Talk:Lagrangian mechanics

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Definitely needs a rewrite. A lot of the content here overlaps with the content in action (physics), but the derivation of the Euler-Lagrange equations differs. Some consolidation is probably in order, and I think I prefer the one here to the one in action (physics). There's definitely a notational issue, since this page uses r' and the other uses r-dot.

Taral 08:13, 19 Jun 2004 (UTC)

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Kinetic energy relations

We are just about to calcuate a derivative of kinetic energy, I just wondered if it wouldn't be easier to put this part in a following way, without the need of further explaining "vanishing" 1/2 factor and so on:

$\frac {\partial T}{\partial \dot{q}_j} = \frac 1 2 \cdot \frac {\partial \sum_{i=1}^n m_i \mathbf {v}_i \cdot \mathbf {v}_i}{\partial \dot{q}_i} = \frac 1 2 \cdot \frac {\partial \sum_{i=1}^n m_i \mathbf {v}_i^2}{\partial \dot{q}_i} = \frac 1 2 \cdot \sum_{i=1}^n \frac {\partial m_i \mathbf {v}_i^2}{\partial \dot{q}_i}$

Now, according to $\, {\frac {\partial}{\partial x} f^2(x) = 2 f(x) \frac {\partial}{\partial x} f(x)}\,$, we have:

$\frac {\partial T}{\partial \dot{q}_j} = \frac 1 2 \cdot \sum_{i=1}^n 2 \cdot m_i \mathbf {v}_i \frac {\partial \mathbf {v}_i}{\partial \dot{q}_i} = \sum_{i=1}^n m_i \mathbf {v}_i \frac {\partial \mathbf {v}_i}{\partial \dot{q}_i}$

Ender2101 10:50, 14 Feb 2009 (CET)

I have trouble understanding this bit below:

More generally, we can work with a set of generalized coordinates and their time derivatives, the generalized velocities: {q_j, q′_j}. r is related to the generalized coordinates by some transformation equation:

$\mathbf{r} = \mathbf{r}(q_1 , q_2 , q_3, t). \,\!$

What is q

What is this equation? $\mathbf{r}(q_1 , q_2 , q_3, t). \,\!$

The above equation makes no sense what so ever.

I cleared it up a bit, I hope, by reordering that sentence and adding a really simple example. Laura Scudder 00:08, 7 Mar 2005 (UTC)

Great article. But since the biggest clincher is generalized coordinates, perhaps there should be separate discussion on the matter elsewhere? --Rev Prez 13:10, 28 May 2005 (UTC)

"Generalized coordinates", according to Landau & Lifshitz, refers to any s quantities $q_1, q_2, ..., q_s$ which completely define position in a system with s degrees of freedom. "Generalized velocities" are the associated velocities. You can think of them as vectors. For example, in a 3D system, which has three degrees of freedom, the usual way to think about the q variables are x, y and z. The Lagrangian works in spherical and cylindrical coordinate systems as well, which may be why the "generalized" label is used. - mako 30 June 2005 00:25 (UTC)

The generalized coordinates are really about picking the coordinates that make your life easiest. For instance, simple problems often have lower dimensional motion embedded in 3 dimensions, so you don't actually need 3 generalized coordinates. It all depends on whether the constraints on the motion are holonomic or not. So the generalized reminds you that your coordinates may need to be a totally non-traditional system, like the length along a wire bent into a weird shape (perhaps a bead is moving along the wire). --Laura Scudder | Talk 30 June 2005 01:16 (UTC)

generalized coordinates

The link generalized coordinates links to this page (Lagrange mechanics), I believe it would be nice to have either a larger discussion of generalized coordinates on this page or perhaps its own article. Article could give examples (such as how they are used in cartesian or spherical-polar coordinates) and discuss the relation to degrees of freedom. Perhaps a discussion on the related generalized momenta. Ideas? 71.131.37.89 02:37, 21 December 2005 (UTC)

gen. coord. added

Generalized coordinates now have their own article, which is not 100% yet, but is a good start. I'll continue to work on it and parse the content between this page and that over the next week.Jgates 03:13, 25 December 2005 (UTC)

• As pointed before, nice article, however i find this lack on several "rare" but sometimes common examples involving Lagrangians:

$L(q,q',q'',t)$ (Lagrangian with an acceleration term)

and the usual Hamiltonian Mechanics, i think in this case Hamiltonian is given by:

$xp'+x'p''-H(q,p,p')=L$

these Hamiltonians happens in the euler-Lagrange equations for GR see "Ray D'inverno: Introducing Einstein's Relativity" (university course in Cosmology )

overloaded symbols

Symbols on this page like $\mathbf{r}$, $\mathbf{v}$, $\mathbf{a}$ are overloaded to denote multiple functions. I realize this is a common practice is physics; stating which function a symbol refers to at key points, however, can make the article easier to follow.

Let's say $A$ is the map that relates the coordinates $q_i$, $t$ to $\mathbf{r}$, ie $\mathbf{r} = A(q_1, \dots, q_m, t)$, and $f_i$ is a function that gives an object $i$'s coordinates $q_j$, $t$ with time $t$, $\left\langle q_1, \dots, q_m, t \right\rangle = f_i(t)$. Frustrating ambiguities arise when the article discusses $\mathbf{v}$'s partial derivatives. In expression $\mathbf{v}_i = \sum_{j=1}^m \frac {\partial \mathbf{r}_i}{\partial q_j} \dot{q}_j + \frac {\partial \mathbf{r}_i}{\partial t}$, $\mathbf{v}_i$ could be regarded as a function of only $q_j$, $t$ where $\frac {\partial \mathbf {v}_i}{\partial q_j}$ has one meaning (and you're expected to assume the operator in $\dot{q}_j$ commutes with $\frac {\partial}{\partial q_j}$) or $\mathbf{v}_i$ could be regarded as a function of $\dot{q}_j$, $q_j$ (hidden in $\mathbf{r}_i$), $t$ where $\frac {\partial \mathbf {v}_i}{\partial q_j}$ has a completely different meaning or $\mathbf{v}_i$ could be regarded as simply $(A \circ f_i)^\prime = (A^\prime \circ f_i) f_i^\prime$ (the proper definition, a function of only 1 variable, $t$), among other possibilities. It seems to be the second one.

My point is the reader has to guess and shouldn't. --L0mars01 23:17, 7 November 2007 (UTC)

stationary vs. minimum action integral

While the action must be stationary for the entire path, Landau and Lifschitz argue that for sufficiently small displacements it needs to be a minimum (I think 2nd page of 3rd ed. ). I do not exactly see why this helps them, other than allowing them to prove that mass is positive (don't take my word for this though).. Could this possibly be worth a mention in the article? —Preceding unsigned comment added by 163.1.62.20 (talk) 13:55, 5 March 2008 (UTC)

New/old lagrange equations

What is the difference between the sections "lagranges equations" and "old lagranges equations"? Has it just been rewritten? Then the old one should be removed. If they have independently valuable content, then perhaps the titles should be changed, or at least a summary explaining what the differences between these lengthy derivations are. Feyrauth (talk) 01:35, 23 April 2008 (UTC)

Yes, why are two sections to derive the same equations? 189.179.243.82 (talk) 00:53, 16 January 2011 (UTC)

Lagrange multipliers and Cartesian Coordinates

The aspect of (undetermined) Lagrange multipliers for the Lagrangian mechanics is missing in this article. Stressing this aspect would lead how to deal with systems witch still contains constraints e.g. using cartesian coordinates. I think this could be a start: http://electron6.phys.utk.edu/phys594/Tools/mechanics/summary/lagrangian/lagrangian.htm —Preceding unsigned comment added by 217.229.106.204 (talk) 08:57, 4 May 2008 (UTC)

Radical rewrite or a new article to the many already existing about more or less the same subject?

The basic form of the Lagrangian equations of motion valid also for non-conservative forces is:

$\dot{\overbrace{\frac{\partial T}{\partial \dot{q_i}}}} = Q_i + \frac{\partial T}{\partial q_i}$

where

$T = T(q_1,...,q_n,\dot{q_1},...,\dot{q_n},t)$

is the kinetic energy expressed in generalized coordinates

$q_i= q_i (x_1,...,x_n,t)$

where $x_1,...,x_n$ the cartesian coordinates of the many mass points involved and $t$ is time

and

$Q_i=\sum_{j=1}^n F_j \cdot \frac{\partial x_j}{\partial q_i}$

are the generalized force components.

This relation is in my opinion easiest and most understandably derived by straightforward variable transformation (i.e. invariance under transformation) without reference to any variation principle!

If the forces are conservative, i.e

$F_j=-\frac{\partial V}{\partial x_j}$

one gets that

$Q_i=-\sum_{j=1}^n \frac{\partial V}{\partial x_j} \cdot \frac{\partial x_j}{\partial q_i} = -\frac{\partial V}{\partial q_i}$

and

$\dot{\overbrace{\frac{\partial T}{\partial \dot{q_i}}}} = \frac{\partial T}{\partial q_i}- \frac{\partial V}{\partial q_i}$

One should also point out that $\frac{\partial T}{\partial \dot{q_i}}$ are the generalized momenta

$p_i$

and that the first order differential equation system to integrate is

$\dot{p_i} = \frac{\partial T}{\partial q_i}+ Q_i$

$\dot{q_i} = \operatorname{F_i}(p_1,...,p_n,q_1,...,q_n)$

where $F_i$ are the "inverse functions" to $p_i=\frac{\partial T}{\partial \dot{q_i}}$ obtained by solving for $\dot{q_i}$

And examples how this differential equation system is derived, i.e. some concrete examples of explicitly derived functions

$\frac{\partial T}{\partial q_i}+ Q_i$

and

$\operatorname{F_i}(p_1,...,p_n,q_1,...,q_n)$

should be included, possibly also with the result of a numerical integration of this system of first order differential equations

This would make the subject more praxis oriented and of use for a larger audience!

Stamcose (talk) 12:26, 17 July 2008 (UTC)

Scalar potential vs. potential energy

Hello everyone, I am a bit confused about that generalised force derivation, Q_j, in conservative field, given by scalar potential. The quantity V, this is evidently the potential energy, although the author denotes that as scalar potential and subsequantly use it as potential energy. Please, if there is anybody involved, explain it to me. Thanks a lot!

Pavol, Slovakia —Preceding unsigned comment added by 147.175.85.38 (talk) 19:41, 5 November 2009 (UTC)

Proposed change, at another editor's request. Requesting feedback

Hello, a new editor (User:Desfuscay) requested to make the following changes in this section. Since I don't know enough of the subject, can someone look it over and see if it is correct and okay to add?

indicating the presence of a constant of motion. Performing the same procedure for the variable $\theta$ yields:

$\frac{\mathrm{-d}}{\mathrm{d}t}\left[ m( \dot x \ell \cos\theta + \ell^2 \dot\theta ) \right] + m (\dot x \ell \dot \theta + g \ell) \sin\theta = 0;$

Thanks--Obsidi♠n^Soul 06:10, 24 February 2011 (UTC)

Removed last sentence from the intro

I removed the following sentence from the end of the intro: This does not appear to be true, since using standard Newtonian mechanics one would ignore an analysis of forces and use conservation of energy which follows directly from Newton's laws; alternatively one can eliminate the constraint force by resolving tangentially to the groove. It did not fit in and seemed like a topic of discussion. Hamsterlopithecus (talk) 11:20, 5 December 2011 (UTC)

Clean up

I made various changes as shown in the edit summary. I intend to clean up more but too busy for now. Also - what is "Old Lagrange's equations" supposed to mean when the result is identically stated in the article as the Euler-lagrange equations? Should no-one object "old" will be eventually removed. -- F = q(E + v × B) 13:45, 17 February 2012 (UTC)