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Article 2424 of comp.ai.philosophy:
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>From: long@next3 (Richard Long)
Newsgroups: comp.ai.philosophy
Subject: Re: Ignore QM and be happy
Message-ID: <1991Dec27.213037.4038@cs.ucf.edu>
Date: 27 Dec 91 21:30:37 GMT
References: <61072@netnews.upenn.edu>
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In article <61072@netnews.upenn.edu> weemba@libra.wistar.upenn.edu (Matthew P  
Wiener) writes:
> In article <1991Dec25.061455.29709@news.media.mit.edu>, minsky@media (Marvin  
Minsky) 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.
> 
> The counterintuitiveness of quantum effects for such a large object has
> an OBVIOUS EXPLANATION[caps mine].  The synapse transition is unusually fast,  
about
> a thousand times faster than, say, vesicle transport down the axon.  If
> I remember correctly, that slower speed is roughly the same as cellular
> component thermal diffusion.  Somebody fix the slug in me...in an ideal
> gas, 300K vesicles would have thermal velocites ~~ 1m/s.
> -- 
> -Matthew P Wiener (weemba@libra.wistar.upenn.edu)

	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).  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], 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?

Richard Long: long@next1.acme.ucf.edu


