On 04 Mar 2012, at 18:52, John Clark wrote:

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On Sat, Mar 3, 2012 at 3:43 PM, Bruno Marchal <marc...@ulb.ac.be>wrote:>> The probability is 100% that if you receive sights and soundsfrom Moscow and not Washington you will become the Moscow man andnot the Washington man.> But the question which was asked is avoided here. What if you knowin advance that 3-you will be both in M and in W, knowing that withcomp the 1-you [...]I said that much of the problem is one of language and this is aperfect example of that because you just made a grammatical error;its plural, you should have said "the 1-yous".

`You still don't get the point. Below I suggest another experience to`

`help you. You are confusing the third person projection of 1-you(s) on`

`another people, which can be plural indeed, and the 1-you that you`

`might feel to be from your 1-pov, which is never plural, for the`

`people doing the experience (assuming comp, and the correctness of the`

`level chosen, etc.).`

`You consider the 1-you of the duplicate person, from a 3p outsider`

`point of view, without ever putting yourself in the place of each one,`

`to see their 1-view from their 1-view points of view.`

> [...] will not feel to be at both place.If I'm avoiding the question its because I don't understand exactlywhat the question is, let me try to cover all the bases. What is theprobability that both the Washington and the Moscow man willremember being the Helsinki man? 100%.

That's correct.

What is the probability the Helsinki man will receive signals fromMoscow turning him into the Moscow man? 100%.

`That's ambiguous. If you say 100%, it means that you are talking on`

`the first person that you can attribute to different people. But the`

`question is practical and concerns the expected 1-you, as considered`

`by the future 1-people after the duplication. In that case we get a`

`paradox if you say that it is 100% for both Moscow and Washington.`

What is the probability the Helsinki man will receive signals fromWashington turning him into the Washington man? 100%.

Idem.

What is the probability the Helsinki man will receive signals fromneither Washington nor Moscow and thus leaving him as the Helsinkiman? 100%.

`In the protocol considered the Helsinki guy is annihilated. So the`

`probability is zero, unless you attribute a first person sensation to`

`ashes. Duplication without annihilation of the "original in Helsinki"`

`is handled in step 5. We are in step 3.`

What is the probability the Helsinki man will feel like the Moscowman? 0% because if he felt like the Moscow man he wouldn't be theHelsinki man anymore.

`In that case, the probability to survive, in the usual clinical sense,`

`a teleportation experience is 0, and you should say no to the doctor,`

`which assure you that IF comp is true, and if the level is right, then`

`you survive without trouble (= with probability one, that's step one,`

`which is an equivalent definition of comp as the one I give usually).`

What is the probability the Moscow man will feel like the Washingtonman? 0% because if he felt like the Washington man he wouldn't bethe Moscow man anymore.

`I guess the last "Moscow" should be replaced by "Helsinki". Again, if`

`you are correct, you can no more use a teleportation device and you`

`make comp false.`

What is the probability that a third party in all this will see aperson in Helsinki and Washington and Moscow with all 3 having aexactly equal right to call themselves John K Clark? 100%.

`The guy in Helsinki is annihilated, in step 3. I insist that the case`

`with the non annihilation of the 'original' will be handled in step 5.`

`Then you are correct (except for Helsinki where the third party will`

`see only ashes after the experiment).`

Where is the indeterminacy in all this and what question have Iavoided?

`You have avoided the question, asked in Helsinki to you: "where can`

`you expect to be from a personal, first person point of view, after`

`the duplication is done?".`

`You cannot answer in W and in M, because you will not write, after the`

`experience, in your diary "I feel to be W and I feel to be in M",`

`because any "I" cannot feel to be in both place at once, except no`

`relevant pathologies.`

`You cannot answer "I will feel to be neither in W nor in M", because`

`this would make comp false.`

`So you will feel from any first person point of view to be in W, or in`

`M. And that "or" is easily shown to be non constructive, and that's`

`the first person indeterminacy.`

> To say, when imagining yourself in Helsinki, that you neitheragree or disagree with the indeterminacy illustrates that you areindeterminate on the outcome you will live.It illustrates that the question is gibberish. This entire businessreminds me of Thomas Nagel's famous essay "What is it like to be abat?" that for reasons I've never understood was the vogue incertain circles a few years ago. Personally I've never had much usefor it. Nagel doesn't want to know what it would be like if he werea bat, he want's to know what it's like for a bat to be a bat. Theonly way to do that would be to turn Nagel into a bat, but thenNagel still wouldn't know because he'd no longer be Nagel, he'd be abat. Like all self contradictory tasks Nagel concludes that this onecan not be done and so demonstrates a keen grasp of the obvious, butI don't know what deep philosophic insight is supposed to be gainedfrom it.

`The question is just hard, if not impossible, for the bat, which is`

`very different from us. Yet it makes some sense to ask some question`

`on qualia, due to the bat peculiar use of sound. (we can come back`

`later on this).`

`Nagel points on the 1-view as seen from the 1-view, indeed, like in`

`the case of the question asked to you in Helsinki.`

`The deep thing to gain is a more precise formulation of the mind-body`

`problem, the discovery of a subjective or first-person indeterminacy`

`in the case of self-multiplication, and of its important fundamental`

`consequences.`

`Note that something akin to that first person indeterminacy is used`

`implicitly in the QM-without collapse theory:`

`If you observe, with an {up, down}-discriminating device, a particle`

`prepared in state`

1/sqrt(2)[up + down],

`then the probability is given by (1/sqrt(2))^2 = 1/2. (in both`

`Everett and Copenhagen)`

`Everett supposes that the physicist obeys to QM. So he applies QM to`

`the couple made from you (the physicist) together with the particle.`

`This is, supposing You represent the physicist's quantum state:`

(You X 1/sqrt(2)[up + down]) ("X" = the tensor product)

`This state is equivalent (before you make the observation) with the`

`state 1/sqrt(2) [(You X up) + (You X down)], by the linearity of the`

`tensor product "X".`

`Then by the linearity of the SWE, this evolves into 1/sqrt(2) [(You-up`

`X up) + (You-down X down)], when you decides to observe the particle;`

`with You-up (resp. You-down) represents the quantum state of the`

`physicist with the memory of having observed the particle in the state`

`up (resp. down).`

`For Copenhagen the wave mysteriously reduces into either (you-up X`

`up), or (you-down X down), by a process which is non linear and can't`

`be described by QM.`

`Everett suggests that the wave doesn't collapse, and that the entire`

`system just obeys QM. This is indeed well justified because the final`

`state explains well why the guy will believe that the wave did`

`collapse, despite it did not. The interaction has just split or`

`differentiated the observer into two components of the wave, with each`

`observer being in front of a definite result.`

`This is strikingly similar to the comp duplication, and in both case,`

`we have a global 3p view which justifies the use of the probability`

`for the relative state of the observers.`

`In fact, the goal is to show that such a similarity is not a`

`coincidence, for the global indeterminacy brought by the UD.`

>> comp can never be proved > Yes. >> or disproved > False. By UDA.But as the proof of UDA critically relies on something you call"first person indeterminacy" the proof is not valid.

`The proof critically relies on the 1-person indeterminacy indeed. (But`

`not in AUDA, except for the motivation).`

`I think that you are just not doing the thought experiment completely.`

`You keep the 3p-view, instead of listening to what the resulting`

`persons will say, or putting yourself at their place.`

`The guy in W feel to be in W, and has no reason justifying why he is`

`the one in W. He could not have predicted this in advance with`

`certainty, because this would single out arbitrary W from {W, M}, and`

`the guy in M, would feel to have been wrong.`

`If the guy in Helsinki has enough cognitive ability to put himself at`

`the place of each resulting people, he can understand that the`

`personal feeling of being in "this precise city" , after the`

`duplication, is not predictable.`

* * *

`Let me propose you now a slightly different self-multiplying`

`experience which has often help people to understand more easily the`

`"1-view of the 1-view", on which the probability question is bearing.`

`(as opposed to your 3-view on the 1-views)`

`For further reference I call it the Random-Movie Experience (not to`

`confuse with the movie graph, or filmed graph experience of the step 8).`

`Consider a giant screen composed of 16180 x 10000 black and white`

`pixels.`

`There are 2^(16180 x 10000) possible images that can be done on that`

`screen. OK?`

`Now, here is the self-multiplying protocol. I multiply you in 2^(16180`

`x 10000) exemplar, in front of each of the possible screen image ...`

`... and I iterate that experience, meaning that I re-multiply all the`

`resulting persons again by 2^(16180 x 10000), putting them again in`

`front of each possible screen, and this 24 times per second, during`

`1h30 hours (= 90 minutes).`

`You can see that the number of people getting out of the lab will be`

`2^[(16180 x 10000) x (60 x 90) x 24], given that the candidate is`

`multiplied 24 times per second, and that there is (60 x 90) seconds in`

`1h30. OK?`

`Again the question is asked to the guy (you) before the experiment`

`begin. What question? This one: what experience do you expect to live.`

`If you believe in comp, you will clearly not predict that you will see`

`more than one movie, given that each final examplars will agree having`

`lived the seeing of one particular movie. So you can predict with`

`certainty that you will live the experience of seeing one movie, just`

`one.`

`Let me make the question more precise: what living experience do you`

`expect, before the experience begin, to be more probable among:`

- "I will see Monty Python's Flying Circus with italian subtitles" - "I will see a black screen movie" - "i will see a white screen movie"

`- "I will see a random movie (white noise), but looking again at the`

`detail, the pixels of the successive`

images describe the digit of Pi in binary" - "I will see a random movie (white noise).

`If you want, I can still make everything 3p in that question, in the`

`following manner. I make a genuine sample of 1000 persons among the`

`2^[(16180 x 10000) x (60 x 90) x 24] resulting persons, by selecting`

`them with a random coin, or whatever choice reasonable enough for not`

`biasing the statistics. I ask them the same question, including "did`

`you expect to see the movie you did see?".`

`In that last case, it is just an exercise in combinatory analysis to`

`show that most of them will acknowledge that they were unable to`

`predict the movie that they have seen, and this is close to a 3p`

`definition of the 1p-indeterminacy we live in such self-multiplication`

`exercise.`

Does this helps?

`It might be relevant to compare this with the following quasi feasible`

`experiment, and certainly feasible soon or late.`

`You are invited to see a quantum random movie, where each pixels is`

`black or white, according to the state up or state down of some photon`

`sent to that pixel.`

`And I send photons, prepared in the state 1/sqrt(2)[up + down] on the`

`pixels, 24 times per second, for a lps of time equal again to 1h30.`

`I ask exactly the same question. What is your answer, and what is your`

`justification?`

`And a last question-experiment protocol. This time I don't tell you in`

`advance if I will do the experiment with the iterated self-`

`multiplication, or with the quantum pixels. But I do the`

`self)multiplication experience, without you knowing if it is the comp`

`multiplication or the quantum experience. I chose randomly one`

`exemplar among the 2^[(16180 x 10000) x (60 x 90) x 24], which I`

`suppose separated (so they have no obvious clues of what choice of`

`experiment I have done). Can that random person have a clue of what`

`has been going one? Can that person distinguish about the comp-`

`multiplication, and the quantum differentiation?`

What do you think?

`Here the "1-view on the 1-view" is implicitly captured by the word`

`"live" in "what experience do you expect to live", because a "personal`

`life" is always a sequence of 1-views on (preceding) 1-views. The`

`question is really simply "what do you expect to live in those`

`experiment, made possible in principle in the comp theory.`

`You can also define the first person indeterminacy by the statistics`

`inferred by the majority of the resulting people (here, among the`

`2^[(16180 x 10000) x (60 x 90) x 24] one).`

`My (real) director of the thesis was an opponent to such notion of`

`probability so I was asked to define them in term of bets. This can be`

`done by multiplying, not one person, but two persons, together, in the`

`same read-annihilation box. They can bet with each other, and that is`

`what I call the first person plural. Likewise, in Everett QM, we can`

`share the indeterminacies because we are collectively multiply, when`

`you apply QM on us.`

`Tell me if this helps. I work in the axiomatic way. I don't define a`

`term more precisely than I need for the reasoning.`

Bruno http://iridia.ulb.ac.be/~marchal/ -- You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to everything-list@googlegroups.com. To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/everything-list?hl=en.