Talk:Isotropic radiator

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[edit] Comment from T - Antenna Theory Section

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)

[edit] Comment from JR Thorpe - Antenna Theory Section

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.

[edit] Antenna Theory Section - Inaccuracies

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 k2 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)

[edit] Comment from K T McDonald

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)