Talk:Angular momentum/Archive 1

Archive 1

Archive 2

this is just too nerdy... way too nerdy

Seriously, this is an important topic for someone trying to grasp quantum mechanics and it just not explained well at all. I saw below that someone had removed the laymans description - for fair reasons, it was shit, - but really please give us a new one. What the hell is this thing describing?

I think there's probably a good case for an article called something like Introduction to angular momentum.- Wolfkeeper 04:50, 6 February 2010 (UTC)

Old, unsectioned comments

The reason the proof of angular momentum was originally attached to Torque as a subpage, was because I considered the proof trivial, and not interesting to the majority of users interested in the article. In fact, I thought it would scare aware the math-phobic general public. I wrote it shortly after I rewrote Torque. The previous version of that article contained the misconception that the derivative of angular momentum is equal to torque only in special cases. I wrote Torque/Angular momentum proof to justify my alteration, to explain to whoever wrote the article why I'm right, and of course to increase the amount of information in Wikipedia by a tiny, trivial amount. I disagree with its inclusion in Angular momentum because the inclusion of this trivial, apparently random factoid is inelegant, confusing, and makes the article overly mathematical. -- Tim Starling 01:52 Apr 28, 2003 (UTC)

Only, it's non-trivial. To make it trivial requires certain quite reasonable assumptions - but a rigorous demonstration is far more awkward than (say) energy conservation. Essentially, in general you are dealing with an infinite number of particles. Then you have to assume that some terms in an infinite sum go to zero, which isn't obvious. PML.

That depends on your definition of trivial, doesn't it? There's trivial as in the maths is easy, and there's trivial as in List of songs which have the word Song in title or lyrics. I think it's both -- it's trivial mathematically for someone competent in the field, and also trivial in the sense that most people reading angular momentum or torque don't care about how to prove this identity. But I'm getting off the point. Even if it were non-trivial mathematically, it still shouldn't be in the article, due to the confusion factor. Since confusion factor increases with non-triviality, your statement seems to support my main point. -- Tim Starling 03:02 Apr 28, 2003 (UTC)

I would say it supports the reverse: since there is something to be said, it is better to note it than to slide past it. Your remarks, while accurate as far as they go, don't lead to us leaving it out but to us making the editing draw attention to the fact that there is more for interested people to follow up - without distracting casual browsers. The principle of a good encyclopaedia. Now, how to achieve both? If in doubt, I'd rather leave it in. PML.

How do we achieve both? Simple, by leaving it how it was when I wrote it, i.e. on its own page. That's the whole point -- I'm responding to Looxix suggesting merging Proof of angular momentum (which started out as Torque/Angular momentum proof) with Angular momentum, an act which I disagree with. Sorry if I didn't make that clear. -- Tim Starling 05:44 Apr 28, 2003 (UTC)

No problem for me, it was only an suggestion/question. In fact, until we have the possibility to link within a specific part of a page, I often prefer to have well linked small pages than one BIG page having all the sub-subject linked to the main article. -- Looxix 20:47 Apr 28, 2003 (UTC)

Tesla coil

I removed the following paragraph, which seems a bit dubious:

A good model of angular momentum is the inventor Nikola Tesla's Tesla coil, where due to angular momentum of the field, a energy force can be pulled off at 90 degrees to the spin of the field on the coil, called phase,see also polyphase coils.

This might be true, but it's not very clear. What is an 'energy force'? -- Heron 19:33, 4 Jul 2004 (UTC)

Definition

I'm not happy about the definition used at the beginning of this article. I think we ought to be able to improve on this. I'd like to suggest the following. Any comments / improvements, please? Ian Cairns 22:37, 2 Oct 2004 (UTC)

"Angular Momentum is the tendency of an object orbiting an origin to continue orbiting. This is the rotational analog(ue) of linear momentum, and reflects the moment of the linear momentum of the object orbiting the origin. A body rotating about an axis can be considered to be the sum of its individual particles. Likewise the body's angular momentum can be considered as the sum of the individual angular momenta of all the particles' The SI unit of angular momentum is the kilogram metre squared radian per sec (kg m² s^-1)."

seconded if no one speak against, i'll implement it next week. 12 Dec 2004

I do not think it is a good definition, since it conveys the idea of some kind of inertia that will keep a particle in orbit. This is not true, since a force is ´´´always´´´ required for the particle to follow the orbit (although a torque may not be necessary) . TomasG

the current def. is really poor, (the one above seems better, but do see a problem with orbit, AM is usually about rigid bodies, orbits usually considering LM); "In physics, the angular momentum of an object rotating about some reference point is the measure of the extent to which the object will continue to rotate about that point unless acted upon by an external torque." reasons; . "in physics", what else could it be? . "reference point" is a point from which something else is measured, not as used here, . a spinning object can move its center of rotation to another point, if that point is very close by, very easily, this is not whats written here . "external torque" understanding this term requires at least as much physics knowledge as the definition is trying to put accross, so does not help understanding. Asplace 14:29, 24 February 2007 (UTC)

I think what's really necessary is a definition that doesn't mention rotation in order to encompass the generality of the concept of angular momentum. Objects moving in straight lines can have angular momentum. I know you're thinking that that would make it unintuitive or something, but in the real world, that's what it is: angular momentum is not something intuitive. It really is just an abstract vector quantity that just happens to be conserved. RobertRHannah89 (talk) 16:20, 15 January 2009 (UTC)

Thank you all for listening to my comments about the meaning of angular momentum! Take a look at the link that I posted (and wrote incidentally) abouf the conservation of angular momentum (in "see also"). It makes what I said a little clearer RobertRHannah89 (talk) 1:15, 25 April 2009 (GMT+10)

If a proof is needed, reference a published one, rather than your own. And don't put external links into the see also section. Dicklyon (talk) 16:43, 24 April 2010 (UTC)

He is right though. Angular momentum is defined by the equation which is just to do with the sum of relative motions multiplied by their perpendicular distances; whereas you only get rotation when there's additionally centripetal forces between parts that are in relative motion.- Wolfkeeper 17:30, 24 April 2010 (UTC)

It's just a matter of how you define rotation. Striaght motion not through a point is rotation from the point of view of that point. Dicklyon (talk) 21:11, 24 April 2010 (UTC)

Nah ;-) A rotation is typically defined in terms of cos and sin and wt stuff. You can't do that with straight lines (unless the rotation axis is at infinity, with infinite radius, which is really dubious.- Wolfkeeper 22:47, 24 April 2010 (UTC)

Really angular momentum is a measure of shear; it's like I once saw somebody lay down on top of a pair of straight escalator handrails that were moving in opposite directions, their body turned around in a flat rotation. It's because shear and rotation are deeply linked... but not the same thing. In that case there's shear from the handrails.- Wolfkeeper 22:47, 24 April 2010 (UTC)

Of course you can do that with straight motion. Just take the radial from the point in question and work with the changing angle. It doesn't matter whether the motion is pure rotation, or straight, or shear; it all gives the right answer if you apply the right equations. It's called angular because it's most easily conceptualized with pure rotation, or with rotation of the radials to the point masses for general motions. Dicklyon (talk) 01:40, 25 April 2010 (UTC)

Dicklyon expresses the matter admirably. Xxanthippe (talk) 02:34, 25 April 2010 (UTC).

diagrams

Diagrams!!! We need diagrams! Otherwise it is very hard for young students to understand what is being talk about in this article. For example the angles. Without diagrams the young students are merely guessing where the angle is measured from.

Post requests to Wikipedia:Requested pictures, if you will. —Josh Lee 02:39, May 6, 2005 (UTC)

Cutting editorial comment from the article page here for discussion

I cut the editorial comment below for someone to integrate as text in the article page.

• • • • Comment from somebody else: this only defines "torque" as being the time derivative of angular momentum - when is the torque zero? Torque does not appear in any of Newton's laws, and should therefore not be used to explain under which circumstances angular momentum is conserved. The time derivative of the angular momentum is not a law in analogy to Newton's second law, but a consequence of Newton's law: simply take the time derivative of the angular momentum as defined above, and plug in Newton's laws. It then follows that a particle in a central force field has its angular momentum conserved. This was shown explicitly in a previous entry on angular momentum, which has been removed. *******

The above editorial comment was added in this edit. ---Rednblu | Talk 20:58, 6 Jan 2005 (UTC)

Relativistic angular momentum

Does anyone know if there is a symmetry under spacetime rotations in quantum field theory?—Preceding unsigned comment added by 212.202.0.66 (talk) 12:10, 18 July 2005

Kinetic energy

I think it would be useful to show the relatationship between kinetic energy (I*omega^2) and angular momentum. What do you think?

I added the link rotational energy.--Patrick 22:31, 1 November 2005 (UTC)

Kinetic energy in terms of angular momentum is (L*omega/2). Kenny56 02:02, 2 November 2005 (UTC)

When a rotating object is required to store an increase in kinetic energy within its system of rotation, it stores it in increased units of angular momentum.WFPM (talk) 18:36, 13 February 2011 (UTC)

@WFPM: I disagree with you. The units of angular momentum are not the same as the units of energy. (They are not even the same as the units of linear momentum.) It would not even be true to say whenever the kinetic energy of a rotating object increases, the angular momentum also increases. To illustrate this, imagine a rotating system with a variable polar moment of inertia I. Imagine the polar moment of inertia halving in magnitude and the angular velocity doubling. Angular momentum does not change. Kinetic energy is a function of the square of the angular velocity, so the kinetic energy of this system doubles. Dolphin (t) 06:38, 14 February 2011 (UTC)

The matter within a rotating material system contains kinetic energy, but it's not evenly distributed amongst the constituents of the material system. And when we examine the distribution of kinetic energy amongst the constituents, we find that each element has a circular motion around an axis of rotation, with the value of the motion of each element being equal to its vr (velocity times radius) value. And thus the total resultant motion of the rotating material system is the Mvr value. And if you double the amount of kinetic energy contained within a system (with the same mass value) it will result in a doubling of the sum of the vr values of the constituents.WFPM (talk) 15:47, 14 February 2011 (UTC) However, if I reduce the constituents of the rotating material system to only 1 unit that is rotating at a radius r with a velocity v and examine its kinetic energy versus angular momentum relationship, I note that if I double the kinetic energy content, the v value would only increase by the factor of the square root of 2, as also would the vr value. So maybe the angular momentum value relationship is to the square root of the kinetic energy value and , if so, I apologize and thank you for the input. I'll think more about it.WFPM (talk) 16:45, 14 February 2011 (UTC)

Thanks for your prompt and thoughtful response.

Angular momentum doesn't have a square root relationship with kinetic energy. If a rotating system has polar moment of inertia I and angular velocity ω its angular momentum is:

Iω and its kinetic energy is:

½ I (ω)²

The relationship between kinetic energy KE and angular momentum L is:

$KE=L*{\frac {\omega }{2}}$

Dolphin (t) 22:36, 14 February 2011 (UTC)

Let's please stick with my single rotating object for further analysis in this matter. My Robeson says that its moment of inertia would be equal to its mass M times the square of its radius r. And its contained kinetic energy value would be equal to one half of the square of its velocity, which would be one half of the square of its (omega times r) value. So we're looking for the relationship between the kinetic energy (omega squared times r squared)/2 value and the angular momentum (vr or omega times r squared) value, and we find the KE/Mvr ratio to be proportional to one half of the v/r value. But let's fix the value of r, for the single example, or for any solid object, and then the KE/Mvr value becomes related to just one half of the v value, which increases as the square root of the kinetic energy. So if we double the KE the increase in the mvr value increases in proportion to the square root of the KE increase ratio. However I can see that if you want to change around the configuration of the rotating material mass, so as to change its radius of rotation value, you could then change the r value in the equation.WFPM (talk) 15:51, 15 February 2011 (UTC)

It is true that the moment of inertia of a particle rotating about an axis at distance r from the axis is the mass of the particle times r². However, for the more general case of multiple particles or a rigid body, there is no single value for r and the moment of inertia is based on the square of the radius of gyration. If you consider the radius of gyration to be a constant, your results will be true for that special case, but they won't be true for systems of particles in general, or rigid bodies in general. Dolphin (t) 20:54, 15 February 2011 (UTC)

I guess I'm leaving out the part about how a rigid material body would uniformly acquire the same additional ratio of increased kinetic energy. Because if it kept the same mass value and radius of gyration value then only the v value would change and the math would work out. Isn't that correct?WFPM (talk) 21:38, 15 February 2011 (UTC)

So we can say that the value of angular momentum is related to the storage of kinetic energy in rotating objects, and varies in proportion to the square root of the stored kinetic energy (around a specific axis) in rigidly configured objects. In freely moving (orbiting) objects the stored angular momentum remains a constant, unless disturbed by a force.WFPM (talk) 15:32, 24 February 2011 (UTC)

Sounds plausible. Are you proposing adding that connectin between kinetic energy and angular momentum to the article? If so, based on what source? I'd be happy to review. Dicklyon (talk) 15:47, 24 February 2011 (UTC)

Gee! I'm just responding to the question about what happens when you try to store kinetic energy in a rotating body. To keep track of it in kinetic energy values is hard because of the variable amount stored within the constituent parts, So I imagined the problem of twirling a baton and worked out the distribution of energy values in the different lengths of the baton and came to that equivalent energy storage value. And the source is obviously Newton's laws of physical motion as related to the kinetic energy values.WFPM (talk) 19:35, 24 February 2011 (UTC) An interesting point on this matter that I tried to make to Martin on his talk page is that although a rotating object stores kinetic energy as angular momentum, it is still possible to extract kinetic energy from it like from a baseball bat by letting it supply a linear force on a baseball and then redistributing the remaining kinetic energy internally. But he didn't like my physics.WFPM (talk) 21:31, 24 February 2011 (UTC)

Oh, I didn't realize you were answering the 2005 query just for fun; I thought you were talking about the article. Never mind. Dicklyon (talk) 00:22, 25 February 2011 (UTC)

The article is typical about the subject matter in that it departs almost immediately into terminology requiring elements of knowledge not relevant to the principles of operation of the subject matter under discussion, and so I have to shop around in the talk section to find a section that pursues the ideas about the subject matter that I'm interested in. And the relative importance of the ideas is, of course, debatable, but I will argue, in accordance with Newton's rules of Philosophy, that the simplest one is probably the best.WFPM (talk) 19:54, 25 February 2011 (UTC)

@WFPM: On 24 February 2011 you wrote (see above) a rotating object stores kinetic energy as angular momentum. This is the same as your comment above on 13 February 2011 where you wrote that a rotating object stores it [kinetic energy] in increased units of angular momentum. This is not a scientific viewpoint. There is no reason why an object can't store kinetic energy as kinetic energy! As I explained above on 14 February 2011, angular momentum doesn't even have units of energy. In the SI system, angular momentum has units of kg.m².s^-1 and these are not the same as Joules. (Joules are the same as kg.m².s^-2) Dolphin (t) 02:26, 25 February 2011 (UTC)

Well certainly it contains kinetic energy! But the distribution of the kinetic energy within the mass constituents is uneven. And so we need a system of measurement and analysis of some property of a rotating system that can be better measured and analyzed as a means of keeping track of the dynamics of the situation. And we've come up with its property of angular momentum, which can be mathematically related to its kinetic energy content value and also used to determine how best to add to or subtract from the kinetic energy contained within the rotating system.WFPM (talk) 16:46, 25 February 2011 (UTC)

Please note that some systems of rotation have centers of mass concentration that can be dimensionally determined such as to permit an almost direct effect on the contained kinetic energy of the system. Others do not. But we can hypothetically overcome this difficulty by dreaming up some torque or other nonlinear force application method of changing the kinetic energy contained in the rotating systems. And we can also get involved in situations where the energy distribution system leads to indeterminable high velocity and resultant kinetic energy values as occurs in the operating characteristics of the whip.WFPM (talk) 17:30, 25 February 2011 (UTC)

Sorry to be so long winded on this subject matter and appreciate having my contributions defended.WFPM (talk) 03:14, 4 March 2011 (UTC)

Angular momentum in relativistic mechanics

Although this section is very short (and very interresting), I think that it is VERY advanced! Quite frankly, I am not sure that the average reader of this page would find it very usefull!

There should be a way to separate the basic introduction of the subject to the more advanced material.

Alain Michaud 04:19, 19 February 2006 (UTC)

Units

This page needs to mention the units of angular momentum.

Operator

As far as im aware, the $L^{2}$ operator is missing a factor of minus h-bar squared in the front there.

Ice Skater, Big Bang, KE goes up!

I was once asked, Where do stars get their energy from? Where does the energy for the spinning earths core come from? Nuclear fusion for the first case, but what started out the process? My answer was "Conservation of Angular momentum". As a spinning rotating body contracts due to gravitation, Work is done: W=FxD. Ice skaters expend energy to bring their arms and legs to the center. In a primordial dust cloud this energy is converted to heat. Rotating contracting dust clouds get hotter until plants form and fusion is initiated in the center. All thanks to "Conservation of Angular momentum" and realizing that the original source of the energy comes from resigual potential / kinetick energy left over from the big bang.

Is this a fair answer and is it worthy of including as an interesting discussion for the general public.

--Tbmorgan74 18:59, 6 June 2006 (UTC)

The explanatory power of angular momentum

In the article it is stated:

The conservation of angular momentum explains the angular acceleration of an ice skater as she brings her arms and legs close to the vertical axis of rotation. By bringing part of mass of her body closer to the axis she decreases her body's moment of inertia. As angular momentum is constant in the absence of external torques, the angular velocity (rotational speed) of the skater has to increase.

In my judgement, the logic of this statement is flawed. My intention is to replace the flawed statement with a correct one. In an article that I wrote for my own website, you can read what in my judgement the correct presentation is. Conservation of angular momentum (the sections 'a centripetal force doing work' and 'causality') --Cleonis | Talk 09:11, 14 July 2006 (UTC)

Smoke rings

Angular Momentum is a cross product, it is not a vector. Consider the Angular Momentum of a smoke ring.

• That's a question of definitions. OK it doesn't "transform like a vector", but the only author I've read who used that definition is Einstein. One modern definition of a vector is as an element of a vector space. The other main one is of an element of a three-dimensional vector space with extra operations (dot product, cross product). According to both definitions angular momentum is a perfectly reasonable vector. If you want to define vector differently and change every entry in the wikipedia to reflect that then go ahead. If I had my way angular momentum would be a skew-adjoint linear transformation, but I tried that on Angular velocity once and I admitted defeat fairly quickly. Buster79 15:26, 8 October 2006 (UTC)

What is the Angular Momentum of the Universe? Is it a zero tensor? Cave Draco 01:14, 7 October 2006 (UTC)

• What's your point? Buster79 15:26, 8 October 2006 (UTC)

If there is only 1 universe, logic says it would have Zero Angular momentum. However, if there are more than one, then they could individually have individual amounts, with the sum being zero.WFPM (talk) 21:15, 12 March 2011 (UTC)

Angular Momentum in layman's terms

"In layman's terms, it describes the "fasterness" of moving an object after changing its "closerness" to its "centerness" of rotation. A person in a spinning chair, who moves his legs and head inward toward the "centerness" of rotationality, will therefore, spin with more fasterness."

This doesn't make sense, rotationality isn't even a word. Someone might like to have another go at adding a layman's definition in there, though personally I don't see what's not to understand :) -TeeEmCee 21:23, 21 November 2006 (UTC)

We could use a better one, but that does make the most sense for a layman than anything else i've read. Something a bit easier to visualize than the anonymous "particle" would help. -RikiTikiTavi

L=Iw, vector or scalar

In the equation L=Iw, L is bold and I ad w are italic. Bold means vector and italic means scalar. This should be fixed so that both sides of the equation have the same types. --Alanic 23:08, 8 January 2007 (UTC)

Conservation of Angular Momentum

In the section on the conservation of angular momentum, this article states that if the external torques add to zero, then the angular momentum is constant. However, this fact is NOT obvious because we have not considered internal torques. In fact, Newton's laws do NOT allow us to assume that internal torques always add to zero. The Kleppner and Kolenkow textbook on mechanics has a long discussion on this, where it concludes that it is not possible to prove mathematically that angular momentum is conserved and is an empirical fact. Does anyone with more expertise on this topic care to comment and maybe even change the article? -maxx

Now I'm wondering if MY question is the same as yours Maxx, excepting that I'm asking it from the point of view of a non-specialist. But it has to do with how the "preservation of angular momentum in the Earth / Moon system works. I have posted a 10 point question at the talk page on Moon, and it's basically asking whether the Moon would still recess in order "to preserve combined angular momentum" if the tidal effects were insignificant. I should think not. I am not going to reprint the whole thing here, but here's the link Talk:Moon. And go to Point 11 on the list. I would certainly appreciate any answer either here or there--Myles325a 14:30, 11 April 2007 (UTC)

I've not read Kleppner and Kolenkow, and perhaps their discussion goes deeper than most, including relativistic considerations, etc. However, what I do know is that it's a fairly simple matter to show in the context of Newtonian mechanics that if the total external torque acting on a system is zero, then the total angular momentum of that system is conserved (see Corben and Stehle "Classical Mechanics", pp. 36-38, Dover, 2nd Ed.). Another way of stating this fact is that total angular momentum is conserved if the Hamiltonian (or energy function) describing the system is rotationally invariant. This is a special case of Noether's theorem, which says that, for every symmetry of a system, there corresponds a conserved quantity (and her theorem yields a prescription for computing these quantities as well).Milez (talk) 00:18, 14 April 2008 (UTC)

Conservation is an inescapable consequence of the veracity of Newton's laws. It might seem as if Newton's laws don't imply conservation of angular momentum until you look at the STRONG FORM of law number three. The proof can be completed from there in less than a quarter of a page. Isotropic space does not imply conservation of angular momentum, the nature of forces in general (Newton's third law) does. I can expand on that if you're curious. RobertRHannah89 (talk) 15:30, 15 January 2009 (UTC)

Robert, you say "isotropic space does not imply conservation of angular momentum". Are you familiar with Noether's theorem? If the laws of physics don't change when the universe is rotated -- and everyone believes they don't -- then conservation of angular momentum has to be true by Noether's theorem. The proof you like that uses Newton's laws is less general than the proof from Noether's theorem, because the Newton's laws are only true in classical mechanics while Noether's theorem applies immediately to classical systems, quantum systems, the standard model of particle physics, etc. Also, Noether's theorem I think gives a deeper understanding, by relating it to an intuitively-obvious symmetry of physics. :-) --Steve (talk) 16:44, 15 January 2009 (UTC)

Hamiltonian is not gauge invariant

The Hamiltonian given in the article is invariant only under gauge transformations that do not depend on time. Xxanthippe 03:14, 5 April 2007 (UTC)

O.K. Now give the people something they can read....

rant
The following discussion has been closed. Please do not modify it.
This is not an encyclopedia article really. It is of almost no use to anyone who would be reading Wikipedia. Try to keep the context in mind. The person who reads this is not doing Physics at Uni. No, they have come upon some term and would like to knnw a little about it, just the JIST of the matter. Consider that they are disinclined to go to a reference library, and are too cheap to shell out for a sub to Britannica. No, what do you think a punter like that would make of the hieroglyphs you have got there? Absolutely nothing. And yet, there is a fair bit the average mortal COULD know about angular momentum without dedicating themselves to a life of being a physicist. After all, there are academic treatises on anthropological issues which the lay reader could never understand, and does not want to understand. But in these fields, there are quite a few reference sources which will enlighten average Joe Blow as to what is happening in the Moggabogga tribe without weighing them down with a great deal of jargon, and material which is of interest primarily to the specialist. August and widely respected intermediary publications like New Scientist and Scientific American (and I read both) manage to bridge the gap between cocooned boffins and an interested and reasonably well-educated LAY audience. If they can do it, why is it that a bunch of amateurs, working for free, and writing for a lot of school students who can't / won't get to a library, feel impelled to dump this stuff on them. What, is it a "Look at me, I'm way, way, way ahead of you lot? Watch the master at work" syndrome"? I suppose that before I started looking up stuff in WP, I had the idea that an Encyclopedia written BY the masses FOR the masses, would of course be far more accessible than one written by experts. Then I realised that that was not the case. At first, it seemed counter-intuitive. But then I began to see score upon score of articles, many on subjects far less forbidding than this one, but with a common problem. They were not accessible to the great unwashed, many were deliberately obscurantist, many had talk pages in which authors made open avowals that they would never compromise on their "work" by trying to "oversimplify" their material, that the Joe Blow who had stumbled on their pronouncements just had to realize the wonderful, incredible complexity of this material. Or otherwise, they should just take up macrame, or something. I then came to an understanding. You see, in my working life, I spent a fair bit of time compiling manuals and tutorials for novices in computing areas. Most of my time was not spent in demonstrating the august splendour of which I had such mastery. In fact, most of my time and effort was spent in doing the very opposite. The real skill in my job was telling people something in as sweet, plain and simple a language as I could. I never burdened them with some aspect of computing that they did not need to know to do their jobs. I never used a technical term that had a plain English parallel that would do. I spent a fair bit of time grading the material, so that those who did not NEED to know the more complex stuff did not have to wade thru it to get to what they did need. But it was only until I came upon the phenomena of Wikipedia, and thought about it some, that I began to understand that the PRIMARY skill in any kind of explanatory writing lies in COMMUNICATION. In Communication, NOT in expertise in the subject. It takes a good communicator to make the difficult simple, it takes a COMMUNICATOR to assess from first principles - BEFORE they begin to write - WHAT people they are writing for, WHY these people are liable to be reading this stuff, WHAT they intend to do with it, HOW MUCH they know already, WHAT can I spare them, HOW can I grade the material so that it does not become dispiriting? HOW can I provide safety nets so that if they do lose the way, they can find the path again before they lose hope? And then, while I was thinking of these matters, I would also have to think about how I could make a bridge BETWEEN what the trainee had experienced, what he/she had seen, was familiar with and even liked, to the more abstruse, foreign world I was trying to introduce them to. I spent a lot of time thinking about how I might do these things, a lot of time talking to the people who would be reading and learning from these manuals. I spent about 90% of my writing time on doing that and 10% of it on doing the research. Now I didn't think that much of it back then. I guess I sort of keep my light under a bushel a bit. But it was only when I started becoming familiar with the Wikipedia world, that I bagan to realise just how good I was. I was a TEACHER, I was a COMMUNICATOR. I was not primarily an expert. And just as well, because many experts are rotten communicators. Of course, I had to be at least a second tier expert, and I guess that's why I was being paid. I had to have a useful mix of two skills not often found living happily together - the communicator who facilitates the education of people with the minimum of tears and the maximum of wonder and enjoyment, and the expert, who knows enough to at least give a reasonably accurate and useful account of the matter in question. Now, a while back, I came upon an article in WP which purported to be NOT an article on SPECIAL RELATIVITY, but an Introduction to that subject. In other words, an introduction to an introduction to special relativity. And what did I find? Within 3 PARS of the start, we are into the kind of mathematical scribbling we see in YOUR article. Here is the VERY FIRST sentence of that "introduction". 'Everything started with the Michelson-Morley experiment to determine the absolute speed of the Earth through the theorised "luminiferous aether."' Why, I wondered, was it necessary to broach the concept of "luminiferous aether" in the first sentence of a introduction to an introduction to a brief entry on Special Relativity designed for people who won't shell out for a Britannica? And why mention "luminiferous aether" at all, considering that is not only a forbidding technical term, but an antiquated one to boot? Why indeed? Eventually, I came up with the idea that the author did not know - or simply did not care - if people understood him or not. Rather like you. After all, neither he nor you, nor about a thousand others out there, are being PAID to write comprehensible English. I guess that's the difference. I WAS heing paid to write same, and I would have been out of a job had I not made every effort to get as MANY PEOPLE AS POSSIBLE to understand what it was I was saying, and so become proficient and happy workers. As it is, my work was widely regarded, by people who had a real vested interest in getting their staff to understand new technology, understand it well, and even - hey why not! - actually enjoy it. Now I imagine that if you were being paid a couple of dollars for every poor uneducated sod who read your material and came away a little wiser, then you would, I am sure, place a greater premium on clear and intelligible writing for a lay audience. And further, were you ACTUALLY BEING PAID a reasonable return, and you were menaced with dismissal if those depending on your tuition did NOT learn what they needed to know, then I feel sure that this would concentrate your mind wonderfully, and you would go to great lengths to write articles of the most lucid and transparent simplicity and beauty. But as it is, you are not, and you will be damned if you are going to make concessions for the stupid plebs. Now, where was I? As yes, ANGULAR MOMENTUM. Well it's not the Bemuda Triangle, sure. It's not Paris Hilton. But it does have potential. Want to know the first thing I noticed when I opened the page. First off, all I saw were all those equations which would require college level maths to understand, and which therefore are a completely closed world to me and all but the most minute fraction of the human race. The second was the "ice skater". I always look for her when anybody tries to explain "angular momentum". And there she was, but instead of being in the first par, as the sort of illustration of angular momentum in action that everyone has seen, everyone enjoys, and thus is a PERFECT lead in to explaining what is going on as she spins faster by pulling in her arms, she is well into the body of the text, and the attendant description does not use HER to explain angular momentum, but utilizes the image replete with all the jargon. Now, how did I get to the ANGULAR MOMENTUM article, anyway? You probably don't care by now, and I don't blame you. Once again, in my erstwhile job of making difficult things easy to understand rather than the reverse, I made sure I knew WHY people needed to know the stuff I was about to teach them. If our positions were reversed and it was me writing on "angular momentum" I might, for example, be interested in how people arrived at the article in the first place. You know, "Did you hear about the WP angular momentum article from your friends, your colleagues, or did you read about it in magazine, or did you simply lose a bet?" Well, this kind of information is of the UTMOST importance to any writer with a claim to being a purveyor of knowledge. Why do people read - or at least had that intention at the start - an article on angular momentum? Well, in my case, I had been reading, and proofreading the WP article on the Moon, and noticed that the material there on the Moon's recession from the Earth, due to angular momentum concepts, was less than perfectly clear. I had pondered in the past as to how the Moon might "sense" Earth's rotation across the space that separates them - and slow down accordingly. The whole question of the rate of the Moon's recession - whether it would stop, or go on forever, or stablize - was one which interested me. There was a link from there to angular momentum, so I took it, and I did not profit thereby. Now I am not underestimating the difficulty of this subject. But, I would suggest it incumbent on the writer in such cases to make a concerted effort to take what IS EXPLICABLE concerning the subject and ISOLATE it. Then, using all the good educator's techniques of moving from the particular to the general, moving from the reader's everyday experience to experiences less familiar and so on, make an effort so that a diligent reader will have an "AHA" moment. Then a bit more slog, and then another AHA, and by the end, there will be a bit of light in that reader's head. They will know next to nothing about "angular momentum" when you figure how very much more YOU know about it. But they don't HAVE to know much about it, and if they did, they wouldn't be studying it there in WP. As it is, I'm not a deadhead, and I can tell you, I most certainly did not get a AHA moment out of your article. No, there was no glimpse of the Promised Land. I came in confused and left even more confused. Everything was and remains complete darkness. And I can tell you that I am not champing at the bit to get a eyeful of your "Torque" article. No sir, I've had enough for one day. But I can tell you, that if you working for me, then you wouldn't be working for me. I guess that's part of the lesson of Wikipedia. It's an old one, and many would say, a cynical one, but true nonetheless. You get what you pay for. Still, even so, it is astonishing at times just how good a lot of the stuff is. I've noticed that with many articles, the Discussion page throws just as much light - if not more - on the subject, than does the main entry. And it has the sort of anarchic, to-and-froing, disputational flavour that is actually a lot closer to how knowledge is won from the mud of apathy than does the calf-bound, hide-bound, letter perfect productions of the big publishing houses. But not here. Turning to the Discussion Page, I found the meditations of a very-like minded coterie of physicists. All of them are genuinely interested in the subject, and at a pretty advanced level, which is to be admired. But not one of them had the slightest interest in the issues I have raised. It was as if they were hermetically sealed off from all the readers who had appraoched the topic because THEY KNEW NOTHING ABOUT IT. WP is not designed as a common room in an ivy-league university. It is meant to be a repository of accessible knowledge for a PEOPLE'S ENCYCLOPEDIA. And this is not an article which falls within that ambit. So, how can you do better? Well, I won't repeat myself. Hey, you probably disagree with everything I wrote here. You probably hate it. But you READ it. And I'll bet you understood every word. Myles325a 19:36, 8 April 2007 (UTC) Actually, while I appreciate what you say (at length), I'm a fourth year undergrad who's brain isn't working and was looking for alternative angular momentum definitions (for an accretion disk problem where packets of mass are falling onto a central star), and I suspect that Angular momentum is an article that is only ever going to looked at by people studying it, albeit quite probably at a school level. I agree the article looks a bit busy. GBM 23:18, 10 April 2007 (UTC) Well, that was kinda my point. Wikipedia is not really here to cater to fourth year undergraduates. I would imagine that there would be any number of academic sites you could have googled, not to mention the material in your university library, where there is probably a whole floor of stuff for you to look at. This site’s first priority should be to serve the non-specialist, the general public in other words. I thought I had dealt with that in my piece. And no, lots of people run into the concept of angular momentum, often, as in my case, coming from such relatively non-technical topics like the Earth-Moon interaction. I also dealt with this in my piece. I would have suggested in these cases at aiming for people who are reasonably well-educated, have an amateur interest in physics, cosmology, astronomy and so on, but only rudimentary maths. The task is to provide a non-technical background explanation to these topics for a lay readership, only moving to the technical stuff at the end. If Scientific American and New Scientist can do it, then THIS organisation can. I also dealt with that, as you point out, at length. While I wrote at length, I can’t say your perusal was any more than cursory. --Myles325a 01:05, 11 April 2007 (UTC) If Scientific American and New Scientist can do it As a former reader of those magazines as a teenager and later a Physics student I can absolutely tell you. They Can't. The article is not complicated, learning takes effort and unless you have some concrete suggestions to the article you should just stop rambling. 89.181.97.102 (talk) —Preceding comment was added at 01:25, 12 December 2007 (UTC) Wow. And I thought I was wordy! :o) But while I've lost a lot of respect for New Scientist since their recent 'Darwin Was Wrong' nonsense (24 Jan 09), I have to otherwise agree with Myles. I like to think I have a reasonably good amateur's grasp of physics. I can visualise what I think are probably some reasonably complicated ideas - for example, I think I understand (to some extent) notions like gravitation, spacetime, time dilation, the expansion of the universe, and so on. Certainly I wouldn't claim expertise, but I've had conversations with people properly trained in physics and they didn't give me any reason to believe I was going hugely off-target. But for me these things are a bit like what pilots call seat-of-the-pants flying. Or, better still, like catching a ball. I can see the ideas, and I can see what they do, but I couldn't in a month of Sundays sit down and write out the maths behind them. And similarly, while I can grasp the aforementioned skater analogy, I have absolutely no clue at all what the bulk of this article says. This is how much I understood: "In physics, the angular momentum of a particle..." The anonymous contributor from 89.181.97.102 claims that the article is not complicated. But I suspect that that ties in to Myles's point: it probably isn't if you're already equipped to understand the mathematics involved. Anyone who could argue - with a straight face - that the article as it stands is simple for the lay person is obviously completely out of touch with the real world. If you have the foundation training in equations and such, then I've no doubt that reading the mathematical glyphs of which this article is primarily composed is no different from reading text. And I've seen the claim on here before that "learning is supposed to be hard" - but that misses the point of Wikipedia to some degree. Of course there's no reason why this encyclopaedia shouldn't go into infinitesimal detail about a subject if its authors wish - but it should also be borne in mind that it serves a non-expert readership as well. There's no reason why a clear, simplified explanation for the non-expert couldn't be followed up by all the equations you guys can think of. And I don't think Myles was 'rambling', as the anonymous contributer so dismissively said. I think the point was well made: some people will come here because they've seen the term and want a brief, rough-and-ready idea of what angular momentum is and what it does. The skater is buried in the middle of this article, but she's doing her thing where a lot of people won't see her, because they'll already have quit. And people may come here for many reasons. I found my way to this article from the website for an online space game I'm considering signing up to. One of the features I was interested in was what flight model it would use - I was hoping it would incorporate Newtonian flight mechanics. It mentioned angular momentum and linked here. So please don't assume that only people studying physics are going to turn up here wanting to know what angular momentum is. 89.181.97.102 demands a concrete suggestion for improving the article. Since it would take an expert to properly decipher and translate what exists on it already, I can't do anything but add a prominent link to the Simple English Wikipedia equivalent article, and good luck to the reader. - Laterensis (talk) 12:10, 4 February 2009 (UTC)

Relevance

Hoaglang, Richard C. and Bara, Mike, Dark Mission: The Secret History of NASA, (2007), Feral House ISBN 978-1-932595-26-0. What has this book got to do with angular momentum? Xxanthippe (talk) 21:52, 8 February 2008 (UTC).

That appears to be some possibly weak justification for what sounds an awful lot like nonsense inserted below AM in ED. --Altered7th (talk) 11:05, 9 February 2008 (UTC)

Thanks. I have deleted it. Xxanthippe (talk) 11:41, 9 February 2008 (UTC).

Picky suggestiong? remove the word massive

2nd to the last sentence in the 2nd paragraph - "... the kinetic energy stored in a massive rotating object such as a flywheel is proportional to the square of the angular momentum." I think that the adjective 'massive' is mis-leading as this will be the case with any rotating object, even one that is of, relatively, low mass. Not being a subject matter expert, I hesitate to directly edit, but I do suggest this for your consideration.

Also, the discussion on the usefulness of this page to the layman is quite interesting in itself. I agree that the page takes some effort to read. Looking around the web, I don't find anything easy that also provides meaningful content. 75.35.217.119 (talk) 16:08, 10 February 2008 (UTC)

With regards to "massive": perhaps in this context 'massive' is being used to indicate the property of simply having mass - not necessarily having a relatively large amount of mass. Nevertheless it isn't the pinnacle of clarity and I agree with 75.35.217.119 that revision would be helpful. The lay reader would benefit more by the removal of the word - anyone predisposed to wonder whether the statement can be extended to the massless will likely already know the answer. --67.160.66.72 (talk) 13:56, 11 February 2008 (UTC)

INTRODUCTION nit-picking

In both paragraphs of the introduction, the author states that the angular momentum of a closed system can't be altered unless acted upon by an external TORQUE. Now, I know this isn't wrong since any force applied to the rotating mass at any given distance from the axis of rotation is equivalent to a TORQUE, but the wording just bothers me. In any classical Newtonian physics text relating to momentum, they will stress that an object will not change its momentum unless acted upon by an external FORCE.

Another issue I have is with the use of a "point" instead of an axis to define the rotation of a body. Using a point as a reference may have some relevance in advanced topics, however for an introduction, it seems counter-intuitive. With an axis (IE: a line), a lay person can visualize an object rotating with a given direction (which experts in the field go on to define with the right hand rule) without even realizing it. However, by saying "a point" instead, I personally first picture a sphere rotating in an given direction whithout rhyme or reason.

In the intent of keeping at least the introduction of this subject as simple as possible, wouldn't it be better to state:

"In physics, the angular momentum of an object rotating about some reference AXIS is the measure of the extent to which the object will continue to rotate about that AXIS unless acted upon by an external FORCE. In particular, if a point mass rotates about an axis, then the angular momentum with respect to the axis is related to the mass, the velocity and the distance of the mass to the axis.

Angular momentum is important in physics because it is a conserved quantity: a system's angular momentum stays constant unless an external FORCE acts on it. When that FORCE is acted upon the object at a given distance from the axis of rotation, the result is a TORQUE. Torque can also be described as the rate at which angular momentum is transferred in or out of the system.

When a rigid body rotates, its resistance to a change in its rotational motion is quantified by something called the Moment of Inertia. This Moment of Inertia is a concept that has alot to do with the shape of the object. In essence, it accounts for all the mass of the object and its distance from the axis of rotation. Follow the link to the Moment of Inertia page for more information.

Angular momentum is an important concept in both physics and engineering, with numerous applications. For example, the kinetic energy stored in a rotating object with a given mass, such as a flywheel, is proportional to the square of the angular momentum. Knowing how much kinetic energy something like a flywheel has, allows engineers to use that knowledge to design systems that can change the orientation of a sattelite in space, or store electrical power to be used later, or smooth out the rotation of a crankshaft in a traditional combustion engine.

Conservation of angular momentum also explains many phenomena in sports and nature."

To me, that has a better flow with fewer confusing "jumps" in knowledge. Please feel free to revise if I got some of the concepts wrong. 141.119.184.10 (talk) 18:24, 22 February 2008 (UTC)

I think the first sentence should remain as it is. First, it is incorrect to state that angular momentum remains constant unless acted on by a force. There are a lot of ways to apply a force that will not result in a change in angular momentum. It is a net torque that is necessary to change the angular momentum of an object, not just a force. But thanks for thinking about it and trying to make improvements to the article. PhySusie (talk) 18:37, 22 February 2008 (UTC)

Adding that clarification "...by an external FORCE resulting in a net TORQUE.", would still convey my point, and clarify the issue.141.119.184.10 (talk) 17:57, 25 February 2008 (UTC)

Angular momentum vs rotation

I'm sorry, but a subtle point of angular momentum is that it doesn't have to involve any rotation at all (and this isn't OR).

For example, if I have a mass moving at 1m/s downwards, 5 cm to the right of my finger, it has an angular momentum about my finger. If the mass moves down a bit (we'll ignore gravity) and then tightens a string, it will then swing around my finger in circular motion, with that same angular momentum.

It's not usually discussed very much, but it is true, and it's not my OR.

Hope this helps.- (User) WolfKeeper (Talk) 23:57, 15 July 2008 (UTC)

I am afraid it does not help at all. Your contribution does not add usefully to the article and the source you cite is worthless. I commend the article on Noether's theorem for a good discussion of the principles underlying angular momentum and other conserved quantities. If you want to write an article on your own interpretation of angular momentum I suggest you start a new one. Xxanthippe (talk) 03:20, 16 July 2008 (UTC).

Um... Which bit of the equation for angular momentum don't you understand? Angular momentum is simply instantaneous linear speed times orthogonal instantaneous distance from the point you're measuring angular momentum about. Absolutely nothing needs to rotate at all. Rotation occurs when objects with angular momentum interact by any kind of forces, but the angular momentum exists in the system (and is conserved) before, during and after they interact. I mean, how the heck do you think it works if a body goes past a planet in an escape orbit? At great distance there's negligible rotation rate around the planet, during the pass, it's rotating around, afterwards the rotation decays to zero again. That's gravity which is a long-range force. Other forces, particularly mechanical ones have only short ranges. Where do you think the angular momentum comes from if it only appears when something is physically rotating if I drop a weight onto the end of a ruler so that it sticks to it? Afterwards, it's spinning. Before no spin.- (User) WolfKeeper (Talk) 03:51, 16 July 2008 (UTC)

This is not my OR, this is just you not understanding the topic.- (User) WolfKeeper (Talk) 03:51, 16 July 2008 (UTC)

If you're going to be rude, you could atleast be CORRECT. Wolfkeeper is perfectly right about the fact the angular momentum need not have anything to do with rotation. What you don't seem to understand "Xxanthippe" is that angular momentum is merely convenient in the study of rotation -- these two topics are only related. The quantity is far more general than some "tendency of an object orbiting an origin". Look at the situation with Newton's three laws (+strong 3rd) in more detail. And now I hope you atleast get THIS connection when I say that a little knowledge is a dangerous thing my friend... RobertRHannah89 (talk) 16:01, 15 January 2009 (UTC)

Angular momentum notation is an effort to tell you that there is some motion going on in a system of matter that is on an offset line from a parallel line through the center of gravity by a certain distance. The result of this is that the kinetic energy distributed amongst the constituents of the system will not be evenly distributed, and that a condition of rotary motion of the matter is going on.WFPM (talk) 16:59, 3 March 2011 (UTC)

Engineering Relevence

Numbers and math are useless without a real world application of the concept. We could spend a hundred thousand man hours describing in detail the composition and properties of a carpenters hammer, but information of 'what is' is as infinite as the universe. 'What is usefull' is how that hammer can be used, and to what goal. Perhaps someone should right a section citing enegineering uses for the concept. For example the increased stablitity of a motorcycle. Or the use to measure change of motion. —Preceding unsigned comment added by 96.2.117.153 (talk) 05:13, 14 September 2008 (UTC)

I wonder how long it took for mankind to learn that a rock is a better tool when it's equipped with a handle?WFPM (talk) 17:03, 3 March 2011 (UTC) That's because with a handle it can be given more angular momentum and thus deliver more energy to the impact area.WFPM (talk) 17:02, 4 March 2011 (UTC)

accessibility

Like a lot of WP articles on science, this one IMO lacks accessibility. It starts out with a definition in terms of the vector cross product. Oops, sorry, 99% of WP readers aren't conversant with the vector cross product. We need some basic, simple examples without math. WP articles are supposed to be intelligible to the general reader; sections in this article such as Relation to spherical harmonics are not accessible to anyone without at least an undergraduate degree in physics.--76.167.77.165 (talk) 04:04, 5 February 2009 (UTC)

This is the reason for the existence of the Simple English Wikipedia. Xxanthippe (talk) 05:57, 5 February 2009 (UTC).

That's one approach, certainly. But, as I said further up, I'm not sure it's entirely in keeping with the purpose of Wikipedia, which is to make information accessible for as wide a readership as possible. Clearly there are people who enjoy making articles as technically precise and detailed as they possibly can. That's not limited to the science articles, by the way: the same tendency occurs on, say, articles about movies that basically go through the entire plot frame by frame. It's a sign that people want to share their knowledge, and that's good. But when those people assume a similar level of understanding on the part of every reader then you have articles like this one, which basically slams the door shut in the face of anyone who doesn't have a thorough grounding in maths and physics.

Packing readers off to an alternative version of Wikipedia is one solution (I added the link to Simple English myself), but I think it will just compound that impression for a lot of people: "You're not smart enough to read the proper article, so go here where we explain it in words of one syllable". And even then, the Simple English version isn't all that Simple. I think what's required on this particular article is a more realistic approach. Explain it in simple - by which I mean plain English - terms first, and by all means put in all the equations you want after that. Just leave the door open for a paragraph or two. - Laterensis (talk) 12:40, 5 February 2009 (UTC)

I think you make an excellent point, why don't you put it in place? -- Talkstosocks 17:17, 12 April 2009 (UTC) —Preceding unsigned comment added by 130.15.161.187 (talk)

Maybe the article should cover angular momentum in 2D and 3D separately. The 2D version is just a scalar, and should be fairly easy for most people.- (User) Wolfkeeper (Talk) 17:41, 12 April 2009 (UTC)

The connection with Kepler's law of areas

The current version of the angular momentum article deals exclusively with a particular case of angular momentum: the angular momentum of a solid that is spinning around its own center of mass.

However, angular momentum has a wider range of applicability than that. For instance, there is also the angular momentum of a planet in its orbit around the sun. That angular momentum is conserved, and this conservation principle has played a seminal role in the history of physics. Kepler's law of areas is a conservation law. In retrospect we see that Kepler's law of areas anticipated the principle of conservation of angular momentum.

In the Principia the very first proof (Book I, Proposition I, Theorem I), is a derivation of Kepler's law of areas from the laws of motion.

If you google with the combination of Kepler and conservation of angular momentum you get oodles of pages pointing out the connection. --Cleonis | Talk 23:38, 12 April 2009 (UTC)

And the periodic exchange of angular momentum around the sun between the earth and the moon is involved in explaining the orbits of both the sun and the moon.WFPM (talk) 18:43, 13 February 2011 (UTC)

New introduction to article

It is wikipedia policy that scientific articles have a buildup structure. The opening paragraphs should be accessible for as wide an audience as possible. As the article proceeds it can move on to increasingly deeper discussion. The reader can stop reading when he has reached his saturation point.

The current version of the article fails to implement that policy. For example, in the second paragraph the reader is slammed with Noether's theorem, a highly advanced subject.

I have written a new starting section for this article. It contains four images, two of them are existing images, two are uploaded by me to wikipedia. Please review the Sandbox version of the new introduction. I will merge this with the article, replacing what I think should be replaced. I am announcing this to give you time to form an opinion. Please respond to the new starting section when it has been merged with the article.

Wolfkeeper mentioned that it is perhaps helpful to first discuss a 2D version of angular momentum and then proceed to the fully 3D version. That is what I did. --Cleonis | Talk 04:10, 30 April 2009 (UTC)

I agree that this article has many shortcomings and I commend the substantial effort you have put in to improve it. I particularly like your notion of a gradual increase in conceptual complexity throughout the article. In the past 100 years the idea of the angular momentum (and energy and momentum) of fields has superseded that of particles alone. That is why a mention of Noether's theorem as the overarching principle is appropriate.

Unfortunately I cannot support the changes that you propose as I do not think that they improve upon the admittedly imperfect present version. For example, I think that your first diagram is too complicated for a neophyte to understand, and the introduction of the notion of a gimbal early in the piece is confusing. A practical suggestion I have to make is that you simplify the conceptual level of your piece, keep it to two dimensions, and use it as a contribution to Simple English Wikipedia. The present state of that article is even worse than that of this one. Xxanthippe (talk) 10:44, 30 April 2009 (UTC).

I get the impression that understanding of angular momentum is a very personal experience. On one hand you suggest that Noether's theorem is not too complicated to present to novices, and on the other hand you suggest that the geometrical representation of angular momentum (which does not require prior mathematical knowledge) is too difficult.
Communicating the concept of angular momentum presents special challenges. Why are both the behavior of gyroscopes and orbital dynamics - in appearance so dissimilar - examples of conservation of angular momentum? I believe that postponing or even avoiding that aspect is a disservice.
The starting section of an article should be as accessible as possible, but there is no point in dummifying the picture to the point of violating the physics, or ignoring half of it.
Angular momentum is inherently a more complex concept than momentum. For instance, to understand the information about angular momentum it is necessary to be familiar with the concept of angular velocity. That necessity can't be circumvented. --Cleonis | Talk 11:17, 30 April 2009 (UTC)

Agree that Noether's theorem is too high up. Yes, there is an advanced connection with rotational symmetry, but in a more straightforward introductory way angular momentum is just a special case of linear momentum (applied to the constituents of rigid bodies). There should be a separate section that details advanced concepts such as the relation to symmetry, quantum mechanics, etc. 124.168.201.193 (talk) 11:26, 25 May 2009 (UTC)

The Sandbox version of the proposed new introduction is still there, but I'm undecided what to do with it. For instance, if it is the intention to introduce angular momentum step by step, then what should be the first and subsequent steps?
As for me, I see the angular momentum of planetary orbits as the best starting point. But contributor 124.168.201.193 regards the motion of constituents of a (spinning) solid as the first step. What angular momentum is, is to some extent in the eye of the beholder.

The current version of the article treats the subject in a very abstract way (thus avoiding subjectivity). Advantage: the math and physics is correct and undisputed. Disadvantage: the content is inaccessible for people who seek a first introduction. --Cleonis | Talk 19:20, 26 May 2009 (UTC)

I personally don't have a problem with the abstraction so much as the vector notation.

I think the article really needs to minimise the amount of talking about rotation (without going crazy about avoiding it). The relationship between rotation and angular momentum is surprisingly complex.

But I think we need to cover the 2-d case upfront.- (User) Wolfkeeper (Talk) 19:48, 26 May 2009 (UTC)

Angular momentum in the case of a spinning gyroscope

In the current version of the article the very first picture is one of a spinning gyroscope, with the caption saying: "This gyroscope remains upright while spinning due to its angular momentum."

That statement remains suspended in the air. How is angular momentum involved? The article doesn't say.

So let's get into that.
If you position a spinning gyroscope on a surface, and you position it as close to perfectly upright as you can, then the gyroscope will appear to remain unwaveringly upright.

Next I will look at a precessing gyroscope. For denoting change of orientation I will borrow some terminology from aviation. When an upright object starts to tip over I will call that motion 'pitch', as the pitch of an aeroplane. Tipping perpendicular to the pitching I will call 'roll'.

If you position a gyroscope on a surface, not upright, then when you release the gyroscope the Earth's gravity is exerting a torque upon the spinning gyroscope, and the gyroscope will tend to pitch. In fact it will pitch, just a fraction, and in doing so the gyroscope's angular momentum changes. As the gyroscope pitches (a fraction), gyroscopic precession comes into play. The pitching motion gives rise to roll. That roll reorientates the gyroscope, so now the gravity's torgue does not act in the same direction as before.

The point should be clear by now: while gravity exerts a torque at all times, gravity doesn't get the opportunity to topple the gyroscope over because the gyroscope reorients itself all the time. The reorienting is because of gyroscopic precession, and the gyroscopic precession occurs because of change of angular momentum.

It is physically wrong to suggest that a spinning gyroscope remains upright due to conservation of angular momentum. It's the very opposite: in the case of the spinning precessing gyroscope angular momentum is not conserved.
--Cleonis | Talk 20:12, 26 May 2009 (UTC)

Introduction without vector notation and the 2-d case

Hi wolfkeeper, let me give some more thoughts

I propose as first step in introducing angular momentum two point masses, one much heavier than the other, with the two point masses exerting an attractive force upon each other.

In such a scheme the angular momentum of a solid object, such as a gyroscope, is thought of as a compound property; a compounded angular momentum arising from the many atoms that the solid consists of.

Angular momentum has both magnitude and direction.
- In the case of planetary orbits it is the conservation of (angular momentum) magnitude that catches the eye.
- In the case of a spinning gimbal mounted gyroscope it is the property of keeping the same (angular momentum) direction that catches the eye.

What the two cases have in common is that a torque is required to change either magnitude or direction of angular momentum. The gyroscope example is very unsuitable for introduction: the distance of mass to the axis of rotation is fixed; a dynamic degree of freedom is missing in the gyroscope example.

Question: what demarcates cases with 2 spatial degrees of freedom and cases with 3 spatial degrees of freedom? (in short: the 2-d case and the 3-d case) If one says: the 2-d case is when the motion is confined to a single plane then change of angular momentum direction is not possible. it seems to me that case is too restrictive to be of interest.

I propose that introducing angular momentum should involve the possibility of a torque changing angular momentum direction.
On the other hand, a useful restriction, I think, is to have the reference point of the angular momentum coinciding with the common center of mass of the masses involved. Later on this restriction can be lifted to move on to a more generalized, more abstract concept.

Newton's derivation of the area law using geometric means.

As I have argued several times: many textbooks and websites (including wikipedia) mention that Kepler's second law is in fact a case of angular momentum magnitude conservation. I think it is very appropriate that Newton's geometrical derivation of Kepler's second law from the laws of motion is presented in the angular momentum article.

Newton's geometrical derivation:
In the absence of a force the object moving along BCDE would proceed along the interval Bc in an equal interval of time. The actual motion BC is the vector sum of the motion at point B and added motion due to an impulse received at point B
All triangles have the same area. In the limit of infinitisimally small intervals of time the line BCDE approaches infinitely close to the actual planetary orbit.

It looks as if the above might fall into the category of original research (WP:OR "Articles may not contain any new analysis or synthesis of published material that serves to advance a position not clearly advanced by the sources.") If so it will not be acceptable in WP. Xxanthippe (talk) 12:03, 30 May 2009 (UTC).

- The original diagram in the Principia: facsimile page. Service provided by Carnegie Mellon University, Pennsylvania, USA

- Discussion using Newton's original diagram for angular momentum derivation, University of Sheffield, England.

- Proof that angular momentum is proportional to area swept out by orbiting body, Western Washington University, USA

In my opinion the diagram that I uploaded advances exactly the same position as the one that is advanced by the sources. I did not make a facsimile copy of Newton's original diagram because in my opinion the reasoning can be illustrated in a more accessible way. --Cleonis | Talk 12:25, 30 May 2009 (UTC)

You need to keep it really simple, particularly in the introduction. You're still wayyy overcomplicating.- (User) Wolfkeeper (Talk) 13:01, 30 May 2009 (UTC)

The introduction to this article has some problems. For example, what do the words "In two dimensions..." refer to? Two dimensional motion, I think, but as far as I'm aware the definition of angular momentum doesn't depend on whether the object's motion is restricted to a plane or not. Therefore, I'm unclear as to why the introduction differentiates between the 2D and 3D situations. Angular momentum is angular momentum, and it's definition for a point particle is ${\overrightarrow {L}}={\overrightarrow {r}}\times m{\overrightarrow {v}}$, the magnitude of which is, $L=rmvsin\theta$, and the direction of which is given by the right-hand-rule. I'd be happy to try to revamp the intro in a way that's suitable for the lay reader but accurate.Andonee (talk) 18:12, 6 August 2009 (UTC)Andonee

Here's my stab at an intro below. I don't think the first two paragraphs would scare off the average lay person, but I think some mention of the formal definition needs to be made.

---

Angular momentum is a quantity that is useful in describing the rotational state of a physical system. For a rigid body rotating around an axis of symmetry (e.g. the fins of a ceiling fan), the angular momentum can be expressed as the product of the body's moment of inertia and its angular velocity ($\mathbf {L} =I{\boldsymbol {\omega }}$). In this way, angular momentum is sometimes described as the rotational analog of linear momentum.

Angular momentum is conserved in a system where there is no net external torque, and its conservation helps explain many diverse phenomena. For example, the increase in rotational speed of a spinning figure skater as the skater's arms are contracted is a consequence of conservation of angular momentum. The very high rotational rates of neutron stars can also be explained in terms of angular momentum conservation. Moreover, angular momentum conservation has numerous applications in physics and engineering (e.g. the gyrocompass).

Fomally, the angular momentum of a point object is defined as the cross product of the object's position vector and linear momentum vector ($\mathbf {L} ={\boldsymbol {r}}\times m{\boldsymbol {v}}$). Angular momentum is a pseudovector whose magnitude is given by $L=rmv\sin \theta$ where $\theta$ is the angle between the object's position vector and velocity vector. The direction of the angular momentum can be determined by applying the right-hand rule. The angular momentum of a system of particles (e.g. a rigid body) is the sum of angular momenta of the individual particles.

---

Let me know what you think.Andonee (talk) 20:45, 6 August 2009 (UTC)Andonee

I changed the intro. I don't think it's perfect by any stretch, but I do think it reads better and is more accessible (at least the first two paragraphs) than the previous version. Angular momentum is a difficult subject for many physics students--I hope the new intro is enough to satisfy the educated lay person while remaining true to the physics. Let the discussion continue!Andonee (talk) 15:18, 9 August 2009 (UTC)Andonee

It looks like you started right after a strange uncommented anonymous edit that used non-wiki math notation and such. I reverted to before that, as I hate to see big uncommneted anon changes. I didn't study the various versions in details; perhaps you'll want to put some or all of that back, but if so please use normal wiki math notation. Dicklyon (talk) 16:51, 9 August 2009 (UTC)

I've changed the intro, and the math is now LaTex.Andonee (talk) 19:33, 9 August 2009 (UTC)andonee

"spin" of rigid body

In the "Angular momentum in the centre of mass frame" section, a rigid body rotating about its centre of mass is identified as a "spin" (with a link to the spin (physics) article). I find this misleading. According to the spin (physics) article, spin is considered an intrinsic property of subatomic particles, not something that you get by (e.g.) twirling a pencil. I'm not sure how to edit the sentence to remove the reference to spin without just removing the reference to the rigid body case. The article claims that the expression for angular momentum is simplified in the case of a rigid body, but this is unsubstantiated. Perhaps someone could explain the simplification? And4e (talk) 20:53, 21 January 2010 (UTC)

I reworded and deleted the link. OK now? --Steve (talk) 05:17, 22 January 2010 (UTC)

Merge with Introduction to angular momentum

The following discussion is closed. Please do not modify it. Subsequent comments should be made in a new section. A summary of the conclusions reached follows.

The Introduction to angular momentum article was proposed to be merged with this article in July 2010. The result of this discussion is no consensus. – PIE ( CLIMAX )  20:35, 14 December 2011 (UTC)

---

User:FilipeS has posted the Merge banner on Angular momentum and Introduction to angular momentum. I am opposed to a merger of these two articles. On Wikipedia there are many articles that are unavoidably complex, especially for young people and newcomers to science. A number of these complex articles are supported by a Introduction to ... article that provides a simple introduction. A good example is Introduction to entropy that supports the more complex article Entropy. A complete listing of these Introduction to ... articles is available HERE. The article Introduction to angular momentum is an ideal supporting article for Angular momentum and the two should remain separate. Dolphin (t) 23:39, 13 July 2010 (UTC)

Yes, although:

• stats.grok.se/en/201006/Angular_momentum
• stats.grok.se/en/201006/Introduction_to_angular_momentum

Speaking as the guy that actually made this into a full article, it doesn't look like the new article has been successful, nor do I see any evidence that it will pick up.- Wolfkeeper 01:09, 14 July 2010 (UTC)

Intro articles don't seem to be read nearly as much as the main article no matter what the subject. I've checked several, and the traffic on the intros is always relatively low. Since the Merge template has been removed from this main article, I shall remove it from the intro article, as well, and then close this merge proposal. It began in July of 2010, and there doesn't appear to be much interest, let alone a consensus, to merge. – PIE ( CLIMAX )  20:35, 14 December 2011 (UTC)

The articles on entropy actually contain a whole lot of redundancy amongst themselves. They would also benefit from some mergers. FilipeS (talk) 16:46, 15 October 2011 (UTC)

The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

Angular Momentum Commutation Relations

I've seen this incorrect formula in several places on Wikipedia:

$[L_{l},L_{m}]=i\hbar \varepsilon _{lmn}L_{n}$.

Without a summation on the index n, the commutation relations become multivalued. (For example, for $l=1$, $m=2$ we could have either $[L_{1},L_{2}]=0$ (corresponding to the case $n=1$ or $2$) or $[L_{1},L_{2}]=i\hbar L_{3}$ (corresponding to $n=3$.) Therefore, the correct formula is

$[L_{l},L_{m}]=i\hbar \sum _{n=1}^{3}\varepsilon _{lmn}L_{n}$.

I took the liberty of making the correction for this article. —Preceding unsigned comment added by Fourier Transform (talk • contribs) 12:06, 23 May 2011 (UTC)

Fair enough, but the convention is often adopted of summing over repeated indices. Xxanthippe (talk) 13:02, 11 June 2011 (UTC).

Misleading definition

discusion about the "truth"
The following discussion has been closed. Please do not modify it.
"In physics, angular momentum, moment of momentum, or rotational momentum is a conserved vector quantity that can be used to describe the overall state of a physical system." Overall state ??? What about linear momentum ? I think the "can be used to describe the overall rotational state" should be better. "... where r is the particle's position from the origin" From the origin ? Coordinate origin ? I think "from the center of mass" should be correct. Center of mass represents the position related to linear momentum. Rotation is complementary ("orthogonal") balanced motion around the center of mass. Rotation of independent system around an other point than the center of mass is not physically possible. In other words, rotation cannot disturb the linear momentum of the center of mass. Softvision (talk) 23:28, 10 June 2011 (UTC) Both questioned statements are correct. Angular momentum is independent of coordinate origin only when linear momentum is zero. Xxanthippe (talk) 23:57, 10 June 2011 (UTC). It is clear, that the reference point of the r vector cannot be placed anywhere. Acording to the angular momentum conservation, reference point of the vector r must conserve angular momentum. The center of mass meets this criterium. If the coordinate origin is identical with center of mass, requirement is fulfiled, of course. However, I am convinced, that the general definition must use general, physically distinct, concepts (reference points, vectors, formalism), valid in general circumstances, coordinate systems. Equation $\mathbf {L} =\mathbf {r} \times \mathbf {p}$ satisfies these requirements, if the vectors and quantities in equation have valid physical meaning. Softvision (talk) 10:28, 11 June 2011 (UTC) No, you can measure the angular momentum about any origin anywhere, and angular momentum is conserved about it; it's just that the lowest angular momentum is found about the centre of mass.There's a simple, well known relationship between the angular momentum about any point and the distance of that point or origin from the centre of mass of the system of particles. -Rememberway (talk) 10:33, 11 June 2011 (UTC) If $\mathbf {r} _{c}$ is the cetre of mass referenced vector, and the $\mathbf {r} _{d}$ is constant displacement vector of the other arbitrary reference point from the centre of mass, than $\mathbf {r} ^{\prime }=\mathbf {r} _{c}-\mathbf {r} _{d}$ , and $\mathbf {L} ^{\prime }=\mathbf {r} ^{\prime }\times \mathbf {p} =\left(\mathbf {r} _{c}-\mathbf {r} _{d}\right)\times \mathbf {p} =\mathbf {r} _{c}\times \mathbf {p} -\mathbf {r} _{d}\times \mathbf {p} =\mathbf {L} -\mathbf {r} _{d}\times \mathbf {p}$. Considering the $\mathbf {L}$ is constatnt (conserved), then $\mathbf {L} ^{\prime }\neq \mathbf {L}$ , and it is not constant (conserved), because $\mathbf {r} _{d}\times \mathbf {p} =0\Longleftrightarrow \mathbf {r} _{d}\parallel \mathbf {p}$ and $\mathbf {r} _{d}\times \mathbf {p} ={\mbox{const}}\Longleftrightarrow \mathbf {p} ={\mbox{const}}$, but generally ${\mbox{not}}\left(\mathbf {r} _{d}\parallel \mathbf {p} \right){\mbox{and}}\left(\mathbf {p} \neq {\mbox{const}}\right)$. Softvision (talk) 11:22, 11 June 2011 (UTC) True, different origins have different angular momentum, but angular momentum is conserved with respect to any particular origin. -Rememberway (talk) 13:44, 11 June 2011 (UTC) The idea that angular momentum, linear momentum or other conserved quantities are independent of the choice of frame of reference is in general quite wrong. For example, conservation of energy- the total energy in a stationary frame is different from the energy of the same system of particles from a differently moving frame (because even with Galilean relativity kinetic energy is frame-dependent, NOT frame independent.) -Rememberway (talk) 13:51, 11 June 2011 (UTC) As you can see in equations above, the $\mathbf {L^{\prime }}$, derived from an arbitrary reference point ("origin"), is not conserved, and $\mathbf {L}$ is conserved, because rotation of rigid body around the center of mass results in circular motion - $\mathbf {r} _{c}\perp \mathbf {p}$, and conservation holds also in relation to the elyptical orbits around the center of mass (...). That means, the definition - formulation and reference to arbitrary origin, is not valid. Believe me, I have expected the relativistic arguments :-), but here we are talking about general definition of angular momentum, for all inertial reference frames, linear motions. Angular momentum is so say "orthogonal" to the linear momentum. That means they do not overlap, and their superposition generates overall inertial motion, and this fact is directly related to the individual conservation of both quantities. The definition of angular momentum must correspond to these facts. The center of mass is identical in any inertial reference frame. This is fundamental theoretical stone :-). Angular momentum in relation to the center of mass is therefore invariant in context of inertial reference frames. This represents the valid, distinct and physically consistent quantity of angular momentum - related to the center of mass of independent system. Arbitrary "origin" is just mathematically and physically invalid statement. PS: Einstein's Theory of Relativity is invalid, as the world will see soon, so do not try arguments based on ETR. Softvision (talk) 20:20, 11 June 2011 (UTC) Sorry, that's incorrect, even in Newtonian mechanics. I liked your math, but it's not even wrong. You can calculate angular momentum about any point in space and even from a frame of reference that is moving. Any galilean transformation preserves conservation of angular momentum just fine; it would be completely useless if it didn't. But it doesn't preserve the numerical amount. The numerical angular momentum is different between different frames, but in any given inertial frame it's constant, as with any other conserved quantity. For example linear momentum is also conserved, but it's not the same between non comoving frames. Angular momentum is similar. -Rememberway (talk) 20:46, 11 June 2011 (UTC) (edit conflict) I agree with Rememberway: L is conserved, and L' is also conserved, and L≠L'. None of those statements contradict each other and they're all true. Both L and L' are valid "angular momentum"s. I do agree with Softvision that the first sentence of the article ("...a conserved vector quantity that can be used to describe the overall state of a physical system) could use improvement. The problem is, a reader could (mis)interpret it as "completely describe the overall state." Obviously it takes more than one vector to describe every aspect of a physical system! :-) I also agree that the word "origin" should be explained in the article, although of course it should be explained correctly (not "center of mass") :-) --Steve (talk) 20:51, 11 June 2011 (UTC) I have expected relativistic bias in argumentation. In equation above $\mathbf {r} _{d}$ represents the point in the same reference frame - constant (static) position from the center of mass - arbitrary "origin" in the same refernce frame. That means, in the same reference frame there are two different values of angular momentum $\mathbf {L^{\prime }} \neq \mathbf {L}$, according to the definition in article - the quantity is not distinct. If there is no specific dependence of definition related to the reference frame, definition must hold in all reference frames, including "static" :-), which is in current theory identical with the "same" reference frame, because current theory does not recognize the "static" reference frame by other modus. Softvision (talk) 21:26, 11 June 2011 (UTC) Look, L is a linear function of the positions of all the particles. So you can write it as L = sum(f(r_n, v_n)). So transforming you get L'=sum(f(r_n+r0,v_n))=sum(f(r_n,v_n)+f(r_o,v_n))=L+sum(r_0, v_n). That's all. -Rememberway (talk) 22:01, 11 June 2011 (UTC) L'=sum(f(r_n+r0,v_n))=sum(f(r_n,v_n)+f(r_o,v_n))=L+sum(f(r_0, v_n)). sum(f(r_0, v_n)) is not constant. $\mathbf {r} _{d}\times \mathbf {p}$ is not constant. Calculus is Calculus. Equivalence. Understand ? Proof. This is not the fight of ideas, this is necessity of truth. I appreciate all submisions. Truth is not obvious. Because truth is noncontradiction, and noncontradiction is not obvious. :-) Softvision (talk) 22:33, 11 June 2011 (UTC) No, you're still not right, the thing is there's nothing internal to a system of particles which changes the angular momentum after the initial state. Each action-reaction internal force pair acts oppositely along a common line, and so the resultant accelerations cannot change the angular momentum, and that's true whether or not the angular momentum was initially calculated around the centre of mass. -Rememberway (talk) 23:09, 11 June 2011 (UTC) Ok. Result above is quantitative (mathematical, calculus) contradiction of the article definition. However, let's trace the logic of physical reality. We are deeper in the question. I appreciate revelation of the concept of acceleration, in the context of angular momentum. There is no angular momentum without interaction, because without interaction there is no acceleration. Curved paths without acceletation are not possible, therefore angular momentum without interaction is not possible. If we talk about rotation, in general meaning of cyclic bound in region of space, we must take conservation of linear momentum into account, we must take into account the system's center of mass. The rest of the motion, not overlaping (<=> orthogonal) with the linear momentum (the linear geometrical path of the system), is curved motion that maintains the "particles" around the center of mass, the angular motion, forming the quantity of angular momentum, not overlaping (<=> orthogonal) the linear momentum of any particle in the system. Intuitively you can understand, that the angular (curved) motion, not overlaping (<=> orthogonal) the linear motion, can only be the motion in the reference frame of the center of mass. Any point with constant position to the center of mass is in the same reference frame. However, the quantitative (mathematical, calculus) result does not confirm the same value of angular momentum quantity for every inertial "origin" of r. The physical science is based on quantitative consistency, because any measurement generates the value of physical quantity, defined or proposed by the theory, determining the measurement method. That's where the "numbers" and units of our conscious domain are from, and how they are related to the physical reality, on the level (limit) of current knowledge. That means, numerical relations are fundamental part of the physical science. Numbers above do not confirm the article definition of the r vector reference point. Softvision (talk) 23:58, 11 June 2011 (UTC) Don't even bother trying to put this OR into the article, unless you've referenced it, and you won't be able to anyway, because it's wrong. -Rememberway (talk) 00:57, 12 June 2011 (UTC) Reconciliation : Rememberway has pointed correctly that symmetry makes "origin" irrelevant. If $\mathbf {r} _{c}$ is the cetre of mass referenced vector, and the $\mathbf {r} _{d}$ is constant displacement vector of the other arbitrary reference point from the centre of mass, taking the symmetrical body, symmetrical pairs of elements, than (simplified) $\mathbf {r} _{1}^{\prime }=\mathbf {r} _{c}-\mathbf {r} _{d}$, $\mathbf {r} _{2}^{\prime }=-\mathbf {r} _{c}-\mathbf {r} _{d}$ , then $\mathbf {L} ^{\prime }=\mathbf {r} _{1}^{\prime }\times \mathbf {p} +\mathbf {r} _{2}^{\prime }\times \left(-\mathbf {p} \right)=\left(\mathbf {r} _{c}-\mathbf {r} _{d}\right)\times \mathbf {p} -\left(-\mathbf {r} _{c}-\mathbf {r} _{d}\right)\times \mathbf {p} =\left(\mathbf {r} _{c}-\mathbf {r} _{d}-\left(-\mathbf {r} _{c}-\mathbf {r} _{d}\right)\right)\times \mathbf {p} =2\mathbf {L}$. Under these conditions, the $\mathbf {r} _{d}$ displacement vector of the other arbitrary reference point "origin" is irrelevant. As far as rotation around the centre of mass must preserve the linear momentum, the centre of mass symmetry must be implicitly general. The "origin" of r can be arbitrary, when considering the system as a whole, when integrating the angular momentum. Whether we are talking about the integral angular momentum ${\frac {1}{V}}\int \mathbf {L} \mathbf {d} V$, or individual (discrete) angular momentum $\mathbf {L} =\mathbf {r} \times \mathbf {p}$, is significant difference, according to current definition. If the integral angular momentum is distinct, independently of the "origin", than if we want the individual angular momentum to be distinct physical quantity, we must reference the origin of r distinctly, using the centre of mass, the distinct physical point of the system, directly and distinctly related to the linear momentum. Maybe this could be improved in article. This is the meaning of my discussion - the distinctness of the physical quantity. Current definition says : $\mathbf {L} =\mathbf {r} \times \mathbf {p}$ can be any value, because $\mathbf {r}$ is arbitrary. Real physical quantity cannot be any value. Softvision (talk) 10:32, 12 June 2011 (UTC)

Just for fun : What is angular momentum of Solar system at 19:49, 12 June 2011 (UTC) ? Softvision (talk) 19:49, 12 June 2011 (UTC)

Classical Derivation Using Center of Mass

I think "Angular momentum simplified using the center of mass" is a nice derivation, but it uses ambiguous terminology. To be specific, the derivation seems to show that if you want the angular momentum of a collection of point particles about an ARBITRARY point, it turns out you can just use the angular momentum of their COM about that same ARBITRARY point plus the angular momentum of the point masses about their COM.

My issue is with the intro to this section as well as with the end where the final result is given as "the total angular momentum with respect to the center." Somebody not reading close enough might become confused. In fact, I might be confused. What do you guys think?

Scaram25 (talk) 03:40, 9 August 2011 (UTC)

Extra images

Just thought some of the equations could do with more illustrations - I added a couple in the Angular momentum in classical mechanics section, in the Fixed axis of rotation and Conservation of angular momentum subsections. Hope they're clear enough.

Maschen (talk) 23:36, 5 September 2011 (UTC)

Conservation of Angular Momentum

The conservation and corresponding symmetries are well expounded in Landau Lifshitz. I'll add the page nos subsequently. It is well written below in the section angular momentum#Conservation of angular momentum. The lead section does not need to be "sourced". What I wanted to remove was the ambiguous statement that it was a "conserved vector quantity" (which too is unsourced). --Dhatsavan 02:44, 17 December 2011 (UTC)

When we talk about a conserved (constant in time) vector, we need to say something about the space in which it exists. Take the figure skater for instance. Her angular momentum is conserved when spinning but how did she start spinning in the first place? And how will she stop spinning? The answer is the following. She uses the ground in both cases. The frictional force is not rotationally symmetric about any axis. So there is no conservation of angular momentum. When she is spinning with negligible friction, she can indeed be considered a closed system on whom no external force (not just torque) is acting. But the conservation law holds more generally. Even if there is a force field, as long as it is rotationally symmetric, (like a central force) angular momentum is still conserved. Consider the figure skater from a rotating frame of reference. The centrifugal force will be circularly symmetric and hence angular momentum conserved, but only if the axis of rotation is collinear with her own axis of rotation (or the "angular momentum vector"). Saying bluntly that angular momentum is a conserved vector quantity without qualification at the very beginning implies more generality and is misleading. This is for instance different from energy (scalar) conservation which relies on time translation symmetry. Dhatsavan 03:55, 17 December 2011 (UTC)

All that needs to be said (and it has been said many times before) is that angular momentum (as well as linear momentum and energy) is conserved for an isolated system. Xxanthippe (talk) 05:09, 17 December 2011 (UTC).

Archive 1

Archive 2