E Classical Mechanics 1273
Taking the time derivatives of Eqs. (E-28) and (E-29), we can write the Lagran-
gian asL
1
2
m 1(
.
xc−m 2.
x
M) 2
+
1
2
m 2(
.
xc+m 1.
x
M) 2
−V(x)1
2
m 1⎡
⎣
.
xc
2
− 2 m 2.
xc.
x
M+
(
m 2.
x
M) 2 ⎤
⎦+^1
2
m 2⎡
⎣
.
x
2
c+^2 m^1.
xc.
x
M+
(
m 1.
x
M) 2 ⎤
⎦−V(x)(E-30)
The terms containing.
xc.
xcancel, so thatL
1
2
(m 1 +m 2 ).
x
2
c+1
2
m 1 m 2 (m 1 +m 2 )
M^2.
x
2
−V(x)1
2
M
.
x^2 c+1
2
μ.
x^2 −V(x) (E-31)Thereduced massof the pair of particles is denoted byμand is defined byμm 1 m 2
m 1 +m 2m 1 m 2
M(E-32)
The kinetic energy isK
1
2
M
.
x
2
c+1
2
μ.
x
2
(E-33)Since the variables are separated in the Lagrangian, we obtain separate equations of
motion forxcandx:M
d.
xc
dt∂L
∂xc0 (E-34)
μd.
x
dt∂L
∂x−
dV
dx(E-35)
If there are no external forces, the center of mass of the two particles moves like a
particle of massMthat has no forces acting on it, while the relative coordinate changes
like the motion of a particle of massμmoving at a distancexfrom a fixed origin
and subject to the potential energyV(x). The motion of two particles moving in the
xdirection has been separated into two one-body problems. The motion in theyand
zdirections is completely analogous, so that in three dimensions we can assert:The
fictitious particle of massμmoves around the origin of its coordinate in the same way
that particle 1 moves relative to particle 2, while the center of mass moves like a free
particle of massM.
The separation is the same in Hamiltonian mechanics. In three dimensions, the
kinetic energy isK
1
2
M(
.
x
2
c+.
y
2
c+.
z
2
c)+1
2
μ(.
x
2
+.
y
2
+.
z
2
) (E-36)