Talk:Bessel beam

Say what?

I was led to this article via a BBC News item today. However, the second sentence of the article states "It is particularly important to note that the fundamental zero-order Bessel beam has an amplitude maximum at the origin, whereas a high-order Bessel beam (HOBB) possesses an axial phase singularity at the transverse origin where the amplitude vanishes as expected from the mathematical descriptive nature of the high-order Bessel function of the first kind". Pardon? What is this in English? Perhaps some editor can improve this lead as per WP:MoS General principles, clarity --Senra (Talk) 15:23, 3 March 2011 (UTC)[reply]

It's saying the zero-order beam is the only mode that has an intensity peak at the center of it's axis. For all higher-order modes, the same intensity cross-section is donut shaped. 162.246.139.210 (talk) 14:16, 29 August 2023 (UTC)[reply]

Diagrams

Added diagrams of Bessel beam etc. Removed reqdiagram tag. Egmason (talk) 10:20, 3 June 2011 (UTC)[reply]

Nice diagrams. Thanks.--Srleffler (talk) 04:05, 4 June 2011 (UTC)[reply]

Is it just me or is the first maximum off-center in the cross-section? Laura Scudder | talk 17:02, 1 August 2013 (UTC)[reply]

It seems to be an artifact. I see it off-center in the thumbnail, but correctly centered in the full image.--Srleffler (talk) 01:22, 2 August 2013 (UTC)[reply]

Needs math

The article needs a formula and derivation of the Bessel beam. 178.39.150.251 (talk) 16:04, 11 June 2015 (UTC)[reply]

A formula, but not a derivation. Wikipedia is not a textbook.--Srleffler (talk) 02:07, 12 June 2015 (UTC)[reply]

Controversial and lacking evidence

The notion of a "non-diffracting" beam suggests that there is direct experimental evidence of a propagating beam that is proven to not lose any energy (compared to already very small proportion lost by a diffraction-limited beam, over long enough distances to be reliably measured. Where is this experimental evidence? Wikibearwithme (talk) 08:52, 3 February 2018 (UTC)[reply]

Diffraction doesn't cause any energy loss. It just causes a beam to spread out as it propagates. A perfect plane wave also has no diffraction in free space. Neither a perfect plane wave nor a perfect Bessel beam can be made in practice, however, because either would require an infinite expanse and an infinite amount of energy. We can make finite approximations to both, however. I haven't looked into it, but I presume that a finite approximation to a Bessel beam exhibits much lower diffraction than a similarly-sized finite plane wave. That is not particularly surprising.

You ask "where is this evidence?" Reference 2 looks like it would answer your question. Many of the other 27 references cited in the article would likely do so as well.--Srleffler (talk) 15:55, 3 February 2018 (UTC)[reply]

I wondered about why this is called a "beam" at all. The beam is conic, and what is referred to as a "Bessel beam" only occurs at a particular intersection point. It should be called a Bessel [interference] pattern rather than a "beam". 162.246.139.210 (talk) 14:18, 29 August 2023 (UTC)[reply]

All beams are conic. There is no such thing as a beam of finite width that propagates through space without expanding. Such a thing would violate fundamental physical law. --Srleffler (talk) 23:09, 2 September 2023 (UTC)[reply]

That has nothing to do with my point. A Gaussian beam diverges, yes, but it is always a Gaussian beam. It doesn't become something else as we travel along the beam. We could say it is longitudinally stable. There is no such Bessel beam. The properties referred to as "Bessel beam" only exist within a finite window of a focused beam with circular cross-section, where opposite sides of the circle interfere. Therefore it's a Bessel interference pattern, not a "beam" in any sense of the word. 162.246.139.210 (talk) 19:34, 6 June 2024 (UTC)[reply]