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I have this conceptual question: In Landau's book of classical mechanics, about the principle of least action, it's written:

$$\left. \delta S =\frac{\partial L}{\partial v} \delta q \right\rvert_{t_1}^{t_2} + \ \int\limits_{t_1}^{t_2} \ dt \left(\frac{\partial L}{\partial q}\ - \frac d {dt}\frac {\partial L}{\partial v}\right) \delta q \ =0 ,$$

where $q=q(t)$ is the position function, $v=v(t)$ velocity function, $S$ the action, and L the Lagrangian of the system.

There is the condition $\delta q(t_1)=\delta q(t_2)=0$. So the first term is zero, and then it says " there remains an integral which must vanish for all values of $\delta q$. This can be so only if the integrand is zero identically.

Well I can't understand why

$$\left(\frac{\partial L}{\partial q}\ - \frac d {dt}\frac {\partial L}{\partial v}\right) =0 \;.$$

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  • $\begingroup$ Because you are minimizing the action, that whole expression must be zero in order for it to be a minima. Since t1 and t2 are arbitrary, the integrand has to be zero. $\endgroup$ – Señor O Apr 15 '17 at 0:56
  • $\begingroup$ But if the result is x^2 it also vanishes everywhere, isn't it? $\endgroup$ – santimirandarp Apr 15 '17 at 0:59
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    $\begingroup$ The mathematical statement you seem to be asking about is known as the "fundamental lemma of variational calculus". $\endgroup$ – Adomas Baliuka Apr 15 '17 at 1:02
  • $\begingroup$ What do you mean "if the result is x^2"? What result is x^2? $\endgroup$ – Señor O Apr 15 '17 at 1:06
  • $\begingroup$ Oh, i was saying something silly. I am sorry. But i still wonder if there is a better way to justify that the integrand is zero. $\endgroup$ – santimirandarp Apr 15 '17 at 1:07
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Consider the following integral $$\int_{t_1}^{t_2}F(t)\delta q(t) dt=0,$$ for every possible $\delta q$. Then choose a function $\delta q$ which has a large value but is different from zero only in an infinitesimal neighborhood of a point $t_0\in[t_1,t_2]$. Then, $$0=\int_{t_1}^{t_2}F(t)\delta q(t) dt=\int_{t_0-\epsilon}^{t_0+\epsilon}F(t)\delta q(t) dt=F(t_0)\int_{t_0-\epsilon}^{t_0+\epsilon}\delta q(t) dt.$$ In the last equal sign we took $F$ out of the integral because the function is approximately constant in the infinitesimal interval $2\epsilon$. Now, the last integral above is different from zero, so $F(t_0)$ has to vanish. Repeat this argument for all $t\in[t_1,t_2]$ and you obtain that $F(t)$ is identically zero. This is basically the fundamental lemma of variational calculus.

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  • $\begingroup$ Okey, i think it is fairly good.. $\endgroup$ – santimirandarp Apr 15 '17 at 1:38
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The point is that if $\delta q(t_1) = \delta q(t_2) = 0$ then

$$ 0 = \int_{t_1}^{t_2}{\rm d}t \left(\frac{{\rm d}}{{\rm d}t}\frac{\partial L}{\partial v} - \frac{\partial L}{\partial q}\right)\delta q $$

and this must be true regardless of the selection of $t_1$, $t_2$ or $\delta q$. Imagine that $q$ live in a $n$-dimensional space, all the variations are independent of each-other, this quadrature can be interpreted as a linear combination of such variations. Since they are independent, and add up to zero, this means that the coefficients must vanish, that is

$$ \frac{{\rm d}}{{\rm d}t}\frac{\partial L}{\partial v} - \frac{\partial L}{\partial q} = 0 $$

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  • $\begingroup$ Sorry but i am a poor chemist, is there a simpler way to undestand it? $\endgroup$ – santimirandarp Apr 15 '17 at 1:02
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When a particle travels from point 1 to point 2, the fundamental law is $F = ma$. One way to find the path a particle follows is to start with the velocity and acceleration of the particle and point 1, and use $F = ma$ to predict the new position after a small time step. Then find the velocity and acceleration at the new place, and follow it another small step. Repeat. Newton used this approach.

Lagrange had a different approach. Start with the initial and final positions and velocities. Consider all paths that connect them. The particle only follows one of these paths. Find a condition that is true for the actual path, and false for all the others. The condition is that $F = ma$ for all points of the path. This turns out to be equivalent to the condition that the integrand is $0$ on the path.

You can find a good derivation of this and a discussion of the benefits of this approach at David Tong's website.

Landau uses a different approach. He chooses the principal of least action as the fundamental law instead of $F=ma$. This is fine. You can derive one from the other.

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Maybe this approach will help:

Since the integral $ \ \int\limits_{t_1}^{t_2} \ dt \left(\frac{\partial L}{\partial q}\ - \frac d {dt}\frac {\partial L}{\partial v}\right) \delta q \ $ is zero for any $\delta q$ , it must also be zero for $$\delta q = \epsilon \left(\frac{\partial L}{\partial q}\ - \frac d {dt}\frac {\partial L}{\partial v}\right), $$ with $\epsilon$ a "small enough" real positive number. In this specific case, the action principle requires that, $$ \epsilon \ \int\limits_{t_1}^{t_2} \ dt \left(\frac{\partial L}{\partial q}\ - \frac d {dt}\frac {\partial L}{\partial v}\right)^2 =0. $$ The only way a non-negative integrant to yield an integral equal to zero, is for the integrant to be strictly zero, hence $$ \frac{\partial L}{\partial q}\ - \frac d {dt}\frac {\partial L}{\partial v}=0. $$

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protected by Qmechanic Apr 15 '17 at 4:43

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