LinkTest 2005-08-16 20:33:20 2005-08-17 00:19:48 Definition % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here direction cosine electric charge Euler angle action simple harmonic oscillator Euler 123 Sequence Euler 212 Sequence Euler 121 Sequence Euler 131 Sequence insulator I am trying to kill the database. Need to see if Mysql server has gone away error reappears!!!!!!!! So far so good, but who knows. It just keeps going and going .... Perl Perl, P is for PeRl!.!.!.! 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 \begin{equation} \vec{f_i} = \sum_j \vec{F_ij} \: \: \: where i \ne j \end{equation} So when we go to sum up all the forces we notice 'force pairing' such that \begin{equation} \sum_i \vec{f_i} = \sum_i \sum_{j>i} \vec{F_{ij}} + \vec{F_{ji}} \end{equation} Plugging this into the virial of Claussius yields \begin{equation} \sum_i \vec{f_i} \cdot \vec{r_i} = \sum_i \sum_{j>i} (\vec{F_{ij}} \cdot \vec{r_i} + \vec{F_{ji}} \cdot \vec{r_j}) \end{equation} Now we need to take the gradient of the potential energy to get the force. This is a tedious calculation which can be found here (insert link) \begin{equation} \vec{F_{ij}} = -\nabla_i \frac{G m_i m_j}{r_{ij}} = - \frac{G m_i m_j(\vec{r_i} - \vec{r_j})}{r_{ij}^3} \end{equation} Inserting this into (13) gives \begin{equation} \sum_i \vec{f_i} \cdot \vec{r_i} = \sum_i \sum_{j>i} (- \frac{G m_i m_j(\vec{r_i} - \vec{r_j}) \cdot \vec{r_i}}{r_{ij}^3} - \frac{G m_i m_j(\vec{r_j} - \vec{r_i}) \cdot \vec{r_j}}{r_{ij}^3} \end{equation} The two keys in understanding the above equation is to note that \begin{equation} r_{ij} = \vert \vec{r_i} - \vec{r_j} \vert \end{equation} and that \begin{equation} \vert \vec{r_i} - \vec{r_j} \vert^2 = (\vec{r_i} - \vec{r_j}) \cdot (\vec{r_i} - \vec{r_j}) \end{equation} So now we add the numerators of (15) to get \begin{equation} (\vec{r_i} - \vec{r_j}) \cdot \vec{r_i} - (\vec{r_j} - \vec{r_i}) \cdot \vec{r_j} = (\vec{r_i} \cdot \vec{r_i} - \vec{r_i} \cdot \vec{r_j} + \vec{r_j} \cdot \vec{r_j} - \vec{r_i} \cdot \vec{r_j}) \end{equation} Next we expand (17) to see that it is equal to (18) and this cancels a power of 2 in the denominator of (15) to finally yeild the expression for potential energy \begin{equation} \sum_i \sum_{j>i} - \frac{G m_i m_j}{r_{ij}} = -U \end{equation} So going back to (8) we see \begin{equation} \frac{d^2I}{dt^2} = 4T - 2U \end{equation} From the energy equation of the system, E = T - U and the key to finish up the virial theorem is to note that the momemt of inertia does not change on average with time ("After one dynamical timescale, the time derivative of I is constant so the second derivative is zero \begin{equation} \frac{d^2I}{dt^2} = 0 \end{equation} This leads to \begin{equation} T = \frac{1}{2}U \end{equation} plugging this into the energy equation gives us the result of the virial theorem which states that the total energy of a stationary system (no significant dynamical evolution) is one-half the potential energy of the system. \begin{equation} E = \frac{1}{2}U - U = - \frac{1}{2}U \end{equation}