From: toms@fcs260c2.ncifcrf.gov (Tom Schneider)
Newsgroups: bionet.info-theory
Subject: New Questions
Date: 4 Sep 92 23:24:24 GMT
Organization: Frederick Cancer Research and Development Center

Hi Chowder folks,

I received some email from Kenneth Tolman (he goes by the name Tom,
tolman@mindblow.cs.utah.edu, he has a funny .plan, check it out!) who had some
questions.  He allowed me to post them here, where we perhaps can hold a
discussion.

> Actually, I am doing research for my thesis, which is an unusual and
> somewhat radically different approach to information.  In your papers you
> make some unusual comments (which I am not sure I agree with).  (But for that
> matter, I am not sure I agree with my own theory, and certainly do not
> agree with some of the claims of Szilard and others)
> 
> Let me briefly explain my peculiar viewpoint....
> 
> I am attempting to understand the complexity (or information content) of
> the universe as a whole.
> Or in other words, I am attempting to understand information change within
> a closed system.....  Can it change, or can it not change?
> 
> According to this theory,  information can change, and it can only change
> through nondeterminism.  For in a reversible deterministic system, all
> that is here now was dictated by what was there before, and so on.
> Basically, information changes when the system makes a "choice"
> (which you have associated with a required dissipation, but I am not sure
> you are using the word choice to mean what I do)
> (Also, I think that jumping to kTln(2) is a little hasty, for I believe the
> foundations of statistical mechanics should fall out of the foundations
> of information, not the other way around)
> 
> Anyways, thats enough for now...
> Tom

First, it is important to have a clear definition of information, and also to
have a clear idea of the system one is applying it to.  In the case of a
molecular machine, it is not deterministic because the machine is surrounded by
a thermal bath which the machine itself cannot predict.  Interestingly, the
entire bath (including the machine) could be deterministic!  But a little
ribosome cannot predict the impacts of surrounding water molecules, so must act
in such a way that it deals with them as random noise.

The Second Law makes it rather clear that the entropy of the universe as a
whole in increasing, so the total information has to be going down toward
``heat death''.  This does not prevent local pockets such as us from having
entropy decreases, so plenty of interesting things can still happen!

I do agree that there is an information gain of a system when that system makes
choices.  In other words, information is a decrease of the entropy of a
system.  This necessitates a heat dissipation to the surroundings, according to
the Second Law, dS >= dq/T.  If you look at Schneider.edmm you will see that,
of the two ways to get to the formula you allude to, one starts with dS >=
dq/T.  You may have it the other way if you prefer the work of Jaynes and
Tribus.  I'm not sure why you considered it hasty.  I'm not trying to rebuild
physics.  dS >= dq/T is pretty well established, so I started there.  One ends
up with the EQUATION Emin = kTln(2) (or E >= kTln(2) joules per bit).  So this
equation IS the Second law (under isothermal conditions).  For some reason this
was missed in the literature.  Always write it as an equation, and keep those
units.

...
> Anyways, let me ask here more specifically, you write:
> "The energy of the photon must be dissipated for the molecule to choose
> between [bathorhodopsin] and the original unexcited state, just as a 
> ribosome must dissipate energy to bind to its binding site"
> Now, here are some questions:
> 1) What is the energy of the reaction moving in either direction?
> Is it exothermic to produce [bathorhodopsin], or endothermic, or is it
> neutral?
> 2) What does dissipate energy mean, exactly?  I presume that the photon
> overcomes the [activation] energy, and then it falls into equilibrium upon
> dissipation.  Is this manifest by spitting out some black body energy
> photons, or by translating it into vibrational energy, or giving it
> out as kinetic energy?  Do I have the right idea at all here?
> 3) What are the relative probabilities of going to the [bathorhodopsin] form
> versus going to the original unexcited state?
> 
> Thanks a lot for any comments you might have,
> 
> Tom (my self chosen name)

1) To go from the 'before' high energy state to the low energy 'after' state,
clearly the molecular machine must loose energy to the surroundings.  Hence it
is exothermic.  The priming step (for eg. ribosomes) goes the other way and so
is endothermic.  Beware that these are microscopic steps.  In a large solution
of water RNA and ribosomes (+tRNA-f-met), the individual ribosomes will be
binding to the sites and coming off them even when the entire solution is at
macroscopic equilibrium.  So when the individual ribosome binds, a tiny bit of
heat moves away into the bulk; when it comes off, a little bit of heat had to
have gone into the local region of the ribosome.

2) A wonderful book is:

@book{Atkins1984,
author = "P. W. Atkins",
title = "The Second Law",
publisher = "W. H. Freeman and Co.",
address = "N. Y.",
year = "1984"}

I strongly recommend it.  Atkins shows clearly with lots of nice pictures how
EVERYTHING runs by dissipation.  Dissipation is the spreading out of energy or
matter.  Basically, when things are bouncing around by thermal motion, they
tend to get more spread out or disordered.  (Do you hear the grumbling in the
background?  Some people disagree with that statement for reasons I still don't
understand!  Or is it my stomach rumbling?)

> I presume that the photon overcomes the [activation] energy, and then it
> falls into equilibrium upon dissipation.

No, the photon is ABSORBED by rhodopsin.  At this point the molecule is
bouncing around a lot more than before.  If it doesn't do something quickly, it
will just lose that energy to the surroundings.  The molecule settles down only
upon dissipation.

> Is this manifest by spitting out some black body energy
> photons, or by translating it into vibrational energy, or giving it
> out as kinetic energy?  

It could be any of these.  The vibrational energy will, however, not last long,
as the rhodopsin will hit the water molecules near by and most of the energy
will travel away as heat.  The impacts are roughly at the speed of sound (!) in
the medium, but since they keep going forwards and backwards, the energy of
heat does not travel at the speed of sound.

> 3) What are the relative probabilities of going to the [bathorhodopsin] form
> versus going to the original unexcited state?

To bathorhodopsin:  70%, back to rhodopsin:  30%.     ;-)

  Tom "letting the mystery lie" Schneider
  National Cancer Institute
  Laboratory of Mathematical Biology
  Frederick, Maryland  21702-1201
  toms@ncifcrf.gov
Newsgroups: bionet.info-theory