36 Classical mechanics
the trajectory, the conditionσ(r) = 0 will be satisfied. The latter condition defines
a surface on which the motion described by eqns. (1.10.1) must remain. This surface
is called thesurface of constraint. The quantity∇σ(r) is a vector that is orthogonal
to the surface at each pointr. Thus, the second equation (1.10.1) expresses the fact
that the velocity must also lie in the surface of constraint, hence it must be perpen-
dicular to∇σ(r). Of the two force terms appearing in eqns. (1.10.1), the first is an
“unconstrained” force which, alone, would allow the particle to driftoff of the surface
of constraint. The second term must, then, correct for this tendency. If the particle
starts from rest, this second term exactly removes the component of the force perpen-
dicular to the surface of constraint as illustrated in Fig. 1.10. This minimal projection
Constraint surfaceFF||Fig. 1.10Representation of the minimal force projection embodied inGauss’s principle of
least constraint.
of the force, first conceived by Karl Friedrich Gauss (1777-1855), is known asGauss’s
principle of least constraint(Gauss, 1829). The component of the force perpendicular
to the surface is
F⊥(r) = [n(r)·F(r)]n(r), (1.10.2)
wheren(r) is a unit vector perpendicular to the surface atr;nis given by
n(r) =∇σ(r)
|∇σ(r)|. (1.10.3)
Thus, the component of the force parallel to the surface is
F‖(r) =F(r)−F⊥(r) =F(r)−[n(r)·F(r)]n(r). (1.10.4)If the particle is not at rest, the projection of the force cannot lieentirely in the surface
of constraint. Rather, there must be an additional component ofthe projection which
can project any free motion of the particle directed off the surface of constraint. This
additional term must sense the curvature of the surface in orderto affect the required
projection; it must also be a minimal projection perpendicular to thesurface.
In order to show that Gauss’s principle is consistent with the Lagrangian formula-
tion of the constraint problem and find the additional projection when the particle’s
velocity is not zero, we make use of the second of eqns. (1.10.1) anddifferentiate it
once with respect to time. This yields:
∇σ· ̈r+∇∇σ··r ̇r ̇= 0, (1.10.5)