Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I am reading Gelfand's Calculus of Variations & mathematically everything makes sense to me, it makes perfect sense to me to set up the mathematics of extremization of functionals & show that in extremizing a certain functional you can end up with Newton's laws (i.e. you could extremize some arbitrary functional $L$ & in examining the form of Newton's laws you see that one should define $L$ as $T - V$) & this way of looking at things requires no magic, to me it seems as though you've just found a clever way of doing mathematics that results in Newton's laws.

However in books like Landau one must assume this magical principle of least action using the kind of thinking akin to Maupertuis, & I remember that every time I read some form of justification of this there's always a crux point where they'd say 'because it works'. I'm thinking that there may be a way to explain the principle of least action if you think of extremizing functionals along the lines Euler first did & use the method of finite differences (as is done in the chapter on the Variational Derivative in Gelfand, can't post a link unfortunately), i.e. because you're thinking of a functional as a function of n variables you can somehow incorporate the $T - V$ naturally into what you're extremizing, but I don't really know... I'm really hoping someone in here can give me a definitive answer, to kind of go through what the thought process is in this principle once and for all!

As far as I know, one can either assume this as a principle as in Landau & use all this beautiful theory of homogeneity etc... to get some results, or else once can assume Newton's laws & then use the principle of virtual work to derive the Euler-Lagrange equations, or start from Hamilton's equations & end up with Newton's laws, or one can assume Newton's laws & set up the calculus of variations & show how extremizing one kind of functional leads to Newton's laws - but I'm not clear on exactly what's going on & would really really love some help with this.

share|improve this question
    
Possible duplicates: physics.stackexchange.com/q/9/2451 , physics.stackexchange.com/q/15899/2451 and links therein. –  Qmechanic Jun 15 '13 at 21:55
    
Thanks very much, maybe I should have indicated that I'd read a ton of threads on this & many other forums & none seem to address my concerns unfortunately :( –  bolbteppa Jun 15 '13 at 22:48
    
I noticed your interesting comment about equations of motion having no action principle in one thread, a comment like that is exactly the kind of thing this thread is enquiring about, hopefully you'd take a moment to expand on this in light of the question asked if you have the time, if not no problem ;) –  bolbteppa Jun 15 '13 at 22:52
1  
@bolbteppa (2 comments up) yes, definitely. If you've looked at similar questions but they don't address what you want to know, you should mention that in your question. Something like "I looked at [other question] but it doesn't explain [thing], and also at [other other question] but it doesn't explain [other thing]." Tell us what research you've done so we don't duplicate it. –  David Z Jun 16 '13 at 1:23
1  
Dimension could you stop troll-editing my post, thank you. –  bolbteppa Sep 9 '13 at 22:04
show 8 more comments

3 Answers 3

I) OP's questions about the variational principle (e.g. the derivation to and from Newton's laws) are good and interesting questions, but most have been asks before, see e.g. this and this Phys.SE posts and links therein. Lagrangians not of the form Kinetic Energy minus Potential Energy are discussed in this Phys.SE post. The question of whether (or not) a set of equations of motion has an action principle is a duplicate of this Phys.SE post and links therein. Another possible helpful Phys.SE post concerning calculus of variation is this one.

II) The only new ingredient in OP's question seems to be the finite-difference method. Gelfand's use of the finite-difference method seems mostly to be an alternative way to gain insight into the Euler-Lagrange formula for the functional derivative via discretization. It doesn't add anything new to the continuum theory.

For instance, let us for simplicity consider point mechanics (as opposed to field theory). A continuum formulation for point mechanics here refers to a continuous time variable $t$ (as opposed to a discrete time variable). Let there be given a continuum Lagrangian $L(q,v,t)$ that depends on three arguments: position $q$, velocity $v$ and time $t$. Let $\frac{\partial L}{\partial q}$ and $\frac{\partial L}{\partial v}$ denote the derivative wrt. the first and second argument, respectively, and so forth.

Let us approximate the continuum action functional

$$\tag{1} S[q]~:=~\int_{0}^{T} \!\! dt~L(q(t),\dot{q}(t),t)$$

via the following discretized expression

$$\tag{2} S(q_0, \ldots q_{N})~:=~\Delta t\sum_{n=0}^{N-1} L(q_n,v_n,t_n), $$

where

$$\tag{3} \Delta t~:=~ \frac{T}{N}, \qquad t_n ~:= n \Delta t, \qquad q_n~:=~q(t_n),\qquad v_n~:=~\frac{q_{n+1}-q_n}{\Delta t}.$$

In eq. (3) we have chosen the forward difference for $v_n$. Obviously other choices are possible. Then

$$\frac{1}{\Delta t}\frac{\partial S(q_0, \ldots q_{N})}{\partial q_n} $$ $$\tag{4}~=~ \frac{\partial L(q_n,v_n,t_n)}{\partial q} -\frac{1}{\Delta t}\left[\frac{\partial L(q_n,v_n,t_n)}{\partial v}-\frac{\partial L(q_{n-1},v_{n-1},t_{n-1})}{\partial v} \right], $$

where $n\in\{1,2, \ldots,N-2, N-1\}$. Eq. (4) is a discretized version of the Euler-Lagrange formula

$$\tag{5} \frac{\delta S[q]}{\delta q (t)}~=~\frac{\partial L(q(t),\dot{q}(t),t)}{\partial q(t)} - \frac{d}{dt} \frac{\partial L(q(t),\dot{q}(t),t)}{\partial \dot{q}(t)} $$

for the functional derivative$^1$. The functional derivative (5) satisfies by definition the following infinitesimal variation formula

$$\tag{6} \delta S[q] ~:=~S[q+\delta q]- S[q]~=~ \int_{0}^{T} \!\! dt~\frac{\delta S[q]}{\delta q (t)}\delta q(t). $$

--

$^1$ The existence of functional derivative depends on the appropriately imposed boundary conditions, which we have not discussed. In plain English: We need to integrate by part at some point.

share|improve this answer
add comment

Action principles generalize to situations where Newton's laws are no longer valid--quantum mechanics and general relativity being the most obvious examples. They are therefore a more solid fundamental assumption to make than Newton's Laws, which are a subset of the possible laws of dynamics allowed by action principles.

share|improve this answer
    
On the other hand, many dissipative problems have no variational formulation, differing from the extremal of a functional by a total derivative. So I'm not sure how accurate it is to say that Newton's laws are a subset of the possible laws of dynamics allowed by action principles. A large intersection, maybe, but a strict subset, probably not. (That is, if you mean "action principle" in the strict sense of extremizing a local functional.) –  Chay Paterson Jun 17 '13 at 17:47
    
fair enough, but from the perspective of fundamental physics, those processes don't exist--they're emergent statistical phenomena. –  Jerry Schirmer Jun 17 '13 at 18:50
add comment

The principle of stationary action can be motivated from quantum mechanics. In quantum mechanics you don't have one outcome, but a whole set of them with different probabilities. The contribution of a path to the probability of an outcome is related to the path integral of something that depends on the action. In some sense, the classical principle of stationary action picks out the "most likely" path rather than considering all of them like we do in quantum mechanics. Philosophically this justification makes sense because classical mechanics is not reality; quantum mechanics is reality and classical mechanics is just an approximation. Therefore it makes sense that the justification for the principle of stationary action comes from an approximation of quantum mechanics (that is, considering only 1 path instead of all paths). However, using this justification it remains a mystery that historically the principle of stationary action came before quantum mechanics...

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.