Deriving Hamilton's Principle from Maupertuis' Principle

Question

I'm trying to understand the derivation of Hamilton's Principle from Maupertuis' Principle:

In this answer, they say "$E$ is a constant". Which is consistent with:

Maupertuis's principle requires that the two endpoint states $q_{1}$ and $q_{2}$ be given and that energy be conserved along every trajectory.

But, if, to prove Hamilton's Principle from Maupertuis' principle I have to assume that, I can't then have

By contrast, Hamilton's principle does not require the conservation of energy, but does require that the endpoint times $t_{1}$ and $t_{2}$ be specified as well as the endpoint states $\displaystyle q_{1}$ and $q_{2}$ holding true?

What am I missing?

Answer 1

Before I can discuss Maupertuis' principle I must first address some historical aspects.

In the original presentation by Maupertuis there was no provision to handle cases where there is interconversion of potential energy and kinetic energy. That means that phenomena such as falling and the motions of the planets in their orbits is out of scope. Euler provided the means to handle dynamics.

Available on wikisource: translations of the original texts by Maupertuis

Euler's foray into application of Calculus of Variations in Dynamics

The effort by Euler is informative for the question: is it possible to go from Maupertuis' principle to Hamilton's stationary action?

The book in which Euler introduced calculus of variations is for brevity referred to as 'Methodus invenientes'. Wikisource offers a rendering of 'Additamentum II' of Methodus Invenientes. It is in that 'Additamentum II' that Euler treats several cases of applying calculus of variation in dynamics.

Important to know: the translator has gone through great lengths to make the text accessible for modern readers. It is very much not a literal translation. That is why I refer to it as a 'rendering'.

Euler opens with the simple case of an object that is launched with a horizontal velocity, proceeding to be accelerated vertically by gravity.

(As we know, in the case of gravity we have that inertial mass and gravitational mass are equivalent, so that the mass of the object drops out of the calculation.)

The translator has Euler stating:
"The external force produces an acceleration described by $dv^2=2gdy$ which integrates to $v^2=v_0^2+2gy$"

From today's point of view we recognize that $v^2=v_0^2+2gy$. Euler is using what today is seen as application of the work-energy theorem: with a uniform gravitational force $mg$ we have: $mgy=\tfrac{1}{2}mv^2-\tfrac{1}{2}mv_0^2$.

After a string of manipulations Euler arrives at an expression for the height as a function of the horizontal coordinate (which of course is a parabola).

In section 11. of the Additamentum II Euler adresses the case of an object acted on by a central force, using as coordinates radial distance and arc-of-a-unit-circle.

The translator has Euler stating:
"the applied force results in an acceleration $dv^2=-2F_{r}\,dr$, from which we obtain $v^{2}=v_{0}^{2}-\int 2F_{r}\,dr$" (Euler is treating the force as an acceleration factor.)

Again the work-energy theorem.
So we see that Euler had already recognized the utility of the work-energy theorem. (The work-energy theorem reached its modern form a century later, around 1850.)

We see a pattern here:
Euler is treating cases in dynamics; cases with continuous interconversion of potential energy and kinetic energy.

Maupertuis' original concept does not have the means to handle such interconversion. And that means: in order to do dynamics you have to supply the work-energy theorem. That is what Euler is doing in the treatments in that 'Additamentum II'.

By contrast: Hamilton's stationary action is self-contained. There is no need to supply Hamilton's stationary action with the work-energy theorem; the work-energy theorem is already built into Hamilton's stationary action.

Another way of saying the same thing:
We have: when the force that is acting is a conservative force then throughout the motion the sum of kinetic energy and potential energy is a conserved quantity. In order to go from Maupertuis action to Hamilton's stationary action it is necessary to supply the constraint of sum-of-kinetic-and-potential-must-be-conserved.

What I expect:
I any presentation that goes from Maupertuis action to Hamilton's stationary action: in order to reach the goal an extra element is supplied. It can be supplied in the form of the work-energy theorem, or in the form of the constraint sum-of-kinetic-and-potential-must-be-conserved, that is equivalent.

I submit: without supplying that extra element there is no way to go from Maupertuis' action to Hamilton's stationary action.

About the reason there is no practical application for Maupertuis' action.

In the case of orbiting motion: that treatment was set up using as coordinates the radial distance and arc-of-a-unit-circle. Euler arrives at an expression for the radial distance as a function of the arc-along-the-unit-circle. That is, the expression is for one spatial coordinate as a function of another spatial coordinate. That is unpractical: what you want is an expression for the position of the object as a function of time.

With an expression that gives you one spatial coordinate as a function of another spatial coordinate you have to do an additional computation to convert to an expression that gives all the spatial coordinates as a function of time.

About interderivability

In expositions of Hamilton's stationary action the customary approach is to posit Hamilton's stationary action, and to show that $F=ma$ can be recovered from it.

Now, in physics it is often the case that derivations can be run in both directions. $F=ma$ and Hamilton's stationary action is an instance of that.

It is possible to go from $F=ma$ to Hamilton's stationary action in all forward steps. That path consists of two stages:

• Derivation of the work-energy theorem from $F=ma$
• Demonstration that in cases where the work-energy theorem holds good Hamilton's stationary action will hold good also.

In Juli of this year (2024) I went all the way back to the first question about Hamilton's stationary action (which was posted in 2010) and I posted an answer that presents the path from F=ma to Hamilton's stationary action .

Answer 2

I'm posting another answer. In this answer I will address the two quotes in your question from the wikipedia article about Maupertuis's principle.

Repeating the quote from further down in the wikipedia article about Maupertuis' action:

Maupertuis's principle requires that the two endpoint states $q_{1}$ and $q_{2}$ be given and that energy be conserved along every trajectory.

(1) is how Euler restated Maupertuis' proposition in integral form.

$${S}_0 = \int_{q_1}^{q_2} \mathbf{p} \cdot d\mathbf{q} \tag{1} $$

The thing is: this expression is in terms of (generalized) velocity only. (1) does not provide the means to express interconversion of potential energy and kinetic energy.

In order to have the capability to handle dynamics (1) must be augmented with the work-energy theorem

Well: the work-energy theorem is in an of itself the most powerful statement in dynamics. (Note: it follows from the work-energy theorem that in any interconversion of potential and kinetic energy the sum of potential and kinetic is a conserved quantity.)

To solve a dynamics problem the following two factors are sufficient:
-An expression for the potential as a function of (generalized) position coordinate(s)
-The work-energy theorem

So it's not clear what (1) is supposed to be contributing.

The other quote:

By contrast, Hamilton's principle does not require the conservation of energy, but does require that the endpoint times $t_{1}$ and $t_{2}$ be specified as well as the endpoint states $\displaystyle q_{1}$ and $q_{2}$

Well, the condition of conservation of energy is built into Hamilton's stationary action.

The true trajectory has the property that everywhere along the trajectory the derivative of the kinetic energy matches the derivative of the potential energy.

Hamilton's stationary action is about comparing derivatives: at the point in variation space where the derivative of the kinetic energy matches the derivative of the potential energy the derivative of Hamilton's action is zero.

That raises the question: what is it that Maupertuis' concept brings to the table?

As is well known, Maupertuis was inspired by Fermat's stationary time.

Fermat's stationary time:
In refraction a variation can be defined: apply variation to the angle of incidence and angle of refraction by moving the refraction point along the interface between the two optical media. A path for which the derivative of the total time is zero is a path that satisfies Snell's law.

Stated differently: at the point in variation space such that there is a match of the derivative-of-the-incident-ray-time and the derivative-of-the refracted-ray-time the path satisfies Snell's law.

There is a recurring theme: matching derivatives.

As we know, Huygens had shown that it is possible to account for refraction in terms of properties of wave propagation.

Suppositions:
Light propagates as a wave phenomenon
In an optically denser medium light propagates slower
While propagating slower the frequency must remain the same, so there will be a correspondingly shorter wavelength

That is, Huygens' principle can be thought of as expresssing a conservation concept; conservation of frequency of a propagating wave.

What Maupertuis had in mind for refraction of light

In Maupertuis time it was clear that in order to account for Snell's law it must be assumed that propagation of light is different depending on the medium; there is a ratio of speed of light.

But that ratio could be either way; the mathematics works out for both. Maupertuis went with the supposition that Descartes had offered: that in the denser medium light propagates faster than in air.

For light Maupertuis' action is the product of velocity and distance, which comes out as time. In the variation space: the true refraction is where the derivative-of-the-incident-ray-time matches the derivative-of-the-refracted-ray-time. At the point in variation space such that those two derivatives are matching the derivative of the total time is zero.