Talk:Angular momentum

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Introduction section[edit]

In the body of our article Angular momentum, there is comprehensive coverage of the fact that it is a vector cross product. However, there is some disagreement about how the subject should be summarised in the lede section. For some time, the lede has included the following:

the angular momentum can be expressed as the magnitude of its linear momentum, mv, multiplied by the radial distance, r. Thus, the angular momentum L of a particle with respect to some point of origin is
\mathbf{L} = m\mathbf{v}r
where r is the particle's distance from the origin and mv is its linear momentum.

On 11 July User:Essex.edwards amended this text to say the following. See the diff.

the angular momentum can be expressed as its linear momentum, mv, crossed by its position from the origin, r. Thus, the angular momentum L of a particle with respect to some point of origin is
\mathbf{L} =  \mathbf{r} \times m\mathbf{v} \, .

There has since been a sequence of reversions. In the interests of avoiding an edit war it is important that we now determine whether the definition of angular momentum as a vector cross product is appropriate for the lede section, or whether the lede should avoid the complexity of the vector cross product.

My preference is that lede sections should summarise the content of an article in terms as clear and simple as possible so that the lede is accessible to as many readers as possible, not just those with a high-level education. The comprehensive detail can come later in the body of the article. For that reason, I am in favour of the lede explaining angular momentum without resorting to the vector cross product. Dolphin (t) 01:43, 12 July 2012 (UTC)

Independent of whether or not the cross product is appropriate in the introduction, the formula without the cross product is completely wrong. It should not be there. It's not the formula for any common physical quantity. If the cross product is too complicated, then this formula should just be removed entirely.

I agree that the cross product expression is only understandable by someone with a higher-level math education. It would be better to have an introduction that can be understood more generally - so long as this doesn't mean resorting to factual inaccuracies.

In my opinion, the intro doesn't lose much by removing that formula and paragraph altogether. The cross-product formula is the very first thing in the definition anyway. Essex.edwards (talk) 02:12, 12 July 2012 (UTC)

Thanks for joining this discussion. You have written the formula without the cross product is completely wrong. It should not be there. It's not the formula for any common physical quantity. Please explain. (Why is it completely wrong without the cross product?) Dolphin (t) 02:40, 12 July 2012 (UTC)
mvr is only correct in the special situations where the velocity v is perpendicular to r. It's not true for orbits for example.Teapeat (talk) 02:54, 12 July 2012 (UTC)
Thanks. That makes sense. Dolphin (t) 12:53, 12 July 2012 (UTC)
That's not entirely correct either. r and v are vectors. You can't write rv without specifying whether or not your're using the cross product, dot product, or something else. It's not a mathematically well formed statement. As the equation was written, r had been replaced with a scalar r - which is inconsistent with the rest of the page. With that substitution, mrv is well-formed, but wrong. It will be a vector of the correct magnitude only when r is perpendicular to v, as mentioned above. However, it will always have the wrong direction. It will be parallel to v, while the angular momentum vector is always perpendicular to v. You could replace r and v both with their scalar magnitude, and then say that the magnitude of the angular momentum is mrv - and this will be correct for the special case that r and v are perpendicular. But that's not the angular momentum, it's just the magnitude of the angular momentum. Essex.edwards (talk) 01:03, 13 July 2012 (UTC)

────────────────────────────────────────────────────────────────────────────────────────────────────Essex.edwards makes sense too. What do people think about the following explanation as an attempt to say what has to be said, but without introducing the complexity of the cross product?

the angular momentum of a particle has magnitude equal to its linear momentum, mv, multiplied by the perpendicular distance, r, from some point of origin to the vector representing the linear momentum. Thus, the magnitude of the angular momentum L of a particle with respect to some point of origin is:
\mathbf{L} = rm\mathbf{v}
where r is the particle's perpendicular distance from the origin to the linear momentum vector, and mv is its linear momentum.
The direction of the angular momentum vector is perpendicular to both the vector representing the linear momentum and the vector representing the perpendicular distance from the origin, as given by the right hand rule. Dolphin (t) 02:46, 13 July 2012 (UTC)

All of that perpendicular distance stuff sounds rather confusing to me. It's not actually clear to me what you're trying to say with that. For one thing, the linear momentum vector only defines a direction, not a line. It's not obvious to me what the correct interpretation is, or if there is one. To be totally general, and avoid the cross product, you can just bring in the definition of the cross product, like so ...

the angular momentum of a particle has magnitude equal to its linear momentum, mv, multiplied by the the distance to the origin, and further scaled by \sin(\theta), where \theta is the angle between the particle's velocity vector, and the line to the origin.
L=mvr\sin(\theta)
The direction of the angular momentum vector is perpendicular to both the vector representing the linear momentum and the vector representing the perpendicular distance from the origin, as given by the right hand rule.

Or, you could discuss just circular motion.

For the special case of a particle moving in a circle around an origin point, it's angular momentum points along the axis of rotation, and has magnitude equal to its linear momentum (speed times mass) multiplied by the radius of the circle it's moving on. If the particle's direction were changed, but its speed and position left fixed, then its angular momentum shrinks in proportion to \sin(\theta) where \theta is the angle between the particle's heading, and the direction from it to the origin.

That is ambiguous about the sign of the vector, but for a qualitative and correct intro piece, I would argue that that's okay. It's ambiguous, rather than being incorrect. Essex.edwards (talk) 03:33, 13 July 2012 (UTC)

Perpendicular distance? Wikipedia has two articles on the subject: Perpendicular distance, and Distance from a point to a line.
I like both Essex.edwards' suggestions. At this stage I propose we go with the first one:
the angular momentum of a particle has magnitude equal to its linear momentum, mv, multiplied by the distance to the origin, and further multiplied by \sin(\theta), where \theta is the angle between the particle's velocity vector, and the line to the origin.
L=mvr\sin(\theta)
The direction of the angular momentum vector is perpendicular to both the vector representing the linear momentum and the vector representing the perpendicular distance from the origin, as given by the right hand rule.
Dolphin (t) 06:59, 14 July 2012 (UTC)

Relationship to the kinetic energy of motion[edit]

A nondimensional particle in uniform directional motion has a kinetic energy of motion equal to the mass value times the square of the speed. Also a nondimensional particle in uniform circular motion (around a fixed point) has an equal amount of kinetic energy of motion the value of which is now called angular momentum. However if the particle has components with a varying radius dimension around the point of rotation, the angular motion (and the kinetic energy of motion) are reduced in accordance to the integral of the value of the angular momentum (and the kinetic energy of motion of the components.WFPM (talk) 21:48, 18 November 2012 (UTC)

Angular momentum in electrodynamics seems to be wrong.[edit]

The expression  \mathbf{p} -\frac {e \mathbf{A} }{c} should be replaced by  \mathbf{p} +\frac {e \mathbf{A} }{c}. The way it is stated now contradicts the

at the beginning of the paragraph. But since I am not sure whether my assumptions are correct I thought I would write in the talk section. Positrino (talk) 15:16, 6 October 2012 (UTC)

Actually it is correct, although confusing, since
  • in this article no symbol is used to differ between the canonical momentum \frac{\partial L}{\partial \dot{q}} (generalized momentum in Lagrangian and Hamiltonian mechanics) and kinetic momentum - both use lower case p,
  • in that other article the kinetic momentum mv is lower case p and canonical is capital P.
I tried to clarify by making canonical momentum capital P. Ok? Maschen (talk) 16:20, 6 October 2012 (UTC)
Also, FYI, some people define e as the absolute value of (a.k.a. negative of) the charge of an electron and others define e as just the charge of an electron. There could be inconsistencies arising from that. --Steve (talk) 14:19, 7 October 2012 (UTC)
There shouldn't because it just says "e = electric charge of particle", which could be positive or negative depending on the particle. The see also link above given by Positrino should clarify why the "potential momentum" eA is removed from canonical momentum rather than added on. Maschen (talk) 18:50, 7 October 2012 (UTC)

Animation on the right is upside down.[edit]

Preceding unsigned comment (just the title) by 79.184.72.28.

So it was. I reverted. M∧Ŝc2ħεИτlk 19:37, 27 December 2012 (UTC)

Suggest merge[edit]

Introduction to angular momentum covers the same topic as this article and could usefully be merged with it. If the introduction to this article is too technical, it should be rewritten, not split off into a separate "introductory" article. This is not break-the-sky bleeding edge physics, so should be possible to treat in a unified and compact way here. --Wtshymanski (talk) 16:47, 22 April 2013 (UTC)

Good idea. In the process should we reduce the quantum angular momentum section to a briefer paragraph firmly linking to angular momentum operator? M∧Ŝc2ħεИτlk 17:28, 22 April 2013 (UTC)
Taking a look at the history of the intro article, the original motivation was to provide an exposition that wasn't based on vectors, and probably not calculus either. If the intro section of the angular momentum article could retain the same mathematical simplicity, without sacrificing the more general approach later in the article, this could work. To do this, the lead of the article would need to be rewritten to excise vector notation as well. I would guess that the intro section would go before the Angular momentum in classical mechanics section and the illustrations in the article would need to be rearranged as well. --Mark viking (talk) 18:16, 22 April 2013 (UTC)
Oppose – The angular momentum article is 26k in length, the intro article is 9k. Let's keep the intro in the simplest form so that more elementary readers, such as myself, are not scared away by an even longer one. Merging an introductory formated article into the main article, and saving the main concepts presented in an introductory format, would result in a duplication of material within the single article. Also, I see 20 other "Introduction to" articles and I consider the idea to be most helpful. A sort of "World Book" level encyclopedia vs the Britannica level that more advanced readers might expect. (Such introduction to articles should be a step above the Simple English Wikipedia.) – S. Rich (talk) 04:30, 26 April 2013 (UTC)
Comment by nominator A two-column list of "see also" articles suggests that this article is not a comprehenisive encyclopediac overview of the subject, no matter how much pretty maths typography we have in it. Perhaps less redundant maths notation and more explanation is in order. If you understand the vector notation, you don't need to look up a concept as fundamental as "angular momentum" here. Very few of our 3 million articles need separate "introductory" articles, and I don't think this concept warrants an exception. --Wtshymanski (talk) 13:48, 26 April 2013 (UTC)
  • In the last 90 days, we have 219412 page views for this page and 6216 for Introduction. The question is what will be helpful to the readers, especially those 6,216 less knowledgeable ones? I submit that those people should be considered, and not discarded as a small minority of people interested in the subject. What does guidance say? I submit that WP:CRITERIA is met in all regards by the Intro article. Moreover, WP:USE has the goal of making WP "aims to make Wikipedia easier for everyone to use." An "Introduction to ..." article is a simple and effective step towards that goal. (BTW: 5 of the see alsos have been removed as duplicates.) – S. Rich (talk) 14:52, 26 April 2013 (UTC)

────────────────────────────────────────────────────────────────────────────────────────────────────Maybe we need a better link to the introductory article. Martinvl (talk) 12:43, 6 May 2013 (UTC) The ratio of the main article to the introductory one is about 40:1. There is a similar ration between the articles Metric system and Introduction to the metric system (80251 and 2038 hits respectively in the last 30 days). Would it help if the hatnote relating to introductory articles were reformatted as shown below:

This hatnote could be made more general to include "outline" articles as well. Martinvl (talk) 20:43, 6 May 2013 (UTC)

Yes, the box presentation is an excellent idea. Dolphin (t) 00:13, 7 May 2013 (UTC)
Strong oppose - Type "Introduction to" into the search box at the top right of the page and you will see Wikipedia has a whole series of Introductory articles, including Introduction to angular momentum. It is possible to make diminishing remarks about any one of these, and then propose it be merged with the parent article, as has been done here. We can assume that the majority of people who read Wikipedia don't have a comprehensive understanding of either momentum or angular momentum. For this majority, momentum, and especially angular momentum, are out there at the leading edge of their understanding of physics and an Introductory article is likely to reassure them that they can aspire to assimilating the concept. It's not a case of finding someone willing to write an Introductory article - someone has already done the hard work to create the article so I see no benefit in burying it within the main article. Dolphin (t) 13:47, 6 May 2013 (UTC)

Center of Mass Wording[edit]

Under the section Angular momentum simplified using the center of mass it says

 It is very often convenient to consider the angular momentum of a collection of particles about their center of mass

Then after the equation,

 where ri is the position vector of particle i from the reference point

Shouldn't "from the reference point" say "from the center of mass" since the basis of simplifying the equations is that the reference point is the same as the center of mass? Or am I missing something? Infogulch1 (talk) 06:36, 30 November 2013 (UTC) Original: 06:30, 30 November 2013 (UTC)

Misunderstanding[edit]

@Xxanthippe:, I've noticed you've reverted an edit of mine suggesting a misunderstanding. There would have to be a good number of things I don't understand but I'm not sure about this case. There were a number of edits I made at once so I'm not sure which of them were due to a misunderstanding. Was it all of them? I kind of doubt it could be all of them since part of it was to delink Sun which shouldn't have been linked for an irrelevant page like this. Perhaps I ought to get in the habit of doing thing one at a time. Jimp 09:42, 5 June 2014 (UTC)

I think that For the case of an object that is small compared with the radial distance to its axis of rotation flummoxed me as it seemed an unneeded restriction. Best wishes. Xxanthippe (talk) 10:06, 5 June 2014 (UTC).
Actually it's quite important. There is a component of the angular momentum due to the spin of the object itself. If we take the ball on the string, we'll note that the ball is spinning with the same angular velocity as it's being whirled (since it's tied on). The angular momentum we get is the following (assuming a ball of uniform density & a string of negligible mass).
L  = Iω
= m(r2 + 12R2)ω
Where R is the radius of the ball, r = |r| and everything else is defined as on the page. The mr2ω is the r×mv mentioned but the 12mR2ω is due to the ball's own spinning. We can ignore the 12mR2ω if and only if we have a long string relative to the radius of the ball (r>>R). Planets are a bit of a different case because they aren't tied to the Sun by string but are free to rotate anyhow they feel like. The angular rotational speed of the Earth, for example, is 365.242... times its angular orbital speed and the angular velocities are not in the same direction, but since the square of the distance to the Sun is about 550 million times the square of the Earth's radius we can still use the approximation. Nevertheless it is an approximation and so I'd imagine it would best be explicitly described as one so as to avoid confusion. As for my other edits: the expression crossed didn't seem of encyclopædic tone, in fact it seemed a bit clumsy, nor was it very clear (especially if you print the page and thus can't go to the wikilink); origin likewise was completely unclear (undefined and jargonistic); the rest was punctuation and the delinking I mentioned. I hope this clears thing up. I'll be reverting the revert, I hope you don't mind. Jimp 11:09, 10 June 2014 (UTC)