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From: OsherD on 27 Apr 2010 22:40 From Osher Doctorow Let X, Y be continuous random variables, and let {a_i}, {b_i} with a_i < = b_i be sequences of nonnegative real numbers such that a_i --> 0+ and b_i --> 0+ as i--> infinity, and similarly for pairs {c_i, d_i}. Then by continuity of the cumulative distribution function, we have: 1) P(a_i < = X < = b_i)P(c_i < = Y < = d_i) --> 0+ as i --> infinity which contradicts the Heisenberg Uncertainty Principle (HUP) since each of the probabilities in (1) is an analog of variances Var(X), Var(Y) respectively in the Heisenberg Uncertainty Principle: 2) Var(X)Var(Y) > = k, where k is a positive linear function of h (Planck's constant). The only counter-argument that defenders of HUP can give is that position (X) and momentum (Y) are discrete or not continuous random variables. Readers can mull that over, but it won't work in terms of approximations and in terms of "leaps of faith" about discreteness. Osher Doctorow
From: OsherD on 28 Apr 2010 00:41
From Osher Doctorow For simplicity, I considered above that X and Y are nonnegative with minimum zero. If they have other minima, similar results would be obtained with replacing 0 by the other minimum or minima. Notice, by the way, that with momentum p and position x and frequency v, at the quantum level: 1) p = hv/c 2) px = hvx/c 3) px = hvx/c > = h iff vx > = c With x = 0 or sufficiently near 0 and v fixed, (3) is false, since 0 > = c is false, so that even without introducing probability the idea of the Probable Causation/Influence (PI) analog gives a 0 (false) result for the analog of the HUP. Osher Doctorow |