Talk:Action (physics)

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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.

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[edit] ≤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 r2 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)

[edit] 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)

[edit] 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, Jk is a single variable. Moreover, the Jk 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 S0 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 Jk, since the former is integrated over a generic trajectory, whereas Jk 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 q0 = q(t0), 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 Jk 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 S0 (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 Sk(qk) (still a function)
  • the constant of motion Jk, which equals the change in Sk(qk) as qk 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)

[edit] 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)

[edit] 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 (talk • contribs).

[edit] Intro

Can any one simplify the intro for a dummy like me to understand? —The preceding unsigned comment was added by Tugjob (talk • contribs) 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

[edit] 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)

[edit] 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)