Mij q j + Vij q j

’ i=

0= ¨ (1.72)

‚ qi

dt ™ ‚q j=1

for each i.

1.3.4 Continuous Systems

The action principle can be extended to ¬eld theories and to continuum me-

chanics. Here one has a continuous in¬nity of dynamical degrees of freedom,

either one for each point in space and time or one for each point in the mate-

rial, but the extension of the variational derivative to functions of more than

one variable should possess no conceptual di¬culties.

Suppose we are given an action S depending on a ¬eld •(xµ ) and its ¬rst

derivatives

‚•

•µ ≡ µ . (1.73)

‚x

18 CHAPTER 1. CALCULUS OF VARIATIONS

Here xµ , µ = 0, 1, . . . , d, are the coordinates of d + 1 dimensional space-time.

It is traditional to take x0 ≡ t and the other coordinates spacelike. Suppose

further that

S = L dt = L(•, •µ ) dd+1 x, (1.74)

where L is the Lagrangian density, in terms of which

L ddx,

L= (1.75)

where the integral is over the space coordinates. Now

‚L ‚L

dd+1 x

δS = δ•(x) + δ(•µ (x))

‚•(x) ‚•µ (x)

‚L ‚ ‚L

dd+1 x.

’µ

= δ•(x) (1.76)

‚•(x) ‚x ‚•µ (x)

In going from the ¬rst line to the second, we have observed that

‚

δ(•µ (x)) = δ•(x) (1.77)

‚xµ

and used the divergence theorem,

‚Aµ n+1

Aµ nµ dS,

d x= (1.78)

‚xµ

„¦ ‚„¦

where „¦ is some space-time region and ‚„¦ its boundary, to integrate by

parts. Here dS is the element of area on the boundary, and nµ the outward

normal. As before, we take δ• to vanish on the boundary, and hence there

is no boundary contribution to variation of S. The result is that

δS ‚L ‚ ‚L

’µ

= , (1.79)

δ•(x) ‚•(x) ‚x ‚•µ (x)

and the equation of motion comes from setting this to zero. Note that a sum

over the repeated coordinate index µ is implied. In practice, however, it is

easier not to use this formula, but instead do the variation explicitly as in

the following examples.

1.3. LAGRANGIAN MECHANICS 19

The Vibrating string

The simplest continuous dynamical system is the vibrating string. We de-

scribe the string displacement by y(x, t).

y(x,t)

0 L

Let us suppose that the string has ¬xed ends, a mass per unit length of ρ, and

is under tension T . If we assume only small displacements from equilibrium,

the Lagrangian is

1 1

L

2

dx ρy 2 ’ T y .

L= ™ (1.80)

2 2

0

The variation of the action is

L

dtdx {ρyδ y ’ T y δy }

δS = ™™

0

L

dtdx {δy(x, t) (’ρ¨ + T y )} .

= y (1.81)

0

To reach the second line we have integrated by parts, and, because the ends

are ¬xed, and therefore δy = 0 at x = 0 and L, there is no boundary term.

Requiring that δS = 0 for all allowed variations δy then gives the equation

of motion

‚2y ‚2y

ρ 2 ’ T 2 = 0. (1.82)

‚t ‚x

This is the wave equation for waves with speed c = T /ρ. Observe that

from (1.81) we can read o¬ the functional derivative of S with respect to the

variable y(x, t) as being

δS

= ’ρ¨(x, t) + T y (x, t).

y (1.83)

δy(x, t)

In writing down the ¬rst integral for this continuous system, we must

replace the sum over discrete indices by an integral:

‚L δL

’L ’ ’ L.

E= qi

™ dx y(x)

™ (1.84)

‚ qi

™ δ y(x)

™

i

20 CHAPTER 1. CALCULUS OF VARIATIONS

When computing δL/δ y(x) from

™

121

L

2

ρy ’ T y ,

L= dx ™

2 2

0

we must remember that it is the continuous analogue of ‚L/‚ qi , and so, in

™

contrast to what we do when computating δS/δy(x), we must treat y(x) as

™

a variable independent of y(x). We then have

δL

= ρy(x),

™ (1.85)