Talk:Angular momentum
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[edit] 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)
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- 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.
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- 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)
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- 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)
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[edit] 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)
[edit] 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
[edit] 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)
[edit] 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.
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- 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. *******
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The above editorial comment was added in this edit. ---Rednblu | Talk 20:58, 6 Jan 2005 (UTC)
[edit] relativistic angular momentum
Does anyone know if there is a symmetry under spacetime rotations in quantum field theory?
[edit] 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)
[edit] 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)
[edit] Units
This page needs to mention the units of angular momentum.
[edit] Operator
As far as im aware, the L2 operator is missing a factor of minus h-bar squared in the front there.
[edit] 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)
[edit] 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)
[edit] 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)
[edit] 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)
