# Virial ratio behaviour

I think I didn't understand something about virial theorem for an $$N$$-body system, for instance the behaviour of virial ratio $$T/\Omega$$, with $$T$$ kinetic energy and $$\Omega$$ gravitational potential energy. As far as I understood, if the system is stationary, than virial ratio is $$0.5$$. This means on the other hand that the system is not expanding nor collapsing. But the behaviour of virial ratio is something like the plot in this question. Here the system starts from $$T/\Omega < 0.5$$, so it starts collapsing and than reach a maximum. I would expect this maximum to be at $$T/\Omega \sim 0.5$$, since once the collapse stops the system is not accelerating nor decelerating, but it's about $$0.75$$. If the system is stationary at $$0.5$$ how can that maximum (and the next minimum too) be explained? I don't understand how virial ratio has a single value since it actually oscillates until converging to a specific value.

• What does $T$ and $\Omega$ mean? The question becomes clearer if that is explained. Apr 29 at 14:41
• @AndersSandberg Presumably those are the kinetic and potential energies. Apr 29 at 15:56

The ratio $$T/\Omega$$ tells you about the acceleration of the system - or more specifically, the second derivative of its moment of inertia - it does not tell you about the velocity.
If the system collapses because it has $$T/\Omega<0.5$$, then when it reaches $$T/\Omega=0.5$$ it stops accelerating. That doesn't mean it stops collapsing. It overshoots in the same way that if you compress a spring it goes beyond the equilibrium position before bouncing back once the deceleration becomes large.
• Yes I understood the system can be considered as an oscillating spring, but I was confusing something about the acceleration/deceleration part, you're right. Ok, so supposing the system is collapsing, after $T/\Omega$ reaches $0.5$ from negative values, the system decelerate (so acceleration sign changes) until it stops collapsing when $T/\Omega$ reach the peak right?