James1238765 said:
TL;DR Summary: What is the significance of the T - V Lagrangian of a system?
As you point out: we have that
over time the sum of potential energy and kinetic energy remains at the same value.
$$ E_{potential} + E_{kinetic} = E_{total} \tag{1} $$
(From here on I will abbreviate as ##E_p## and ##E_k## )
This can be expressed in differential form. As an object is moving along, the derivative with respect to time of the total energy is zero
$$ \frac{d(E_k + E_p)}{dt} = 0 \tag{2} $$
As we know: we can understand that as follows: as potential energy and kinetic energy are being exchanged they both change at the same rate, in opposite direction. That is: (2) can be rearranged to a form that states that the the time derivative of the kinetic energy is equal to the time deriviative of minus the potential energy:
$$ \frac{d(E_k)}{dt} = \frac{d(-E_p)}{dt} \tag{3} $$
You ask about the meaning of the integral with respect to time of the Lagrangian ##(E_k - E_p)##.
I assume you are asking about Hamilton's stationary action, since Hamilton's action is defined as the integral with respect to time of the Lagrangian ##(E_k - E_p)##.
The variation of a trial trajectory is variation of the position coordinate of the trial trajectory.
The importance of that is this: derivative with respect to
variation and derivative with respect to
position coordinate are one and the same thing.
The change that is evaluated when applying variation of a trial trajectory is a hypothetical change. In that hypothetical change you are looking for the sweet spot: the spot where rate of change of kinetic energy matches the rate of change of potential energy.
The following expression differentiates with respect to the position coordinate 's', not with respect to the time coordinate 't'.
$$ \frac{d(E_k)}{ds} = \frac{d(E_p)}{ds} \tag{4} $$
The point is: at the point in variation space where the trial trajectory coincides with the true trajectory we have that (3) holds good, and when (3) holds good (4) holds good also.
Notice that compared to (3) the expression above does not have a minus sign. That is because when you apply variation to the trial trajectory the two hypothetical changes are co-changing. When you make the trial trajectory
steeper the rate-of-change of the hypothetical kinetic energy and rate-of change of the thypothetical potential energy both become larger; those rates-of-change are
co-changing.
(4) can be rearranged as follows:
$$ \frac{d(E_k - E_p)}{ds} = 0 \tag{5} $$
At this point I need to point out a particular property of calculus of variations.
As we know, it was in the wake of the Brachistochrone challenge that Calculus of Variations was developed. Johann Bernoulli had issued the Brachistochrone challenge, and his older brother Jacob Bernoulli was among the mathematicians who was able to solve it. This means Jacob Bernoulli solved the problem without having Calculus of Variations, nor any precursor of it.
Jacob Bernoulli recognized a particular feature of the brachistochrone problem, and he presented that feature in the form of a lemma:
Let ACEDB be the desired curve along which a heavy point falls from A to B in the shortest time, and let C and D be two points on it as close together as we like. Then the segment of arc CED is among all segments of arc with C and D as end points the segment that a heavy point falling from A traverses in the shortest time. Indeed, if another segment of arc CFD were traversed in a shorter time, then the point would move along AGFDB in a shorter time than along ACEDB, which is contrary to our supposition.
(
Acta Eruditorum, May 1697, pp. 211-217)
This property that applies in the case of the Brachistochrone problem also applies in the case of Hamilton's stationary action.
You want to evaluate a trajectory from a particular start point ##t_1## to an end point ##t_2##
Divide the total time interval into sub-intervals. Every single sub-interval is an instance of evaluating a trajectory from a start point to an end point. In order for Hamilton's action to be stationary over the entire interval it is necessary
and sufficient that Hamilton's action is stationary over each
sub-interval concurrently.
So:
The property of being stationary propagates from the sub-intervals to the entire interval. This propagation from sub-intervals to the entire interval obtains at every scale, down to arbitrarily small sub-intervals.
That means:
The process of ascertaining that the action is stationary for the entire interval is actually a process of ascertaining that the action is stationary along every sub-interval, down to infinitisimally small sub-intervals
There is a more extended discussion of Hamilton's stationary action by me on this forum, I posted that one in december 2021:
https://www.physicsforums.com/threads/stationary-point-of-variation-of-action.1009770/#post-6571395
That extensive discussion is illustrated with diagrams. Those diagrams are animated GIF's; the frames of those GIF's are screenshots of interactive diagrams. The interactive diagrams are on my own website. (Link to my own website is available on my physicsforums profile page.)
Summerizing:
- In the case of Hamilton's stationary action the derivative with respect to variation and derivative with respect to position coordinate are the same thing.
Therefore:
At the point in variation space where the trial trajectory coincides with the true trajectory we have:
$$ \frac{d(E_k - E_p)}{ds} = 0 $$
- In the case of the true trajectory this differential relation holds good on every infinitisimal subsection concurrently. From there the property of being stationary (along the true trajectory) propagates to the integral.