# Very short ($\ell \ll \lambda$) dipole as a receiving antenna

I'm trying to understand how a very short dipole of length $\ell \ll \lambda$ works for receiving radiation. (It is center-fed, and has two thin perfectly conducting arms each of length $\ell/2$ separated by a negligible distance.)

Say I have such a dipole oriented along the direction of the spatially uniform incident electric field $\vec{E}\left(t\right)$, and at all times in the region where the antenna is places, we have $$\vec{B}\left(t\right) \perp \vec{E}\left(t\right)$$ $$\left|\vec{B}\left(t\right)\right| = \dfrac{\left|\vec{E}\left(t\right)\right|} {c}$$

What will be the open circuit voltage $V\left(t\right)$ seen across the dipole? Do I need to specify $\vec{E}\left(t\right) = \vec{E}_0 \cos \omega t$ for a meaningful answer, or will a general $\vec{E}\left(t\right)$ do?

My first guess would be $\displaystyle V\left(t\right)=\int_{\left(0,-\ell/2\right)}^{\left(0,+\ell/2\right)} \vec{E}\left(t\right)\cdot d\vec{r} = \ell\;\left|\vec{E}\left(t\right)\right|$, but I have two confusions:

1. The integral above would be the potential difference between the tips of the dipole, but given that the electric and magnetic fields are time varying, I understand that the scalar potential itself becomes meaningless. Is this expression still usable for EMF anyway? Or does the fact that $\ell \ll \lambda$ make the potential difference still usable--and if so, why?

2. Given that the arms of the dipole are perfect conductors, is the potential difference between the two ends of any arm zero? Should I therefore look at the potential difference between the midpoints of the two arms rather than the tips? I will then get $$\int_{\left(0,-\ell/4\right)}^{\left(0,+\ell/4\right)} \vec{E}\left(t\right)\cdot d\vec{r} = \frac{\ell} 2 \;\left|\vec{E}\left(t\right)\right|$$ Or does the radiation resistance of the arms come into play somehow?

Thanks ...

• "the potential difference between the two ends of any arm should be zero". Reconsider that conclusion. Oct 18, 2013 at 11:16
• Well I'm a little confused as to why one would use a dipole antenna, where the electrical length was much shorter than the wavelength. By inference, the antenna is being used way below its design frequency. Typically very short antennas, can be considered to have a very low radiation resistance, so they appear to be simply a capacitance connected in series with the arriving field Voltage, and the free space impedance (377 Ohms) For example, the Loran-C signal at 100 kHz, has a wavelength of 3,000 meters, so practical antennas are all short.
– user26165
Oct 18, 2013 at 22:27
• @GeorgeESmith -- I am not actually trying to use it for any reception. This was just a kind of theoretical exercise to satisfy my own chain of thought. First get the open-circuit voltage, then using the radiation resistance plus an equal load resistance to obtain the maximum power drawn, then dividing it by the Poynting vector to get the effective aperture. Oct 19, 2013 at 3:35
• Well it really doesn't matter what the application is, from a radio telescope, to EM radiation from two atoms in collision; the same principles apply. Efficient dipole antennas are typically one half wavelength long, tip to tip. So if the antenna is very short compared to a half wavelength, it is very inefficient, either to receive or transmit, and it will look like a capacitance, rather than a radiation resistance. If it is longer than a half wave it will start to look inductive.
– user26165
Oct 20, 2013 at 3:25

For your question 1, the Hertzian dipole is assumed to be much shorter than the wavelength, so that around the antenna itself, the situation is essentially electro / magnetostatic.

So the "EMF" is the expression you have given for $V(t)$. Although the situation as you rightly point out is not truly static, one can still interpret your $V(t)$ as an source voltage driving whatever your antenna is connected to. This is much the same as the interpretation of the EMF of a generator that I explain here. The reason the nett potential difference along the arms is nought happens analogously to that answer too. Charges build up to cancel the electric field inside the conductor.

It can be shown that (see here for example, andI recall R. E. Colins, "Antennas and Radiowave Propagation" being an excellent source) that the antenna as a receiver works as voltage source with the $V(t)$ in your formula, and the voltage source has a source impedance equal to the radiation resistance (of the antenna as a sender rather than receiver antenna). Intuitively what this means is that as the antenna receives, the motion of its charges means that it reradiates some of its power as a transmitter antenna. So the maximum power transfer from the antenna to whatever load you connect to it happens when the load is equal to the radiation resistance, when half the incident power is re-radiated.

For receiver antennas, the wonted calculation method makes use of the Reciprocity theorem (Lorentz lemma), which can show (as in the Colins reference I gave) that:

$$G = \frac{4\pi}{\lambda^2} A_{eff}$$

where $G$ is the "gain" of an antenna as a transmitter, $\lambda$ the wavelength and $A_{eff}$ the effective aperture. This formula is altogether general, for ANY antenna. $G$ is calculated for the orientation that your antenna is in relative to the incoming field. It defines the ratio of the transmitted intensity to the intensity that would happen were the antenna isotropic (radiating power equally in all directions). So once you know the gain of the antenna as a sender in the particular orientation it is in, you can find $A_{eff}$. Then you simply calculate the Poynting vector of the incoming field, and multiply the intensity by this $A_{eff}$ to find the power delivered to a matched antenna, i.e. one with a load equal to the antenna's radiation resistance.

So you can use either of the above techniques to find the current through and voltage across your load for each Fourier component of the incoming field. The small antenna approximation means that the Thévenin equivalent voltage for the system is in-phase with the incoming electric field. These observations will let you extend your analysis to an arbitrary $E(t)$.

Here is another reference you might find useful:

http://whites.sdsmt.edu/classes/ee382/notes/382Lecture34.pdf

• so will $V\left(t\right)$ be $\displaystyle \ell\;\left|\vec{E}\left(t\right)\right|$ or will it be $\frac{\ell} 2 \;\left|\vec{E}\left(t\right)\right|$? Thanks... Oct 18, 2013 at 16:20
• @Avijit That will depend on the loading on the ends of the antenna. The Hertzian dipole is wontedly assumed to have capactive loading on the ends so that the current is constant along its length when a transmitter. In this case the expression will be $\ell\,|\vec{E}|$. However, if there is zero capacitance (an idealisation very hard to realise), the charges bunch up along the wires so that you get a current distribution that decreases from a maximum at the feeders to nought at the ends. In this case the right expression as a receiver will be $\ell |\vec{E}|/2$. .... Oct 19, 2013 at 12:41
• @Avijit The receiver's behaviour mirrors that of the transmitter. So you may have to do some experiments to find out the antenna's radiation pattern. You'll probably just have to assume some factor $\alpha |\vec{E}| \ell$ with $\alpha$ between 0.5 and 1. Oct 19, 2013 at 12:43
• I meant the one with capacitive loading. This antenna has a gain of $\frac 3 2$ and a radiation resistance for that antenna is $\displaystyle R=\frac{2\pi}{3}Z_0\left(\frac{\ell}{\lambda}\right)^2$, where $\displaystyle Z_0 = \sqrt{\frac{\mu_0}{\epsilon_0}}$, right? Now for a radiative electric field $E\left(t\right)$, the Poynting vector should be $\displaystyle \frac {E^2\left(t\right)}{Z_0}$ (contd...) Oct 19, 2013 at 16:10
• So if I connect a matched resistive load equal to $R$ across the terminals of the antenna, the power absorbed by the antenna [the part dissipated + the part re-radiated] should be $\displaystyle \frac {V^2\left(t\right)}{2R} = \frac {E^2\left(t\right)\;\ell^2}{2R}$? And then the effective aperture would be $\displaystyle \frac {V^2\left(t\right)}{2R} = 3\frac {\lambda^2}{4\pi}$, which is twice as much as I expect. So what am I doing wrong here? Oct 19, 2013 at 16:11