Rigid body motion 39CQ S
POθn rr’Fig. 1.12 Rotation of the vectorrtor′about an axisn.a massm. We shall assume that the motion occurs in thexyplane. Let the Cartesian
positions of the two atoms ber 1 andr 2 and let the molecule be subject to a potential
of the formU(r 1 −r 2 ). The Lagrangian for the molecule can be written as
L=1
2
mr ̇^21 +1
2
m ̇r^22 −U(r 1 −r 2 ). (1.11.4)For such a problem, it is useful to transform to center-of-massR= (r 1 +r 2 )/2 and
relativer=r 1 −r 2 coordinates, in terms of which the Lagrangian becomes
L=1
2
MR ̇^2 +
1
2
μr ̇^2 −U(r), (1.11.5)whereM= 2mandμ=m/2 are the total and reduced masses, respectively. Note
that in these coordinates, the center-of-mass has an equation of motion of the form
MR ̈= 0, which is the equation of motion of a free particle. As we have already seen,
this means that the center-of-mass velocityR ̇is a constant. According to the principle
ofGalilean relativity, the physics of the system must be the same in a fixed coordinate
frame as in a coordinate frame moving with constant velocity. Thus,we may transform
to a coordinate system that moves with the molecule. Such a coordinate frame is called
abody-fixed frame. The origin of the body-fixed frame lies at the center of mass of the
molecule, and in this frame the coordinates of the two atomsr 1 =r/2 andr 2 =−r/2.
It is, therefore, clear that only the motion of the relative coordinaterneeds to be
considered. Note that we may use the body-fixed frame even if thecenter-of-mass
velocity is not constant in order to separate the rotational and translational kinetic
energies of the rigid body. In the body-fixed frame, the Lagrangian of eqn. (1.11.5)
becomes
L=1
2
μr ̇^2 −U(r). (1.11.6)In a two-dimensional space, the relative coordinateris the two-component vector
r= (x,y). However, if the distance between the two atoms is fixed at a valued, then