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)

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)

Diagrams

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

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

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

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)

Needs math

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

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

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)

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)

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)

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)