Cm2 Solutions

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Physics 31600 R. WaldClassical Mechanics Autumn, 2002

Problem Set II Solutions

1) Let L(q, q; t) be a Lagrangian [where, as in class, q stands for ( q1 , . . . , qn )].Suppose we introduce new coordinates ( Q1 (q), . . . , Q n (q)) on congurationspace. Relate the new momenta, P , to the old momenta p and show that

i P i Qi = i pi qi . (For the purposes of this problem, it is convenient to viewall quantities as functions of the independent variables ( q, q).)

The momentum has the form:

pi =L qi

=L Q j

Q j qi

+LQ j

Q j qi

(1)

(There is an implied sum over double indices in the above equation and thoseto follow.) But the second term is zero because the Qi s are functions onlyof qi . Because of this we also have that:

Q j =Q jqi

qi (2)

So: Q j qi

=Q jqi

(3)

Thus the new momentum P i is related to the old by the formula:

pi =L Q j

Q jqi

= P jQ jqi

(4)

We also immediately have:

pi qi = P jQ jqi

qi = P j Q j (5)

Recall that there is an implied sum over double indices.

2) (a) Show that the Euler-Lagrange equations for the Lagrangian

L =12

mdxdt

2

e +ec

A dxdt

yield the usual Lorentz force equations of motion of a charged particle in anelectromagnetic eld.

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(b) Obtain the corresponding Hamiltonian formulation of the problem. Writeout Hamiltons equation of motion and show explicitly that they also areequivalent to the usual Lorentz force law.

(a) First calculate the conjugate momenta:

L x i = mx i +

ecAi (6)

The force term is:Lx i

= ex i

+ec

x Ax i

(7)

So the Euler-Langrange equations are:

ddt

(mx i +ec

Ai ) = ex i

+ec

x Ax i

(8)

Subtracting the time derivative of A from both sides and using the chain rulewe have:

mxi = [ex i

ec

A it

] +ec

x Ax i

ec

( x )Ai (9)The rst term in brackets is eE i , where E is the electric eld. The secondterm can be calculated as follows:

x j i A j x j j Ai = x j lAk (li kj lj ki ) = x j lAk ( lkm ijm ) (10)But lAk lkm = Bm and x j Bm ijm = ( x B )i , so:

m x = e E +ec x B (11)

which is the Lorentz Force Law.(b) From equation 6 above we know that:

p = m x +ec

A (12)

Thus,

H = p x L = p p ec Am [

m2

( p ec Am

)2 e +ec

A ( p ec Am

)] (13)

which gives

H = (1 12

) p2

m+ ( 1 + 1 1)

e p Amc

+ ( 12

+ 1)(eA)2

mc2+ e

=p2

2m e p A

mc+

(eA)2

2mc2+ e (14)

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Hamiltons equations are:

xi =H p i

=pim

eAimc

(15)

pi = H x i

= +e pmc

Ax i

e2

mc2 A

Ax i e

x i

(16)

Using the equation for x we can write the p equation as:

ddt

(mxi +eAi

c) = x

ec

Ax i e

x i

(17)

This is the same as equation 8 above; thus Hamiltons equations also reduceto the Lorentz Force Law.

3) In the context of special relativity, it is much more in keeping with thecovariant nature of the theory to treat all four spacetime coordinates

(t ,x,y,z ) on an equal footing, and thus to describe particle motion as apath t(), x(), y(), z() in a 4-dimensional conguration space (with anarbitrary parameter along the path) rather than as a curve x(t), y(t), z(t) in a3-dimensional conguration space (with t the time coordinate of a particularglobal inertial coordinate system).(a) Show that the Lagrangian

L = mdtd

2

dxd

2 12

yields the correct equations of motion for a free particle. (In keeping with theabove remark, treat ( t ,x,y,z ) as the degrees of freedom and as time.)(b) Show that the conjugate momenta satisfy the relation

p2t ( p2x + p

2y + p

2z ) = m

2

and, thus, are not independent, i.e. one cannot eliminate the qs in favor of ps.(c) Nevertheless, obtain a (constrained) Hamiltonian formulation for the freerelativistic particle by the procedure described in class, with = dt/d .

(a) First, lets dene some simpler notation:

q = u = (dtd

,dxd

,dyd

,dzd

) (18)

q = u = (dtd

, dxd

, dyd

, dzd

) (19)

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where q are the coordinates and u are the rst derivatives with respect to. Then the Lagrangian is now expressed as:

L = m[u u ]12 (20)

The momenta are:

p =Lu = mu [u

u ]

12

(21)Since the coordinates do not explicitly appear in L, the equation of motionis:

dd

(mu [u u ]

12 ) = 0 (22)

which says the term in parenthesis is a constant, which we will call P .

P = mu [u u ]

12 (23)

Note that this expression is independent of the parametrization along

the world-line. If I choose a new = f () then P is:

P = mu [u u ]

12 = mu [u u ]

12 (24)

where = dd . So we can choose equal to the proper time and equation23 becomes:

P = mdqd

(25)

which is the familiar expression for the (constant) momentum of a relativisticparticle.

The parameter is affine if the function f () = 0 in the following equa-tion:

u u = f ()u (26)

Carrying out the derivative in equation 22, we will get (after dividing by

m):u [u u ]

12 u [u

u ]

32 u u = 0 (27)

This leads to:u = [

u u

u u ]u (28)

After noticing that the left-hand side of equation 24 is u , we see that thefunction in brackets is f (). Thus, choosing affine gives u = 0, whichallows us to set u u to a constant (which is 1 if is the proper time). It iseasy to show that if is not affine, one can choose a new that is affine.

(b) Using equation 21, we see that:

p p = m2 u u [u u ]

12 [u u ]

12 = m 2 u u [u u ]

1 = m 2 (29)

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(c) Since the p are not independent, we cannot eliminate all of the u infavor of them. We choose one of them ( = dtd ) to serve as a non-dynamicalconstraining variable. Thus we have the Hamiltonian:

H = p0 + pi ui L(, p i ) (30)

where i runs over the spatial variables. Dene = [u

u ]

12

(note that thishas no dependence.) Thus:

H = p0 pi pi

m[u u ]

12 + m[u u ]

12 = p0

pi pi

m

+ m

(31)

Note that pi pi = p2 and that:

(m )2 =m2

1 v2= m 2

1 v2 + v2

1 v2= m 2 + m 2 ( v)2 = m2 + p2 (32)

Thus:H = p0 + p2 + m2 (33)

So Hamiltons equations read:

qi =H p i

=pi

p2 + m2 =pi

m(34)

pi = H qi

= 0 (35)

H

= p0 + p2 + m2 = 0 (36)

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