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What is the closest antenna ever made to an isotropic antenna? That is, has the same gain in every direction in free space?

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  • $\begingroup$ Without some definition of "closeness" to isotropy, this question is not answerable. $\endgroup$ – Phil Frost - W8II Jun 2 '14 at 1:27
  • $\begingroup$ @PhilFrost I think the question does define "closeness": closeness to uniform radiation gain in every direction (and it also does specify free space). Slightly more technically, I suppose the same thing could be phrased as that the strength of the far field at a fixed distance from the antenna approaches uniform with regards to elevation and azimuth from the antenna. $\endgroup$ – user Jun 2 '14 at 9:38
  • $\begingroup$ Does it actually need to be designed to radiate a significant fraction of the RF input? Because otherwise, I'd expect a dummy load to come pretty close to isotropic; it accepts RF, it radiates some (very small) fraction of it, and it's likely to radiate about equally in all directions. $\endgroup$ – user Jun 2 '14 at 9:40
  • $\begingroup$ @MichaelKjörling it most definitely does not. Which is "closer" to isotropic: an antenna with a pattern in azimuth of $\sin \theta$, or one of $\sin(12\theta)$? $\endgroup$ – Phil Frost - W8II Jun 2 '14 at 11:45
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    $\begingroup$ As an example of figures you might consider, see directivity, beamwidth, or front-to-back ratio. All of these quantify "closeness to isotropy" in some way, but they are all different. To answer your question specifically, you need to pick one. $\endgroup$ – Phil Frost - W8II Jun 3 '14 at 12:13
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An isotropic antenna can not exist. Asking how "close" we can get doesn't really make sense. It's like asking how close we can get to any other impossible thing, like creating perpetual motion. Can we say one thing is closer to violating the laws of thermodynamics than another? No: all things ever observed obey these laws, no exceptions. Likewise, all antennas ever observed obey Maxwell's equations, so none can be isotropic. Not even a little bit. No exceptions known to science.

How do Maxwell's equations forbid isotropic antennas? I'll explain by analogy. Imagine a ball with some hair on it. Is there any way to comb this hair flat on the ball such that there is not at least one tuft?

hairy ball

There isn't. This is the hairy ball theorem, which states "there is no nonvanishing continuous tangent vector field on even-dimensional n-spheres."

Now, imagine that the hairs are the electric or magnetic field radiating from your antenna. We are looking for a vector field on a sphere, so the "tuft", where the hair sticks straight out, is a vector of zero magnitude. This is a point where there is no radiation.

Why does the hairy ball theorem apply? Because electromagnetic waves are transverse waves. This is why the tufts count as vectors of zero magnitude and not as vectors sticking out: the only direction the lines of force can go in a field that you want to radiate away from that sphere are tangential to the sphere.

Contrast this with sound waves, which are longitudinal waves. For these, the lines of force must be perpendicular to the sphere, and an isotropic radiator is easily realized: just make the hair stick straight out everywhere.

So, the consequence of this is that any antenna must not radiate it at least one direction. Beyond that, arbitrary radiation patterns are possible, but none of them are close to isotropic.

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    $\begingroup$ You comment about thermodynamics does not quite work. The thermodynamical laws are formalized as saying that the derivative of entropy is positive. It is still reasonable to ask how close to zero it can be made. Similarly there are metrics that can be used to compare smooth vector fields to nonsmooth vector fields. In particular I am thinking of a Hilbert metric for this. Your comments on the OP are correct though that the method of comparison does need to be specified. $\endgroup$ – BSteinhurst Jun 6 '14 at 3:14
  • $\begingroup$ @BSteinhurst I think it does work, until you define a way to quantify "closeness to free energy". You just defined it as "highest derivative of entropy". Without such a rule comparisons are meaningless though, because nothing is a free energy machine, not even a little bit. The only difference seems to be that there's pretty much one obvious way to quantify perpetual motion, while there are very many ways to quantify isotropy. $\endgroup$ – Phil Frost - W8II Jun 6 '14 at 12:50
  • $\begingroup$ the difference is that the way that the laws of thermodynamics are written (not their hand wavy popularizations) included the quantified notion of entropy and make claims about that. So a serious discussion of thermodynamics has wedded to it a particular way to quantify and measure entropy. This is not the same situation when talking about smooth vector fields since the notion of smooth vector field does not have bound to it a way to measure the difference between two vector fields. $\endgroup$ – BSteinhurst Jun 6 '14 at 14:40
  • $\begingroup$ You begin your answer with Asking how "close" we can get doesn't really make sense., and end it with but none of them are close to isotropic. (!) So your final conclusion makes no sense, based on your initial assertion? $\endgroup$ – Mark Adler Dec 26 '19 at 20:59
  • $\begingroup$ By the way, you can have something quite close to an isotropic antenna (in the sense of directivity, let's say), if you relax a key constraint that you did not mention, which is that the waves are all linearly polarized. See arxiv.org/pdf/physics/0312023v1.pdf . $\endgroup$ – Mark Adler Dec 26 '19 at 21:05

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