# Why is classical mechanics determinism based on position and momentum only and not forces and scattering rules?

Consider a closed system (say a box) of $n$ particles. There is a well-known idiom/meme/law in classical mechanics that says that the position and momentum of those $n$ particles is all that is needed to determine the future and past configuration of the system. (A minor question, does this idiom have a name?)

Why do we not consider two other pieces of information to be as important as the position and momentum of the $n$ particles?

• The force field (force as a function of space, and preferably of time too).

• The particle interaction rules (When two particles touch each other, do they whiz by or do they scatter? if they scatter what are the rules for determining angles?)

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I guess by force field I mean external forces, and by scattering rules I mean internal ones (particle-particle interaction). So in the end I guess it's just one piece of information, forces, not two. –  user3236443 Feb 20 '14 at 1:51
As a note, the closest named statement I know of is the Newton-Laplace Principle of Determinacy, which says that a particle's position and velocity at a given time completely determines the particle's trajectory. –  EtaZetaTheta Feb 20 '14 at 2:20

The information about the forces is just as important in predicting the time evolution of a classical mechanical system.

When people say that the initial position and momenta are "all" that's needed to predict the evolution of the configuration of the system, they do not mean this in the strict sense that you're indicating. In particular, the statement they're really making is is along the lines of the following:

For a given set of interactions (forces) between particles in a system, and for a given set of initial positions and momenta of all of the particles, one can (modulo some somewhat pathological examples) determine the positions and momenta of the particles for all times using Newton's Second Law for each particle: \begin{align} \mathbf F_i = m\ddot {\mathbf x}_i \end{align} where $\mathbf F_i$ is the net force on particle $i$.

Notice that if the forces are unknown, then one can't even write down Newton's Second Law, so there are no equations of motion to solve for the time evolution of the system and, in that case, the initial data are essentially useless.

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which are those "somewhat pathological examples" you are talking about? –  silvrfück Feb 27 '14 at 22:07
@silvrfuck Great question. Check this out: physics.stackexchange.com/q/39632 –  joshphysics Feb 27 '14 at 23:57

An experimentalist's answer is that the dynamics of all systems are described by second order differential equations, either classical or quantum mechanical.

The classical case is called deterministic because all one needs for describing the trajectories in space and time of the particles in the system are solutions of the differential equations . The forces and fields entering the problem exist in the setups of the differential equations.

Once one has the solutions to the equations, the general form of the trajectories, the values of the position and momentum at one time, t=t', define uniquely the evolution of the system, basically by construction.

In the quantum mechanical case even though the differential equations of the dynamics of the problem are second degree, and the potentials are included, the solutions define probabilities for observing the evolution of the system, not definite trajectories, and thus the system cannot be deterministic in the classical sense.

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