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Article 2450 of comp.ai.philosophy:
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>From: weemba@libra.wistar.upenn.edu (Matthew P Wiener)
Newsgroups: comp.ai.philosophy
Subject: Re: Ignore QM and be happy
Message-ID: <61196@netnews.upenn.edu>
Date: 30 Dec 91 20:50:02 GMT
References: <61072@netnews.upenn.edu> <1991Dec27.213037.4038@cs.ucf.edu>
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Reply-To: weemba@libra.wistar.upenn.edu (Matthew P Wiener)
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In-reply-to: long@next3 (Richard Long)

I found the complete reference: J C Eccles FRS "Do mental events cause
neural events analogously to the probability fields of quantum mechanics?",
PROC R SOC LOND B 227, 411-428 (1986).

In article <1991Dec27.213037.4038@cs.ucf.edu>, long@next3 (Richard Long) writes:
>> 	presynaptic membrane width:	5e-9  m
>> 	synapse emission time:		1e-3  s
>> 	neurotransmitter vesicle mass:	3e-17 g

>> So a vesicle's worth of released neurotransmitter has momentum roughly
>> 3e-17*5e-9/1e-3=1.5e-22 gm/s.  Using this figure for order of magnitude
>> in momentum uncertainty, we use the uncertainty relations to identify a
>> positional uncertainty (6.6e-31 gmm/s)/(1.5e-22gm/s)=4e-9 m, about the
>> width of the presynaptic membrane.

>I have two complaints about your back-of-the-envelope calculation.
>First, you have calculated the momentum of the vesicle from the
>presynaptic membrane width and the synapse emission time.  But the
>synapse emission time is not the time it takes a vesicle to move a
>distance equal to the membrane width, it is the time required for the
>visicle to merge with the presynaptic membrane, thus expelling
>neurotransmitter (I may be mistaken about this definition in detail,
>but not, I think, in broad outline).

So let's get down to details.  The vesicle membrane itself stays behind
and neurotransmitter itself is emitted.  The mass of 10k acetylcholine
molecules is about 10% of the vesicle mass, but the size of the vesicle
is about 10x the membrane, so the relevant distance is on average 5
times larger than I used above.  The emission time across the membrane
is, says Eccles, citing Katz et al, "many tenths of a millisecond"--say
half of the 1e-3s above?  I get the same momentum!

>						  Your calculated
>velocity is then (5e-9m/1e-3s)=5e-6 m/s!  Since delta x=h/(delta p)
>[it should be further divided by 4pi, but that only decreases it by
>an order of magnitude anyway],

The Heisenberg relations are an inequality.  The minimal uncertainty is
achieved only for a Gaussian.  So the back-of-my-envelope tells me that 
this buys me back that order of magnitude.

>			        no wonder your position uncertainty
>was so large, since your velocities are unreasonably small, which
>brings me to my second complain, which is one of principle.  Mr.
>Wizard tells us that we can have a 10kg bowling ball with an
>uncertainty in position of 1 kilometer.  Want to see how?  Just give
>is a velocity of 6.6e-38 m/s!  then, by your calculation, delta x is
>(6.6e-31 gmm/s)/(6.6e-38 m/s*1e 4 g)=1e 3 m.  What's wrong with this
>argument?

Easy.  Since when did you have a bowling ball with such a precisely known
velocity?  A rough upper bound on momentum uncertainty is the momentum
itself  (Ie, if a back-of-the-envelope calculation gives a momentum p,
than if non-bogus, you know the momentum is maybe 2p but not 10p, ie,
the uncertainty is of order of magnitude at most p.)  This is what I
used above.  It won't work for bowling balls.
-- 
-Matthew P Wiener (weemba@libra.wistar.upenn.edu)


