™

leading to

121

L

2

E= dxρy + T y .

™ (1.86)

2 2

0

This, as expected, is the total energy, kinetic plus potential, of the string.

Exercise: Consider an action of the form

dd+1 xL(•, ‚µ •)

S= (1.87)

which does not depend explicitly on xµ . Generalize the Noether derivation

of the energy conservation law to one exploiting variations of the form

µ

δ• = (x)‚µ •, (1.88)

where depends on space and time, and hence show that

‚µ T µν = 0, (1.89)

where

‚L

T µν = µ

‚ν • ’ δν L (1.90)

‚(‚µ •)

is known as the canonical energy-momentum tensor .

Exercise: Apply the results of the previous exercise to the Lagrangian of

the vibrating string, and so establish the two following local conservation

equations:

‚ ρ2 T 2 ‚

{’T yy } = 0,

y+ y +

™ ™ (1.91)

‚t 2 2 ‚x

and

‚ ‚ ρ2 T 2

{’ρyy } +

™ y+ y

™ = 0. (1.92)

‚t ‚x 2 2

1.3. LAGRANGIAN MECHANICS 21

Verify that these are indeed consequences of the wave equation.

The two equations obtained in the last exercise are “local” conservation

laws because they are of the form

‚q

+ · J = 0, (1.93)

‚t

where q is the local density, and J the ¬‚ux, of the globally conserved quantity

Q = q dd x. In the ¬rst case, the local density q is

ρ T

T 00 = y 2 + y 2 ,

™ (1.94)

2 2

which is the energy density. The energy ¬‚ux is given by T 10 ≡ ’T yy , which

™

is the rate of working by one piece of string on its neighbour. Integrating

over x, and observing that the ¬xed-end boundary conditions are such that

L‚

{’T yy } dx = [’T yy ]L = 0,

™ ™0 (1.95)

0 ‚x

gives us

d L ρ2 T 2

y+ y

™ dx = 0, (1.96)

dt 0 2 2

which is the global energy conservation law we obtained earlier.

The physical interpretation of T 01 = ’ρyy , the locally conserved quan-

™

tity in the second case, is less obvious. If this were a relativistic system,

we would have no di¬culty in identifying T 01 dx as the x-component of the

energy-momentum 4-vector, and therefore T 01 as the density of x-momentum.

Our transversely vibrating string has no signicant motion in the x direction,

though, so T 01 cannot be the string™s x-momentum density. Instead, it is

the density of something called pseudo-momentum. The distinction between

true and pseudo- momentum is best understood by considering the corre-

sponding Noether symmetry. The symmetry associated with Newtonian mo-

mentum is the invariance of the action integral under an x translation of

the entire apparatus: the string, and any wave on it. The symmetry asso-

ciated with pseudomomentum is the invariance of the action under a shift,

y(x) ’ y(x ’ a), of the location of the wave on the string ” the string

itself not being translated. Newtonian momentum is conserved if the ambi-

ent space is translationally invariant. Pseudo-momentum is conserved if the

string is translationally invariant ” i.e. if ρ and T are position independent.

A failure to realize that the presence of a medium (here the string) requires us

to distinguish between these two symmetries is the origin of many paradoxes

involving “wave momentum.”

22 CHAPTER 1. CALCULUS OF VARIATIONS

Maxwell™s Equations

Faraday and Maxwell™s description of electromagnetism in terms of dynam-

ical vector ¬elds gave us the ¬rst modern ¬eld theory. D™ Alembert and

Maupertuis would have been delighted to discover that the famous equations

of Maxwell™s Electricity and Magnetism (1873) follow from an action princi-

ple. There is a slight complication stemming from gauge invariance but, as

long as we are not interested in exhibiting the covariance of Maxwell under

Lorentz transformations, we can sweep this under the rug by working in the

axial gauge, where the scalar electric potential does not appear.

We will start from Maxwell™s equations

·B = 0,

™

—E = ’B,

™

—H = J + D,

·D = ρ, (1.97)

and show that they can be obtained from an action principle. For convenience

we shall use natural units in which µ0 = 0 = 1, and so c = 1 and D ≡ E

and B ≡ H.

The ¬rst equation · B = 0 is non-dynamical, but is a constraint which

we satisfy by introducing a vector potential A such that B = — A. If we

set

™

E = ’A, (1.98)

then this automatically implies Faraday™s law of induction

™

— E = ’B. (1.99)

We now guess that the Lagrangian is

1

d3 x E2 ’ B2 + J · A .

L= (1.100)

2

1™

The motivation is that L looks very like T ’ V if we regard 1 E2 ≡ 2 A2 as

2

being the kinetic energy and 2 B2 = 1 ( — A)2 as being the potential energy.

1

2

The term in J represents the interaction of the ¬elds with an external current

source. In the axial gauge the electric charge density ρ does not appear in

the Lagrangian. The corresponding action is therefore

1 ™2 1

d3 x — A)2 + J · A dt.

A’ (

S= L dt = (1.101)

2 2

1.3. LAGRANGIAN MECHANICS 23

Now vary A to A + δA, whence

¨

d3 x ’A · δA ’ ( — A) · ( — δA) + J · δA dt.

δS = (1.102)

Here, we have already removed the time derivative from δA by integrating