PlanetPhysics/Generalized Coordinates for Free Motion

The [[../DifferentialEquations/|differential equations]] for the [[../CosmologicalConstant/|motion]] of a [[../Particle/|particle]] under any [[../Thrust/|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 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 [[../SimilarityAndAnalogousSystemsDynamicAdjointnessAndTopologicalEquivalence/|system]] of coordinates are easily obtained.

Let $q_1,q_2,q_3$, be the coordinates in question. The appropriate [[../Formula/|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/|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 [[../Bijective/|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/|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

${\delta }_{q_1}W=m[\ddot{x}\delta x+\ddot{y}\delta y+\ddot{z}\delta z]$

if expressed in rectangular coordinates. We need, however, to express ${\delta }_{q_1}W$ in terms of our coordinates $q_1,q_2,q_3$.

${\delta }_{q_1}W=m[\ddot{x}\frac{\partial x}{\partial q_1}+\ddot{y}\frac{\partial y}{\partial q_1}+\ddot{z}\frac{\partial z}{\partial q_1}]\delta q_1$

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

${\delta }_{q_1}W=[\frac{d}{dt}\frac{\partial T}{\partial \dot{q_1}}-\frac{\partial T}{\partial q_1}]\delta q_1$

where

$T=\frac{m}{2}[{\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2]$

and is the [[../KineticEnergy/|kinetic energy]] of the particle.

To get our differential equation we have only to write the second member of (1) equal 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

$\frac{d}{dt}\frac{\partial T}{\partial \dot{q_1}}-\frac{\partial T}{\partial q_1}=Q_1$

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 [[../Lagrangian/|Lagrange's equations]]

$Q_i=\frac{d}{dt}\left(\frac{\partial T}{\partial \dot{q_i}}\right)-\frac{\partial T}{\partial q_i}$

In any concrete problem, $T$ must be expressed in terms of $q_1,q_2,q_3$, and their time derivatives before we can form the expression for the work done by the effective forces. $Q_1\delta q_1,Q_2\delta q_2,Q_3\delta q_3$, the work done by the actual forces, must be obtained from direct examination of the problem.