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.

The Euler-Lagrange equations can be derived from the Principle of Least Action using integration by parts and the fact that the variation is zero at the end points.

This has a mystical air about it, with the system somehow "sniffing" out all possible paths in the distant future and choosing the one that minimizes the action. An infinitesimal version would make the method less mystical in this sense, and so I would like to ask:

Can the Euler-Lagrange equations be derived from an infinitesimal principle of least action?

share|improve this question
3  
Can the "infinitesimal form" be anything else but the differential equations themselfs? If you argue locally, you can't further reduce the information. And you will not be able to keep a meaningful expression in terms of an integral if you bring the endpoints together/arbitrarily close. –  NikolajK Oct 17 '12 at 15:17
1  
Is the mystical sniffing of all possible paths just another way to talk of the consequences (classical limit) of the QM path integral? The reason the classical PLA works is because of quantum mechanics perhaps? –  twistor59 Oct 17 '12 at 15:49
    
Wikipedia: en.wikipedia.org/wiki/Hamilton%27s_principle –  Dimensio1n0 Jul 15 '13 at 16:49
    
@twistor59: Yes. –  Dimensio1n0 Jul 15 '13 at 16:49

3 Answers 3

The differential equations do the same thing: after integration you obtain a solution with "two" arbitrary constants. It means you obtain a whole two-parametric family of possible solutions (curves). Fixing endpoints reduces the solution to one unique curve. It has the property to obey the differential equations (=> minimises the action) and to pass through the given endpoints.

share|improve this answer

Let us just consider point mechanics$^1$

$$ S[q]~:=~\int_{t_i}^{t_f}\!dt~ L, \qquad L~=~L(q,\dot{q},t)$$

to be concrete, where $t_i$ and $t_f$ denote initial and final time, respectively; and we have imposed some appropriate boundary conditions.

I guess OP is asking his question(v1) because of the apparent teleology of the stationary action principle, as also discussed in this Phys.SE answer.

One can get a local stationary action principle as follows: If one assumes that there exists a possibly very small (but finite) constant $\epsilon>0$ (of dimension time), such the stationary action principle holds for all $t_i$ and $t_f$ with $|t_f-t_i|<\epsilon$, then one would still be able to derive the Euler-Lagrange equations (via the standard derivation).

By choosing $\epsilon$ much smaller than all the characteristic time scales of the physical problem, this would for all practical purposes be "an infinitesimal stationary action principle".

But at the philosophical level, if OP felt uneasy about the apparent teleology before, he would probably still feel uneasy about the above $\epsilon$-local stationary action principle, no matter how small $\epsilon$ is chosen.

I agree with Nick Kidman's comment above, that "an infinitesimal stationary action principle", in its manifestly infinitesimal form, would just constitute the Euler-Lagrange equations themselves.

--

$^1$The field-theoretic generalization is straightforward.

share|improve this answer

The infinitesimal case eliminates the apparent tetology because we can predict where the system will be at time $dt$ in the future from $dq = \dot q dt$.

Let's start with the one dimensional case. To vary the path over an infinitesimal time, we need at least two infinitesimal time intervals where $\dot x$ is varied by $\delta\dot x_1$ over $dt_1$, and then in the opposite sense $\delta\dot x_2$ over $dt_2$. Thus it ends up at the predicted fixed position $\dot x(dt_1+dt_2)$

The infinitesimal action is then simply the Riemann sum over $dt_1$ and $dt_2$ $$L_t = L(x,\dot x ,t),\,L_{dt1} = L(x+\dot xdt_1,\dot x,t+dt_1),\, L_{\delta \dot x dt1} = L(x+(\dot x+ \delta \dot x_1)dt_1,\dot x,t+dt_1)$$

$$ ds = L_tdt_1 + (L_t + \frac {\partial L_{dt1}}{\partial x}\dot xdt_1)dt_2$$

$$ ds+ \delta ds = (L_t + \frac {\partial L_t} {\partial \dot x}\delta \dot x_1)dt_1 + (L_{\delta\dot x dt1} + \frac {\partial L_{\delta \dot x dt1}} {\partial \dot x}\delta \dot x_2 + \frac {\partial L_{\delta \dot x dt1}} {\partial x}(\dot x+\delta\dot x_1)dt_1)dt_2$$

Since the variation in position over $dt_1$ must be compensated by that over $dt_2$, then $\delta \dot x_1dt_1 = - \delta \dot x_2dt_2$, so that

$$ \begin{align*}\delta ds &= (\frac {\partial L_t} {\partial \dot x}\delta \dot x_1- \frac {\partial L_{\delta\dot x dt1}} {\partial \dot x}\delta \dot x_1 )dt_1 + \frac {\partial L_{\delta \dot x dt1}} {\partial x}\delta\dot x_1dt_1dt_2\\ &=0\\ &= -\frac{\frac {\partial L_{\delta\dot x}} {\partial \dot x}-\frac {\partial L_t} {\partial \dot x}}{dt_2} +\frac {\partial L_{\delta \dot x dt1}} {\partial x}\\ &= -\frac {d}{dt}\frac {\partial L}{\partial \dot x} + \frac{\partial L}{\partial x}\end{align*}$$

Obviously this generalises to any number of coordinates since the infinitesimal action can be varied independently for each $q_i$.

share|improve this answer
    
looks interesting –  Physiks lover Oct 30 '12 at 22:16

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.