# Indeterminism in systems with infinitely and finitely many degrees of freedom John D. Norton Department of History and Philosophy of Science University.

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Indeterminism in systems with infinitely and finitely many degrees of freedom John D. Norton Department of History and Philosophy of Science University of Pittsburgh 1

Indeterminism is generic among systems with infinitely many degrees of freedom. 2 Source: Appendix to Norton, “Approximation and Idealization…”

Enforced by embedding in still larger solution. Enforced by embedding in larger solution. Solution that manifests pathological behavior for a few components, e.g. spontaneous excitation The mechanism that generates pathologies 3 0 1 2 345 system of infinitely many coupled components …and so on indefinitely.

Expected solution dx n (0)/dt = x n (0) = 0 for all n x n (t) = 0 for all n, all t with initial conditions Masses and Springs 4 Motions governed by d 2 x n /dt 2 = (x n+1 – x n ) - (x n – x n-1 )

Unexpected solution with same initial conditions x 1 (t) = x 2 (t) = (1/t) exp (-1/t) Non-analytic functions needed to ensure initial conditions preserved. Masses and Springs 5 Motions governed by d 2 x n /dt 2 = (x n+1 – x n ) - (x n – x n-1 ) Solve for remaining variables iteratively x 3 = d 2 x 2 /dt 2 + 2x 2 - x 1 dx 3 /dt = d 3 x 2 /dt 3 + 2dx 2 /dt - dx 1 /dt x 4 = d 2 x 3 /dt 2 + 2x 3 – x 2 dx 4 /dt = d 3 x 3 /dt 3 + 2dx 3 /dt – dx 2 /dt etc.

Indeterminism is exceptional among systems with finitely many degrees of freedom. 6

7 The Arrangement The mass experiences an outward directed force field F = (d/dr) potential energy = (d/dr) gh = r 1/2. The motion of the mass is governed by Newton’s “F=ma”: d 2 r/dt 2 = r 1/2. A unit mass sits at the apex of a dome over which it can slide frictionless. The dome is symmetrical about the origin r=0 of radial coordinates inscribed on its surface. Its shape is given by the (negative) height function h(r) = (2/3g)r 3/ 2.

8 Possible motions: None r(t) = 0 solves Newton’s equation of motion since d 2 r/dt 2 = d 2 (0)/dt 2 = 0 = r 1/2.

9 Possible motions: Spontaneous Acceleration For t≤T, d 2 r/dt 2 = d 2 (0)/dt 2 = 0 = r 1/2. For t≥T d 2 r/dt 2 = (d 2 /dt 2 ) (1/144)(t–T) 4 = 4 x 3 x (1/144) (t–T) 2 = (1/12) (t–T) 2 = [(1/144)(t–T) 4 ] 1/2 = r 1/2 The mass remains at rest until some arbitrary time T, whereupon it accelerates in some arbitrary direction. r(t) = 0, for t≤T and r(t) = (1/144)(t–T) 4, for t≥T solves Newton’s equation of motion d 2 r/dt 2 = r 1/2.

10 The computation again For t≤T, d 2 r/dt 2 = d 2 (0)/dt 2 = 0 = r 1/2. For t≥T d 2 r/dt 2 = (d 2 /dt 2 ) (1/144)(t–T) 4 = 4 x 3 x (1/144) (t–T) 2 = (1/12) (t–T) 2 = [(1/144)(t–T) 4 ] 1/2 = r 1/2

11 Without Calculus Imagine the mass projected from the edge. Close…

12 Without Calculus Imagine the mass projected from the edge. Closer…

13 Without Calculus BUT there is a loophole. Spontaneous motion fails for a hemispherical dome. How can the thought experiment fail in that case? Now consider the time reversal of this process. Spontaneous motion! Imagine the mass projected from the edge. BINGO!

What should we think of this? 14

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