On 04 Mar 2012, at 18:52, John Clark wrote:
On Sat, Mar 3, 2012 at 3:43 PM, Bruno Marchal <[email protected]>
wrote:
>> The probability is 100% that if you receive sights and sounds
from Moscow and not Washington you will become the Moscow man and
not the Washington man.
> But the question which was asked is avoided here. What if you know
in advance that 3-you will be both in M and in W, knowing that with
comp the 1-you [...]
I said that much of the problem is one of language and this is a
perfect 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 exactly
what the question is, let me try to cover all the bases. What is the
probability that both the Washington and the Moscow man will
remember being the Helsinki man? 100%.
That's correct.
What is the probability the Helsinki man will receive signals from
Moscow 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 from
Washington turning him into the Washington man? 100%.
Idem.
What is the probability the Helsinki man will receive signals from
neither Washington nor Moscow and thus leaving him as the Helsinki
man? 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 Moscow
man? 0% because if he felt like the Moscow man he wouldn't be the
Helsinki 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 Washington
man? 0% because if he felt like the Washington man he wouldn't be
the 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 a
person in Helsinki and Washington and Moscow with all 3 having a
exactly 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 I
avoided?
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 neither
agree or disagree with the indeterminacy illustrates that you are
indeterminate on the outcome you will live.
It illustrates that the question is gibberish. This entire business
reminds me of Thomas Nagel's famous essay "What is it like to be a
bat?" that for reasons I've never understood was the vogue in
certain circles a few years ago. Personally I've never had much use
for it. Nagel doesn't want to know what it would be like if he were
a bat, he want's to know what it's like for a bat to be a bat. The
only way to do that would be to turn Nagel into a bat, but then
Nagel still wouldn't know because he'd no longer be Nagel, he'd be a
bat. Like all self contradictory tasks Nagel concludes that this one
can not be done and so demonstrates a keen grasp of the obvious, but
I don't know what deep philosophic insight is supposed to be gained
from 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/
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