On 3/23/2022 7:02 PM, Bruce Kellett wrote:

On Thu, Mar 24, 2022 at 12:34 PM Brent Meeker <meekerbr...@gmail.com>wrote:On 3/22/2022 11:05 PM, Bruce Kellett wrote:On Wed, Mar 23, 2022 at 4:01 PM Brent Meeker <meekerbr...@gmail.com> wrote: On 3/22/2022 7:55 PM, Bruce Kellett wrote:Actually, that is where I started. I assumed that Alice and Bob were both able to collect results from N trials before they met. Then there are 2^N copies of each experimenter, and a potential (2^N)^(2^N) pairs when they meet. The trouble to be explained is that there are actually only 2^N pairs in a real experiment, each with inequality-violating correlations. What has happened to all the extra pairings that MWI must produce? (Most of which have correlations violating the quantum predictions.)Don't you mean (2^N)x(2^N). Probably. Both Alice and Bob split into 2^N copies for binary results. So each of the 2^N Alice's splits into 2^N further copies, one for each copy of Bob.Why would Alice split again for each copy of Bob. There's no QM involved.There is. That is what MWI predicts: for every splitting of Bob, thewhole world, including copies of Alice, also splits.

`OK, I see your point. I think that's one reading of it. But if MWI is`

`local, as Bruno and smitra contend, the non-locality of the`

`wave-function can't have the effect of splitting Alice when Bob measures`

`spacelike relative to Alice. As I understood it, they were claiming`

`it's a virtue of MWI that this FTL stuff doesn't happen.`

That is certainly the case if both Alice and Bob measure independentparticles that they prepare themselves. The question is, "Why arethere fewer joint branches when they measure entangled particles?" Ifyou consider that all (2^N)*(2^N) branches with Alice-Bob pairingsexist in the entangled case as well as the independent case, as MWIwould predict if the measurements are truly local and independent,then a large number of the couples are going to find correlations thatviolate the quantum predictions. The fact that all experiments havefound correlations that agree with QM is then hard to explain.Each Alice and each Bob has seen N events and has recorded them as a binary number of N bits in their notebook. They are quasi-classical observers and they're going to meet in the future. If there are not (2^N)x(2^N) meetings there must be some quantum magic that it is eliminating or otherwise weighting them.That is the point that I am making. There must be some fairy dust orother magic at play if the MWI predictions are to agree with QM. As Ihave said before, if some of the branches have zero weight or zeroprobability, then such branches can never have been created. There isno mechanism which could make them vanish only when Alice and Bobmeet. In many of the Everettian apologies, capital is made of the factthat the Alice-Bob meeting is also an interaction. But thatinteraction is not (or need not be) of the sort that could removeunwanted branches.

`Right. That was my point in having them light-hours apart so the`

`meeting is classical comparison of notebooks.`

Brent

For example, if Alice and Bob exchange results by email, there is nointeraction that could possibly eliminate branches, or change data intheir notebooks so that everything worked out. Such notions are in therealm of fantasy and magic.Bruce --You received this message because you are subscribed to the GoogleGroups "Everything List" group.To unsubscribe from this group and stop receiving emails from it, sendan email to everything-list+unsubscr...@googlegroups.com.To view this discussion on the web visithttps://groups.google.com/d/msgid/everything-list/CAFxXSLTbQz5wzXfEZ563Eu_BS2%3DhH%2BUU_XYEm9KHijyOPCnxmQ%40mail.gmail.com<https://groups.google.com/d/msgid/everything-list/CAFxXSLTbQz5wzXfEZ563Eu_BS2%3DhH%2BUU_XYEm9KHijyOPCnxmQ%40mail.gmail.com?utm_medium=email&utm_source=footer>.

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