Picture 1. Graphlet
Potential energy proportional to cube of displacement

How to use this demonstration:

This graphlet is about variational approach to mechanics. The main slider, located at the bottom of the graphlet, executes a global variation sweep.

In the starting configuration the trajectory points have been placed such that they coincide (to a very good approximation) with the true trajectory of the object.

When the object is moving along the true trajectory the rate of change of kinetic energy matches the rate of change of potential energy (work-energy theorem). To emphasize this matching property the graph of the potential energy (green) has been turned upside down. It has been mirrored in the x-axis.

In the upper panel the two large grey dots are draggable points that are adjusters for the trajectory. The upper adjuster morphs the trajectory towards a parabola; the lower adjuster morphs the trajectory towards a triangle shape.

These adjusters allow morphing of the trial trajectory while maintaining that during the entire ascent the velocity never reverses from decreasing to increasing again, which would be unphysical.

Clicking the checkbox 'More controls' sets a row of ten grey draggable points to visible. These are separate adjusters for the the individual trajectory points. As you slide the grey point the corresponding black point is moved by a small amount. These ten adjusters are actually not practical. The only reason for keeping them available is so that the user can personally experience that they are unpractical.

There is an additional draggable point all the way to the right.
This adjusts an attenuation factor. If you slide that point all the way up then the connection between the grey points and the black points is a 1:1 connection.

In the lower left sub-panel: the grey point is draggable, and it drags the entire curve of the kinetic energy (red) with it. By shifting the kinetic energy curve over to the potential energy curve the user can verify that the red and green curve are parallel to each other along the entire trajectory. (The evaluation of the integral uses the unshifted position of the kinetic energy curve.)

In the lower right sub-panel:
· Red/Green point: value of the integral of the red/green curve of the lower left sub-panel
· Blue point: the sum of the values of the red point and the green point

As you sweep out variation of the trajectory: since kinetic energy is a quadratic function: the response of the kinetic energy is quadratic. The response of the potential energy depends on the specific potential. This graphlet is for the case where the potential energy increases with the cube of the displacement.

The purpose of the demonstration

The Energy-Position Equation

As you sweep out variation with the main slider you notice that the blue dot is at a maximum when the variational parameter pv is zero. Generalizing: for any type of potential: for the blue dot it will always be the case that the derivative with respect to variation is zero at the point pv = 0. In the following I explain how that comes about.

We start with the Work-Energy theorem:

\int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2

The Work-Energy theorem is about change of energy. The Work-Energy theorem serves as the definition of potential energy. Given a known force: if the integral of that force acting over distance is defined then the potential energy is the negative of the work done. (Any definition of potential energy that isn't based on a quantitive rule for transformation to another form of energy - in this case kinetic energy - is of no practical use.)

The amount of change of kinetic energy is the negative of the amount of change of potential energy.

\Delta (E_k) = - \Delta (E_p)

For the subsequent manipulations it is practical to move the minus sign inside.

\Delta (E_k) = \Delta (-E_p)

The validity of (1.3) extends down to infinitisimal change:

d(E_k) = d(-E_p)

As stated in the how-to-use, in the starting configuration the trajectory points have been placed such that from start to end the following equation is satisfied; in the lower left sub-panel the red and green curve are parallel to each other everywhere. From start to end:

\frac{d(E_k)}{dt} = \frac{d(-E_p)}{dt}

The concept of energy introduces an opportunity not available in the newtonian formalism. For a given trajectory we attribute a potential energy and a kinetic energy to every point along that trajectory. For a given trajectory the rate of change of energy can be expressed in two ways:
- as the time derivative
- as the position derivative

Next to each other the time derivative and the position derivative:
For the true trajectory the following two are both satisfied:

\frac{d(E_k)}{dt} = \frac{d(-E_p)}{dt}


\frac{d(E_k)}{ds} = \frac{d(-E_p)}{ds}

From here on I will refer to (1.6) as the 'Energy-Position equation', as it evaluates the derivative with respect to position.

(1.6) is the practical way forward. Potential energy is the integral of force over distance, so (1.6), the derivative over distance, recovers Newton's second law straight away.

The variation

In this graphlet the global variation is implemented in the following way: what is added to the curve is a smaller version of the curve itself.

With pv as variational parameter:

(variation sweep) = (1 + pv) * (current trial trajectory)

What that means is that the global variation is linear, which in turn means that the following two are identical:
- derivative with respect to variation
- derivative with respect to position

Integration is a linear operation

In the graphlet you home in on the true trajectory by making the two curves in the lower left sub-panel parallel to each other along the entire length of the trajectory. That is, the lower left sub-panel shows whether (1.5) is satisfied (over the entire length of the trajectory).

We have that when (1.5) is satisfied the Energy-Position equation (1.6) is satisfied too.

When the trial trajectory coincides with the true trajectory the lower right panel shows that Hamilton's action is stationary at that point (stationary as in 'the derivative is zero').

The explanation for that is this: integration is a linear operation. Hence:

For variation between two fixed points:
If, from start to end:

\frac{d(E_k)}{ds} = \frac{d(-E_p)}{ds}

Then, from start point to end point:

\frac{d(\int E_k dt)}{ds} = \frac{d(\int -E_p dt)}{ds}

That is: because integration is a linear operation the property expressed by the Energy-Position equation propagates to the integrals.

Lemma. When two functions are changing at the same rate, one ascending the other descending, then the sum will be at an extremum.

The blue dot displays the value of the sum. As you sweep out variation: when the trial trajectory coincides with the true trajectory (1.7) is satisfied, putting the blue point at an extremum of its curve.

In this graphlet the potential increases with the cube of the displacement. The response of the kinetic energy to variation is of lower order: it is quadratic (kinetic energy being proportional to the square of the velocity). Hence in this graphlet the extremum is a maximum.

Variation of subsections of the trajectory

The master slider at the bottom of the graphlet executes a global sweep of variation. In this graphlet, with potential energy proportional to the cube of dispacement, the extremum that appears is a maximum.

However, when we apply the adjusters at the left side of the upper sub-panel we see that such a subsection sweep increases the sum. That is, in the case of potential energy proportional to the cube of displacement the responses do not line up. Response to local variation on one hand and to global variation on the other hand is in opposite directions.

Low order potentials

Of course, a potential proportional to the cube of displacement is rare in Nature. In most cases the potential is of lower order. Let's take the case of a linear potential. Then the response to local variation sweep will be in the same direction as the response to global variation sweep. That brings me to a lemma that I propose to name 'Jacob's lemma'.

Jacob's Lemma

When Johann Bernoulli had presented the Brachistochrone problem to the mathematicians of the time Jacob Bernoulli was among the few who was able to find the solution independently. The treatment by Jacob Bernoulli is in the Acta Eruditorum, May 1697, pp. 211-217

Jacob opens his treatment with an observation concerning the fact that the curve that is sought is a minimum curve.

Picture 2. Image
Jacob's Lemma

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 ACFDB in a shorter time than along ACEDB, which is contrary to our supposition.

Jacob's lemma generalizes to all cases where the curve that you want to find is an extremum; either a maximum or a minimum. If the evaluation is an extremum for the entire curve, then it is also an extremum for any sub-section of the curve, down to infinitisimally short subsections.

With Jacob's lemma in place I return to the case of a potential that is of lower order than that of the kinetic energy. Then the response to local variation sweep and to global variation sweep will be in the same direction.

Derivation of the Work-Energy theorem

It appears many people learn the Work-Energy theorem as a definition, rather than learning the derivation from F = ma.

The following derivation is specifically structured to show that the entire physics content of the Work-Energy theorem is F = ma. Up to (2.7) no physics is involved yet, only mathematical operations.

The steps in the following derivation are all towards the end goal of obtaining an expression in terms of the velocity v.

In the course of the derivation the following two relations will be used:

ds = v \ dt

dv = \frac{dv}{dt}dt

The integral for acceleration from a starting point s0 to a final point s

\int_{s_0}^s a \ ds

Use (2.1) to change the differential from ds to dt. Since the differential is changed the limits change accordingly

\int_{t_0}^t a \ v \ dt

Rearrange the order, and write the acceleration a as dv/dt:

\int_{t_0}^t v \ \frac{dv}{dt} \ dt

Use (2.2) for a second change of differential, the limits change accordingly.

\int_{v_0}^v v \ dv

Putting everything together:

\int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2

Until this point no physics has been introduced. (2.7) follows from definitions: velocity is the time derivative of position; acceleration is the time derivative of velocity.

combining F = ma with (2.7) gives the Work-Energy theorem:

\int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2

The interactive animation on this page is created with the Javascript Library JSXGraph. JSXGraph is developed at the Lehrstuhl für Mathematik und ihre Didaktik, University of Bayreuth, Germany.

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Last time this page was modified: April 11 2021