From: ethan@emx.cc.utexas.edu (Ethan T. Vishniac)
Newsgroups: sci.physics
Subject: Re: Big Bang Alternates
Date: 21 Jan 1993 10:00:40 -0600
Organization: The University of Texas at Austin, Austin, Texas
NNTP-Posting-Host: emx.cc.utexas.edu

Tom Van Flandern  writes: (>> denotes my previous comments)


>> This still seems wrong.  The ISM of the galaxy won't produce a blackbody
>> unless it is optically thick at the relevant wavelengths.  It isn't.

>     I am aware of this argument, but don't understand how it applies here.
>I understand that an optically-thin medium would ordinarily radiate more
>than it would absorb and so not be thermalized, but starlight sets a 3-
>degree minimum to the temperature of the ISM, keeping it in equilibrium. So
>it must be a blackbody, nes pas?
>
>     For example, in Harwitt's "Astrophysical Concepts" (2nd edition, p.
>286), he mentions two conditions for blackbody radiation: (1) thermal
>equilibrium; and (2) "for this equilibrium to become established, we
>require that the assembly of absorbing particles at constant temperature be
>large enough so that a succession of absorption and re-emission steps occur
>before energy escapes from the surface of the assembly."  That's a little
>vague, and usually would require optical thickness, but I think an
>optically-thin ISM surrounding a local superbubble (to provide a "surface")
>continually heated by the radiation of starlight meets these criteria.
>
>     If it doesn't, I would be glad to receive a clear explanation of my
>error.

I won't swear that this is a sufficiently clear explanation because I'm
not sure I understand what your difficulty here is.  The ISM is detectable
both in emission and absorption.  There is no sign of a significant 
component at 3 degrees in absorption.  The ISM cannot be emitting a blackbody
at 3 degrees *even if it were at that temperature* without fulfilling
condition number 2 of Harwitt.  At best it would supply a diluted
blackbody.  Even that is impossible since the gas is not a perfect emitter
or absorber and the actual radiation from an optically thin gas at 3 degrees
would not be a blackbody.  Condition number 2 cited by Harwitt is equivalent
to saying that distant photons in the relevant energy range would be
absorbed before reaching us, i.e. that the galaxy is optically thick at these
wavelengths. (In fact, quite thick since the quadrapole anisotropy in the background
is a few times 10^(-5) including radiation clearly associated with our own
galaxy). If the radiation is emitted locally then it is 
coming from so close that not even the vertical structure of the galactic disk  
is making an impression.  In our vicinity that's a scale height of about
100 pc.)  We would therefore conclude that nothing at greater distances is
visible at these wavelengths.  It doesn't matter whether you ascribe the
radiation to the ISM or a bubble around us.  I note that there are a large
number of radio sources with optical identifications visible at these
wavelengths, only some of which are quasars.

I don't really want to stretch this thread out, but  I don't think there
is any merit to your position.


-- 
"Quis tamen tale studium, quo ad primam omnium rerum causam evehimur,
tamquam inutile aut contemnendum detractare ac deprimere ausit?"-Bridel
Ethan T. Vishniac, Dept. of Astronomy, The University of Texas at Austin
Austin, Texas, 78712                   ethan@astro.as.utexas.edu
Newsgroups: sci.physics