<p>The ratio <span class="math-container">$T/\Omega$</span> tells you about the <em>acceleration</em> of the system - or more specifically, the second derivative of its moment of inertia - it does not tell you about the velocity.</p>
<p>If the system collapses because it has <span class="math-container">$T/\Omega&lt;0.5$</span>, then when it reaches <span class="math-container">$T/\Omega=0.5$</span> it stops <em>accelerating</em>. 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 <em>deceleration</em> becomes large.</p>
<p>NB: I answered the question (of which your question is a subset) that you linked to in the same way.</p>

CC BY-SA 4.0
2021-04-29T15:21:24.663Z
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?
Zebx
CC BY-SA 4.0
2021-04-29T14:30:39.337Z
2021-04-29T14:30:39.337Z
ProfRob
<p>I think I didn't understand something about virial theorem for an <span class="math-container">$N$</span>-body system, for instance the behaviour of virial ratio <span class="math-container">$T/\Omega$</span>, with <span class="math-container">$T$</span> kinetic energy and <span class="math-container">$\Omega$</span> gravitational potential energy. As far as I understood, if the system is stationary, than virial ratio is <span class="math-container">$0.5$</span>. 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 <a href="https://physics.stackexchange.com/q/362633/">this question</a>. Here the system starts from <span class="math-container">$T/\Omega &lt; 0.5$</span>, so it starts  collapsing and than reach a maximum. I would expect this maximum to be at <span class="math-container">$T/\Omega \sim 0.5$</span>, since once the collapse stops the system is not accelerating nor decelerating, but it's about <span class="math-container">$0.75$</span>. If the system is stationary at <span class="math-container">$0.5$</span> 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.</p>

CC BY-SA 4.0
2021-04-29T14:41:22.453Z
What does $T$ and $\Omega$ mean? The question becomes clearer if that is explained.
Anders Sandberg
CC BY-SA 4.0
2021-04-29T15:56:16.430Z
@AndersSandberg Presumably those are the kinetic and potential energies.
Peter Erwin
<astrophysics><star-systems><newtonian-gravity>
Virial ratio behaviour
Zebx
# Virial ratio behaviour

(Asked by: username_0 on 2021-04-29)

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](https://physics.stackexchange.com/q/362633/). 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.

**Tags:** astrophysics, star-systems, newtonian-gravity

> **Comments:**
> **username_1**: What does $T$ and $\Omega$ mean? The question becomes clearer if that is explained.  
> **username_2**: @AndersSandberg Presumably those are the kinetic and potential energies.  

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## 1 answers

### Answer by username_3 on 2021-04-29. Score: 5 | Accepted answer

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.

NB: I answered the question (of which your question is a subset) that you linked to in the same way.

> **Comments:**
> **username_0**: 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?  

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