This question is based on the description of Longair in his book "Theoretical Concepts in Physics".

He starts by giving some provisions:

  • Conservative force field
  • Fixed times $t_1$ and $t_2$
  • Object moves from fixed point at $t_1$ to fixed point at $t_2$

Then he defines:

  • Lagrangian: $L = K-U$
  • Action: $S = \int_{t_1}^{t_2}Ldt$

He goes on to explain, that the principle of least action means, that an object moves on a path so that $S$ is minimized.

Then he claims that this priciple is equal to Newton's 2nd law of motion, following through with a proof which is beyond my comprehension (which of course is my fault).

After I calculated $S$ for a few examples, I am convinced, that this claim is correct only adding one additional provision (which Longair clearly does not state directly or indirectly):

  • The object moves on a path fixed in space. (Just the speeds at the points is allowed to differ.)

My argument for why this is necessary follows from a counterexample:

  • Assume a central force field with constant force. Setup the object so that its trajectory is a circle. Take time $t_1$ and $t_2$ so that the object is at opposing ends of the circle, describing a half circle. Now change the force field, so that there is no force inside this circle. (This is still a conservative force field and the object moves still in the same circle.) Compare the $S$ of this half circle to the $S$ of the object moving with constant lower speed along the diameter of the circle. For both trajectories the $U$ is the same but the $K$ is lower for the shortcut along the diameter (lower speed). So the shortcut along the diameter has a lower action. Still, with the correct initial speed the object will move the half circle, fully in accordance to Newton's second law of motion.

Since I cannot assume, that I found an error in Longair's standard book, can anyone please explain, what I got wrong.

  • $\begingroup$ Without a force you cannot keep on staying on a circle. $\endgroup$
    – Jon
    Nov 28, 2011 at 15:24
  • $\begingroup$ @jon No force only inside the cirlce, still a central force at the edge. $\endgroup$
    – catalyst
    Nov 28, 2011 at 15:34
  • 1
    $\begingroup$ "Setup the object so that its trajectory is a circle" - You don't get to set up the trajectory beforehand in the principle of least action. You have to fix end-points $r(t_1)$ and $r(t_2)$, and the trajectory you obtain will depend on the initial conditions! $\endgroup$
    – Lagerbaer
    Nov 28, 2011 at 15:54
  • $\begingroup$ @lager The quoted sentence applies to a Newtonian perspective. I think the rest of the paragraph, especially its end, makes this quite clear. And I fixed $r(t_1)$ and $r(t_2)$ in the sentence afterwards as the endpoints of the half circle. What I do not understand is what you mean with "will depend on the initial conditions". What are initial conditions? $\endgroup$
    – catalyst
    Nov 28, 2011 at 16:17

1 Answer 1


The general carelessness with the so-called "principle of least action" it that even in very good and reliable sources it is incorrectly stated that the action must be minimal. While the principle only requires that the action must be stationary, e.g. $\delta S = 0$.

So, more correctly, it should be called a "principle of stationary action". Concerning your example -- both of your trajectories are stationary, therefore both of them might be the true trajectory of your body.

  • $\begingroup$ +1: Sorry this is correct. One should also say that if you choose small enough increments of time, the local solution of Newton's laws is usually a true minimum action path along each increment of time (for ordinary force laws). $\endgroup$
    – Ron Maimon
    Nov 28, 2011 at 17:35
  • $\begingroup$ "Sorry?" For what? And I confirm that for very small increments the action is really minimal. Also as selfish promotion I'd like to give a link to this related answer: physics.stackexchange.com/a/17567/386 $\endgroup$
    – Kostya
    Nov 28, 2011 at 17:41
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    $\begingroup$ Sorry because in a comment 10 seconds before, I wrote that you were wrong about the horizontal solution being stationary. But in the case that the points are antipodal, the action minimum is nonunique. I was in the middle of my own answer. The action is a pure local minimum until you reach a focusing point, which is the half-circle in this case, and it is often a global minimum. $\endgroup$
    – Ron Maimon
    Nov 28, 2011 at 17:46
  • $\begingroup$ Yeah -- that is all correct. Actually I've just found nice reference for that: scholarpedia.org/article/Principle_of_least_action $\endgroup$
    – Kostya
    Nov 28, 2011 at 17:54
  • $\begingroup$ @kostya Thank you very much for your answer. I now read Feynman's lectures on this topic (volume 2, chapter 19 as recommended by Longair) and understand it. Feynman at least mentions after the proof, that it is about S being stationary. I read through Longair's description again and find, that there are some more inaccuracies. Well, Longair himself states that his book is meant as a recapitulation for students who are already familiar with the topic, so my fault again for choosing the wrong book. $\endgroup$
    – catalyst
    Nov 29, 2011 at 20:59

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