16 Classical mechanics
m 2 ̈r 2 =U′(|r 1 −r 2 |)r 1 −r 2
|r 1 −r 2 |, (1.4.34)
a more natural set of coordinates can be chosen. To this end, we introduce thecenter-
of-massandrelativecoordinates defined by
R=
m 1 r 1 +m 2 r 2
M, r=r 1 −r 2 , (1.4.35)respectively, whereM=m 1 +m 2. The inverse of this transformation is
r 1 =R+m 2
M
r, r 2 =R−m 1
M
r. (1.4.36)When eqn. (1.4.36) is substituted into eqn. (1.4.33), the Lagrangianbecomes
L=
1
2
MR ̇^2 +
1
2
μr ̇^2 −U(|r|), (1.4.37)whereμ=m 1 m 2 /Mis known as thereduced mass. Since∂L/∂R= 0, we see that
the center-of-mass coordinate is cyclic, and only the relative coordinate needs to be
considered. After elimination of the center-of-mass, the reduced Lagrangian isL=
μr ̇^2 / 2 −U(|r|) which gives a simple equation of motion
μ ̈r=−U′(|r|)
r
|r|. (1.4.38)
Alternatively, one could transformrinto spherical-polar coordinates as described in
the previous subsection, and solve the resulting one-dimensional equation for a single
particle of massμmoving in a central potentialU(r).
We hope that the reader is now convinced of the elegance and simplicity of the
Lagrangian formulation of classical mechanics. Primarily, it offers a framework in which
the equations of motion can be obtained in any set of coordinates. Beyond this, in
Section 1.8, we will see how it connects with a more general concept,the action
extremization principle, which allows the Euler–Lagrange equations to be obtained
by extremization of a particular mathematical form known as theclassical action, an
object of fundamental importance in quantum statistical mechanics to be explored in
Chapter 12.
1.5 Legendre transforms
We shall next derive the Hamiltonian formulation of classical mechanics. Before we
can do so, we need to introduce the concept of aLegendre transform.
Consider a simple functionf(x) of a single variablex. Suppose we wish to express
f(x) in terms of a new variables, wheresandxare related by
s=f′(x)≡g(x) (1.5.1)withf′(x) = df/dx. Can we determinef(x) at a pointx 0 given onlys 0 =f′(x 0 ) =
g(x 0 )? The answer to this question, of course, is no. The reason, as Fig. 1.6 makes