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Suppose a body is moving directly upward against gravity. Its initial velocity is given as $100 m/s$ and acceleration due to gravity is $10 m/s^2$.

Then, $v(t)=100-10t$

$x(t)=100t-5t^2$

We are interested in the time period between $t=0s$ and $t=10s$.

Now, according to the principle of least action, for any body following Newton's law, $S=\intop(K-P)dt$, the value of $S$ is minimum. Here $K$ is Kinetic energy and $P$ is potential energy of the body.

In the above situation, $S=\intop_{0}^{10}(\frac{(100-10t)^2}{2}m-10(100t-5t^2)m)dt$

Assuming $m=1$,

$\Rightarrow S=100\intop_{0}^{10}(t^2-20t+50)dt=-16666.6$

velocity withbrespect to time graph

Now, let us define a new path for the particle which will bring a lower value of $S$.

(Edit: PLEASE NOTE THAT THE CREST AND TROUGH IN THE CASE 2 OF THE GRAPH ARE OF EQUAL AREA, HENCE MAKING THE DISTANCE TRAVELLED UNCHANGED.)

  1. $v(t)=100-10t$ for all $t$ except the time interval $t\in[2,3]$ and $t\in[5,6]$,
  2. In, $t\in[2,3]$, $v(t)=90-10t$,
  3. and in $t\in[5,6]$ $v(t)=110-10t$.

When I integrate for $S$, I get a value less than -16666.6. I really don't know what is happening!

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    $\begingroup$ The "Principle of least action" doesn't state that the action is minimized (as the name may suggest, though this name was given for other historical reasons), but rather extremized (minimum or maximum point). The use of this tool should also be in consistency of its assumptions - that is, the target function (the solution) is smooth (and your newly defined function is not). $\endgroup$
    – Alexander
    Oct 6, 2016 at 0:11
  • $\begingroup$ @Alexander but in the situation i am discussing, the function can be forcefully made continous in a steep way without changing the answer. Making it continous will only make the answer a little difficult to find, but will change nothing. $\endgroup$
    – Prem
    Oct 6, 2016 at 6:05
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    $\begingroup$ It has to be not only continous but also differentiable (after all it's a solution to 2nd order differential equation). There may be many other continuous solutions (some of them may have "better" action values), but unless those are differentiable solutions - calculus of variations won't find them. $\endgroup$
    – Alexander
    Oct 6, 2016 at 6:12
  • $\begingroup$ @marka I just found that the equation I had written was not describing the situation correctly, although the graph I drew is correct. Thanks for pointing out the mistake. The correct equation to describe the situation i want to describe has been updated in (2) and (3). My mistake! I really intend the total distance travelled in both situations to be same. But the problem still remains. Actually the mistakes in the question are typographic. $\endgroup$
    – Prem
    Oct 6, 2016 at 11:05
  • $\begingroup$ "THAT THE CREST AND TROUGH IN THE CASE 2 OF THE GRAPH ARE OF EQUAL AREA, HENCE MAKING THE DISTANCE TRAVELLED UNCHANGED.)" distance refers to the particle, and for the particle the distance is changed, it is not a vector. $\endgroup$
    – anna v
    Oct 7, 2016 at 4:24

2 Answers 2

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I'm afraid you have made a mistake in the calculation. If we generalize so that

  • $v(t) = 100-d+10t$ for $2<t<3$ and
  • $v(t) = 100+d+10t$ for $5<t<6$

then the total action is $S[d]=-50000/3 + d^2$, i.e. the action is minimized when $d=0$, which is the classical path.

More generally, the principle of least action only requires that the classical paths be stationary points of the action, which is a statement about the action of nearby paths. So even if the action of $S[d]$ was lower for some $d>0$ that still wouldn't be a violation of the principle of least action. To violate the principle you would need to show $\lim_{d\to 0} S[d]/d \neq 0$. The $d\to 0$ part is what is meant by 'nearby' paths.

It is possible for a given system to have more than one classical path with the same start and end points (the famous one is the path which orbits the Earth once v.s. the one which goes straight up and comes straight down) in which case a classical path might not minimize the action globally. In your example here though the classical path is unique, and therefore it must be a global minimum (that's why everyone who commented is so confident you have made a mistake).

Although the derivation of the principle of least action (from the usual statements of Newton's laws) assumes the path to be differentiable, it is easy to show it holds for piecewise differentiable paths too, such as what you have here. (If you try to extend this to general continuous paths though then things start to get a bit tricky, because continuous paths can be weird.)

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  • $\begingroup$ Thank you for answering. Yes, there was an error in my calculation. And yes, Hilbert curves are interesting! $\endgroup$
    – Prem
    Oct 7, 2016 at 8:39
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The principle of least action states that among all paths ($t \mapsto x(t)$) that bring a body from $x_1$ at time $t_1$ to $x_2$ at time $t_2$, the physical path is the one that minimizes the action. Therefore it only makes sense to compare actions for paths with identical boundary conditions. Your piecewise-built test path should thus verify $x(10) = 500$ and I do not think this is the case.

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  • $\begingroup$ Thanks for pointing out the mistake! Actually, it is a typographic error. I intend the distance travelled in both the situations to be same. Please see the case 2 in the graph. I intend the crest and the trough in the v(t) graph to be of equal area. $\endgroup$
    – Prem
    Oct 6, 2016 at 11:13

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