I'd like to start with some remarks about the general question of how we categorize our physics concepts into 'fundamental' and 'derivable-from-first-principles'. After that I will go into the stationary action concept specifically.
Two examples:
In the case of refraction of light there was initially Snell's law. Then Huygens proposed the theory that is now commonly known as 'Huygens' principle'. If you grant that light propagates as a wave phenomenon and that in a denser medium light propagates slower, then Snell's law follows as a theorem.
Entropy and Thermodynamics.
Initially entropy was used in a way that is independent of whether matter consists of atoms or not. Entropy was conceived in terms of bulk thermal properties of matter. The development of statistical mechanics moved the description of entropy to a deeper level.
Being able to move the description to a deeper level is progress. In that sense there is a hierarchy.
At any stage of development of physics the deepest available level consists of propositions that must be granted in order to formulate the theory at all. It would appear: there is no way of knowing in advance for which level of description no deeper level exists.
The level of description that is currently the deepest known level is what is referred to as 'first principles'.
Hamilton's stationary action
Hamilton's stationary action is stated in terms of kinetic energy and potential energy. We want to understand the relation with F=ma
There is a theorem that provides the connection between force and acceleration on one hand, and potential energy and kinetic energy on the other hand: the work-energy theorem.
In order to derive the work-energy theorem I first derive the kinematic relation that enables the work-energy theorem.
With:
##s## position coordinate
##v## velocity
##a## acceleration
In the course of the derivation the following relations will be used:
## ds = v \ dt \qquad (1) ##
## dv = a \ dt \qquad (2) ##
The integral for acceleration from a starting point ##s_0## to a final point ##s##
## \int_{s_0}^s a \ ds \qquad (3) ##
## \int_{t_0}^t a \ v \ dt \qquad (4) ##
## \int_{t_0}^t v \ a \ dt \qquad (5) ##
## \int_{v_0}^v v \ dv \qquad (6) ##
Overview of the above steps:
in going from (3) to (4) the differential was changed from ##ds## to ##dt##, in accordance with (1), with corresponding change of limits.
From (4) to (5) was a change of order.
From (5) to (6) the differential differential was changed from ##dt## to ##dv##, in accordance with (2), with corresponding change of limits.
So we have:
$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \qquad (7) $$
Note that (7) is a mathematical relation; it follows from (1) and (2).
To derive the work-energy theorem we start with the force-acceleration principle:
$$ F = ma \qquad (8) $$
Next: for both sides: specify integration with respect to position coordinate:
$$ \int_{s_0}^s F \ ds = \int_{s_0}^s ma \ ds \qquad (9) $$
We use (7) to process the right hand side of (9):
$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \qquad (10) $$
(10) is the work-energy theorem.
We have that potential energy is defined as the negative of the integral of force over distance.
$$ E_p = -\int_{s_0}^s F \ ds \qquad (11) $$
Hence:
$$ \Delta E_k = -\Delta E_p \qquad (12) $$
Of course, (11) is subject to the constraint that is is applicable only if the force that is involved has the property that the outcome of the integration is independent of
how the motion proceeds from start point to end point. That is, in order for (11) to be a valid expression the force that is involved must be a conservative force.
Below is a group of three statements. I present these three statements as a unit to emphasize the tight interconnection; while the statements are different mathematically, the physics content of these three is the same.
$$ \begin{array}{rcl}
F & = & ma \\
\int_{s_0}^s F \ ds & = & \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \\
-\Delta E_p & = & \Delta E_k
\end{array} $$
The work-energy theorem gives us the explanation for why we define kinetic energy as the quantity: ##\tfrac{1}{2}mv^2## The reason that the expression for kinetic energy features that factor ##\tfrac{1}{2}## is that the expression for kinetic energy is the result of an
integration.
The work-energy theorem gives us the following constraint:
During motion the rate of change of kinetic energy must match the rate of change of potential energy. That is: the derivative of the kinetic energy must match the derivative of the potential energy.
Hamilton's stationary action expresses that: the true trajectory corresponds to a point in variation space such that the derivative of the kinetic energy matches the derivative of the potential energy.
At the point in variation space such that the derivative of the kinetic energy matches the derivative of the potential energy the derivative of Hamilton's action is
zero.
The operation to evaluate Hamilton's action is differentiation with respect to variation.
The variation that is applied is exclusively variation of position coordinate. That is: differentiation with respect to variation is a form of differentiating with respect to position coordinate.
Now take a look at the Euler-Lagrange equation, in its form for classical mechanics. The operation that the Euler-Lagrange equation performs on the potential energy is straightforward: differentiation with respect to position coordinate.
So here we see a relation with the work-energy theorem.
To obtain the work-energy theorem: perform
integration with respect to position coordinate.
The Euler-Lagrange equation (for classical mechanics) performs
differentiation with respect to position coordinate.
As we know, we have the fundamantal theorem of Calculus: the operations of integration and differentiation are each other's inverse.
That relation is key to understanding Hamilton's stationary action.
(It's important to be aware that it was only around the 1850's that the physics community recognized the significance of the work-energy theorem. The definition of kinetic energy as ##\tfrac{1}{2}mv^2## dates from the 1850's. Hamilton's work was published in 1834, so Hamilton did not have the work-energy theorem to guide his understanding.)
I'm active on physics.stackexchange
In July of 2024 I went back to the earliest question physics stackexchange about Hamilton's stationary action, and I posted an answer with discussion of how to proceed
from the work-energy theorem to Hamilton's stationary action.
