Circular orbits aren't minimas of lagrangian for Kepler problem?

[thalesofmiletus]

TL;DR Summary

Circular orbits are solution of Euler-Lagrange's equation for lagrangian in Kepler problem. But this is only necessary condition, not sufficient. Another condition is that second variation of lagrangian needs to be positive definite. And if my calculations are correct, it is not. But otherwise in Gelfands and Fomin's "Calculus of variations" there is footnote indicating that it should be (with all other theorems).

My point of view is bases on CALCULUS OF VARIATIONS by I. M. GELFANDS. and V. FOMIN. I would assume that you are familiar with the topic. I've found this book online, If needed I can provide link, but for now I don't know if it's legal in this site, so I won't.
(by the way, I'm new on this site, so if latex isn't working I will try to repair it, but I'd appreciate help in such case)

So basically junctional for Kepler problem is defigned as such

$$\int\left(\frac{1}{2} m(\dot r^2+r^2 \dot \varphi^2) +\frac{GMm}{r} \right)dt$$

From which using E-L equations you can derive solution

$$r=\frac{p}{1+\varepsilon \cos{(\varphi-\varphi_0)}}$$

where $$p=\frac{L^2}{GMm^2}$$ , $$\varepsilon=\sqrt{1+\frac{2EL^2}{G^2M^2m^3}}$$ , E is total energy $$E=\frac{1}{2} m\dot r^2+\frac{L^2}{2mr^2} -\frac{GMm}{r}$$ and L is angular momentum $$L=mr^2 \dot \varphi$$. (I can provide my full solution if you need)

Before we consider special solution of circular orbits, let's derive formula for second variation in general case. Formula is given as

$$\delta^2J[h]=\frac{1}{2} \int_a^b \left( \sum_{i,k=1}^n F_{y_iy_k}h_ih_k + 2\sum_{i,k=1}^n F_{y_iy'_k}h_ih'_k + \sum_{i,k=1}^n F_{y'_iy'_k}h'_ih'_k\right) dx$$

(Note that you cannot use (Ph',h')+(Qh,h) is this case, becouse that require $$F_{yy'}$$ to be simetrical, which IS NOT for this example)

where for junctional in Kepler problem partial derivatives are

$$F_{rr}=m \dot \varphi^2+\frac{2GMm}{r^3}$$
$$F_{r\varphi}=F_{\varphi r}=F_{\varphi\varphi}=0$$
$$F_{r \dot r}=0$$
$$F_{r\dot\varphi}=2mr\dot\varphi$$
$$F_{\varphi \dot r}=F_{\varphi \dot \varphi}=0$$
$$F_{\dot r\dot r}=m$$
$$F_{\dot r \dot \varphi}=0$$
$$F_{\dot \varphi\dot \varphi }=mr^2$$

Now let's asume that $$\varepsilon=0$$ (the solution is circle). Let $$r\equiv R$$, using angular momemtum formula and E-L equation with respect to r we can derive $$L=\sqrt{GMm^2R}$$. Substituting that to partials we get (ignoring ones that equal 0)

$$F_{rr}=m \dot \varphi^2+\frac{2GMm}{R^3}=\frac{3GMm}{R^3}$$
$$F_{r\dot\varphi}=2mR\dot\varphi=2\sqrt{\frac{GMm^2}{R}}=2\sqrt{\frac{GMm}{R^3}\cdot mR^2}$$
$$F_{\dot r\dot r}=m$$
$$F_{\dot \varphi\dot \varphi }=mR^2$$

So under integral we get

$$\frac{3GMm}{R^3}h_1^2+4\sqrt{\frac{GMm}{R^3}\cdot mR^2}h_1\dot h_2+mR^2\dot h_2^2+m\dot h_1^2$$

which can be rearanged to

$$\left(\sqrt{\frac{3GMm}{R^3}}h_1+\sqrt{mR^2}\dot h_2\right)^2+\left(4-2\sqrt{3}\right)\sqrt{\frac{GMm^2}{R}}h_1\dot h_2+m\dot h_1^2$$

what can be negative, for example if we take $$\dot h_2=-\sqrt{\frac{3GM}{R^4}}h_1$$ then the squared bracket is 0 and you can easily come on up h_1 that makes second variation negative.

So to conclude, there are (h_1,h_2) for which $$\delta^2J[h]$$ are negative, so it cannot be positive definite. What means that circular orbits are not minimas of lagrangian. What means that they should not be present in world when we assume least-action principle. Well clearly there is something wrong in here. I hope some one might help me.One last remark. In Gelfands and Fomin's book there is footnote (page 119) that theorems regarding Jacobi equations are true even when $$F_{yy'}$$ is not simetrical. As far as I understand the book, from those theorems we deduce sifficient conditions for minimas of juntional. And if we ignore that $$F_{yy'}$$ is not simetrical and check Legendre and Jacobi conditions as it were simetrical, we get that both holds, even in strengthened versions. So we conclude that circular orbits are minimas of lagrangian in kepler problem? The footnote refers to (H. Niemeyer, private communication), how can I tell if this is valid?

 

[Orodruin]

The principle of least action is a misnomer. The appropriate name would be the principle of stationary action. The important thing is not whether the action takes a minimal value - it is if it takes a stationary value, ie, maxes and saddles are also relevant.

 

[thalesofmiletus]

The principle of least action is a misnomer. The appropriate name would be the principle of stationary action. The important thing is not whether the action takes a minimal value - it is if it takes a stationary value, ie, maxes and saddles are also relevant.

Thanks, that helpfull. However the mathematical conflict with footnote from the book still remains. Perhaps I should post it somewhere on math forum.

 

[Cleonis]

As pointed out by contributor Orodruin, the content of Hamilton's stationary action is: the true trajectory is the point in variation space such that the derivative of Hamilton's action is zero. Whether the condition of derivative-of-Hamilton's-action-is-zero corresponds to a minimum or a maximum is not material.

Phrased differently: the business end is the criterion: the point in variation space such that the derivative of the action (wrt variation) is zero. It's just that if you have a curve, and it goes from sloping one way to sloping the other way, and it has a single point where the derivative is zero: that single point is mathematically constrained to be either a minimum or a maximum.

Of course, just the word 'minimization' feels enticing to a physicist. The temptation to latch onto it is pretty much irresistible. So there we are.

As to when a minimum or a maximum is encountered:
I have looked at cases with a single spatial degree of freedom. There is a crossover point, in the sense of a range of cases, and traversing that range of different cases you go from the true trajectory being a minimum of Hamilton's action to the true trajectory being a maximum of hamilton's action; the determining factor is to how the force that acts is a function of distance. (Examples: there are inverse square laws, but there are also interactions such that the force increases with distance.) In case you are interested: identifying that crossover point is part of a discussion of Hamilton's stationary action that is available on my own website.
Incidentally, the first to offer the expression 'stationary action' rather than 'least action' was Hamilton himself.

David R. Wilkins, Trinity College Dublin, has made transcriptions of many of Hamilton's papers. Among those papers is one titled: 'On a general Method of expressing the Paths of Light, and of the Planets, by the Coefficients of a Characteristic Function' (Listed as being published in the 'Dublin University Review', which presumably only had a small circulation.)

[...] its pretensions to a cosmological necessity, on the ground of economy in the universe, are now generally rejected. And the rejection appears just, for this, among other reasons, that the quantity pretended to be economised is in fact often lavishly expended. In optics, for example, though the sum of the incident and reflected portions of the path of light, in a single ordinary reflexion at a plane, is always the shortest of any, yet in reflexion at a curved mirror this economy is often violated.

That is why earlier in the same paper Hamilton had proposed:

From this Law, then, which may, perhaps, be named the LAW OF STATIONARY ACTION, it seems that we may most fitly and with best hope set out, in the synthetic or deductive process, and in search of a mathematical method.