View Full Version : how are OOM calculations done?

2006-Aug-30, 01:35 AM
How do astronomers determine the distribution of energy emitted (for example) by the nuclear jet of the galaxy M87....across the electromagnetic spectrum?

I'm interested to understand both the total energy emitted, and the relative amount of energy, in each waveband.

2006-Aug-30, 01:41 AM
have a look at this thread (http://www.bautforum.com/showthread.php?t=46270).

2006-Sep-02, 01:35 AM
Each waveband has its own methods, but they are all, essentially, photon counting.

It's clearest in the x-ray (and gamma) bands - the detectors record each photon.

In the UV and optical, CCD and photomultipliers are also, essentially, counters of photons.

Of course, no detector has a 100% efficiency, so calibration is essential (there is a vast literature on this subject; suffice it to say that astronomers rarely make mistakes wrt calibration).

But between detector and astronomical object (the M87 jet, say) there are many veils - the Earth's atmosphere, the interplanetary medium, the inter-stellar medium, ... Through decades of patient, often exceedingly tedious, work, astronomers think they now have a good handle on how to estimate the number of photons 'lost' on their way from source to detector.

Having done all this work, the last part - converting estimated number of photons emitted to energy emitted, per band, is quite easy - simply multiply by the energy of the photons!

There remain, however, one or two niggles, especially for things like 'relativistic jets'. The Sun, Earth, Moon, the stars, galaxies, ... all radiate their electromagnetic energy essentially isotropically (equally in all directions).

But relativistic jets are different - they almost certainly do NOT radiate isotropically (or even close to it), and even if they do (in their own frame), they are certainly not seen as isotropic emitters, by distant observers, moving relativistically.