The Representation of Time in Discrete Mechanics

Abstract

The starting point of the chapter is a twofold observation:

• (i) most current physical theories make use of a continuous parameter t that plays the role of time;

• (ii) the current practice of modeling, namely, of computational modeling, makes use of a discrete parameter t _k (k = 1, 2, …, n) that plays the role of time, because computers cannot handle continuous quantities.

A “parameter” here is a symbol that does not play the same role as the other symbols in the physical theories or models: whereas the other symbols represent physical quantities, t (or t _k ) just plays the role of that within which other quantities evolve. It is “time”, but in a rather shallow sense.

Appendices

Appendices

1.1 Appendix 1: Discrete Mechanics and Some Applications

Let us call A _d the action in Discrete Mechanics (for details see (D’Innocenzo et al. 1987; Marsden and West 2001, Chap. 5)):

$$ {A}_d=\sum_{k=0}^{N-1}{L}_d.\left({t}_{k+1}-{t}_k\right)\kern5.5em $$

(9.15)

with L _d the discrete Lagrangian. The principle of least action states that: δA _d  = 0. It results in the following twofold discrete Euler-Lagrange equations (DEL). The first discrete Euler-Lagrange equation is:

$$ \left({t}_k-{t}_{k-1}\right)\frac{\partial {L}_d\left({t}_{k-1},{q}_{k-1},{t}_k,{q}_k\right)}{\partial {q}_k}+\left({t}_{k+1}-{t}_k\right)\frac{\partial {L}_d\left({t}_k,{q}_k,{t}_{k+1},{q}_{k+1}\right)}{\partial {q}_k}=0 $$

(9.16)

The second discrete Euler-Lagrange equation is:

$$ \frac{\partial }{\partial {t}_k}\left[\left({t}_k-{t}_{k-1}\right){L}_d\left({t}_{k-1},{q}_{k-1},{t}_k,{q}_k\right)\right]+\frac{\partial }{\partial {t}_k}\left[\left({t}_{k+1}-{t}_k\right){L}_d\left({t}_k,{q}_k,{t}_{k+1},{q}_{k+1}\right)\right]=0\kern1em $$

(9.17)

Let us follow d’Innocenzo et al. for the choice of the discrete Lagrangian:

$$ {L}_d\left({q}_k,{q}_{k+1},{h}_{k+1}\right)=\frac{1}{2} m{\left(\frac{q_{k+1}-{q}_k}{h_{k+1}}\right)}^2- V\left(\frac{q_{k+1}+{q}_k}{2}\right) $$

with h _k+1 =t _k+1 −t _k . Under these conditions, let us solve (i) the free particle system, (ii) the one dimension falling body problem, and (iii) the one dimension harmonic oscillator system.

(i) The free particle is the system with V=V ₀ . The DEL become:

$$ m\frac{v_{x, k+1}-{v}_{x, k}}{h_{k+1}}=0\kern2.5em $$

(9.18)

$$ \frac{1}{2} m{v}_{x, k+1}^2=\frac{1}{2} m{v}_{x, k}^2\kern3.25em $$

(9.19)

with ${v}_{x, k}=\frac{x_k-{x}_{k-1}}{t_k-{t}_{k-1}}$. The solution of the equations is:

$$ {x}_k={v}_{x, i}{t}_k+{x}_i\kern3em $$

(9.20)

with the initial conditions t ₀ =0, x _i =x ₀ , v _x,i =(x ₁ −x ₀ )/h ₁ .

(ii) The one dimension falling body problem is the system where $\left({z}_k,{z}_{k+1}\right)= mg\frac{z_{k+1}+{z}_k}{2}$. Thus, the DEL are:

$$ {v}_{z, k+1}-{v}_{z, k}=-\frac{g}{2}\left({t}_{k+1}-{t}_{k-1}\right)\kern1.5em $$

(9.21)

$$ \frac{1}{2} m{v}_{z, k+1}^2+ mg\frac{z_{k+1}+{z}_k}{2}=\frac{1}{2} m{v}_{z, k}^2+ mg\frac{z_k+{z}_{k-1}}{2}\kern0.5em $$

(9.22)

with ${v}_{z, k}=\frac{z_k-{z}_{k-1}}{t_k-{t}_{k-1}}$. We follow the resolution of d’Innocenzo et al. (1987) with different notations. The solution of the equations is t _k  − t _k − 1 = h and:

$$ {z}_k=-\frac{1}{2} g{t}_k^2+\left({v}_{z, i}+ gh\right){t}_k+{z}_{i\ }\kern1.25em $$

(9.23)

with the initial conditions t ₀ =0, z _i =z ₀ , v _z,i =(z ₁ −z ₀ )/h ₁ .

(iii) The one dimension harmonic oscillator system is the system where $V\left({x}_k,{x}_{k+1}\right)=\frac{1}{2} K{\left(\frac{x_{k+1}+{x}_k}{2}\right)}^2$. Thus, the DEL are:

$$ m\left({v}_{x, k+1}-{v}_{x, k}\right)=-\frac{K}{4}\left[\left({x}_k+{x}_{k-1}\right){h}_k+\left({x}_{k+1}+{x}_k\right){h}_{k+1}\right]\kern1.25em $$

(9.24)

$$ \frac{1}{2} m{v}_{x, k+1}^2+\frac{1}{2} K{\left(\frac{x_{k+1}+{x}_k}{2}\right)}^2=\frac{1}{2} m{v}_{x, k}^2+\frac{1}{2} K{\left(\frac{x_k+{x}_{k-1}}{2}\right)}^2 $$

(9.25)

Following d’Innocenzo et al. (1987) with different notations (in particular see (Cieslinski and Ratkiewicz 2006), we have:

$$ {x}_k={x}_0 \cos \left({\omega}_d{t}_k\right)+\frac{x_1-{x}_0 \cos \left({\omega}_d h\right)}{ \sin \left({\omega}_d h\right)} \sin \left({\omega}_d{t}_k\right) $$

with ${\omega}_d=1/ h\ { \tan}^{-1}\left(\frac{\omega h}{1-{\omega}^2{h}^2/4}\right) a n d\ \omega =\sqrt{K/ m}$.

1.2 Appendix 2: Trajectory of a Falling Body in DM

Since x _k  = v _{x, i} t _k  + x _i  (see Eq. (9.20) in Appendix) then, ${t}_k=\frac{x_k-{x}_i}{v_{x, i}}$. Hence, put it in the Eq. (9.23) in Appendix, we derive the equation of the trajectory:

$$ {z}_k=-\frac{1}{2} g{t}_k^2+\left({v}_{z, i}+ gh\right){t}_k+{z}_i = -\frac{g}{2{v}_{x, i}^2}{\left({x}_k-{x}_i\right)}^2+\frac{v_{z, i}+ gh}{v_{x, i}}\left({x}_k-{x}_i\right)+{z}_{i\ } $$

Now, we can deduce the highest position of the body. It is the position where the partial derivative z _k with respect to x _k vanishes: $\frac{\partial {z}_k}{\partial {x}_k}\left({x}_k^{\ast}\right)=0$. Hence,

$$ {x}_k^{\ast }={v}_{x, i}\left({v}_{z, i}+ gh\right)/ g+{x}_i\kern1em and\kern0.75em {z}_k^{\ast }={z}_k\left({x}_k^{\ast}\right)={\left({v}_{z, i}+ gh\right)}^2/\left(2 g\right)+{z}_i $$

1.3 Appendix 3: Ge-Marsden Theorem

We report here the Ge-Marsden theorem and its proof as they are formulated in the original paper:

We recall that there are algorithms which exactly preserve energy, some of which also preserve other quantities […]. However, these algorithms cannot be symplectic , according to the following result of Ge:

Let H be a Hamiltonian which has no conserved quantities (in a given class $\mathcal{H}$, for example analytic functions) other than functions of H. That is, if {F,H}=0, then F(z)=F ₀ (H(z)) for a function F ₀ . Let Φ_Δt be an algorithm which is defined for small Δt and is smooth. If this algorithm is symplectic , and conserved H exactly, then it is the time advance map for the exact Hamiltonian system up to a reparametrization of time. In other words, approximate symplectic algorithms cannot preserve energy for nonintegrable systems.

This result is in fact easy to prove. The algorithm being symplectic , is generated by a dependent function F(z,t), which we assume belong to $\mathcal{H}$. Since Φ_Δt preserves H, and H is assume to be time independent, F commutes with H, and so F(z)=F ₀ (H(z)). It follows that the hamiltonian vector fields of F and H are parallel, so their integral curves are related by a time reparametrization. (Ge and Marsden 1988, p. 135).

As far we understand this result, while the Hamiltonian system is assumed to be nonintegrable – i.e that an exact solution cannot be constructed – if the symplectic  algorithm exactly preserves energy, the solution of the algorithm would be the exact solution of the system modulo a reparametrization. This contradicts the assumption according to which the system is nonintegrable (see also (Ge 1991, p. 380; Marsden 2009, p. 178)). However, as Marsden (2009, p. 179) emphasizes, if the time step is variable , the previous result does not hold and symplectic algorithms can exactly preserve energy.