X-Message-Number: 3360
From: Brian Wowk <>
Date: Thu, 27 Oct 94 01:51:28 CDT
Subject: SCI.CRYONICS Brain Scanning, reply to Yvan Bozzonetti

Yvan Bozzonetti:
 
>I come to the number of photon estimate:
>I take as an example a one dm^3 cube scanned with a 0.1 micrometer 
>resolution. On one cube face there will be 10^12 squares .1 micron on a 
>side. The information is encoded in tiny phase shifts on each wave, there 
>must be one wave (one photon) for each square. Doing that on three 
>perpendicular sides produces both, all the informations requested to define 
>the content of each .1 micrometer cell and a redundant data for result 
>checking. 3 x 10^12 have been used, even if one in ten only was not 
>absorbed (this is plainly wrong), 3 x 10^13 photons would travel in the 
>brain. If each one has a 1 keV energy, the total deposited energy is at 
>most:
(calculation deleted)
 
        1 keV x-rays will be completely absorbed by photoelectric
interactions with the K shells of oxygen, carbon, and nitrogen within
the first 1 mm of brain tissue.  To penetrate a 100mm wide brain you  
will need at least 50keV x-rays, and even then 90% will be absorbed
before reaching the other side.  >>>> This absorption will occur even
if your x-rays are so monochromatic that they have coherence length
of kilometers. <<<<  The photoelectric effect doesn't care about
coherence.  (*Please* consult any elementary health physics or
nuclear physics textbook to learn about the photoelectric effect.
You will also find there information about partial absorption windows 
that exist near K edges.)
 
        There is an even more serious problem with your calculation.
The number of photons you have estimated is at least 6 orders of
magnitude too small to make a hologram.  A hologram consists of an  
interference pattern recorded on a photographic plate or other
recording medium.  To encode the three dimensional structure of 
a brain to 0.1 micron resolution, the interference pattern must
contain 10^18 bits of information.  How are you going to get an
interference pattern with 10^18 bits of information by recording
only 10^12 photons?
 
        Here is yet another problem: How are you going to stop 
50keV photons in your detector in a short enough distance to record
nanometer-scale interference patterns?  Even if you could magically
stop all 50keV x-ray photons within a nanometer, how would you
detect only the primary ionization event and distinguish it from
the shower of secondary scattered electrons (which would spread up
to 100 microns away from the initial event).  The only solution
would be to enlarge the interfernce pattern by moving the recording
film hundreds of meters away, and using film hundreds of meters wide!
 
        Give it up, Yvan.  Holography with 50keV x-rays is
psuedoscience!
 
--- Brian Wowk

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