Chapter 6
The Lagrangian Method
Copyright 2007 by David Morin, [email protected]
(draft version)
In this chapter, we’re going to learn about a whole new way of looking at things. Consider
the system of a mass on the end of a spring. We can analyze this, of course, by using
F
=
ma
to write down
m
¨
x
=

kx
. The solutions to this equation are sinusoidal functions, as we well
know. We can, however, figure things out by using another method which doesn’t explicitly
use
F
=
ma
. In many (in fact, probably most) physical situations, this new method is far
superior to using
F
=
ma
.
You will soon discover this for yourself when you tackle the
problems and exercises for this chapter. We will present our new method by first stating its
rules (without any justification) and showing that they somehow end up magically giving
the correct answer. We will then give the method proper justification.
6.1
The EulerLagrange equations
Here is the procedure. Consider the following seemingly silly combination of the kinetic and
potential energies (
T
and
V
, respectively),
L
≡
T

V.
(6.1)
This is called the
Lagrangian
. Yes, there is a minus sign in the definition (a plus sign would
simply give the total energy). In the problem of a mass on the end of a spring,
T
=
m
˙
x
2
/
2
and
V
=
kx
2
/
2, so we have
L
=
1
2
m
˙
x
2

1
2
kx
2
.
(6.2)
Now write
d
dt
∂L
∂
˙
x
¶
=
∂L
∂x
.
(6.3)
Don’t worry, we’ll show you in Section 6.2 where this comes from. This equation is called
the
EulerLagrange (EL) equation
. For the problem at hand, we have
∂L/∂
˙
x
=
m
˙
x
and
∂L/∂x
=

kx
(see Appendix B for the definition of a partial derivative), so eq. (6.3) gives
m
¨
x
=

kx,
(6.4)
which is exactly the result obtained by using
F
=
ma
.
An equation such as eq. (6.4),
which is derived from the EulerLagrange equation, is called an
equation of motion
.
1
If the
1
The term “equation of motion” is a little ambiguous.
It is understood to refer to the secondorder
differential equation satisfied by
x
, and not the actual equation for
x
as a function of
t
, namely
x
(
t
) =
A
cos(
ωt
+
φ
) in this problem, which is obtained by integrating the equation of motion twice.
VI1
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VI2
CHAPTER 6.
THE LAGRANGIAN METHOD
problem involves more than one coordinate, as most problems do, we just have to apply eq.
(6.3) to each coordinate. We will obtain as many equations as there are coordinates. Each
equation may very well involve many of the coordinates (see the example below, where both
equations involve both
x
and
θ
).
At this point, you may be thinking, “That was a nice little trick, but we just got lucky
in the spring problem. The procedure won’t work in a more general situation.” Well, let’s
see. How about if we consider the more general problem of a particle moving in an arbitrary
potential
V
(
x
) (we’ll stick to one dimension for now). The Lagrangian is then
L
=
1
2
m
˙
x
2

V
(
x
)
,
(6.5)
and the EulerLagrange equation, eq. (6.3), gives
m
¨
x
=

dV
dx
.
(6.6)
But

dV/dx
is the force on the particle.
So we see that eqs. (6.1) and (6.3) together
say exactly the same thing that
F
=
ma
says, when using a Cartesian coordinate in one
dimension (but this result is in fact quite general, as we’ll see in Section 6.4). Note that
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 Fall '11
 Kennefick
 mechanics, Mass, Richard Feynman, Coordinate system, Polar coordinate system, Lagrangian mechanics, Noether's theorem

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