Magnitude of the variations $\delta q_i$ in the principle of stationary action

Question

To determine the equation of motion using the principle of stationary action, one has to consider the variation of the action due to variations $\delta q_i$ in all the generalized coordinates $q_i$. Do we need that the variations $\delta q_i$s be small in derivation and if so, can we make the notion of smallness precise? Thanks

Answer 1

In this answer I will first point out a general property of the stationary action concept, and with that property laid out I will proceed to your question specifically

The criterium of stationary action expresses the following demand: as an object is moving along some trajectory the rate of change of kinetic energy must match the rate of change of potential energy continuously. (As opposed so any kind of averaging.)

Here is how that 'continuously' comes into play:
Let's say you have obtained a trajectory from coordinates $(q_1, t_1)$ to $(q_3, t_3)$.
You can subdivide that trajectory in multiple subsections, for simplicity I will divide in two concatenated subsections: from $(q_1, t_1)$ to $(q_2, t_2)$ and from $(q_2, t_2)$ to $(q_3, t_3)$

If the action is stationary for the trajectory from $(q_1, t_1)$ to $(q_3, t_3)$, then if you evaluate the sections $(q_1, t_1)$ to $(q_2, t_2)$ and $(q_2, t_2)$ to $(q_3, t_3)$ individually: for each the action will be stationary.

In order for the action to be stationary from $(q_1, t_1)$ to $(q_3, t_3)$ the action must be stationary for each of the subsections. The subdivision reasoning goes down all the way to the limit of infinitisimally short subsections, concatenated. (This property was first pointed out by Jacob Bernoulli, older brother of Johann Bernoulli, in the course of solving the brachistochrone problem.)

So we have that the criterium of stationary action expresses the demand that as an object moves: from instant to instant, down to infinitisimal, the rate of change of kinetic energy must match the rate of change of potential energy.

I will refer to this as 'the concatenated subsections lemma'.

The concatenated subsections lemma is relevant in the following way: I will present a line of reasoning stated in terms of the global trajectory, from the very start to the very end. The concatenated-subsections-lemma informs us that the same reasoning is simultaneously valid for all subsections, down to subdivision into infinitisimally short subsections.

To your question specifically:

Implementation of variation:

Take the simplest case that involves acceleration: there is a uniform gravity, acting vertically. A projectile is launched vertically. The projectile climbs, is decelerated to velocity zero, and then proceeds to be accelerated downward.

As we know: under those circumstances the height of the object as a function of time is a parabola.

The simplest parabola with heigh 'h' as a function of time 't':

$$ h = -t^2 + 1 $$

The simplest variation of that parabola is with a single parameter 'p'.

$$ h = (1 + p)(-t^2 + 1) $$

So that when the parameter 'p' is zero the variation is zero.

The variation of the parabola is variation of the position coordinate of the parabola. To evaluate how the kinetic energy and potential energy respond to the variation you take the derivative of the variation with respect to the position coordinate.

That is: taking the derivative of the energy with respect to the variational parameter 'p' is in effect taking the derivative of the energy with respect to vertical position, since the variation is variation of vertical position.

Work-Energy theorem

In order to express physics taking place in terms of kinetic and potential energy we need the Work-Energy theorem.

• Potential energy is the the integral of force with respect to position coordinate

• Kinetic energy is the integral of mass-times-acceleration with respect to position coordinate

Variation sweep is change of some hypothetical energy

As you perform variation sweep of the trial trajectory: at every position of the trial trajectory that is not the true trajectory the derivative of the hypothetical energy with respect to position is different from the derivative of the true energy with respect to position.

The sweet spot is the point in variation space where the trial trajectory coincides with the true trajectory. At that point the derivative of the hypothetical energy and the derivative of the true energy match.

The true trajectory has the following property:
Everywhere along the trajectory the derivative (with respect to position) of the kinetic energy and potential energy match each other.

The criterium of stationary action is that you find a trial trajectory such that the derivative of the kinetic energy (with respect to position) matches the derivative of the potential energy.

At the point in variation space where you have that match you are at the point where the trial trajectory coincides with the true trajectory.

The Euler-Lagrange equation

The usual derivation of the Euler-Lagrange equation involves treating terms as negligable that are negligable only when the variation sweep is just an infinitisimally small distance away from the true trajectory.

For sure: with a larger amplitude of the variation sweep those higher order terms are actually larger than the ones that are retained. But whether those higher order terms are neglected or not, the outcome is the same anyway: you can tell whether the derivatives match each other or not.

The only thing that matters is whether the derivatives match each other. Mismatch is mismatch; when there is mismatch it isn't necessary to know how large the mismatch is.

So in that sense it isn't necessary for the varation to be small. Large variation sweep or infinitisimal variation sweep: either way there is only one sweet spot, and you have the criterium to tell you when you are at the sweet spot.

For further discussion, illustrated with diagrams, a 2021 answer by me about the reason why Hamilton's stationary action holds good