Principle of Least Action via FiniteDifference Methodby bolbteppa Tags: action, finitedifference, method, principle 

#1
Jun1513, 10:21 PM

P: 124

I have to be honest, the principle of least action seems to me more of a religious claim one takes on complete faith, though of course I'm hoping this is just because I don't understand it. I tried to explain this to a friend suffering through a mechanics class & was literally pushed to say 'one must just assume this magical religiouslike principal, alright...' & I'm not happy to do that again!
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', or worse refer to QFT or Feynman as if any of this foresight was available to Einstein when employing variational principles on [classical] manifolds etc... 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), 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 EulerLagrange 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. Just to be clear I've read a ton of links online, on this site & many others & haven't found an answer to any of my questions unfortunately, thus any help is really really welcome! 



#2
Jun1713, 05:49 PM

P: 1,504

First of all, Landau  Mechanics is very criptic and he doesn't make clear proves of what he finds; I sincerely advise you to start from another book (I started reading that book 30 years ago and I still have a lot of things to understand .)
Second, I am not sure if you are talking of what is usually called the Hamilton's principle (the action, that is the time integral of L is stationary on the real motion q(t)) or if you are talking of Maupertuis' principle which rely on the "reduced action". Third, I believe you should formulate the question in a more direct way, for example: "why we define L as T  V in the context of...?" Fourth, the Hamilton's principle is a real mistery, however its utility is probably greater than what you believe, even in mechanics only. Anyway the real Big powerful theorem, which comes from the Lagrangian approach, is Noether's Theorem (you can cansider it as the "first word of the Physics"). 



#3
Jul313, 11:05 PM

P: 124

A quote from Weinstock's Calculus of Variations book:




#4
Jul413, 08:54 AM

P: 428

Principle of Least Action via FiniteDifference MethodThat being said, if you want to give analogies that will make him/her more comfortable, Fermat's principle is similar and is easier to visualize. I'm not sure that this is your issue though. 



#5
Jul413, 05:36 PM

P: 1,504

 lightarrow 



#6
Jul413, 05:54 PM

C. Spirit
Sci Advisor
Thanks
P: 4,941

I agree with you that the premise of the principle of least action is extremely mysterious in classical mechanics. I have yet to find a satisfactory answer for exactly the question you have posed. At the moment I just accept it because "it works and fits experiments" much like many other things in physics.




#7
Jul413, 09:47 PM

Sci Advisor
P: 8,009

A differential equation is often the solution of a variational problem, and many of those in classical mechanics are.
When does the variational formulation exist? There's discussion of the "inverse problem" of the calculus of variations in some of the papers at http://www.dic.univ.trieste.it/persp...nti/papers.htm, maybe try eg. http://www.dic.univ.trieste.it/persp...TO/Rassias.pdf. Also an interesting comment in the introduction of http://arxiv.org/abs/1008.3177 . Although the differential equation and the variational problem have the same physical content, when one wishes to enforce symmetries, it's often easier to do so via the variational formulation. IIRC, it's also easier to enforce constraints. 



#8
Jul513, 07:59 PM

P: 6

The argument for the stationary action principle is subtle, albeit powerful. I'm sure there might be some way to derive it independent of Newton's Laws, however this particular argument does use it, but it's still illuminating, in my opinion at least.
If you begin by defining L = TV, with V and T as functions of the vectors [itex]\vec{r},\dot{\vec{r}}[/itex], respectively, and t in Cartesian coordinates, it follows easily that [itex]\frac{\partial L}{\partial x} =\frac{\partial U}{\partial x}=F_x [/itex], and likewise for y and z, and that [itex]\frac{\partial L}{\partial \dot{x}}=\frac{\partial T}{\partial \dot{x}}=m\dot{x}=p_x [/itex] etc. Recalling that [itex]F_x=\frac{d}{dt}[/itex] it follows that [itex]\frac{\partial L}{\partial x}=\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}[/itex], and likewise for y and z. These equations take the form of an EulerLagrange equation, which implies (this part might require more proof) δ∫L=0. The crux (and subtle) move here is that if you let any generalized coordinate [itex]q_i[/itex] (e.g. polar or spherical coordinates) be a function of the position vector [itex]\vec{r}[/itex], then, as long as there is a unique value of x,y,z corresponding to each point of the generalized vector [itex](q_1,q_2,q_3)[/itex], then [itex] L(x,y,z,\dot{x},\dot{y},\dot{z},t)\rightarrow L(q_1,q_2,q_3,\dot{q_1},\dot{q_2},\dot{q_3},t)[/itex]. This doesn't change the action integral ∫L in anyway, and so, by going in reverse now, the EulerLagrange equations with respect to the new generalized coordinates take on the exact same form, i.e. [itex]\frac{\partial L}{\partial q_i}=\frac{d}{dt}\frac{\partial L}{\partial \dot{q_i}}[/itex] 



#9
Jul513, 08:03 PM

P: 6




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