Talk:Action (physics)

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1 ≤2005
2 Intuitive definition
3 Problems with definition of action
4 Examples

[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 $r 2$ 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:

$\frac{\partial L}{\partial \phi} = 0$

$\frac{d}{d t} \frac{\partial L}{\partial \dot{\phi}} = \frac{d}{dt} (m r^2 \dot{\phi} )$

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

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

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,

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)

[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).

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