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Rayleigh-Jeans or not Rayleigh-Jeans?

Let $J(\nu,T)$ be the power radiated by a black-body at frequency $\nu$ and temperature $T$. It is always possible to define in a conventional way a Rayleigh-Jeans temperature \ensuremath{T_\ensuremath{\mathrm{}}\ifthenelse{\equal{RJ}{}}{}{^\ensuremath{\mathrm{RJ}}}} such as

...henelse{\equal{RJ}{}}{}{^\ensuremath{\mathrm{RJ}}}}= J(\nu,T).
\end{displaymath} (1)

It happens that \ensuremath{T_\ensuremath{\mathrm{}}\ifthenelse{\equal{RJ}{}}{}{^\ensuremath{\mathrm{RJ}}}} has a direct physical meaning, i.e. $\ensuremath{T_\ensuremath{\mathrm{}}\ifthenelse{\equal{RJ}{}}{}{^\ensuremath{\mathrm{RJ}}}}= T$, when $h\nu<<\ensuremath{k_\ensuremath{\mathrm{bolt}}}\,T$. However, \ensuremath{T_\ensuremath{\mathrm{}}\ifthenelse{\equal{RJ}{}}{}{^\ensuremath{\mathrm{RJ}}}} can always be defined (even when $h\nu >> \ensuremath{k_\ensuremath{\mathrm{bolt}}}\,T$: \ensuremath{T_\ensuremath{\mathrm{}}\ifthenelse{\equal{RJ}{}}{}{^\ensuremath{\mathrm{RJ}}}} just loses its physical meaning. The advantage of doing this is that all the equations of calibrations are considerably simpler to write and to implement in source code using the conventional Rayleigh-Jeans temperature than the black-body formula. The only constraint is to remember to transform physical temperature (in particular hot and cold load temperature) in conventional Rayleigh-Jeans temperature for all the input parameters. This is what is done in GILDAS/TELCAL, in ATM and thus in this memo.

Gildas manager 2014-07-01