generalized coordinates for free motion

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 ¨x = Fx

m ¨y = Fy

m ¨z = Fz

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 mẍ,mÿ,mz, 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 system of coordinates are easily obtained.

Let q₁,q₂,q₃, be the coordinates in question. The appropriate formulas for transformation of coordinates express x,y,z in terms of q₁,q₂,q₃.

x = f1(q1,q2,q3), y = f2(q1,q2,q3), z = f3(q1,q2,q3)

For the component velocity ẋ we have

∂x ∂x ∂x x˙= ---q˙1 + ----˙q2 + ---q˙3 ∂q1 ∂q2 ∂q3

and ẋ,ẏ,ż are explicit functions of q₁,q₂,q₃,q₁,q₂,q₃ linear and homogeneous in terms of q₁,q₂,q₃.

We may note in passing that it follows from this fact that ẋ²,ẏ²,ż² are homogeneous quadratic functions of q₁,q₂,q₃.

Obviously

∂x˙- -∂x- ∂q˙1 = ∂q1

and since

d--∂x- ∂2x- -∂2x--- --∂2x-- dt∂q = ∂q2 ˙q1 + ∂q ∂q q˙2 + ∂q ∂q ˙q3 1 1 2 1 3 1

and

∂-˙x- ∂2x- --∂2x-- -∂2x--- ∂q1 = ∂q21q˙1 + ∂q1 ∂q2q˙2 + ∂q1∂q3q˙3

-d ∂x-- ∂x˙- dt ∂q1 = ∂q1

Let us now find an expression for the work δ_q₁W done by the effective forces when the coordinate q₁ is changed by an infinitesimal amount δq₁ without changing q₂ or q₃. If δx,δy,δz are changes thus produced in x,y,z, obviously from the definition of work

δq1W = m [¨xδx + ¨yδy + z¨δz]

if expressed in rectangular coordinates. We need, however, to express δ_q₁W in terms of our coordinates q₁,q₂,q₃.

[ ] ∂x ∂y ∂z δq1W = m ¨x ∂q--+ ¨y∂q--+ ¨z ∂q-- δq1 1 1 1

Now

∂x d ( ∂x ) d ∂x ¨x ----= -- x˙---- − ˙x-- ---- ∂q1 dt ∂q1 dt ∂q1

but from earlier definitions

∂x--= -∂ ˙x and -d ∂x--= -∂ ˙x ∂q1 ∂ ˙q1 dt ∂q1 ∂q1

Hence

( ) ( 2) ( 2) ¨x ∂x--= -d ˙x ∂-˙x − ˙x-∂ ˙x-= d- ∂--- x˙- − -∂-- ˙x-- ∂q1 dt ∂q˙1 ∂q1 dt ∂ ˙q1 2 ∂q1 2

and therefore

[ ] d ∂T ∂T δq1W = dt∂-˙q-− ∂q-- δq1 1 1

(1)

where

m-[ 2 2 2] T = 2 x˙ + y˙ + z˙

and is the 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 q₁ is changed by δq₁.

If we represent the work in question by Q₁δq₁, our equation is

d ∂T ∂T ------− ----= Q1 dt∂q˙1 ∂q1

(2)

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

( ) d- ∂T- ∂T- Qi = dt ∂q˙ − ∂q i i

(3)

In any concrete problem, T must be expressed in terms of q₁,q₂,q₃, and their time derivatives before we can form the expression for the work done by the effective forces. Q₁δq₁,Q₂δq₂,Q₃δq₃, the work done by the actual forces, must be obtained from direct examination of the problem.

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