# Does Hamilton's principle allow a path to have both a process of time forward evolution and a process of time backward evolution?

This is from Analytical Mechanics by Louis Hand et al. The proof is about Maupertuis' principle. The author seems to say that Hamilton's principle allow a path to have both a process of time forward evolution and a process of time backward evolution. But it's odd to me, because if the each part have different sign of extreme value, that is, some part of positive value, some part of negative value, how can the combined whole path still has the extreme value (sure maybe it's a inflection, but will it be a inflection every time)?

Meanwhile, Eq. 2.109 makes me confused. If I memory is correct, Hamilton's principle uses contemporaneous variation. So how can we use Hamilton's principle to derive Eq. 2.109?

1. OP asks an interesting question, which we rephrase as follows:

Given an action $$S[q]~=~\int_{t_1}^{t_2} \!dt~L(q,\frac{dq}{dt},t),$$ are the stationary action principle (SAP) and Euler-Lagrange (EL) equations modified if we allow zig-zag parametrizations of the time parameter $$t$$, i.e. if we integrate by substitution (IBS) $$t=f(\lambda)$$, where $$f$$ is a not necessarily monotonic function of the free parameter $$\lambda\in[\lambda_1,\lambda_2]$$?

The answer is: Generically it is not modified. Note for starters that IBS is valid for non-monotonic substitutions.

One may show that the Lagrangian formalism behaves covariantly under world-line reparametrizations. It is e.g. an easy exercise to check that the new EL equations for the new Lagrangian $$\widetilde{L}(q,\dot{q},\lambda)~=~\left. \dot{t}L(q,\frac{\dot{q}}{\dot{t}},t)\right|_{t=f(\lambda)}$$ are generically equivalent to the old. Here dot denotes differentiation wrt. $$\lambda$$.

Let us mention for later that we can allow discontinuities in some of the (possibly derivatives of the) Lagrangian $$\widetilde{L}$$ and the path $$q$$, as long as the paths $$q$$ themselves and their momenta $$\frac{\partial\widetilde{L}}{\partial\dot{q}}$$ are continuous, cf. e.g. this Math.SE post.

2. Now let us consider the proof of Maupertuis' principle (MP) in Ref. 1. The MP assumes

• (i) that $$q(t_i)~=~q_i\text{ fixed}\quad\text{and}\quad q(t_f)~=~q_f\text{ fixed},$$ while the initial and final times $$t_i$$, $$t_f$$ are free;

• (ii) that the Lagrangian $$L$$ has no explicit time dependence;

• (iii) and that all the allowed virtual paths $$q$$ have the same constant energy $$E$$.

3. Ref. 1 is considering 2 allowed virtual paths:

• $$q$$ between the points $$AB$$ and
• $$q^{\prime}$$ between the points $$CD$$,

as shown in Fig. 2.15. A piecewise linear zig-zag parametrization of the curve $$ACDB$$ is $$f(\lambda)~=~\left\{\begin{array}{lcl} t_A(1-3\lambda) + t_C3\lambda &{\rm for}& \lambda~\in~[0,\frac{1}{3}],\cr t_C(2-3\lambda) + t_D(3\lambda-1) &{\rm for}& \lambda~\in~[\frac{1}{3},\frac{2}{3}],\cr t_D3(1-\lambda) + t_B(3\lambda-2) &{\rm for}& \lambda~\in~[\frac{2}{3},1].\end{array} \right.$$ We note that the constant segments $$AC$$ and $$BD$$ might not have the energy $$E$$, but that in itself is not disqualifying as we are trying to apply the SAP.

If the CD curve is $$q^{\prime j}(t)~=~g^j(t), \qquad t\in [t_C,t_D],$$ then the ACDB curve is $$q^{\prime j}(\lambda)~=~\left\{\begin{array}{lcl} g^j(t_C)&{\rm for}& \lambda~\in~[0,\frac{1}{3}],\cr g^j(t_C(2-3\lambda) + t_D(3\lambda-1) ) &{\rm for}& \lambda~\in~[\frac{1}{3},\frac{2}{3}],\cr g^j(t_D)&{\rm for}& \lambda~\in~[\frac{2}{3},1],\end{array} \right.$$ whose derivative (the velocity) has discontinuities at $$C$$ and $$D$$. This typically means that the momentum has discontinuities at $$C$$ and $$D$$, which is a problem for the current proof method, cf. above remark.

TL;DR: Yes, OP has a point: The proof of the MP in Ref. 1 is flawed/at best incomplete.

References:

1. L.N. Hand & J.D. Finch, Analytical Mechanics, 1998; Section 2.9 p. 79-80.

As I understand it you are wondering about the following:

What will be the implications when Hamilton's action is evaluated over a time interval where the end point is earlier than the start point?

I take it you will agree that in simple cases the direction of the time coordinate is not material. As we know, when you watch a video of a Kepler orbit then the motion that you see does not give you a clue as to whether the video is played in forward direction or backward direction.

The stationary action concept accomodates that type of independence.

With that out of the way: that's not what you are asking about.

I surmise that you expect that the sign of the extremum will be reversed in the case that the end point is earlier in time than the start point

I guess so. For evaluation within that interval the sign of the extremum makes no difference, either way you are looking for the point in variation space such that the derivative of Hamilton's action is zero. Knowing the sign of th extremum is not necessary.

Your question, as I understand it, is about what happens when you concatenate two such intervals, one time-forward, the other time reversed.

It seems to me: that would be mathematically ill defined.

I assume that combining intervals is well defined only if the intervals are such that the connection is differentiable; there shouldn't be any discontinuity. That excludes the case of having one interval time-forward, and the other interval time reversed.