These formulae were collected from P. Schwering's thesis, from the IRAS explanatory supplements, and from various IRAS reviews

Since there are a number of different items here, I give you a rough navigation list

- Luminosities derived from the IRAS band
- Temperatures measured from the long wavelength part
- Masses of dust

A table summing up the IRAS waveband characteristics can be found here while this will give you the system response tables and graphs.

Luminosities at 12, 25, 60 and 100 µm are given, by definition by the formula:

For a numerical application, let's have:

this results in:

The far infrared flux and luminosity from 40 to 120 µm are defined by:

flux

luminosity and log(3.96 10^{5}) = 5.60

One can also make the ratio of the 12 µm to FIR flux (as in Ryter et al. 1987, A&A 186, 312) which expresses as: and log(19.841)=1.30

Similarly one can define the Mid-IR luminosity by summing over the 12 and 25 µm bands, or the total IR luminosity by summing over the 4 IRAS bands.

which numerically transforms to:

with log(1.61 10^{6}) = 6.21

which numerically gives:

with log(3.14 10^{5}) = 5.50

Alternatively, Sanders & Mirabel (1996, ARA&A 34, 749) have defined an 8 to 1000µm luminosity by:

One can then work toward deriving a temperature and a dust mass provided one has the spectral properties of the dust and that one assumes only one component contribute emission in the 60 and 100 µm, and that furthermore, the IRAS flux densities can be considered as monochromatic (which is far from beeing true if you examine the bandwidths). Typical spectral properties come from Hildenbrandt (1983).

This gives with n=1-1.5

The dust temperature is then obtained by

, with (1.67)^{4.5}=9.96 and (1.67)^{4.0}=7.78

Once the temperature is known, the mass can be derived, provided on is in the optically thin regime (usually the case at 60 and 100 µm). The dust mass is given by:

with the "black-body" function

after frequency normalization this gives a semi-numerical formula, that depends on the spectral index *n*:

then, for different values of n (1.5 or 1) and using either the 100 or 60 µm flux, one has:

*n*=1.5

with log(0.959) = -1.82 10^{-2}

with log(9.63 10^{-2}) = -1. 02

*n*=1.0

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