virial theorem
VirialTheorem
2004-11-19 01:01:34
2004-11-19 01:08:31
Definition
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
We start with the moment of inertia about the origin for the system of particles, which is defined as
\begin{equation}\label{first}
I = \sum_i m_i {r_i}^2
\end{equation}
Differentiate using the chain rule. Note that a vector dotted into itself yields its magnitude square.
\begin{equation}
\vec{r_i} \cdot \vec{r_i} = r_i^2
\end{equation}
This lets us make the connection that
\begin{equation}
I = \sum_i m_i {r_i}^2 = \sum_i m_i (\vec{r_i} \cdot \vec{r_i})
\end{equation}
So after differentiating we get
\begin{equation}
\frac{dI}{ dt} = \sum_i m_i \frac{\vec{dr_i}}{dt} \cdot \vec{r_i} + \vec{r_i} \cdot \frac{\vec{dr_i}}{dt}
\end{equation}
Differentiating again yields
\begin{equation}
\frac{d^2I}{dt^2} = \sum_i m_i \frac{d^2\vec{r_i}}{dt^2} \cdot \vec{r_i} + \frac{\vec{dr_i}}{dt} \cdot \frac{\vec{dr_i}}{dt} + \frac{\vec{dr_i}}{dt} \cdot \frac{\vec{dr_i}}{dt} + \vec{r_i} \cdot \frac{d^2\vec{r_i}}{dt^2}
\end{equation}
In short form
\begin{equation}
\frac{d^2I}{dt^2} = 2\, \sum_i (m_i \dot{\vec{r_i}} \cdot \dot{\vec{r_i}} + m_i\vec{r_i} \cdot \ddot{\vec{r_i}})
\end{equation}
When dealing with a system of particles we found that the Kinetic Energy associated with a system of particles was
\begin{equation}
T = \frac{1}{2} \sum_i m_i \dot{\vec{r_i}} \cdot \dot{\vec{r_i}}
\end{equation}
Plugging in T into (6) gives us
\begin{equation}
\frac{d^2I}{dt^2} = 4T + 2 \, \sum_i m_i\vec{r_i} \cdot \ddot{\vec{r_i}}
\end{equation}
Next we need to tackle the $2 \, \sum_i m_i\vec{r_i} \cdot \ddot{\vec{r_i}}$ term. We first bring in the potential energy through its connection with force. This part is named the Virial of Claussius.
\begin{equation}
\vec{f_i} = m_i \ddot{\vec{r_i}} = \nabla_i U
\end{equation}
This gives us the equality
\begin{equation}
2 \, \sum_i m_i\vec{r_i} \cdot \ddot{\vec{r_i}} = \sum_i \vec{r_i} \cdot \nabla_i U
\end{equation}
Now due to Newton's 3rd law that states for every action there is an opposite and equal reaction we have the forces on the ith particle in our system given by