"Consider: the B and anti-B are entangled. Until one of them decays--thereby "picking" a flavor--the other one doesn't know how it is supposed to decay. The "second" one to decay will subsequently oscillate back and forth between being a Bo and an anti-B meson: effectively, it has two decay rates, one in its role as a Bo and one in its role as an anti-Bo)."
The 2 particles were in causal contact during the the initial decay, with each other and the original Y(4S) they came from. That's the point where the entangled state was established. It's within the light cone, the region where all the info is passed out at. The Y(4S) is a particle containing 2 quarks, b and anti-b. When it decays, each b will get a newly generated d quark eventually, as follows:
Bo = d,anti-b
anti-Bo = anti-d, b
The decay mechanism on one of the particles is stabilized by by raising the activation energy for the normally available decay paths as follows. The 2 components that make up the particle are oscillating back and forth with their own anti-particle with a high enough frequency, that before any path can be taken, it changes to the wrong particle and that path is unavailable then.
The b quark can't oscillate alone, or a net charge would be seen magnets and the "delay clock" could be seen running. The net result is that it is neither Bo, nor anti-Bo. This oscillation is called a CP eigenstate, f. f looks like:
d anti-b <=> anti-d, b
I said one of the particles would begin as an f, but there's no reason, that I know of to constrain it that way. I'm not too familiar with the Std Model. Both particles could begin in that state, with one more likely to have a short life as such. It will become the tag. It seems the flavor symmetry takes some time to develop and One of the particles gets a head start in the region of the original decay. No particle in particular becomes a tag as far as I know.
"you can't say that it was pre-informed about the decay fate of the "first" meson because that doesn't help: such a scheme will necessarily obey Bell's inequality
I looked, but was unable to find, or access any description of the technique and data. Page 4 of the second ref. contains statements I can't construct a picture of. They also name the states differently. In that paper, it seems to me that they're referring to some initial decay, before particle decay begins. Others gave me the impression that they used the first decay to tag both particles ID and then extract curves. I think the 2nd particle's decay curve contains the ttag/decay. ?
Suppose someone pokes their finger in the detector and pokes at random particles giving them enough activation energy to decay. The probability is 1/2, that anyone poked will be an f. The f must decay to the correct particle even though it never saw the tag. The quantum correlations for the finger experiment must match those for the tag run.
Regardless of the details of where ttag/decay ends up in the second particles decay curve. Flavor is a symmetry and if ttag/decay defines to for the second particles decay, then all I see is a phase change(shape) in the fields, Higgs and EM. Phase velocities do exceed c. In this case, the CP eigenstate is maintained by the field between the particles. The finger experiment has the same effect as ttag.
"what about non-EPR correlated nuclear decays?"
A decay here is simply a high energy state of a system falling to a lower energy state with the release of equivalent energy. The system has an inherent stability and an activation energy for any decay path available. Nuclear and particle decays are normally considered with the concept of half life and the equation,
N = Noe-t/t1/2.
Nuclear decays around room temperature are relatively insensitive to temperature, until T times Boltzmann's constant approaches the activation E. In short, in terms of energy, decay(non tunneling) is generally proportional to,
e(-Eact(or delta E)/kT)
, or is the sum of terms containing this factor.
In addition to activation energy driven processes, there is tunneling. One of the first verifications for QM was Gamow, Condon, and Gurney's calc of predicting the number of alpha particles emitted per second in alpha decay. They considered the problem as tunneling through a barrier. The calculated the transmission coefficient and took the number of trials per second, N, as:
N = 1/2 v/R, where v is the velocity of the particle and R is the radius of the nucleus.
The eq for the number of particles/sec, n, is,
n=v/2R*e-2*integral from R to infinity of U,
where U = (2m/hbar2)((zZe2/r-E)dr
E is the energy of the particle. z and Z are the charges of the alpha particle and nucleus respectively. e is the electron charge and m is the mass of the alpha particle.
The following were calculated for n.
n(U238) = 5*10-18 events/sec
n(Po212) = 2*10-6 events/sec
The reference values are.
n(U238) =7.05 *10-18 events/sec
n(Po212) = 3.04*10-7 events/sec
Here's a web calc within a factor of 2. Here's a paper that shows more detail. Stability against decay in isotopes depends on binding energy/nucleon(protons and neutrons). The binding energy represents the height of the potential well. Iron has the highest binding energy.
"Occam's Razor leads us to conclude that all subatomic decays have a random, causeless element, somewhere."
The driving force, or cause, is simply the fact that the system has a path to a lower energy configuration. Even if that path is over an intermediate higher energy hill, or by tunneling. The path over the hill simply requires an activation E. Tunneling simply requires a successful attempt. Even with tunneling, there are underlying fluctuations, that effect "wall height" and particle E. Random just refers to the distribution of events that amount to, "a path has been taken."
I pulled some results from some CLEO B factory analysis here.
The average lifetimes for the 2 particles are, Bo = 1.55ps and anti-Bo = 1.49ps.