In article <1994Jan5.170448.3190 at aber.ac.uk> hrs at aber.ac.uk
(Herbert Martin Sauro) writes:
>What happens if a thermometer is put in a vacuum that is bathed in
>photons? Does the thermometer register a reading or not? (I think it
>must do).
A thermometer, allowed to equilibrate in a vacuum, would indeed
register some temperature or other. Suppose that our thermometer (be it
a thermocouple or whatever) were hot when placed in the vacuum. What
does that mean? I mean by "hot" that the various quantum states of the
device are populated in such a way that it would be in equilibrium with a
conventional heat bath with a high temperature. Such a thermometer, if
placed in a "cold" vacuum (I'll return to this term in a moment) would
lose energy through various radiative processes until it reached radiative
equilibrium with the ambient electromagnetic field. At this point, the
states of the device will be populated in a certain way and we can
associate with this distribution a temperature PROVIDED the distribution
is of the usual Boltzmann shape. (If not, the electromagnetic
field is not "thermal", i.e. it is a nonequilibrium field and the
concept of temperature becomes problematic. In such a situation,
thermometers of different constructions will register different
temperatures. This is well known and explained in many introductory
treatments of thermodynamics.) I can now say what I meant by a "cold"
vacuum: The temperature of the vacuum is the temperature of a
thermometer in radiative equilibrium with it. By a "cold" vacuum I
meant one with a low temperature compared to the initial temperature of
our measuring device.
Marc R. Roussel
mroussel at alchemy.chem.utoronto.ca
