do no net work, never appear. The disadvantage is exactly the same: if we

need to ¬nd the constraint forces “ in this case the tension in the string ”

we cannot use Lagrange alone.

Example: Polar Coordinates

y

a‘

ar

r

‘ x

Consider a central force problem with Fr = ’‚r V (r). The Newtonian

method begins by computing the acceleration in polar coordinates. This

is most easily done by setting z = reiθ and di¬erentiating twice:

™

z = (r + ir θ)eiθ ,

™ ™

™ ™™ ¨

z = (¨ ’ r θ2 )eiθ + i(2r θ + rθ)eiθ .

¨ r (1.43)

Reading o¬ the components parallel and perpendicular to eiθ gives for the

acceleration

™

ar = r ’ r θ2 ,

¨

¨ ™™

aθ = r θ + 2rθ, (1.44)

Newton™s equations therefore become

‚V

™

m(¨ ’ rθ2 ) = ’

r

‚r

d

¨ ™™ ™

m(rθ + 2r θ) = 0, ’ (mr2 θ) = 0. (1.45)

dt

™ ™

Setting l = mr 2 θ, the conserved angular momentum, and eliminating θ gives

l2 ‚V

m¨ ’ =’

r . (1.46)

mr3 ‚r

1.3. LAGRANGIAN MECHANICS 13

(If this were Kepler™s problem, where V = GmM/r, we would now proceed

to simplify this equation by substituting r = 1/u, but that is another story.)

Following Lagrange we ¬rst compute the kinetic energy in polar coordi-

nates (this requires one less derivative than computing the acceleration) and

set

1 ™

L = T ’ V = m(r2 + r 2 θ2 ) ’ V (r).

™ (1.47)

2

The Euler-Lagrange equations are now

d ‚L ‚L ‚V

™

= 0, ’ m¨ ’ r 2 θ2 +

’ r =0

dt ‚r

™ ‚r ‚r

d ‚L ‚L d ™

(mr2 θ) = 0.

’ = 0, ’ (1.48)

™

dt ‚θ dt

‚θ

The ¬rst integral for this problem is

‚L ™ ‚L

’L

E=r

™ +θ

‚r

™ ‚r ™

1 ™

m(r2 + r 2 θ2 ) + V (r)

= ™ (1.49)

2

which is the total energy. Thus the constancy of the ¬rst integral states that

dE

= 0, (1.50)

dt

or that energy is conserved.

Warning: We might realize, without having gone to the trouble of deriving

it from the Lagrange equations, that rotational invariance guarantees that

™

the angular momentum l = mr2 θ will be a constant. Having done so, it is

almost irresistible to try to short-circuit some of the arithmetic by plugging

this prior knowledge into

1 ™

L = m(r2 + r 2 θ2 ) ’ V (r)

™ (1.51)

2

™

so as to eliminate the variable θ in favour of the constant l. If we try this we

get

l2

?1 2

L ’ mr + ’ V (r).

™ (1.52)

mr2

2

14 CHAPTER 1. CALCULUS OF VARIATIONS

We can now directly write down the Lagrange equation r, which is

l2 ? ‚V

=’

m¨ +

r . (1.53)

mr3 ‚r

Unfortunately this has the wrong sign before the l2 /mr 3 term! The lesson is

that we must be very careful in using consequences of a variational principle

to modify the principle. It can be done, and in mechanics it leads to the

Routhian or, in more modern language to Hamiltonian reduction, but it

requires using a Legendre transform. The reader should consult a book on

mechanics for details.

1.3.2 Noether™s Theorem

The time-independence of the ¬rst integral

d ‚L

’ L = 0,

q

™ (1.54)

dt ‚q

™

and of angular momentum

d ™

{mr2 θ} = 0, (1.55)

dt

are examples of conservation laws. We obtained them both by manipulating

the Euler-Lagrange equations of motion, but also indicated that they were

in some way connected with symmetries. One of the chief advantages of a

variational formulation of a physical problem is that this connection

Symmetry ” Conservation Law

can be made explicit by exploiting a strategy due to Emmy Noether. She

showed how to proceed directly from the action integral to the conserved

quantity without having to ¬ddle about with the equations of motion. We

begin by illustrating her technique in the case of angular momentum, whose

conservation is a consequence the rotational symmetry of the central force

problem. The action integral for the central force problem is