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Indeterminismin systems with infinitely and finitely many
degrees of freedom
John D. NortonDepartment of History and Philosophy of Science
University of Pittsburgh
1
Indeterminism is generic amongsystems with
infinitely many degrees of freedom.
2Source: 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 3 4 5
system of infinitely many coupled components
…and so on indefinitely.
Expected solution
dxn(0)/dt = xn(0) = 0 for all n
xn(t) = 0 for all n, all t
with initial conditions
Masses and Springs
4
Motions governed by
d2xn/dt2 = (xn+1 – xn) - (xn – xn-1)
Unexpected solution with
same initial conditions
x1(t) = x2(t) = (1/t) exp (-1/t) Non-analytic functions needed to ensure initial conditions preserved.
Masses and Springs
5
Motions governed by
d2xn/dt2 = (xn+1 – xn) - (xn – xn-1)
Solve for remaining variables iteratively
x3 = d2x2/dt2 + 2x2 - x1
dx3/dt = d3x2/dt3 + 2dx2/dt - dx1/dt
x4 = d2x3/dt2 + 2x3 – x2
dx4/dt = d3x3/dt3 + 2dx3/dt – dx2/dt
etc.
7
The Arrangement
The mass experiences an outward directed force fieldF = (d/dr) potential energy = (d/dr) gh = r1/2.
The motion of the mass is governed by Newton’s “F=ma”:
d2r/dt2 = r1/2.
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)r3/2.
8
Possible motions: None
r(t) = 0solves Newton’s equation of motion
since
d2r/dt2 = d2(0)/dt2 = 0 = r1/2.
9
Possible motions: Spontaneous Acceleration
For t≤T, d2r/dt2 = d2(0)/dt2 = 0 = r1/2.
For t≥T
d2r/dt2 = (d2 /dt2) (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≥Tsolves Newton’s equation of motion
d2r/dt2 = r1/2.
10
The computation again
For t≤T, d2r/dt2 = d2(0)/dt2 = 0 = r1/2.
For t≥T
d2r/dt2 = (d2 /dt2) (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
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!