# Why does an ideal antenna not absorb all incident power?

So, in the optimum case half of the power absorbed by the antenna is immediately re-radiated. Clearly, an antenna which is receiving electromagnetic radiation is also emitting it.

Suppose I have an ideal isotropic radiator radiating $$P_x$$ into empty space. Now every closed surface integral over the Poynting flux enclosing the radiator should give me the $$P_x$$. Take radii $$r_1 < r_2 < r_3$$, all much larger than the wavelength.

Now I place an ideal absorber (a material with wave impedance of $$Z_0=377\Omega$$ converting the incoming radiation into heat??) at $$r_2$$. The closed surface integral at $$r_3$$ should now give less than $$P_x$$ (because a fraction of the energy was absorbed inside). But to my understanding, the closed surface integral at $$r_1$$ should still be $$P_x$$.

Neglecting heat radiation, can anything at $$r_1$$ detect that radiation was absorbed at $$r_2$$?

Now I replace this ideal absorber with an ideal antenna. According to the text above, this antenna would re-radiate half of this power in most ideal conditions.

If I place another antenna at $$r_1$$, this antenna should be able to receive $$\frac{P_x}{L_1} + \frac{P_x}{2 L_2}\frac{1}{L_2-L_1}$$, where $$L_n$$ denotes the ideal free space path loss at radius $$r_n$$. The first term is from the ideal radiator and the second one the re-radiated power from the antenna at $$r_2$$. This implies that the mere presence of a receiving antenna with a perfect match can indeed be detected.

What is the difference between an ideal antenna and an ideal absorber? Why does an antenna re-radiate half of its power? Why can there be an ideal absorber but we cannot use it to detect signals? Why are we instead relying on something that throws away half of the power?

My vague explanation is that the electric field in the wave excites electrons in the conductor which accelerate. This acceleration not only generates the current/voltage (which is detected by the circuit) but also generates electromagnetic radiation. However, with a black-body, it is certainly possible to have something that absorbes 100% of radiation. (I have to note that the most efficient solar cell 47.1%, i.e. nearly half. Is this the same fundamental limit?)

From a wave propagation perspective, an electromagnetic wave can only be reflected if there is a change of impedance. If we assume the load perfectly matched with the antenna, where is this impedance change that causes half of the wave to be reflected? Is this somehow related with a change from $$Z_0=377\Omega$$ to the radiation resistance of the antenna? (mathematically this could make sense for a Hertz dipole where the radiation resistance represents a short as compared to $$Z_0$$ but not so much for a walf-wavelength dipole).

Similarly, assume the antenna is not terminated (open circuit). In this case, only 50% of the incoming power would be reflected by the open circuit?

Any elaborate explanation is much appreciated.

• I don't know much about this, but I note that special cases have been worked out in atomic physics whereby an atom in a cavity manages exactly to absorb a photon that has been prepared in a special state. I wonder if some similar trick would apply to classical e.m. It would involve impedance matching among other things. Aug 21, 2020 at 17:26

I will try to summarize the classic results from 1, 2, 3. In short, a receiving antenna does not scatter as much power as it absorbs except in a few very special cases such as a very short dipole or a small loop. For these small dipoles, when the incident wave induces the same current distribution as would be seen when the antenna is used and driven as a transmitting one, then the scattered field energy is the same as the absorbed energy when used as a receiver.

In general though the scattered and absorbed fields have no such relationship. To see this note that a receiver antenna cannot just be analyzed as an antenna by itself. In fact, its behavior depends both on its orientation and distance to the transmitter antenna. These can be described by a reciprocal 2-pole network and its corresponding $$2\times2$$ impedance (admittance) matrix

$$\mathbf Z =\begin{bmatrix} Z_{11} & Z_{12} \\ Z_{21} & Z_{22} \end{bmatrix}$$

that can be represented as a "T" (or a "$$\Pi$$") network. Assume that the transmitting generator has open-circuit $$emf$$ $$V_g$$ with internal impedance $$Z_g$$ and is connected via a transmission line to the antenna ports $$2$$, while the load impedance is $$Z_L$$ and it is attached to port $$1$$.

The $$Z_{12}=Z_{21}$$ transfer impedance represents the coupling between the antennas through air and its magnitude is proportional to the squared reciprocal distance between the transmitter and receiver antennas and, in practice, it is many orders smaller than either $$|Z_{11}|$$ or $$|Z_{22}|$$. The figures below taken from  shows the impedance matrix representation and, hence, the Thevenin equivalent. (While both can be used if instead a $$T$$ we would use a $$\Pi$$ network then the admittance matrix representation would be more convenient to get the Norton equivalent.)

In the Thevenin equivalent we have $$Z_T = Z_{11}-\frac{Z_{12}^2}{Z_{22}+Z_g}\\ V_{oc}=\frac{Z_{12}}{Z_{22}+Z_g} V_g$$

The Norton equivalent short circuit current is $$I_{sc}=\frac {V_{oc}}{Z_g}$$

The received absorbed and network dissipated powers are $$P_L = \frac{1}{2} \left | \frac{V_{oc}}{Z_L+Z_T} \right |^2 \Re \{Z_L\} \\ P_D ^t = \frac{1}{2} \left \lvert\frac{V_{oc}}{Z_L+Z_T}\right\rvert ^2 \Re \{Z_T\}$$

Interestingly, if we use the Norton equivalent we get the dissipated network power as $$P_D^n = P_D^t \left\lvert \frac{Z_L}{Z_T}\right \rvert^2= \frac{1}{2} \left\lvert\frac{V_{oc}}{Z_L+Z_T}\right\rvert ^2 \Re \{Z_T\} \left \lvert \frac{Z_L}{Z_T} \right\rvert^2$$

This result immediately shows that one cannot and should not interpret $$P_D$$ as scattered power when we use the Thevenin or Norton equivalents to calculate the dissipated power inside the network because they can be *different*. Instead, one may use these equivalent circuits to calculate the power dissipated in the load *only*.

For example, if the receive antenna is open circuited then there is no absorbed power nor internal dissipation, $$P_D=0$$, and get the completely nonsense result that the antenna does not even scatter. On the other hand, in the open circuit case the Norton equivalent shows a complete reflection from the antenna. When the load of the receive antenna is matched to the driving point impedance then $$Z_L = Z_T^{*}$$ and we do have $$P_D^n=P_D^t$$ and it could interpreted for a small dipole as the scattered power, see Collin for details.  Another approach is that of Love who instead of using a Thevenin (or Norton) equivalent circuit that uses a constant voltage (or current) source synthesizes a constant power source representing the incident power wave directly. This is shown below For aperture antennas Love shows that the voltage and current sources of his equivalent circuit depend as in below where $$\eta$$ is the aperture efficiency. His analysis concentrating on aperture antennas, while not generally valid for arbitrary antennas, removes the apparent contradictions and is in agreement with normal physical (engineering) intuition. 1 Silver: Microwave Antenna Theory and Design, chapter 2, sec 10,11 and 12

2 Love, “Equivalent Circuit for Aperture Antennas,” Electronics Letters, 23, 13, June 1987, pp. 708-710.

3: Collin: Limitations of the Thevenin and Norton Equivalent Circuits for a Receiving Antenna, IEEE Antennas and Propagation Magazine, Vol. 45, No. 2, April 2003

• This is really an amazing answer that pointed me to the work of Love and the unterstanding of radiation resistance. I would really Love (pun intended) to mark this as answer but as good as it is, it does not really unswer the actual question(s). Armed with your text above, a good deal of the question can be rephrased as: Can we build an antenna that absorbs power but does not scatter? Why/why not? Looking at it from an optical perspective, such an antenna would be a blackbody, right? In Love's derivation, would this case correspond to $\eta=0$?
– divB
Dec 4, 2020 at 4:22
• Because of the always present edge diffraction there is no way to build a perfect "funnel" for a plane wave. To reduce such scattering effects antennas are coated and/or surrounded around the edges with absorbers but even with ideal absorbers one cannot fully match the edges to a vector EM field. There is no perfect absorbers though designing a better one is a whole industry; look for low observable antennas, i.e., "stealth"! Unfortunately, resistive absorbers on/near a metallic or dielectric antenna contributes to received noise... Dec 31, 2020 at 15:58