Talk:Wavenumber

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The name of it[edit]

I do not agree with the "word" "wavenumber". I think that it ought to be "wave-number" all the way through. The English language does not create new compound words willy-nilly. It is not German.

"Wavenumber" is the established and official name. The word has been around for ages. It isn't up to Wikipedia to decide what to call things. If you don't like a word, take it up with the Queen.199.212.11.110 (talk) 18:41, 10 February 2010 (UTC)

Angular Wavenumber[edit]

I've seen the quantitiy "circular wavenumber" referred to as "angular wavenumber". "Angular" seems to be a bit more descriptive and follows the pattern of "angular frequency". (No one says "circular frequency", but language is not always symmetrical or sensible.) Also, Google finds more pages with "angular wavenumber" than "circular wavenumber". Should we retitle the "circular wavenumber" section and mention both quantity names? Zeroparallax 10:54, 17 March 2006 (UTC)

I think so, that sounds very reasonable. Fresheneesz 06:49, 8 May 2006 (UTC)

Other equation (concerning matter waves)[edit]

The solutions to a physics HW we had involved what it called the "wavenumber" "k", and it said that:

k = \sqrt{2 m E / \hbar}

Does this correspond with anything? I can't find anything about that formula anywhere, and our physics teacher didn't actually teach it to us, although I guess he thinks he did. That and I can't get it to reconcile with de Broiglie's relations. Anyone have any idea what he's talking about? Fresheneesz 06:49, 8 May 2006 (UTC)

If the hbar goes outside the sqrt, then that's the k you get when solving the 1-D Schrödinger equation, for the particle in a box. If you do dimensional analysis on the solution sin(kx), k must have units of inverse length, like the first paragraph says. - mako 11:38, 8 May 2006 (UTC)
Ahh, alright thanks. Do you think that in any way belongs on this page? Fresheneesz 00:30, 9 May 2006 (UTC)
I dunno. sin(kx) comes up a lot, but it doesn't strike me as a particularly important usage of the term. - mako 21:07, 9 May 2006 (UTC)

In summary, the correct relationship is:

k = \frac{\sqrt{2 m E }}{\hbar}.

This relationship defines the angular wavenumber of a matter wave (for example an electron) in terms of its mass, its kinetic energy, and Planck's constant (divided by 2 pi). Another correct relationship is:

k = \frac{p}{\hbar}.

This relationship defines the angular wavenumber of a matter wave in terms of its momentum and Planck's constant (divided by 2 pi). These relationships hold true for a particle in a box (quantized angular wavenumbers) or free particle (continuous angular wavenumbers) because they simply restate the de Broiglie's relations. In fact the page on de Broiglie's relations refers to this article on wavenumber. Therefore the wavenumber article should refer to de Broiglie's relations. The field is quantum mechanics. --John David Wright 22:29, 23 February 2007 (UTC)

Conversion[edit]

This page need to specify the conversion factors between wavenumbers (cm-1) and Energy/angular frequency, preferably in terms of fundamental constants if that is possible. —Preceding unsigned comment added by 82.35.34.20 (talkcontribs)

Done. Han-Kwang 15:16, 4 May 2007 (UTC)

Obscurantism[edit]

It would be easier to understand the concept if we use plain language appropriate to an encyclopedia. I suggest replacing:

Wavenumber in most physical sciences is a wave property inversely related to wavelength, having SI units of reciprocal meters (m−1).

With:

Wavenumber in most physical sciences is a wave property inversely related to wavelength, having SI units of cycles per unit length or radians per unit length, where the unit length is stated and is typically meters, centimeters, nanometers etc. The dimensions are L-1.

Also:

"The energy corresponds to a wavenumber of 300 reciprocal centimeters (or inverse centimeters or per centimeter)"

would be more clearly: "The energy corresponds to a wavenumber of 300 cycles per centimeter"


Any comments? GilesW (talk) 07:30, 22 April 2009 (UTC)

Agreed! To mere mortals (indeed, even to mortals with some physics background) it fails to give a feel for what the concept is, or to give the reader a way of envisaging it. I'll attempt a little reworking. The lead should attempt to explain to a novice the overall idea, so I'll add a sentence there to that effect. Equally the lead should avoid being overloaded with technical detail, so I'll move some of its mathematical detail down into the article itself. Feline Hymnic (talk) 22:29, 12 October 2012 (UTC)

Radial wavenumber[edit]

In the context of RF coils etc., please define "radial wavenumber". It is said to be "all important" [[1]], (see second page between expressions (3) and (4)). I cannot find a definition. GilesW (talk) 07:56, 8 May 2009 (UTC)

Spectroscopy an "oddball field"?[edit]

In this edit, physicist User:Dicklyon threw the whole article around, calling spectroscopy an oddball field. But to hosts of chemists and biologists, wavenumbers has no other meaning than a unit of energy. The lead of this article now ignores that. It only explains this term as angular wavenumber, which is the magnitude of the wave vector, of relevance only to physicists that should already know what this is anyway. The article should aim at the general reader, which is someone who wants to know what it means that a "Raman peak is at 300 wavenumbers". /Pieter Kuiper (talk) 07:41, 24 March 2010 (UTC)

By "oddball" I only meant that it's the minority usage, unique to that field, as opposed to all the fields where wavenumber means what you're calling angular wavenumber. Integrate the alternative into the lead if you like, but don't make it dominate the usual meaning. Dicklyon (talk) 23:48, 24 March 2010 (UTC)
A search like http://www.google.se/search?q=wavenumbers shows that the term is most often used to denote a unit of energy. For chemists and biologists, it is the meaning that they will usually encounter. Look at how common it is to speak of "Raman wavenumbers" or "vibrational wavenumbers". /Pieter Kuiper (talk) 00:07, 25 March 2010 (UTC)

Energy vs frequency[edit]

This edit:

http://en.wikipedia.org/w/index.php?title=Wavenumber&diff=prev&oldid=33599618

seems inconsistent with E=h\nu=\hbar\omega. The equation, as it presently appears, is:

k \equiv \frac{2\pi}{\lambda} = \frac{2\pi\nu}{v_p}=\frac{\omega}{v_p}=\frac{E}{\hbar c},

but should (I think) be:

k \equiv \frac{2\pi}{\lambda} = \frac{2\pi\nu}{v_p}=\frac{\omega}{v_p}=\frac{E}{\hbar v_p}.

Comments? —Preceding unsigned comment added by 70.234.243.222 (talk) 02:14, 11 April 2010 (UTC)

See also[edit]

I was curious as to what the connection is between wave number and curvature. Why would you have a reference to curvature on this page? —Preceding unsigned comment added by 65.199.189.6 (talk) 15:02, 14 July 2010 (UTC)

Other than because wavenumber and curvature both have dimensions of inverse length, I can't think of a relation. Miguel (talk) 08:31, 15 April 2013 (UTC)
Sounds a reasonable question. The curvature article doesn't mention wavenumber, so there's no obvious link. I'll remove the this "see also" item. Feline Hymnic (talk) 22:34, 12 October 2012 (UTC)

wavenumber / energy equivalence in other medium besides vacuum?[edit]

"For electromagnetic radiation in vacuum, wavenumber is proportional to frequency and to photon energy." Ok so what about if we're not in vacuum? Apparently when a photon goes from vacuum (index of refraction 1.0) to glass (index of refraction 1.5) the wavenumber would increase by the same amount (1.5 divided by 1.0). Does that imply its energy increased by the same amount? And when it leaves the glass would it lose that energy?

I can't find answers to these questions anywhere on the internet. Odd...

24.162.242.96 (talk) 13:32, 14 June 2013 (UTC)

Wave vector[edit]

The content of this article is largely redundant with the article Wave vector, and would be even more if that article were as well written as this. Wouldn't it be clearer if the two artilces were merged? 85.23.38.101 (talk) 17:34, 2 March 2014 (UTC)