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When I was reading information about QSOs on a website, it said that Seyfert Galaxies emit less energy than Quasars and QSOs.

Why do Seyfert Galaxies emit less energy than quasars, and how is this energy measured?

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  • $\begingroup$ Welcome to Astronomy SE. Which website were you reading? $\endgroup$ – James K Mar 1 '17 at 22:01
  • $\begingroup$ I was ready from this website: space.com/17262-quasar-definition.html $\endgroup$ – A. Kalyan Mar 2 '17 at 1:44
  • $\begingroup$ The part about Seyfert Galaxies is at the bottom $\endgroup$ – A. Kalyan Mar 2 '17 at 1:46
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The linked website seems to summarise generally accepted theory.

Seyfert galaxies appear less energetic than quasars or blazars. The energy is measured by observations of their brightness in the sky (at different wavelengths of light) combined with their distance (inferred from their red-shift). There are also Radio-galaxies

We think that active galactic nuclei don't release the same amount of energy in all directions, instead they have a "beam". They also have an accretion disc, the inner parts of which are extremely hot If we happen to be looking directly down the beam, they appear exceptionally powerful, and we call them "blazars" If we see them off centre we can see into the extremely hot central part of the accretion disk and they will be a quasar or (if less powerful) a Seyfert galaxy, and if we see them side-on they appear less powerful, but the two lobes of high-energy gas and particles emitted along the beam will be visible, and we will observe a two-lobed radio galaxy.

See these related answers: Quasars and SMBH and Difference between quasar and Active Galactic Nuclei?

http://www.astronomy.ohio-state.edu/~ryden/ast162_9/notes37.html

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  • $\begingroup$ What are the other theories? I was researching quasars, specifically, and I came across the fact that the huge red shift was unprecedented for as the quasars were being observed with no noticeable change in position. Gravitational red shift was also debunked, so how exactly would the energy be measured, knowing that they are very bright objects, but not knowing the cause of the massive energy release, without an inferred "red shift". $\endgroup$ – A. Kalyan Mar 7 '17 at 4:32

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