
#1
Jan1708, 11:32 AM

P: 67

Can somebody explain me how does the derivation for <K>=<delta(U)> goes under the viral theorem?




#2
Jan1708, 02:16 PM

P: 443

1.They define something called called 'scalar moment of inertia around the origin', I. It looks similar to moment of inertia around an axis in mechanics but not quite.
2. The first derivative G = (1/2) dI/dt is called 'virial'. They prove that the first derivative of the virial, dG/dt, i.e. the second time derivative of I , depends on the total kinetic energy T of the system and the potential energy U, when the potential energy is a power law ~ r^n: dG/dt = 2T  nV (n=1 for gravity) 3. They take the time average <...> of the above equation and claim that after long time the system 'virializes' so that the time average <dG/dt> = 0 which gives you an equation between the time averages of kinetic energy and potential energy: <T> = (n/2) <V> 4. If we assume the system is 'virialized' i. e. in a stationary state, equilibrium, so that the total kinetic and potential energy do not change with time, the time averages will equal the energies at any time. When that theorem is applied in astronomy, we measure/observe only the total kinetic and potential energies at a given time. We don't have access to time averaged values because the times involved are millions/billions of years but we assume the system is virialized already so the energy values equal the time averages. All that (except point 4) can be found here http://en.wikipedia.org/wiki/Virial_theorem 



#3
Jan1808, 09:19 AM

P: 67

Thanks!



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