Ignorable Coordinates and Constants of the Motion
Review: Lagrange's Equations: d/dt[T/qk’] - T/qk = Qk
is a generalized equation of motion! [Note that in rectangular coordinates, T/qk = 0 since T
does not depend on the coordinates, only on the velocities.]
Another form of Lagrange's equations can be written IF all the forces are conservative: Qk = V/qk, AND V is not a function of velocities, only positions, so V/qk’ = 0. In this case we
define what is called the LAGRANGIAN: L = T - V .
Then the Lagrange's equation of motion are: d/dt[L/qk’] - L/qk = 0 .
The generalized momentum we defined as: pj = T/qj’ = ΣAkq’ + Bk .
Kinetic Energy:
T = Σi{½mi(x’² + y’² + z’²)} = ΣkΣl{½Akqk’q’} + ΣkBkqk’ + To
where
Ak = Σi{mi[(xi/qk)(xi/q) + (yi/qk)(yi/q) + (zi/qk)(zi/q)]}
Bk = Σi{mi[(xi/qk)(xi/t) + (yi/qk)(yi/t) + (zi/qk)(zi/t)]}
To = Σi{½mi[(xi/t)² + (yi/t)² + (zi/t)²]} .
Generalized Force: Qk = Σi[Fix(xi/qk) + Fiy(yi/qk) + Fiz(zi/qk)]
----------------Ignorable Coordinates: momentum and coordinates
If we can write an L (that is, if we have only conservative forces), and
if qk does not appear explicitly in either T or V, then: L/qk = 0 so
d/dt(L/qk’) - L/qk = 0 becomes
d/dt(L/qk’) = 0 , and so
L/qk’ = constant;
and since V is not a function of velocity (the qk’) , we get:
L/qk’ = T/qk’ =
pk = constant.
This gives a (constant) value for pk, and since pk = ΣAkq’ + Bk = T/qk’ , this expression
provides an equation that can be used to express qk’ in terms of the other coordinates of the
problem, and hence to eliminate it from the other equations of motion. Recall also that qk is not
there either. Therefore, we call this coordinate an ignorable coordinate.
-----------Example:
In spherical coordinates, T = ½ m(r’2 + r2r2sin2()’2). If the forces are radial, then the
potential energies depend on r but not  or . In this case,  would be an ignorable coordinate.
-------------
Time and energy:
What about time? Consider a fixed coordinate system: xi = xi(q1, q2, ... , qn)
but not x1 = x1(q1, q2, ... , qn, t) so that xi/t = 0  Bk = 0 , To = 0.
Then the kinetic energy becomes just: T = ΣjΣ [½Ajqj’q’] .
Now consider:
Σkqk’(T/qk’) = Σkqk’(/qk’)[ΣjΣ{½Ajqj’q’}
= Σkqk’[Σ{½Akq’} + Σj{½Ajkqj’}] = 2T ,
since when j=k in the first sum (Σj) we get the first term above, and when =k in the second sum
(Σ) we get the second term; but both terms are the full expression for the kinetic energy!
Therefore, we have, with L = T - V:
Σkqk’(T/qk’) - L = 2T - (T - V) = T + V = E .
Further recall that T/qk’ = L/qk’, since V does not depend on qk’. Thus we can write:
dE/dt = d/dt[Σk{qk’(L/qk’)} - L]
= Σk{qk’’(L/qk’) + qk’[d/dt(L/qk’)]} - dL/dt .
Now
dL/dt = (1/dt) [Σk{(L/qk’)dqk’ + (L/qk)dqk} + (L/t)dt]
= Σk{(L/qk’)qk’’ + (L/qk)qk’} + L/t .
Continuing with dE/dt, we get:
dE/dt = Σk{qk’’(L/qk’) + qk’[d/dt(L/qk’)] - [(L/qk’)qk’’ + (L/qk)qk’]} - L/t
term 1
term 2
term 3
term 4
term 5
Note that term 1 and term 3 cancel. Note also that term 2 and term 4 express Lagrange's equation
and so equal zero. Thus we have the result:
dE/dt = -L/t .
Note the following relations:
when L/qk = 0  pk = constant (position and momentum are linked),
when L/t = 0  E = constant (time and energy are linked).
EXAMPLE of ignorable coordinate, constrained coordinate and free coordinate: the
SPHERICAL PENDULUM (with θ being the angle with the vertical such that θ=0o is straight up,
θ=90o being horizontal; φ is the angle measured in the horizontal plane) :
T = ½m(r’² + r²θ’² + r²sin²θ φ’²) and
V = m g r cos(θ) .
Since gravity is a conservative force and doesn’t depend on the velocities, we can write a
Lagrangian:
L = T - V = ½m(r’² + r²θ’² + r²sin²θ φ’²) - m g r cos(θ) .
If we don't worry about forces of constraint, our equations of motion are:
d/dt(L/qk’) - L/qk = 0.
If we DO worry about forces of constraint, then:
δW = -m g δz - τ δr = -m g(cosθ δr - sinθ r δθ) - τ δr (spherical coord)
= (-m g cosθ - τ)δr + (m g r sinθ)δθ + 0 δφ = Qrδr + Qθδθ + Qφδφ .
1. Note that φ does not appear in either T or in any of the Qk's (although φ’ does appear in T).
This means that φ is an ignorable coordinate.
2. Note that there is a constraint on the coordinate, r, due to the tension in the string, τ (or the
contact force of the bowl).
3. Thus the only degree of freedom is that of θ.
a) The equation of motion for φ is then:
d/dt(T/φ’) - T/φ = Qφ ,
or d/dt(mr²sin²θ φ’) - 0 = 0,
or:
mr²sin²θ φ’ = pφ = constant, or φ’ = pφ / mr²sin²θ
where pφ is determined from the initial conditions. We can use this expression for φ’ wherever
we encounter it in the other equations.
b) The equation of motion for r is then:
d/dt(T/r’) - T/r = Qr ,
or
d/dt(mr’) - [mrθ’² + mrsin²θ φ’²] = -m g cos(θ) - τ ,
or, since r = constant and r’ = 0, we have for τ, the force of constraint:
τ = -m g cos(θ) + mrθ’² + mr sin²θ [pφ /mr²sin²θ]² ,
or
τ = -m g cos(θ) + mrθ’² + pφ²/[mr3sin²θ]
and once we know θ, we can solve for τ, the force of constraint.
c) The equation of motion for θ is then:
d/dt(T/θ’) - T/θ = Qθ ,
or
d/dt(mr²θ’) - [mr²sinθ cosθ φ’²] = mgr sin(θ) ,
or
[2mrr’θ’ + mr²θ’’] - [m r²sinθ cosθ {pφ / [mr²sin²θ]}²] = mgr sin(θ) ;
but due to the constraint on r we have r’ = 0; there is an m in every term that cancels out; and
dividing through by r² gives:
θ’’ = [cosθ pφ²]/[m²r4sin3θ] + g sinθ / r .
Limiting Cases:
1. If θ = constant, then the φ equation gives:
φ’ = pφ / [m r²sin²θ] = constant = ;
and the θ’’ equation (with θ’’ = 0) gives:
0 = sinθ cosθ φ’2 + g sinθ / r ,
or
 = φ’ = [-g / (r cosθ)]½

½π < θ < (3π/2) ,
which means that the mass can only be below the horizontal for this type of motion.
2. If pφ = 0, then the θ equation gives:
θ’’ = g sinθ / r ,
which is the equation for the pendulum.
IN-CLASS HOMEWORK
Show that the expression above for the first limiting case of θ = constant corresponds to what
should be expected using the rectangular form for Newton's Second Law.