# Principle of Least Action; Newton's 2nd Law of Motion

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.

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Without a force you cannot keep on staying on a circle. – Jon Nov 28 '11 at 15:24
@jon No force only inside the cirlce, still a central force at the edge. – catalyst Nov 28 '11 at 15:34
"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! – Lagerbaer Nov 28 '11 at 15:54
@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? – catalyst Nov 28 '11 at 16:17

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. $\delta S = 0$.