:Wavefront.

Now, in the given equation $$ \vec{E}=\hat{u}_\theta\ p_0 \sin{(\theta)}\frac{k^2}{4\pi\varepsilon_0r}\sin{(kr-\omega t)} $$ the phase is obviously $(kr-\omega t)$. Setting this phase to a constant, you get $$ kr - \omega t = \text{const}$$

By using $\omega = ck$ (with $c$ being the speed of light) you get $$ r - ct = \text{const}$$ which simply means that the wavefront is a sphere with radius $r$ expanding with speed $c$.

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  • $\begingroup$ Am I then correct in assuming the second part of the question, meaning the circular polarization part? Or is it not possible? $\endgroup$ – Henri Södergård Mar 24 '19 at 12:54

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