[Cleonis]

There is obviously a connection between considering arbitrary variation in path taken by a system and just a single variation in some element of position taken by an object at some fixed time during the path.  This was illustrated well in your post about numerical techniques for solving variational problems. 

Numerical analysis implementation

In the course of developing various diagrams I got to think a lot about numerical analysis implementation of variational calculus.

I expect that numerical analysis implementation will be a fruitful way of investigating and demonstrating the relation between differential calculus and variational calculus.

I repeat the link to Preetum Nakkiran's derivation of the Euler-Lagrange equation (classical mechanics). (Spaces inserted; I can't post links yet) As stated before, Preetum uses local reasoning.


Differential calculus
In numerical analysis implementation of differential calculus the smallest unit of evaluation is a pair of points. As we know, the simplest implementation is Euler integration. The integration proceeds from point to point. The current acceleration gives the current velocity, the current velocity gives the distance that will be traversed in moving from the current point to the next point. Rinse and repeat.

Variational calculus
In numerical analysis implementation of variational calculus the smallest unit of evaluation is a triplet of points.
t₀, t₁, t₂
t₁, t₂, t₃
t₂, t₃, t₄

In each consecutive triplet-of-points evaluation the outer points of the triplet are treated as fixed.

For the case of classical mechanics:
As you execute variation of the middle point Hamilton's action is affected in two ways:
1)  The integral of the potential energy is a function of the position of the middle point.
2) The triplet t₀, t₁, t₂ has the following two subsections
t₀, t₁ and t₁, t₂
Lowering the position of the middle point has as effect:
- the velocity along interval t₀, t₁ is lowered
- the velocity along interval t₁, t₂ is raised.
Raising the position of the middle point has the opposite effect.

Schematically:
Differential calculus: smallest unit of evaluation: a pair of points
Variational calculus: smallest unit of evaluation: a triplet of points

The derivation of the Euler-Lagrange equation for the general case demonstrates that any problem stated in variational form can be converted to differential form.

Differential calculus (pair-of-points calculus) has a simpler structure than variational calculus (triplet-of-points calculus)

Conversion of differential form to variational form is referred to as an 'inverse problem'. 

[Cleonis]

There is obviously a connection between considering arbitrary variation in path taken by a system and just a single variation in some element of position taken by an object at some fixed time during the path. 

In the case of differential calculus the smallest unit of operation is a pair of points. Geometrically: you take a pair of points on the horizontal axis, a and (a + h), and then you obtain the following ratio:




The diagram below represents the smallest unit of operation of variational calculus. In numerical analysis implementation this is the unit of operation (concatenated in iteration cycles as necessary).




Horizontal axis: time
Vertical axis: position

Variational calculus: numerical analysis implementation
To find the true trajectory:
- divide the total time in sub-intervals
- adjust the position of the points along the trajectory until all points satisfy the stationary criterium. (Each point evaluated with respect to its adjacent points.)

In the limit of making the time increment infinitely small one obtains the variational counterpart of taking a derivative.

Disclaimer:
This is not writen as an exhaustive derivation. Rather, it is a demonstration of the nature of Hamilton's action. The exposition below is intended to serve as corroborating evidence.


The following exposition is in notation specific for problems in mechanics.
t for 'time'
s for 'position'
v for 'velocity'
a for 'acceleration'

However, the structure of the derivation is not specific to dynamics; it generalizes to all of variational calculus.

S for Hamilton's action
S_k for the kinetic energy component of Hamilton's action
S_p for the potential energy component of Hamilton's action


In order to proceed an auxillary quantity is necessary.
We are trying to find the equation of motion for the trajectory; an equation of motion gives position as a function of time. But in variational calculus what is varied is the position coordinate. Therefore we need a function that assigns a value to each point along the trajectory, allowing us to differentiate that value with respect to position.

In classical mechanics this auxillary quantity function is created by integrating F=ma with respect to position, resulting in the Work-Energy theorem:



The Work-Energy theorem states how potential energy and kinetic energy are related.

We have: potential energy is a function of position; kinetic energy is a function of velocity.

In the diagram δs represents an infinitisimally small variation of position. The potential energy and the kinetic energy each respond to that variation, each responding differently to δs .

The response of the potential energy component of Hamilton's action:



The response of the kinetic energy component of Hamilton's action:
On both side of t₂ the velocity changes; one side increases, the other side decreases.

Before the change:



After adding δs :



The changes of the velocities:



The change of the kinetic energy component of Hamilton's action:





The total time is divided in equal length sub-intervals, so the move from interval (1,2) to the adjacent (2,3) can be rearranged to the form of taking a time derivative:



Combining (5) and (6) we obtain an expression for the change of the kinetic energy component of Hamilton's action as a function of δs:



Combining the two responses:



The variation δs, having served its purpose, can now be eliminated. The stationary-in-response-to-variation condition is satisfied when the following is satisfied:



As we know, in classical mechanics the kinetic energy term in (9 ) simplifies to mass times acceleration:



The above derivation explains why F=ma can be recovered from Hamilton's stationary action.
The Work-Energy theorem is derived from F=ma
Hamilton's stationary action is a corollary of the Work-Energy theorem


Numerical analysis implementation

Numerical analysis implementation is where the rubber meets the road. If variational calculus is implemented numerically the algorithm will proceed according to the diagram. Take a triplet t₁, t₂, t₃, adjust the position of the middle point, then move up one time increment: t₂, t₃,t₄. When you have reached the end point: cycle again over the entire time interval. Over multiple iterations the set of points will converge onto the true trajectory. (Alternatively you could randomize the choice of triplet to evaluate. The algorithm will of course converge onto the same true trajectory.)

(There are Java simulations available on the website of Edwin Taylor where you can actually watch iteration cycles converge rapidly onto the true trajectory. (I don't know whether those simulations still run on recent versions of the Java Virtual Machine.))

[Eternal Student (OP)]

Hi.
Minor printing / typing error:       In equations (7) and (8 )    you have written      where the conventional derivative      should be printed.   This doesn't change the essence of what you have done, it's just a printing error.

Best Wishes.

[Cleonis]

typing error

Yeah, that was a typing error. Then I noticed I had accidentally used (7) twice, so I corrected that too.

More generally: in post #41 the terms E_k and E_p are intended as referring to a value that is a function of the trajectory. I thought about defining some letter of the greek alphabet to represent the trajectory, τ for example, to give indication/reminder that the energy is a function of τ.



So I'm not happy with my notation here, but I'm not sure whether our notation conventions support something better.

[Cleonis]

@eternalstudent

I repeat the diagram from post #41



The diagram illustrates the relation between variational calculus and differential calculus.

The general derivation of Euler-Lagrange equation shows that any problem that can be stated in variational form can be restated in differential calculus form.

The triplet of points of the unit of operation of variational calculus can be thought of as two pairs of points (with the middle point as shared point).


In numerical analysis implementation of variational calculus you start with a seed, and then you iterate. I think it is worthwhile to examine closer how that works out.

The cumulative effect of all the iterations is that the trajectory that you are converging onto satisfies the stationary action criterium along the whole trajectory. In the end result: the convergence of each point along the trajectory has been influenced by the convergences of all the other points along the trajectory.

The points along the trajectory are concatenated, so over the course of the iterations adjustment of any single point propagates to the neighbours of that point; to the neighbours of those neighbours, all the way out.

Because of this propagation-to-neighbours, along the chain, a strictly local evaluation of each unit of operation is sufficient as unit of iteration.

The variation of the unit of operation is linear variation only.



We see the theme of narrowing to linear variation also in the standard derivation of the Euler-Lagrange equation. At the start it is stated that the variation is allowed to be any polynomial. (We have of course that any function can be represented as a Taylor series polynomial.)

At the next stage in the derivation the following is argued:
The variation is taken to be infinitisimally small, therefore when considering the Taylor series expansion all terms quadratic and higher can be dropped.

Now, after dropping quadratic and higher terms what remains is a single linear term; the variation space has been narrowed down to linear variation only.

So we have: in variational calculus you can represent the variation with any polynomial that you want, as long as you agree to drop all terms quadratic and higher.



Hamilton's action is defined as an integral of the Lagrangian ()



So: what is that integral doing there?

The integration serves to conglomorate the set of all infinitisimal units of variation. That is how Hamilton's action assigns a single number to each trial trajectory; that single number is the conglomoration of all of the constituent infinitisimal variational units.

The integration in the definition of Hamilton's action isn't actually doing anything.

The first step of deriving the Euler-Lagrange equation is the step that does the actual job. That first step is to set up the derivative of the action.

Over the course of deriving the Euler-Lagrange equation the integration is removed. (It has to be removed because it isn't doing anything.) What is retained throughout the course of deriving the Euler-Lagrange equation is that first step: taking the derivative with respect to position.



The energy-position equation

The Work-Energy theorem is applicable when there is a well defined potential energy.

It follows from the Work-Energy theorem that the rate of change of kinetic energy will be equal to the rate of change of potential energy, with opposite sign

Hence the following equation, that I refer to as 'the energy-position equation':




As stated earlier in post #26, in classical mechanics it isn't necessary to write the Euler-Lagrange equation with partial derivatives because the potential energy is a function of position only, and the kinetic energy is a function of velocity only.



The way the Euler-Lagrange equation processes the kinetic energy looks different from that of the energy-position equation, but it is the same.

This is what the Euler-Lagrange equation does with the kinetic energy:



Next, here is what the Energy-Position equation does with the kinetic energy:



Both evaluate to because they are actually the same expression.

[Eternal Student (OP)]

  Equation (2)  has a typing error, similar to an earlier post.        should be  .    It is just a typing error, it has no effect on the remainder of the post.
- - - - - - - - - - - - -

    You are spending time to demonstrate something that is reasonably well established for classical mechanics involving one particle moving in a potential V(s).

1.    It might be harder to make the same steps for multiple particles.
Consider, for example, your equation (1) which follows from the Work-Energy theorem.
The Work-Energy theorem is often stated in this form:
    =       showing derivatives with respect to time.
For multi-particle systems this is assumed to be well defined (the Energies are assumed to be differentiable functions of time).
      However, for a multi-particle system, getting from here to your version of the equation, which you call the Energy-Position equation is non-trivial.  It requires a change of variables for the derivatives.

We have    =   provided all the derivatives on the R.H.S. exist

[Change of variables equation]

  where  E = one of the energies of interest E_k or E_p ;     s = position of one particle;    t= time.

   There are some systems where our chosen particle could be at rest for a moment during the path of evolution of our system.  (* See below for an example).   
Then it's velocity = 0 = ,   so that → ∞   and so the change of variables formula doesn't define the quantity  when our chosen particle is at rest.
   This wasn't a problem for a one particle system, since we know in advance that the kinetic energy will remain 0 when the velocity is 0.  Similarly we know that if the particle was at rest then the potential energy doesn't change either.   So we are not solely dependant on the change of variables formula stated above, we can establish the existence of the derivative by more direct methods.
    However, these assumptions are not valid for a multi-particle system.   There could be some other particles in the system that are moving while our chosen particle remains at rest and this could change the total potential energies.  Indeed, if the motion of those other particles is changing, then the kinetic energies change as well.
 
   To paraphrase this,  it is not obvious that we can identify a single particle in a multi-particle and it's corresponding position variable, s,  such that the work energy theorem can be changed from a derivative with respect to time into a derivative with respect to that position variable.

* An example of a system where one particle remains at rest for quite a lot of the time during the evolution of the system is given by a spring or "slinky" when it is extended and then dropped.  See this video if you wish:

2.  I've already said enough and I'll sign off for now.

Best Wishes.

[Cleonis]

However, for a multi-particle system, getting from here to your version of the equation, which you call the Energy-Position equation is non-trivial.  It requires a change of variables for the derivatives.

The thrust of your original question, in post #1 was:
"Why do we see many instances of an action formulation hold good?"

I responded by submitting a discusson of Hamilton's stationary action, showing why Hamilton's stationary action holds good, and I submitted that making Hamilton's stationary action transparent provides insight for the purpose of making all action formulations transparent.


We have: in classical mechanics context Lagrangian mechanics is applied in multiple-particle cases. It's a solved problem.

We have:
- The question why Hamilton's stationary action holds good
- The question of how to apply Lagrangian mechanics in multiple-particle cases

Those two are independent questions.


I submit that the following two equations are mathematically equivalent.





That is:
I submit the following reasoning:
- (1) and (2) are mathematically equivalent.
- (2), the Euler-Lagrange equation (for classical mechanics), is regarded as expressing the foundation of Lagrangian mechanics.
- Since (1) is mathematically equivalent to (2) it follows that (1) equivalently expresses the foundation of Lagrangian mechanics


Is that where you disagree?
That is: do you submit that (1) and (2) are not mathematically interconvertable?


Equations can always be rearranged to accomodate specific circumstances. If circumstances are encountered where some derivative becomes problematic then you work around that with a suitable rearrangement.

So, sure, in the specific case that you describe the equation will have to be rearranged to accommodate the specifics of that case, such that the derivatives used are always defined.

The point is: we know that applying Lagrangian mechanics in multiple particle cases is a solved problem. On the grounds of that it is extremely implausible that the specific example that you discuss would constitute a problem to application of Lagrangian mechanics.