Talk:Isotropic radiator

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Comment from T - Antenna Theory Section[edit]

This looks easily understandable up to "At some point, such coverage must have a discontinuity, i.e. jump in the direction of the vector field." I don't know if there is an easy way to describe this to Joe on the street, but it would help. - T

T, I have made an attempt to correct the description of the hairy ball problem as it relates to the travelling wave and the e and h planes. Kgrr 13:42, 11 August 2005 (UTC)

Comment from JR Thorpe - Antenna Theory Section[edit]

I think that the 3rd paragraph and the 4th paragraph describe the same thing.

"An antenna emits an electromagnetic wave that has two components - the electric and magnetic fields. These are at right angles to each other and also at right angles to the direction of travel of the wave. This presents a problem for a theoretical isotropic radiator since there will be places on the unit sphere where we cannot specify a unique "polarization direction" for the direction of the electric field."

says the same as:

"This is because the electromagnetic wave is made up of two perpendicular components - the electric field E and the magnetic field H. The emitted electromagnetic wave moves perpendicular to the E-plane and H-plane. The wave cannot be lined up so that there is radiation in all directions and that neither the E or H planes cancel each other out. There must be a discontinuity."

Except the second is clearer. I think they should be only one paragraph, preferably the second. (The remark that "An antenna emits ..." is useful though. As is the reminder that perpendicular means at right angles).

I'll change this in a few weeks time if no one objects.

Antenna Theory Section - Inaccuracies[edit]

Some inaccuracies and omissions here (I intend to fix these at some point):

  1. The no isotropic radiator restriction only applies for plane-polarized radiation. If you have two plane-polarized radiators working in quadrature, then in principle it's possible to get isotropic radiation of the total power. One radiator fills in the nulls for the other - though the total radiation is circularly polarized in some directions and plane polarized in others. See Matzner.
  2. It's not satisfying Helmholtz's equation that's the problem (sound waves satisfy the wave equation, and can radiate isotropically), it's the no-divergence thing.
  3. There should be a reference to the hairy ball theorem and a quick explanation. --catslash 11:37, 20 July 2007 (UTC)

I've rewritten it - but reading it back, I see a flaw in my argument. So it still needs fixing. --catslash 16:18, 21 July 2007 (UTC)

This is well worth a read: Scott, W.; Hoo, K.S., "A theorem on the polarization of null-free antennas", IEEE Trans. on Antennas and Propagation, vol. AP-14, no. 5, Sep 1966, pp. 587-590. It reviews the proof of Mathis, and shows that "...that elliptical polarization of all axial ratios, ranging from circular polarization of purely one sense, through linear, to circular polarization of the opposite sense, must exist in the far-field of a null-free antenna". --catslash 14:17, 25 July 2007 (UTC)

Completely false. You cannot have an isotropic radiator, even if it were circularly polarized. It is a violation of the Helmholtz equation. Sound waves are a poor reference, since sound is not a vector, it does not follow the vector Helmholtz equation. --Mr. PIM 17:55, 30 August 2007 (UTC)

You are correct that what I had put was wrong - and I don't have any problem with you reverting it. However, what is there at the moment is equally wrong. Let me take your points in reverse order.

  1. The Helmholtz equation is  ( \nabla^2 + k^2 ) E = 0 , where E could be a vector or a scalar or indeed a tensor, since the Laplacian can operate on any of these (and k^2 is just [multiplication by] a scalar constant). Notice that if E is a vector resolved into x, y and z components, then the operator ( \nabla^2 + k^2 ) does not 'mix' the components - each component of the vector is independently a solution of the scalar Helmholtz equation. Now the acoustics gives us the scalar solution p = \frac{1}{r} e^{- i k r}, so a vector solution is E = (\frac{1}{r}\hat{x} + 0\hat{y} + 0\hat{z})e^{- i k r}. This solution is isotropic since it depends only on r. However it does not satisfy Maxwell's equations, which additionally require E\cdot r = 0 (transverse waves), at least for large r.
  2. I'm not claiming that you can have a circularly polarized isotropic radiator - you can't. I am claiming that you can have a single-frequency (coherent) null-free radiator if you allow it to radiate with a different sort of polarization in each different direction (and it will need to employ all possible polarizations from LHC through linear to RHC). I'm also claiming that such a radiator can achieve at least an arbitrarily small variation in power over the sphere of directions.

I still intend to write something more accurate --catslash 14:32, 31 August 2007 (UTC)

Come to think of it, although the pressure of a sound wave is a scalar field, the velocity (or displacement or acceleration) of the medium is a vector, and it's something like V = \left( \frac{1}{r} + i k\right) \frac{1}{r^{2}} (x\hat{x} + y\hat{y} + z\hat{z}) e^{- i k r} = \left( \frac{1}{r} + i k\right) \frac{1}{r} \hat{r} e^{- i k r} (or at least that's -\nabla p above). Anyway it satisfies the Helmholtz equation, and is purely radial (not transverse) and isotropic, so I reckon that demolishes the Helmholtz equation argument? --catslash 15:37, 31 August 2007 (UTC)

Don't confuse the Laplacian with the Vector laplacian. They are different operators. --Mr. PIM 21:55, 31 August 2007 (UTC)
Different in what sense? --catslash 22:02, 31 August 2007 (UTC)
In Cartesian coords, both are \left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right) are they not? --catslash 22:19, 31 August 2007 (UTC)

That's correct. In cartesian coordinates, the Vector laplacian reduces to three scalar laplacians. However, in spherical coordinates (which is what we are dealing with here), the two operators are completely different, and should not be confused. --Mr. PIM 23:16, 31 August 2007 (UTC)

The Laplacian is a coordinate-independent concept. Nature has no idea what coordinate system you are going to do your calculations in, so you'd better get the same answer in all systems. --catslash 00:13, 1 September 2007 (UTC)
That's correct. The solution is independent of the choice of coordinate system. That is why the fact that the vector laplacian is similar to the scalar laplacian in rectangular coordinates should not confuse you to believe they are the same. The two operators are different. --Mr. PIM 00:34, 1 September 2007 (UTC)
I think you're saying that the r, θ, Ф components of a vector field satisfying the Helmholtz equation, are not three scalar fields satisfying this equation. True. E = (\frac{1}{r}\hat{r} + 0\hat{\theta} + 0\hat{\phi})e^{- i k r}, does not satisfy the equation - but it's a physically different field to E = (\frac{1}{r}\hat{x} + 0\hat{y} + 0\hat{z})e^{- i k r} which does (one field is everywhere radial (and so varies in direction), and one is everywhere parallel to \hat{x} (and so has constant direction)). Conversely E = \left( \frac{1}{r} + i k\right) (\frac{1}{r} \hat{r} + 0\hat{\theta} + 0\hat{\phi}) e^{- i k r} does satisfy the equation, while E = \left( \frac{1}{r} + i k\right) (\frac{1}{r} \hat{x} + 0\hat{y} + 0\hat{z}) e^{- i k r} does not - but again they're physically different things. --catslash 00:56, 1 September 2007 (UTC)
Putting it a different way: If (\nabla^2 + k^2)E = 0 then (\nabla^2 + k^2)(E\cdot v) = 0 for any constant vector v such as v = \hat{x}, but it isn't true for v = \hat{r}. That's because \hat{r} isn't (physically) constant; it points in different directions depending on where you are. Of course, you could tell me that \hat{r} is arithmetically constant in spherical coordinate components; it's everywhere (1, 0, 0), but then you'd have to conclude that the laplacian was somehow different in different coordinate systems, and different for different types of operand. --catslash 01:34, 1 September 2007 (UTC)
Anyway: sound waves: (1) the pressure is a scalar field (in a fluid medium), but the displacement, velocity or acceleration of the medium is a vector field (agree/disagree)? (2) both the scalar pressure field and vector velocity field satisfy the Helmholtz equation (in an ideal medium)(agree/disagree)? (3) it's possible to have an isotropic acoustic source (agree/disagree)? (4) Hence the Helmholtz equation isn't the source of the difficulty in constructing isotropic electromagnetic radiators (agree/disagree)? --catslash 01:54, 1 September 2007 (UTC)

I think where your logic breaks down is that the wave equation for vector fields and for scalar fields should be treated as different equations, even though there is a lot of similarity between the two. I think if you try to plug an isotropic radiation pattern into the Helmholtz equation, you will discover that the equation cannot be satisfied. --Mr. PIM 05:04, 1 September 2007 (UTC)

I'm keen to continue the discussion on the sameness/difference of differential operators applied to scalar and vector fields - but on my talk page, since it isn't strictly relevant here. Since you contend that the distinction matters, I'm happy to continue discussion of the Helmholtz equation strictly with reference to vector fields. ...So, just as a (single frequency) electromagnetic wave can be described by giving either the E or H fields (both vector fields), a sound wave can be described by giving either the pressure field (scalar), or the sound particle velocity field, and the latter is a vector field - or do you disagree? --catslash 15:10, 1 September 2007 (UTC)

You're taking this discussion outside of my league. I'm not an expert on sound waves. I'll just leave saying that if you can come up with additional reasons why an isotropic radiator cannot exist, go ahead and list them. I had only two bones to pick about what you wrote (or more accurately, what I understood from your writing)

  1. An isotropic radiator can satisfy Maxwell's equations
  2. An isotropic radiator can be constructed using two linear dipoles rotated 90° and phased 90° apart to construct a circularly polarized isotropic radiation pattern

Since you do not hold those positions, I have no issues. Go ahead and make whatever changes you feel are necessary, just do not write something that suggests either of those two facts. --Mr. PIM 17:54, 1 September 2007 (UTC)

Sorry you don't wish to continue the discussion - I would be happy to continue without reference to either scalar fields or sound waves should that help. Regarding your two points; in reverse order;
2. A pair of crossed dipoles in quadrature will certainly not constitute an isotropic radiator (since they are in quadrature, they cannot interfere, and so the powers add, and the directivity comes out the same as for a single dipole). However (and this is the difficulty), the quadrature crossed dipoles do evade the oft-quoted hairy ball theorem and produce a pattern with no nulls.
1. Yes and no. When you assert that something is mathematically impossible, you generally need to state the rules very precisely. For example it's false to say simply that angle trisection is impossible; you have to add a number of conditions, (including (i) of an arbitrary given angle, (ii) using classical compass-and-straight-edge methods, and (iii) in a finite number of steps). Unfortunately, the no-isotropic-radiator theorem is most commonly cited as a passing remark in antenna-theory texts, which have no interest in either a precise statement or a proof.
I shan't be writing anything contentious just yet though. --catslash 20:48, 3 September 2007 (UTC)

Comment from K T McDonald[edit]

I am new to Wikepedia talk pages, and I'm not sure this is the right way to add my 2 cents worth, but here goes.

The discussion in the talk page seems to be nearly on track, but that awareness is not yet reflected in the page "Isotropic Radiator" itself.

An isotropic radiator of electromagnetic waves is certainly possible, and two different (theoretical) examples have been given in the note of Matzner and mine that is referenced at the bottom of the Wiki page.

It would be preferable if the text of the Wiki page were updated to reflect to insights contained in the references. I'm not sure how to proceed. I can edit the page if that is appropriate, but maybe it's better if a past editor make changes....

--kirkmcd@princeton.edu --Kirktmcdonald (talk) 20:27, 10 December 2007 (UTC)

Yes this is the right place to comment.
The conventional wisdom is that isotropic radiators are impossible, and that this is a consequence of the hairy ball theorem. I remember being taught this myself (25 years or more ago). Any statement to the contrary is therefore likely to be contentious, and so according to Wikipedia policy must be verifiable by reference to reliable sources. It is not enough that you have proved something yourself (original research), and Arxiv is not considered a reliable source, since it is not peer-reviewed. This policy is vital in order to keep the cranks at bay. Have you published any peer-reviewed papers on this topic?
As to the actual facts of the matter, I reckon:
  1. The hairy ball theorem together with the fact that electromagnetic waves are transverse, proves that at any instant in time, there must be a direction of zero radiation (for some given radius).
  2. By choosing two radiation patterns with non-coincident nulls, and arranging them in quadrature, it is possible to achieve non-zero radiation of power in all directions (that's the time-averaged power (over one cycle)).
  3. As shown by Scott and Hoo (mentioned above), such a null-free radiation pattern must include radiation with all possible polarizations (so it's impossible for one given sort of polarization).
  4. The possibility of a null-free pattern does not imply the possibility of an isotropic pattern, but it does show that the hairy ball theorem does not preclude it.
  5. The u-shaped antenna and the stack of crossed dipoles clearly do give a pattern which is arbitrarily close to isotropic (in time-averaged power). This may be verified analytically or (very easily) with any computational EM software. But do you have any references for peer-reviewed papers which state this?
  6. Is 'arbitrarily close to isotropic' the same as isotropic?
These points should probably be mentioned in the article.
--catslash (talk) 01:09, 12 December 2007 (UTC)

I'm not sure if anyone is watching this page, but these points are correct: an isotropic radiator is NOT excluded if you don't insist on linear polarization, and there are antennas which do not have nulls in their total radiation pattern. In particular, the turnstile antenna (which the WP article mistakenly identifies as being directional). Though I hate to resort to it, I am looking at a RS which says "its three dimensional radiation pattern is nearly omnidirectional." I could compute just how close it is to being isotropic, but that would be OR so I'll skip it. But I'm tagging the claim that a short dipole has the lowest gain possible, since that is wrong, and so is the blanket statement that there must be a null point. EITHER you retract those claims, OR you qualify it as saying "For linearly polarized antennas...." I'll wait for someone already involved with this page to correct it, before I do that myself. Interferometrist (talk) 22:11, 31 March 2011 (UTC)

The antenna shown in the image[edit]

If an Isotropic antenna cannot exist, then what is the device shown in the image? I'm guessing is as some antenna designed as a very close approximation to a isotropic antenna, but if that is the case, it really ought to be mentioned. 74.5.162.102 (talk) 23:34, 22 October 2009 (UTC)

It seems to be three dipoles (or loops) with three receivers, one for each axis. The sum of the three received powers will independent of the polarization and direction of illumination. However any linear combination of the signals from the three loops/dipoles would be zero for some particular illumination, and so no electrical connection of the three would constitute a single isotropic antenna which could be connected to a just one receiver. --catslash (talk) 14:32, 23 October 2009 (UTC)

Why a point source?[edit]

Is there some reason (tradition, definition, ...) why an isotropic radiator must be a point source? Is it so that the radiation can travel an unbounded distance? If not, then why can't it be either of the following?

  1. The inside surface of a closed cavity (of any shape) at a uniform temperature.
  2. An infinite volume of CO2 at a uniform pressure and temperature equal to that at say the surface of Venus, with nothing else in the universe.

The latter has in common with a point source that it is ideal and physically unrealizable, but the former is easily realized and is the basis for constructing excellent approximations to a black body when a small hole is drilled in it. At the center of a spherical cavity the radiation is uniform in all directions, meeting that condition, and unlike a point source there is nothing unphysical about the center of a spherical source.

In the latter, every photon travels a finite distance before being absorbed by a CO2 molecule, and the uniformity of the gas ensures uniformity of the radiation at each frequency. (There is no frequency at which the probability of absorption by a nearby CO2 molecule is exactly zero.) The gas is supercritical and pressure broadening is extreme, with photons whose wavenumber is in the vicinity of 650 cm−1 having mean free paths on the order of millimeters or less.

A spherical parcel of Venusian atmosphere of radius 1 m and altitude 1 km would be an extremely good approximation.

While the latter seems likely to be OR, the former obviously isn't, but the reason for excluding it is not obvious to me, though presumably it's a reasonable reason. --Vaughan Pratt (talk) 00:33, 13 November 2011 (UTC)

Lambert's law[edit]

The last sentence of the lead says "Isotropic radiators obey Lambert's law." If the radiation is coming from a point source (like a star), what is Θ in Lambert's cosine law? This sentence would seem to be contradicted by section 7 of these notes which are in good agreement with what I learnt working and publishing in computer graphics at Sun Microsystems and later teaching computer graphics at Stanford (before we hired Marc Levoy and Pat Hanrahan and I could go back to teaching algebra and logic). --Vaughan Pratt (talk) 05:46, 15 November 2011 (UTC)

Meanwhile it occurs to me that more careful definitions of "point source" and "Lambertian surface" might clarify things. A point source is not literally a point in the geometric sense, but rather merely the limiting case of the distribution of radiation from a radiator as it shrinks to a point (or as the sphere centered on it grows unboundedly). In that sense of "point source" a point could be either flat, or spherical, or some other shape, counter to geometric intuition but in good agreement with optical intuition where "distribution of radiation in the limit" is a concept outside the realm of geometry.

Likewise a Lambertian surface need only be locally flat, meaning comprised of elemental flat surfaces each of which has its own normal. For a smoothly curved surface we still need to take a limit, not of distance from the radiator however but of size of the elemental surfaces.

With the concepts thus clarified (hopefully copacetically) we have the following four examples.

  1. The radiation from a spherical black body is both isotropic and Lambertian.
  2. The radiation from a flat black body is Lambertian but not isotropic.
  3. The radiation from the Sun is isotropic but not Lambertian on account of the phenomenon of limb darkening.
  4. The thermal radiation from a warm flat sheet of chrome is neither Lambertian nor isotropic.

These examples demonstrate the logical independence of isotropic and Lambertian, neither of which need imply the other, though both can hold and neither can hold.

Absent objections I'll modify the last sentence of the lead accordingly, which certainly is problematic as it stands. --Vaughan Pratt (talk) 16:23, 15 November 2011 (UTC)