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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

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

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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 system of infinitely many coupled components …and so on indefinitely.

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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 )

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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.

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

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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.

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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.

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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.

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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

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

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

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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!

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What should we think of this? 14

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