# Talk:Action (physics)

WikiProject Mathematics (Rated B-class, Mid-importance)
This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Mathematics rating:
 B Class
 Mid Importance
Field: Mathematical physics
WikiProject Physics (Rated C-class, High-importance)
This article is within the scope of WikiProject Physics, a collaborative effort to improve the coverage of Physics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
C  This article has been rated as C-Class on the project's quality scale.
High  This article has been rated as High-importance on the project's importance scale.
The discussion of the historical development of this concept, including Latin translations, has been moved to Talk:Principle of least action, because that's where the history of the principle is now documented.

## ≤2005

Two things:

• The Euler-Lagrange equations' box looks really odd. I think it would look fine without the box, and with the text "Euler-Lagrange equations" converted to a wiki link (Euler-Lagrange equations).
• The second E-L equation doesn't look properly derived. I get:
$\frac{d}{dt}\left( \frac{\partial L}{\partial\dot{\phi}} \right) - \frac{\partial L}{\partial\phi} = 0 \qquad \Rightarrow \qquad \ddot{\phi}r^2 + 2\dot{r}\dot{\phi} = 0$

-- Taral

Currently Euler-Lagrange equations is an indirect self link, but Euler-Lagrange equation redirects to Lagrangian b4hand 20:08, 20 Jul 2004 (UTC)

At some point, Euler-Langrange equations might end up its own page, but until then, both plural and singular can point to action (physics), which is the simpler of the pair. -- Taral 02:52, 21 Jul 2004 (UTC)

I think it should go on its own page; after all it's not used just for action. e.g. I've used it to maximise the output of a simplified tidal-power plant (which might make a good example). Kwantus 2005 July 1 17:20 (UTC)

division by $r^2$ yields the result as stated in the article :-) -- Unknown editor

No, it doesn't. That gives $\frac{2}{r^2}\dot{r}\dot{\phi}$, and only if r is nonzero. -- Taral 08:23, 18 May 2005 (UTC)

Here's how you get it:

$\frac{\partial L}{\partial \phi} = 0$
$\frac{d}{d t} \frac{\partial L}{\partial \dot{\phi}} = \frac{d}{dt} (m r^2 \dot{\phi} )$
$\frac{d}{dt} (r^2 \dot{\phi}) = \left(\ddot{\phi} \frac{\partial}{\partial \dot{\phi}} + \dot{r} \frac{\partial}{\partial r} \right) (r^2 \dot{\phi}) = \ddot{\phi}r^2 + \dot{r}2r\dot{\phi}$

so

$r^2 \ddot{\phi} + 2 r \dot{r} \dot{\phi} = 0$

or for nonzero r

$\ddot{\phi} + \frac{2}{r}\dot{r}\dot{\phi} = 0$

as stated in the article. --Laura Scudder | Talk 22:03, 18 May 2005 (UTC)

• There are infinitely many actions which give rise to Newton's laws.
• There are quantum field theories which are NOT "derived" from an action.
• This article only seems to be about the use of the action in classical Newtonian mechanics.

Phys 01:22, 31 Jul 2004 (UTC)

Feel free to fix it.

Taral 21:21, 16 Aug 2004 (UTC)

## Intuitive definition

Is it worth making an attempt to give a physical concept of the action: for example, thus:

L = T - V, but total energy E = T + V. So L is a measure of how much of E is put into kinetic energy L, rather than into potential energy V: very approximately, we can say that it's a measure of the amount of "wiggling" or "change" in the system, expressed in terms of its kinetic energy balance. So, the action is a measure of total "change" over the time period. Then, the principle of least action can be stated informally as "everything must occur with the least possible total amount of change, but no less".

Does this make any sense? Would anyone else like to correct or improve this please? -- The Anome 00:54, 13 February 2006 (UTC)

I agree, such an explanation would be useful. I think we could come up with a better definition, though. In particluar "change" seems like a bad term. I guess the powers that be chose "action" for a reason. I myself am still trying to figure out an intuitive definition for L, but whatever it is it should have some intuitive sense of energy so the integral over time has a sense of energy×time. —Ben FrantzDale 22:36, 4 April 2006 (UTC)

## Problems with definition of action

Some physics and math texts distinguish the "Hamiltonian Principal Function"

$R(q_1, q_2, t) = \int_{t_1}^{t_2} L(q, \dot{q}, t) dt$

from the "action", which is the Legendre transform of R, exchanging t time for energy E:

$S(q_1, q_2, E) = R(q_1, q_2, t) + Et = \int_{q_1}^{q_2} p dq$

This article does not distinguish R and S. Here, E is the energy, p the canonical conjugate momentum:

$p=\frac{\partial L}{\partial \dot{q}}$

Its the first one, R, that is used in classical mechanics, to derive the Euler-Lagrange eqns of motion. Its the second that is used in quantum mechanics, the so-called action-angle coordinates, which is quantized in Bohr-Sommerfeld quantization as

$\oint p dq = nh$

with h Planck's constant of course. (See canonical one-form for a mathematical defn of the action). I'm not sure how to go about clarifying this in this article. linas 17:09, 11 June 2006 (UTC)

Responding to your other comments on my Talk page, I'm not sure if I understand why you think that the second and third meanings are the same? Here's how I understand the differences (which may be imperfect!). In the second meaning, $S$ is a function of many variables; in the third meaning, $J_{k}$ is a single variable. Moreover, the $J_{k}$ are constants of the motion (as typically used), whereas $S$ changes with time, even when energy is conserved.
In the literature (e.g., Landau/Lifshitz, Mechanics, "Maupertuis principle"), there is an abbreviated action $S_{0}$ defined as
$S_{0} \equiv \int \mathbf{p} \cdot \dot\mathbf{q} dt = \int \mathbf{p} \cdot d\mathbf{q}$
this is extremal for variations about the physical trajectory, but only when the energy is conserved. Therefore, it is less general than Hamilton's principle. However, even this abbreviated action (when restricted to one dimension) is not the same as $J_{k}$, since the former is integrated over a generic trajectory, whereas $J_{k}$ is integrated over a closed cycle that may not correspond to any physical trajectory.
Perhaps I've misunderstood something? If so, please let me know, and I'll try to fix it. WillowW 21:01, 13 June 2006 (UTC)
I'm sorry, I was under the impression that you wrote the article Legendre transform, but it seems not, and so probably I'm beating on the wrong horse. Lets stick to one variable, just to keep it easy. In the article Legendre transform, there is a section called "Legendre transform in one dimension", which begins with the equation
$g(y) = y \, x - f(x), \,\,\,\, x = f^{\prime-1}(y)$
Now let y=E, x=t, f=-R and g=S (as I defined R and S above). Perform the transform. What do you get?
Here's a simpler approach: its often said that the Hamiltonian $H=p\dot{q}-L$, right? Integrate both sides:
$\int_{t_0}^{t_1} H dt = \int_{t_0}^{t_1} p\dot{q} dt - \int_{t_0}^{t_1} L(t) dt$
Now the first integral is easy: H, the energy, does not depend on time (the energy is conserved, its a constant of motion), so
$\int_{t_0}^{t_1} H dt = E(t_1-t_0)$
hwere I write E for energy to indicate its a "simple plain-old number". The second integral is also pretty easy:
$\int_{t_0}^{t_1} p\dot{q} dt = \int_{t_0}^{t_1} p\frac{dq}{dt} dt = \int_{q_0}^{q_1} p \,dq$
where $q_0=q(t_0)$, etc. right? The third inetgral requires no change, its immediately recognizable as the "action" (whatever that is), right? So put it all togethr, and we get
$E(t_1-t_0) = \int_{q_0}^{q_1} p \,dq - \int_{t_0}^{t_1} L(t) dt$
or rearranging terms,
$\int_{q_0}^{q_1} p \,dq = \int_{t_0}^{t_1} L(q, \dot{q}, t) dt - E(t_1-t_0)$
which demonstrates that the "action" is the Legendre transform of the "action". These are not "different actions", these are all the same thing. For multiple variables, we replace pdq by the vector dot product pdq. This is all I was trying to get at. Calling the darned thing $J_k$ sort-of obscures the point; actually, J=S. linas 22:13, 13 June 2006 (UTC)

Hi, thanks for your nice derivation. I think our confusion may involve the following six quantities, all called "action" in physics

• the full Hamilton-principle action $S$ (a functional, not a function)
• the abbreviated action $S_{0}$ (a functional, not a function)
• Hamilton's principal function S (a function)
• Hamilton's characteristic function W = S - Et (another function; perhaps this is what you mean?)
• one separated term in Hamilton's principal function $S_{k}(q_{k})$ (still a function)
• the constant of motion $J_{k}$, which equals the change in $S_{k}(q_{k})$ as $q_{k}$ is varied over one circuit.

The quantities you cite above, R and S, are not clear to me -- are they functions or functionals? I'm guessing they're meant to be functions corresponding to S and W but, if so, they're not defined quite correctly.

I still don't see how J=S, since S is either a function or a functional and J is either a scalar or at best a list of the various J variables, e.g., $\mathbf{J} \equiv \left(J_{1}, J_{2}, \ldots, J_{N} \right)$. WillowW 12:16, 14 June 2006 (UTC)

There, that's much better. Six quantities! Yikes! In fact, there's more, if one starts talking about the action in quantum field theory; but lets leave that alone for now. By functional, I assume you mean "a function of the path taken". When you say a function I assume that you mean "the function of its endpoints, assuming a classical trajectory was taken between the endpoints". The quantities R and S that I cite can be taken to be functionals, in that they can be given a value for any path. But if you pick a particular path (the classical trajectory), then they become functions of thier endpoints. I beleive that the relationship between R and S holds only if the path is taken to be the classical path; certainly, my little proof assumed that, e.g. by assuming the "energy is constant", which wouldn't be true for an arbitrary generic path.
FYI, there is a seventh definition, over at tautological one-form#Action which I think is equivalent to a sum over your constants of motion. However, I am confused by this definition when there are no constants of motion, and so will have to study this a bit more. linas 00:11, 15 June 2006 (UTC)

### Disambiguation

Seven meanings, right? Then why does the section titled "Disambiguation" state there are eight meanings, whilst enumerating only these seven? - Yoyo 124.191.50.199 08:50, 10 October 2007 (UTC)

## Examples

The polar coordinate example is ok ... but can anyone demonstrate how to use this stuff to show that objects freely falling in a uniform gravitational field follow parabolic paths? —The preceding unsigned comment was added by Paul Murray (talkcontribs).

## Intro

Can any one simplify the intro for a dummy like me to understand? —The preceding unsigned comment was added by Tugjob (talkcontribs) 23:50, 22 May 2007 (UTC).

OK, I'll try; please let me know if you like the results! :) Willow 14:07, 23 May 2007 (UTC)
Yes its a lot better, but still loses me after the first para. Surely someone with an engineering degree should be able to understand the intro?--Tugjob 23:06, 23 May 2007 (UTC)
OK Willow Ive tried to split the lead to make it digestible (to me). Would tyou care to check for accuracy/sense etc?--Tugjob 23:20, 23 May 2007 (UTC)
More ruthless editing. Please comment if anything is now misleading or incorrect.--Tugjob 15:03, 24 May 2007 (UTC)
I have been trying to enhance Lagrangian by giving some canonical examples of test particle and field Lagrangians. You might want to look at it and perhaps link to it. JRSpriggs 09:44, 25 May 2007 (UTC)
I can only edit from a laymans standpoint, as this stuff is not my speciality. Howeve I will try to simplify where I can.--Tugjob 17:16, 29 May 2007 (UTC)

someone really needs to fix this article. Not me, an experienced professor who knows what they're doing. Tacobake 15:05, 13 July 2007 (UTC)Tacobake

At the time of this comment, the intro is really very bad:

“In physics, the action is a particular [what does that mean] quantity in [perhaps "of" - I don't know how a quantity can be in a physical system] a physical system that can be used to describe its operation [what does this mean]. Action is an alternative to differential equations [Gosh! in all circumstances?]. The action is not necessarily the same for different types of systems [what types of system are there?].”

... and so on. I think the problem is that the intro has been (well-meaningly) simplified to the point of complete uselessness! It needs a complete rewrite. Ideally this should be by someone who:

1. fully understands all the physics (including the historical development of the concept).
2. is a brilliant communicator who understands and sympathises with the wide range of backgrounds and education of Wikipedia readers.
3. is willing (in the introduction) to sacrifice accuracy of detail in order to communicate the basic concept to a non-specialist.

I am (sadly) not such a person. However, I believe that I understand enough to improve the current version, so I will have a try! For future editors aiming to improve my version (please, please do!), do try not to "dumb it down" to the point of meaninglessness, or elaborate it to the point of incomprehensibility to anyone without an advanced physics degree. FredV (talk) 14:42, 9 February 2009 (UTC)

I have just reverted a "correction" to the new intro. The principal reason for this prompt revert is that the "correction" applies only to the quantum mechanical interpretation of action. Since the rest of the article is devoted almost entirely to the action in classical mechanics, this was likely to be confusing, especially to the les well-informed reader (who should, after all, be the primary target audience for the article intro). It is notable that the term "wave function" that was introduced in the proposed "correction" is not mentioned anywhere else in the article - which would surely confuse the keen teenage would-be physicist at whom I have aimed this intro.

I know that there are people who believe that classical mechanics died with the discovery of quantum mechanics — believe me, it did not! For any macroscopic system under normal thermodynamic conditions (that is, wherever the Ehrenfest theorem applies), classical mechanics is the only practical tool for analysis. I would guess that for every physicist applying quantum analysis to a problem, there are at least a hundred engineers applying classical mechanics! ... and of course, that means that there are still loads of people being taught classical mechanics, who may seek help and inspiration from Wikipedia. Many of these may never study quantum mechanics. It is also the case that almost everyone who studies mechanics will encounter the classical action before they are introduced to its quantum interpretations, so the classical action should at least appear first in the article.

With regard to my proposed intro, there are some "deficiencies" of which I am well aware:

• "energy distribution" is an ambiguous term, and would indeed be likely to be misunderstood (as, for example, a Boltzmann distribution) by an informed reader if used in isolation. I use this term to capture the essential physical content of the Lagrangian (that is, the distribution of energy between kinetic and potential) for less informed readers who had not (yet, or perhaps ever) been formally taught about the (classical) Lagrangian. The link to the Lagrangian article is available for anyone who wanted to understand more.
• "state L(t)": of course the Lagrangian is a function of the state (not a state); I could have explained this (perhaps introducing some general state vector as an explicit argument of the function), but I felt this would break the flow of the introduction, and make it less readable by the less well-informed. This simplification is fully elaborated later in the article, for anyone with the education and perseverance to read on that far.

I hope future editors will think carefully about my intentions before "correcting" these "deficiencies".FredV (talk) 10:45, 12 February 2009 (UTC)

Since you admit that you do not fully understand this subject and you admit that your use of "distribution of energy" is misleading, it would behoove you to not simply revert my corrections. There is no simple non-quantum explanation of either action or the Lagrangian. They were arrived at by figuring out what would have to be held stationary in order to give the observed equations of motion. JRSpriggs (talk) 06:53, 13 February 2009 (UTC)
They were indeed "arrived at by figuring out what would have to be held stationary in order to give the observed equations of motion", mainly by Joseph Lagrange and Pierre-Louis Maupertuis during the second half of the eighteenth century, with a final classical polish being given by William Hamilton in the 19th century. The resulting development of the classical Lagrangian, the action, and the stationary action principal are still an important part of the engineering syllabus, and are learned and applied today by many graduate engineers who will never see a quantum wave function (less still try to measure its changes of phase). I repeat - many more people will meet and use the classical action than will ever know what a quantum wave function is. This is reflected by the almost entirely classical content of the body of the article (to which I have made no contributions). Your reversion leaves a broken article, with two terms in the first line ("phase" and "wave function"), presented as essential to the definition of action (wrongly, since the concept survived over 100 years without them), which are not mentioned once in the rest of the article.
Indeed, I am not an expert on the subject (though I do hold a first class honours degree in physics, with a distinction in quantum mechanics). However, the issue here is not knowledge (as I have made clear, I would value an improvement in the presentation of the quantum action in the body of the article), but introductory presentation to the typical Wikipedia readership (who almost certainly have only the haziest ideas about what a "wave function" is, and no idea at all about what its "phase" might be). This is a matter of psychology (in which I also happen to hold a first class honours degree), but can usually be managed by intelligent and sympathetic consideration of the needs of the less expert reader.
However, I am reluctant to become engaged in a petty edit war, although it hurts to leave an article broken in just the way (a garbled and unhelpful introduction) that I had tried to remedy. Perhaps someone will come up with an improvement that meets your standards and mine. FredV (talk) 17:37, 13 February 2009 (UTC)

## A small nonsense

The major section (presently numbered 3) entitled "Examples of systems and their 'action'" contains just the following:

1. The text: "Light: Gravity:"
2. The minor section (presently numbered 3.1) entitled "Mathematical definition"

I suspect that the first is just a pair of placeholder notes for examples some author intended to write, whilst the second should be an independent major section.

Whatever the cause, the effect is confusing. Would someone please:

1. Amplify or remove the "Examples of systems and their 'action'" section.
2. Consider elevating the level of the "Mathematical definition" section.

Thanks! - Yoyo 124.191.50.199 09:03, 10 October 2007 (UTC)

## error

i noticed that the Lagrangian in the action integral doesn't contain time, and this is an error, L = L(q,q',t). I don't know how to edit math formulas, so i'm just throwing it out there. —Preceding unsigned comment added by 99.154.6.85 (talk) 22:58, 16 February 2008 (UTC)

## Newton's law

The following sentence

Although equivalent in classical mechanics with Newton's laws, the action principle is better suited ...

seems to connect the concept to Newton's laws of motion:

Third law: for every action there is an equal and opposite reaction.

If my reading is accurate, making that connection clearer and earlier might help the reader grasp the concept by providing a familiar foundation before delving into the phase of the wave function, the Lagrangian function, etc.

Might Newton's third law be interpreted as "conservation of action"? - Ac44ck (talk) 17:17, 20 March 2009 (UTC)

When Newton used the word "action" in his third law of motion, he meant force which is different from the subject of this article. Newton's third law is a way of stating the conservation of linear momentum (whose time derivative is force). JRSpriggs (talk) 13:21, 21 March 2009 (UTC)

## Units of 'action' in quantum-mechanics

Planck's constant is given by

  h = E / v   (energy / frequency)       joules /  hertz or joules per cycle per second


The 'reduced planck's constant' is

  h / 2 pi   (energy / angular frequency)     joules per radian per second


Dimensional analysis gives them both the dimensions of energy x time i.e. ML2T-1

The units of both of the Planck's constants are conventionally often referred to as joule seconds, unfortunately omitting to mention cycles or radians because they are dimensionless. Surely this is technically WRONG? We do not refer to 100MHz as 100M s-1 !

Unfortunately this error was compounded by giving the units of Planck's constant(s) the nick-name 'action' in quantum-mechanics.

I submit that the concept of 'joules per Hz' is reasonably easily comprehended, 'joules per cycle per second' and 'joules per radian per second' would be even clearer, but calling both of these 'joule seconds' is incorrect (being incomplete) and naming them 'action' is confusing nonsense!

Any comments? GilesW (talk) 13:45, 30 April 2009 (UTC)

Actually, Dirac's constant (the reduced Planck's constant) is a conversion factor between the phase (in radians) of the quantum mechanical wave function and the corresponding classical action (in Joule seconds) discussed in this article. JRSpriggs (talk) 00:47, 1 May 2009 (UTC)
Please read this entry: Planck's_constant#Reduced_Planck_constant. It supports my basic point. GilesW (talk) 07:38, 16 May 2009 (UTC)
As a separate point, I see that the name 'Dirac's constant' has been disputed inTalk:Planck_constant#Who_calls_it_.22Dirac.27s_constant.22.3F. GilesW (talk) 07:38, 16 May 2009 (UTC)

## What is "the bounded part of" the trajectory in the lead?

I think it has to be either defined or deleted. Bakken (talk) 00:32, 12 December 2009 (UTC)

It is not "the bounded part". It is "a bounded part", any bounded part. For example, it might be the part which lies between 1 A.M. and 2 A.M. (UTC) today within 20 meters of the tip of the Eiffel Tower. There must be some temporal and spatial bounds or else the integral will not be a definite integral and thus will (generally) not have a value. JRSpriggs (talk) 01:33, 14 December 2009 (UTC)
That's not a definition, sorry. Could you please give a mathematical definition (with a reference or a cross-link, perhaps) and put it the the lead? As it stands now, the lead does not make much sense. Again, the article on principle of least action has no mention of some mystical "bounded parts" of trajectories. Bakken (talk) 10:22, 14 December 2009 (UTC)
There is nothing mystical about it. Consider the simple example of a non-relativistic particle in a potential well. The action S is given by
$S = \int_{t_i}^{t_f} {m \dot{\mathbf{x}} \cdot \dot{\mathbf{x}} \over 2} - V (\mathbf{x}) d t \,.$
The bounded part means the part between the bounds ti and tf. If you just left it as an indefinite integral or set the limits of integration at infinity, then the action would not have a specific value which is unacceptable.
The reason I did not just mention a start time and end time is that when considering fields one must bound space as well as time. The four dimensional hyper-volume of integration must be bounded by a three dimensional boundary. If you want to get technical, "bounded part" means compact submanifold. JRSpriggs (talk) 16:55, 14 December 2009 (UTC)
I have reformulated your sentence and moved it to the end of the lead. Is it all right with you? I don't think this technicality is of a conceptual importance. Indeed sometimes (e.g. when the fields vanish fast enough at large distances) the action integral can include all space and still be well defined. Bakken (talk) 23:03, 14 December 2009 (UTC)
OK. JRSpriggs (talk) 11:07, 16 December 2009 (UTC)

## Reduced action

Reduced action redirects here, but the term is never defined. -Craig Pemberton 08:30, 30 December 2009 (UTC)

## Introduction not clear

I read the introductory paragraph several times but wasn't able to understand anything about the subject. Could someone re-write it? Whatever may have been the purpose of the author of the current text, I ask that the goal be instead to introduce the subject simply and clearly to the intererested reader. Thanks in advance! Mark.camp (talk) 03:42, 7 January 2012 (UTC)

### Introduction concise

Greetings! What I , as a Physics Student, feel about the introduction is that- it is clear, concise and up to the mark. The introduction puts forth a clear idea of what "Action" is. Nevertheless, I agree with Mark's view that it may not be easy for a beginner to understand the introduction properly as there is required some maturity to grasp the concept fully. To address this issue of not-very-clear introduction, I suggest we create one more page titled "An introduction to Action in Physics", similar to what is been done for Angular Momentum. It also makes sense in doing this since Action is as important a concept in Physics as is Angular Momentum(in fact they have the same SI units , showing that they are derived from same "basic" physical quantities). When looking at the need to merge the page of "Action" with Hamilton's Principle, I feel that it won't be a good idea because the page Action basically talks about the concept of Action in Physics and Mathematics whereas the page Hamilton's Principle, as is clear from the title, is intended to present Hamilton's Principle, which has to do with action as a property of motion. With best regards! — Preceding unsigned comment added by Shivsagardharam (talkcontribs) 10:17, 9 January 2012 (UTC)

Yeah, concise to you, a physics major! But I got here to find out what "action" is so I can understand "Lagrangian". However, what I got from this is: "action is the integration of the Lagrangian across time", which, being circular, tells me nothing other than that I'm even stupider than I already knew I was. GIVE EXAMPLES please—even obscure ones, if that's the best you can do for an obscure topic.
Hey, I'm just sayin'...
HelviticaBold 06:31, 10 February 2012 (UTC)
You have to figure it out from the context. Asking for examples of action is like asking for examples of an electron. It just is what it is. It is not made of anything simpler. It can only be understood by how it relates to other things. Read the article on Lagrangians which has more specific examples than this article does. JRSpriggs (talk) 12:24, 10 February 2012 (UTC)

### Intro is still useless

It still needs re-writing, how the hell is anyone supposed to get anything from this??

• "Physical laws are most often expressed as differential equations, which specify how a physical quantity varies over infinitesimally small changes in time, position, or other independent variables in its domain. A solution to a differential equation provides the value of the physical variable at any point in its domain (e.g., $u = f( \mathbf{x}, t )$ ), given both initial conditions and boundary conditions." <--- This is strangely worded (at least in my veiw), though looks pretty much correct. I suggest:
"Physical laws are frequently expressed as differential equations, which describe how physical quantities such as position and momentum change continuously with time. Given the initial and boundary conditions for the situation, the solution to the equation is a function describing the behaviour of the system (positions and momenta of the particles) at all times and all positions within the set boundaries."
• "In analytical dynamics, the action represents the final form obtained by working backwards from classical Newtonian mechanics to achieve an integral minimization expression in the form of a variational statement." <--- What will a reader get from reading this statement? I can't imagine they can, and have no clue what this means... If the implication is that:
"because the E-L eqn from the calculus of variations
$\frac{d}{dx}\frac{\partial F}{\partial y'}=\frac{\partial F}{\partial y},\quad y'=\frac{dy}{dx}$
has a corresponding functional in the form of an integral,
$\int F(y, y',x)dx$
suggestively Lagrange's equations in Lagrangian mechanics
$\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}=\frac{\partial L}{\partial q},\quad \dot{q}=\frac{dq}{dt}$
has a corresponding functional in the form of an integral AKA the action,
$\int L(q,\dot{q},t)dt = S[q]$
and action satisfies a variational principle." Or words to this effect?
• "The statement is profound, simple, and elegant but comes at the cost of several simplifying assumptions." <--- ... also its prosey and flowery (what happened to neutrality?).
• "The integral form espoused here can only be applied to conservative holonomic mechanical systems and to do otherwise can yield incorrect results." <--- should SIMPLIFY to:
"The integral form used here can only be applied to mechanical systems which are conservative and holonomic; beyond this application it yields incorrect results."
• "The equivalence of these two approaches is contained in Hamilton's principle, which states that the differential equations of motion for any physical system can be re-formulated as an equivalent integral equation." <--- There is no mention of what "these two approaches" are, the sentence randomly starts this way. Presumably referring to Lagrangian and Hamiltonian (or Newtonian??) mechanics right? say so...

This is just my suggestion. If no one else does I'll try soon, busy right now...

-- F=q(E+v×B) ⇄ ici 02:34, 10 March 2012 (UTC)

I re-wrote it. Any better/worse?-- F=q(E+v×B) ⇄ ici 13:16, 10 March 2012 (UTC)
My typos are so bad I don't know what to say... Thanks for fixing that JRSpriggs. Not sure if there is a "vibrational principle"...-- F=q(E+v×B) ⇄ ici 09:37, 12 March 2012 (UTC)

## I understood

I see lots of comments arguing about whether the content of this page is confusing. Well I just wanted to say that I understand this shit and I did not understand it before. It's def one of the best-written articles I've ever read: give yourselves a break. -Keith (Hypergeek14)Talk 10:31, 27 June 2012 (UTC)