GeneralizedCoordinatesForFreeMotion 2008-07-17 16:36:23 2008-07-17 20:21:10 Topic % 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 The differential equations for the motion of a particle under any forces when we use rectangular coordinates are known from Newston's laws of motion $$ m \ddot{x} = F_x $$ $$ m \ddot{y} = F_y $$ $$ m \ddot{z} = F_z $$ where $F_x, F_y, F_z$ are the components of the actual forces on the particle resolved parallel to each of the fixed rectangular axes, or rather their equivalents $m \ddot{x}, m \ddot{y}, m \ddot{z}$, are called the \emph{effective forces} on the particle. They are of course a set of forces mechanically equivalent to the actual forces acting on the particle. The equations of motion of the particle in terms of any other system of coordinates are easily obtained. Let $q_1, q_2, q_3$, be the coordinates in question. The appropriate formulas for transformation of coordinates express $x,y,z$ in terms of $q_1,q_2,q_3$. $$ x = f_1(q_1,q_2,q_3), \,\,\,\, y = f_2(q_1,q_2,q_3), \,\,\,\, z = f_3(q_1,q_2,q_3) $$ For the component velocity $\dot{x}$ we have $$ \dot{x} = \frac{\partial x}{\partial q_1} \dot{q_1} + \frac{\partial x}{\partial q_2} \dot{q_2} + \frac{\partial x}{\partial q_3} \dot{q_3} $$ and $\dot{x},\dot{y},\dot{z}$ are explicit functions of $q_1,q_2,q_3,\dot{q_1},\dot{q_2},\dot{q_3}$ linear and homogeneous in terms of $\dot{q_1},\dot{q_2},\dot{q_3}$. We may note in passing that it follows from this fact that $\dot{x}^2,\dot{y}^2,\dot{z}^2$ are homogeneous quadratic functions of $\dot{q_1},\dot{q_2},\dot{q_3}$. Obviously $$ \frac{\partial \dot{x}}{\partial \dot{q_1}} = \frac{\partial x}{\partial q_1} $$ and since $$ \frac{d}{dt} \frac{\partial x}{\partial q_1} = \frac{\partial^2 x}{\partial q_1^2} \dot{q_1} + \frac{\partial^2 x}{\partial q_2 \partial q_1} \dot{q_2} + \frac{\partial^2 x}{\partial q_3 \partial q_1} \dot{q_3} $$ and $$ \frac{\partial \dot{x}}{\partial q_1} = \frac{\partial^2 x}{\partial q_1^2} \dot{q_1} + \frac{\partial^2 x}{\partial q_1 \partial q_2} \dot{q_2} + \frac{\partial^2 x}{\partial q_1 \partial q_3} \dot{q_3} $$ $$ \frac{d}{dt} \frac{\partial x}{\partial q_1} = \frac{\partial \dot{x}}{\partial q_1} $$ Let us now find an expression for the work $\delta_{q_1}W$ done by the effective forces when the coordinate $q_1$ is changed by an infinitesimal amount $\delta q_1$ without changing $q_2$ or $q_3$. If $\delta x, \delta y, \delta z$ are changes thus produced in $x,y,z$, obviously from the definition of work $$ \delta_{q_1}W = m\left [ \ddot{x} \delta x + \ddot{y} \delta y + \ddot{z} \delta z \right ]$$ if expressed in rectangular coordinates. We need, however, to express $\delta_{q_1}W$ in terms of our coordinates $q_1, q_2, q_3$. $$ \delta_{q_1}W = m\left [ \ddot{x} \frac{\partial x}{\partial q_1} + \ddot{y} \frac{\partial y}{\partial q_1} + \ddot{z} \frac{\partial z}{\partial q_1} \right ]\delta q_1$$ Now $$\ddot{x} \frac{\partial x}{\partial q_1} = \frac{d}{dt} \left ( \dot{x} \frac{\partial x}{\partial q_1} \right ) - \dot{x} \frac{d}{dt} \frac{\partial x}{\partial q_1} $$ but from earlier definitions $$ \frac{\partial x}{\partial q_1} = \frac{\partial \dot{x}}{\partial \dot{q_1}} \,\,\,\,\, and \,\,\,\,\, \frac{d}{dt} \frac{\partial x}{\partial q_1} = \frac{\partial \dot{x}}{\partial q_1} $$ Hence $$ \ddot{x} \frac{\partial x}{\partial q_1} = \frac{d}{dt} \left ( \dot{x} \frac{\partial \dot{x}}{\partial \dot{q_1}} \right ) - \dot{x} \frac{\partial \dot{x}}{\partial q_1} = \frac{d}{dt} \frac{\partial}{\partial \dot{q_1}} \left ( \frac{\dot{x}^2}{2} \right) - \frac{\partial}{\partial q_1} \left ( \frac{\dot{x}^2}{2} \right ) $$ and therefore \begin{equation} \delta_{q_1} W = \left [ \frac{d}{dt} \frac{\partial T}{\partial \dot{q_1}} - \frac{\partial T}{\partial q_1} \right ] \delta q_1 \end{equation} where $$ T = \frac{m}{2} \left [ \dot{x}^2 + \dot{y}^2 + \dot{z}^2 \right ] $$ and is the kinetic energy of the particle. To get our differential equation we have only to write the second member of (1) equatl to the work done by the actual forces when $q_1$ is changed by $\delta q_1$. If we represent the work in question by $Q_1 \delta q_1$, our equation is \begin{equation} \frac{d}{dt} \frac{\partial T}{\partial \dot{q_1}} - \frac{\partial T}{\partial q_1} = Q_1 \end{equation} and of course we get such an equation for every coordinate. Even though we derived this differential equation for a single particle in free motion, it is the same for a systems of particles, except the kinetic energy is for all the particles in the system, which brings us to Lagrange's equations \begin{equation} Q_i = \frac{d}{dt} \left ( \frac{ \partial T}{\partial \dot{q_i}} \right ) - \frac{\partial T}{\partial q_i} \end{equation}