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Consider a particle starting at rest in a potential well, the problem is restricted to one dimension. The lagrangian is $L=T-V$, set $V$ to 0 in the starting location, we get $L(t=0)=0$.

If the particle remains at rest, the action is of course 0. If the particle moves down the well, it gets kinetic energy $T>0$, and loses potential energy $V<0$, and so $L=T-V>0$.

We know that the particle WILL fall down the potential well, but this will result in a greater value for the action than staying at rest, so why does it?

Maybe nature "knows" that the particle will end up hitting something that will change it's movement and balance the action, but this seems non-local.

Qmechanic
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The Catalyst
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2 Answers2

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The principle of least action is poorly named, because often the classical solution corresponds to a saddle point, and not actually a minimum of the action. The condition that you use to derive the Euler-Lagrange equations is that the first order variation under a small change in path is zero, and that doesn't guarantee the solution will be a global minimum.

Andrew
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About change of motion: When an object is in a situation where there is a gradient in the potential then the object will accelerate down the potential gradient.

Given that acceleration: the stationary action criterion expresses what the acceleration profile will be.


You raise the question: "We know that the particle WILL fall down the potential well, but this will result in a greater value for the action than staying at rest, so why does it?"

The stationary action criterion does not relate to whether an object starts moving or not: if there is a gradient in the potential then the object will undergo change of velocity, down the gradient.

Let me put it this way: in order to formulate a theory of motion we have to grant as a fundamental law of motion: if there is a gradient in the potential then the object will undergo change of velocity, down the gradient.


Let's say an object has a bit of static charge, and there is another statically charged object and we have Coulomb force interaction going on. Then the inertial mass of each object is a factor in the resulting acceleration of the respective objects.

To find the actual acceleration profile: that is where the stationary action criterion is applicable.

The true trajectory has the property that the derivative of Hamilton's action is zero. The kinetic energy is proportional to the inertial mass. If the objects have a lot of inertial mass then the amount of change of velocity will be small.





The name 'least action' is most unfortunate, because it suggests something that isn't actually the case.

The criterion that always works is: the true trajectory has the property that the derivative of Hamilton's action is zero.


There are also classes of cases such that the true trajectory corresponds to a point in variation space where Hamilton's action is at a maximum.

An example of that is a situation where the potential increases with the cube of displacement.

We have that the expression for kinetic energy is a quadratic expression. But with a expression for the potential energy that is cubic the response of the potential energy to variation will be larger than the response of the kinetic energy.

And of course, that extends to quartic potential, quintic potential, and so on.

So you get a inversion: when the expression for the potential energy is to a power that is larger than the quadratic of the expression for the kinetic energy then the point-in-variation-space-where-the-derivative-is-zero will not be a minimum, but a maximum.

Of course, cases where the expression for the potential energy is cubic or higher are rare. Cases with the power of the potential energy expression lower than quadratic are the most common. (In the cases of Newtonian gravity and the Coulomb force the power of the potential is $-1$; the potential is proportional to $\tfrac{1}{r}$

Regardless of what is most common: it is important to be aware that Hamilton's stationary action is not about minimizing. Minimum or maximum is simply not relevant.

The true trajectory has the property that the derivative of Hamilton's action is zero. That is necessary, and sufficient.

Further reading:
In July this year I went back to the earliest question here on physics stackexchange about Hamilton's stationary action, and I posted an answer, with an exposition of Hamilton's stationary action My aim with that exposition is to make the subject transparent,

Cleonis
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