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Universal quality of virial theorem

by giann_tee
Tags: quality, theorem, universal, virial
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giann_tee
#1
Mar23-08, 02:05 PM
P: 133
"Distances among galaxies in clusters on average are no greater than their diameter".

Luminosity, color, and other qualities are used to obtain distances - and these values are connected to masses that are observed by motions. If a system is in equilibrium then the theorem is:
2T+U=0
Mv^2/2 - GM^2/2R = 0
M=2Rv^2/G
If rotation curve represents a known function in form of a third Kepler's law for a concentration of mass near the core, then one set of motions is given by that function of mass and distance. Further away from galactic center, the function behind curve is not known, then mass and speed and not connected with known law.
However the theorem of virial still works no matter the scales? ...and the clusters are in equilibrium regardless of rotation curves?
The mass/luminosity ration M/L grows steadily per factor of scale, 20-30 per 50kpc and 200 per 1Mpc.
Luminosity can't be diminishing at large distances? I'm presuming that it is not diminishing at those rates, so the galaxies and clusters are rotating faster at large scales?
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oldman
#2
Mar24-08, 02:31 AM
P: 622
Quote Quote by giann_tee View Post
"Distances among galaxies in clusters on average are no greater than their diameter".

If a system is in equilibrium then the theorem is:
2T+U=0
Mv^2/2 - GM^2/2R = 0
M=2Rv^2/G....

.... the theorem of virial still works no matter the scales? ...and the clusters are in equilibrium regardless of rotation curves?
Yes, this is so. The individual galaxies in a cluster are treated as if they were particles in virial equilibrium. As for rotation curves --- the relation 2T +U = 0 applies to orbiting masses as well, for example to the Earth in orbit around the sun. So it also applies to individual objects in orbit around the centre of any spherically symmetric mass distribution.


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