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From: Jacques M. Mallah <jqm1584.domain.name.hidden>

Date: Mon, 23 Aug 1999 19:21:30 -0400 (EDT)

On Wed, 18 Aug 1999, Christopher Maloney wrote:

*> I should have qualified my zombie view in the post before, but I forgot.
*

*> If you re-read it, you'll notice that I always used the term "near-
*

*> zombies". I certainly do find the concept of zombies absurd. But "near-
*

*> zombies", as I was using the term, are physical SAS's that have a very
*

*> low measure of existence. That is, any random conscious entity would
*

*> have a low probability of finding itself to be that one relative to
*

*> others.
*

The near-zombie is physically identical to a normal human. I see

nothing to make them less absurd than true zombies.

*>[JM wrote]
*

*> > In the QS claim, that is. T and H can still label the branches.
*

*>
*

*> Okay as long as T and H are just labels, and there is no coin toss,
*

*> right?
*

*>
*

*> [JM wrote]
*

*> > The two experiments are actually rather different and I should
*

*> > have made the distiction clearer in my post. I assume you refer below to
*

*> > the case with two copying events.
*

*>
*

*> No, I thought I was very clear. All of my six "options" are alternative
*

*> ways of interpreting the original experiment, with a coin toss, and one
*

*> copying event if the toss lands heads. There may be a problem with
*

*> terminology here, because if you assume the MWI, then the coin toss is
*

*> a copy event. I do usually assume MWI, so I may have confused things.
*

A coin toss is not the same as a copy event in the MWI. In a copy

event, my view is that measure is doubled. In an MWI coin toss, measure

is conserved, as was explained in my other posts.

*> > > The copy of Jane who sees the blue card is a new person, who
*

*> > > was just "born" at the instant the copy was made, even though
*

*> > > she has all the same memories as the original.
*

*> >
*

*> > That is NOT my position, though of course I think 'consciousness
*

*> > flowing to and being distributed among continuations' is nonsense. I make
*

*> > no distinction between a copy and the original; 'identity' is not a
*

*> > fundamental concept. Each has the same amount of measure. For practical
*

*> > purposes the distinction is useful, however. It's just a matter of
*

*> > terminology in the practical use.
*

*>
*

*> Then I wish you would describe what you do believe. I am stymied in my
*

*> attempts to make sense of it, and I have tried.
*

*>
*

*> Consciousness "flowing" is perhaps a bad image. I mean nothing more than
*

*> that certain observer moments are related by some sort of identity function
*

*> to other observer moments.
*

I have described my view. I see no random identity functions.

The only thing that even resembles an identity function in my view is that

an extended computation may occur which gives rise to several observer

moments, but in that case all copies are equally related to the original.

Your so called identity function sounds like hidden variables,

related to the many-minds type.

[re: Tegmark's formula]

*> It's very simple, although I guess I haven't written it out in gory
*

*> detail before. Tegmark's formula "for any mutually exclusive and
*

*> collectively exhaustive set of possibilities Bi, the probability of
*

*> an event A is given by
*

*> P(A) = Sum over i [ P(A|Bi) P(Bi) ]
*

What is the meaning of this formula? If P(Bi) is the effective

probability that an observer would see Bi, then Bi is a label for all the

things an observer could see. 'A' is then a particular characteristic

that an observer-moment can have. But then P(A|Bi) is either zero or one,

which is not what I think you meant. Explain.

*> I want to computer P(H, t3). Let B1 be (H, t1) and B2 be (T, t1), then
*

*>
*

*> P(H, t3) = P(H,t3 | H,t1) P(H,t1) + P(H,t3 | T,t1) P(T,t1)
*

*>
*

*> We assume in this option (4) that P(H,t1) = P(T,t1) = 1/2.
*

I guess your 'identity function' is coming into play. Otherwise

the above makes no sense ... unless you are not using an MWI coin flip.

Without the MWI, either the H branch exists or the T branch but not both.

In any case you have changed the game when bringing time into the picture.

Am I now to understand that what Tegmark meant to say is that

P(A,t2) = Sum over i [ P(A,t2|Bi,t1) P(Bi,t1) ]

which not the same at all as what you quoted him as saying.

*> Now, I
*

*> maintain, and I really don't see how it could be otherwise, that
*

*>
*

*> P(H,t3 | H,t1) = 1 and
*

*> P(H,t3 | T,t1) = 0.
*

*>
*

*> Some people have claimed that these relations do not hold, but I don't
*

*> see what they could possibly be talking about. These statements say
*

*> nothing more than that my memories are consistent -- that I can expect
*

*> them to remain constant. If we abandon that, then all hope is lost!
*

*> I wish those who claim this (Hal?, Wei?) would explain themselves better.
*

I claim that P(H,t3|H,t1) does not mean anything useful. Neither

does P(H,t3 and H,t1).

Let me remark again that an extended computation can, and often

does, give rise to two such observer moments. In the sense that "you"

are an extended computation over time, your memories are consistent

over time. You can then say that "'P(H3,t3|H,t1)' = 1" *in the sense of*

the effective probability of an extended computation giving rise to (H3,t3)

at time t3 being 1 if it gave rise to (H1,t1) at time t1. But be

careful!!! This is a very weak sense of the word "you". If copying

occurs, the measure of the first moment does not equal the measure of the

second moment, so this quantity can NOT be inserted into the Tegmark-like

formula. Instead one must multiply each observation by its measure, then

divide by the total measure - the absolute SSA.

I would also say that English has an advantage over French in that

the same word 'you' must be used for both singular and plural in English.

*> [JM said]
*

*> > remember something non-random, that shouldn't cut your measure. If you
*

*> > remember a random bit, it cuts the total measure of each type of you in
*

*> > half but now there are twice as many types. By total measure I mean, as
*

*> > always, the number, so this is consistent with the SSA and leads to no
*

*> > zombies.
*

*>
*

*> But you said it yourself, if I remember a random bit, "it cuts the total
*

*> measure of each type of [me] in half." I don't know what you mean by
*

*> "type". I'd say it cuts the measure of me in half. That's exactly my
*

*> point.
*

The bit is either 0 or 1. If you don't know the value, let your

measure be M_d. If you know it's 0, you have measure M_0; if you know

it's 1, M_1. So M_d = M_0 + M_1. Suppose there are two guys, Dummy and

Ben. Dummy is ignorant and has M_d. Ben knows the bit and the versions

of him that saw 1 have M_1, the versions that saw 0 have M_0. A randomly

chosen observer is equally likely to be Dummy or to be one of the Bens.

*> [JM wrote]
*

*> > The Omega Point CRAP is disproven because the universe is open.
*

*> > (CRAP=causally retroactive anthropic principle)
*

*>
*

*> Cute! But I thing the jury is still out. I probably shouldn't use the
*

*> term "Omega-Point", since that refers to a specific theory of the infinite
*

*> future evolution of the universe. I sometimes use it to refer in general
*

*> to the infinite future in which life continues.
*

Life will continue but with decreasing measure. Still it seems

that you can make a refutable prediction: namely, that the universe we are

in is not optimised for us to be here, but is optimised to give you a long

lifetime. Basically you are saying that what the measure ratio (say,

between two universes) will be in the future affects the measure ratio in

the present. For example a universe in which lives decay polynomially

would be favored over one in which they decay exponentially.

- - - - - - -

Jacques Mallah (jqm1584.domain.name.hidden)

Graduate Student / Many Worlder / Devil's Advocate

"I know what no one else knows" - 'Runaway Train', Soul Asylum

My URL: http://pages.nyu.edu/~jqm1584/

Received on Mon Aug 23 1999 - 16:35:02 PDT

Date: Mon, 23 Aug 1999 19:21:30 -0400 (EDT)

On Wed, 18 Aug 1999, Christopher Maloney wrote:

The near-zombie is physically identical to a normal human. I see

nothing to make them less absurd than true zombies.

A coin toss is not the same as a copy event in the MWI. In a copy

event, my view is that measure is doubled. In an MWI coin toss, measure

is conserved, as was explained in my other posts.

I have described my view. I see no random identity functions.

The only thing that even resembles an identity function in my view is that

an extended computation may occur which gives rise to several observer

moments, but in that case all copies are equally related to the original.

Your so called identity function sounds like hidden variables,

related to the many-minds type.

[re: Tegmark's formula]

What is the meaning of this formula? If P(Bi) is the effective

probability that an observer would see Bi, then Bi is a label for all the

things an observer could see. 'A' is then a particular characteristic

that an observer-moment can have. But then P(A|Bi) is either zero or one,

which is not what I think you meant. Explain.

I guess your 'identity function' is coming into play. Otherwise

the above makes no sense ... unless you are not using an MWI coin flip.

Without the MWI, either the H branch exists or the T branch but not both.

In any case you have changed the game when bringing time into the picture.

Am I now to understand that what Tegmark meant to say is that

P(A,t2) = Sum over i [ P(A,t2|Bi,t1) P(Bi,t1) ]

which not the same at all as what you quoted him as saying.

I claim that P(H,t3|H,t1) does not mean anything useful. Neither

does P(H,t3 and H,t1).

Let me remark again that an extended computation can, and often

does, give rise to two such observer moments. In the sense that "you"

are an extended computation over time, your memories are consistent

over time. You can then say that "'P(H3,t3|H,t1)' = 1" *in the sense of*

the effective probability of an extended computation giving rise to (H3,t3)

at time t3 being 1 if it gave rise to (H1,t1) at time t1. But be

careful!!! This is a very weak sense of the word "you". If copying

occurs, the measure of the first moment does not equal the measure of the

second moment, so this quantity can NOT be inserted into the Tegmark-like

formula. Instead one must multiply each observation by its measure, then

divide by the total measure - the absolute SSA.

I would also say that English has an advantage over French in that

the same word 'you' must be used for both singular and plural in English.

The bit is either 0 or 1. If you don't know the value, let your

measure be M_d. If you know it's 0, you have measure M_0; if you know

it's 1, M_1. So M_d = M_0 + M_1. Suppose there are two guys, Dummy and

Ben. Dummy is ignorant and has M_d. Ben knows the bit and the versions

of him that saw 1 have M_1, the versions that saw 0 have M_0. A randomly

chosen observer is equally likely to be Dummy or to be one of the Bens.

Life will continue but with decreasing measure. Still it seems

that you can make a refutable prediction: namely, that the universe we are

in is not optimised for us to be here, but is optimised to give you a long

lifetime. Basically you are saying that what the measure ratio (say,

between two universes) will be in the future affects the measure ratio in

the present. For example a universe in which lives decay polynomially

would be favored over one in which they decay exponentially.

- - - - - - -

Jacques Mallah (jqm1584.domain.name.hidden)

Graduate Student / Many Worlder / Devil's Advocate

"I know what no one else knows" - 'Runaway Train', Soul Asylum

My URL: http://pages.nyu.edu/~jqm1584/

Received on Mon Aug 23 1999 - 16:35:02 PDT

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