Suppose all of the particles in the universe should happen to come to rest at the same time, in positions so arranged that all of the forces on every particle balance to zero at that time. What would happen next? Or rather, what does Newtonian mechanics say will happen next? Preface Inertia and Stasis 2.1 Stating the Law of Inertia more precisely 2.2 The stasis scenario Indeterministic Examples 3.1 Abstract example 3.2 Second example Non-Lipschitz Forces and Determinism Beyond the Stasis (...) Scenario CiteULike Connotea Del.icio.us What's this? (shrink)

In talking about the compatibility of quantum observables, discussions often center on the question of whether the corresponding operators commute—even though commutativity is a coarse-grained notion that largely fails to capture the salient “nonclassical” features of quantum theory. Often, too, such discussions involve the issue of whether the operators in question satisfy a Heisenberg-like inequality, of the form ΔA·ΔB≥r>0—even though such inequalities are specific to unbounded operators and (for this and other reasons) are typically not a useful way to discuss (...) joint uncertainty in quantum mechanics. In the present paper we emphasize a simpler dichotomy, in which operator pairs (A, B) are classified according to whether or not states can be found with arbitrarily small dispersions in both A and B. If A and B cannot be made arbitrarily dispersionless simultaneously, then we call A and B an uncertainty pair. Otherwise, we call A and B a certainty pair. An interesting feature of uncertainty pairs in particular is that they are stable, in the sense that if A and B form an uncertainty pair, then slight enough perturbations of A and B must also form an uncertainty pair. This stability, obvious in the finite-dimensional case, follows in general from an operator inequality derived herein. A consequence of this inequality is that perturbed position and momentum operators X+δX and P+δP cannot share an eigenvector unless |δX|·|δP|≥ℏ/2. (Here vertical bars denote the operator norm.) This, despite the fact that (as we show) arbitrarily slight perturbations of X and P can fail to satisfy a Heisenberg inequality—a fact which raises interesting measurement issues in its own right. (shrink)