Chapter 4

Symmetries and Conservation Lawsxx

Legendre transformA system is described by a function f with x, y as independent variables.

f = f (x, y)

d f =∂f∂x

d x +∂f∂y

dy = ud x + vdy

Now we want to describe the system in terms of x, v.

Since d(vy) = vdy + ydv,

d(vy − f ) = − ud x + ydv

The new function is g = g(x, v) = vy − f.

∂g∂x

= − u; ∂g∂v

= y

HamiltonianLagrangian L = L(qμ, ·qμ, t)

pμ =∂L∂ ·qμ

Legendre’s transform to go from ·qμ to pμ as a independent

variable. The Hamiltonian

H = ∑ pμ·qμ − L,

Hence

dH = ∑ (pμδ ·qμ + ·qμδpμ −∂L∂qμ

δqμ −∂L∂ ·qμ

δ ·qμ) −∂L∂t

δt

dH = ∑ (pμδ ·qμ + ·qμδpμ − ·pμδqμ − pμδ ·qμ) −∂L∂t

δt

dH = ·qμδpμ − ·pμδqμ −∂L∂t

δt

Hence

·qμ =∂H∂pμ

; ·pμ = −∂H∂qμ

; ∂H∂t

= −∂L∂t

Section 1

Hamiltonian

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Note: dLdt

=∂L∂qμ

·qμ +∂L∂ ·qμ

··qμ +∂L∂t

δt = ·pμ·qμ + pμ

··qμ +∂L∂t

dHdt

= −∂L∂t

Homogeneity of time implies that for an isolated system, the potential function is not an explicit function of time. For a pair charge particles, U depends only the distance between the charged particles. Hence, for an isolated system, ∂L /∂t = 0. Hence

dHdt

= 0,

or the total energy of the system is a constant. This is the statement of conservation of energy, which is derived using the homogeneity property of the space.

Exercise

1. Argue that the energy is conserved for spring-mass system and planetary system.

2. An oscillator of mass m and spring constant k is forced with F0 cos(ωf t). Write down the Lagrangian for the same. Is the

energy conserved for this system?

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Conservation of Linear MomentumHomogeneity of space: If each particle of an isolated system is shifted by a constant distance ε, then the system’s Lagrangian should not change. Again think of a collection of charged particle. Interaction potential is function only of the distances between the particles. Here

δL = ∑a

∂L∂ra

⋅ ϵ = 0

Since ε is arbitrary,

∑a

∂L∂ra

= 0.

Since

∂L∂ra

=ddt

∂L∂va

,

we obtain

∑a

∂L∂ra

= ∑a

ddt

∂L∂va

=ddt ∑

a

∂L∂va

= 0.

Therefore, the total linear momentum P = ∑a

∂L∂va

of the system is

conserved. It is derived from the homogeneity of space.

On many occasions, a system may be symmetric about some translations, but not arbitrary translations. For example for a line charge aligned along the z axis, the system is invariant under any

translation along z. Therefore only Pz = ∑a

∂L∂va,z

is conserved for

such system.

If the Lagrangian is not an explicit function of qi, then the generalized force Fi = ∂L /∂qi = 0, and the generalized momenta

pi =∂L∂ ·qi

is conserved.

Section 2

More Conservation laws

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Conservation of Angular MomentumIsotropy of space: The system or Lagrangian of an isolated system is invariant under the rotation of the whole system by an arbitrary angle.

Let us rotate the system by angle δφ about an axis. The vector δφ is a vector aligned along the axis, and its magnitude is δφ. Under this operation, each particle is shifted by

δra = δϕ × ra and δva = δϕ × va.

Therefore, the change in the Lagrangian under the above operation is

δL = ∑a

∂L∂ra

⋅ δra +∂L∂va

⋅ δva = 0 or

δL = ∑a

·pa ⋅ δϕ × ra + pa ⋅ δϕ × va = 0

Therefore,

δϕ ⋅ddt ∑

a

ra × pa = 0.

Since δφ is arbitrary, the total angular momentum of the system about the origin

L =ddt ∑

a

ra × pa

is a constant. This is related to the isotropy of space.

If the system is invariant under rotation about an axis, the angular momentum about the axis is conserved.

The above three symmetries (homogeneity and isotropy of space, and homogeneity in time) have never been broken. So far, we have not observed any violation of conservation laws of energy, linear momentum, and angular momentum.

Robust conservation

Example:

Galilean invariance:

Vr is the relative velocity between the two inertial frames. For a set of particles,

L = ∑a

12

mav2a and L′� = ∑

a

12

ma(va + Vr)2

Hence

δL = L′�− L = ∑a

δ [ 12

ma(va + Vr)2] = [∑a

mava] ⋅ Vr = PCM ⋅ Vr

for small Vr.

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We do not set δL zero like in earlier cases because it will lead to PCM = 0, a special case. We set

δL =dFdt

=ddt

[RCM ⋅ Vr]

Comparing the above we obtain

PCM =d RCM

dt,

which is an identity.

Self similarity

Exercise:1. Construct conserved quantities for two masses coupled by a

spring.

2. A particle moves in the field for the following system. Construct the conserved quantities of the charged particle: Infinite charged plate, infinite line charge, two charges of equal magnitude, infinite half-plane charge, charged helix, charged torus.

3. Construct conserved quantities for the system of Exercise 1 of Section 3.2 as well as a system of two masses m1 and m2 coupled by a spring of a spring constant k.

4. What are the conserved quantities for a particle moving on the inner side of the cone?

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Nonrelativistic particle dynamics

Time Reversal Symmetry: U(−t) = U(t),

hence the Lagrangian is symmetric under time reversal.

We cannot distinguish between the forward and time-reversed dynamics.

Parity or Mirror Symmetry

r → − r = [(x, y, z) → (x, − y, z)] + [(x, − y, z) → (−x, − y, − z)]

Parity = mirror (about xz plane) + rotation about the y axis by π.

symmetry of mirror reflection.

(a)(b)

Can’t figure out who is the real gun man!

g g

(a)(b)

Can’t figure out the real projectile motion

Section 3

Discrete Symmetries

39

Proper and pseudo vectors

When a mirror is placed as an xz plane, the symmetry of mirror reflection is mathematically expressed as

z′� = z, y′� = − y, x′� = x; t′ � = t

Hence,

v′�x = vx, v′�y = − vy, v′�z = vz

a′�x = ax, a′�y = − ay, a′�z = az

A vector that transforms as above is called proper vector or real vector.

A cross product of two proper vectors transforms very differently under mirror transformation. Let us consider two proper vectors A and B. A cross product of these two vectors is

C = A × B

= (AyBz − AzBy) x + (AzBx − AxBz) y + (AxBy − AyBx) z (8.8.4)

Since A and B are proper vectors, the rules of mirror reflections of Eq. (8.8.1) yield

C′�x = − Cx, C′�y = Cy, C′�z = − Cz (8.8.5)

A vector that transforms like C is called pseudo vector or axial vector. Thus,

Cross product of two proper vectors is a pseudo vector.

Two corollaries of the above statements are

A cross product of a proper vector and a pseudo vector is a proper vector.

A cross product of two pseudo vectors is a proper vectors.

Examples of pseudo vectors

(1) Angular momentum L = r × p since the linear mo-mentum p = mv is a proper vector. Also, spin angular momentum of a body.

(2) Angular velocity Ω, which is defined using v = Ω × r.

(3) Magnetic field B since

B =μ0

4π ∫ Id l × rr3

40

F

B B

V

F

V(a)(b)

Can’t figure out real experiment!

Does the mirror symmetry hold for all experiment?

Ans: No! In Beta decay

v

S S

v

(b) (a)

e e

Wu and coworkers discovered that that the electrons (beta) are emitted preferentially opposite to the direction of the spin of the Cobalt-60 nucleus (see Fig. (a)). Note that the spin vector perpendicular to the mirror does not change sign under mirror reflection, but the velocity vec-tor does. Hence in the mirror image, the electrons would move preferentially in the direction of the spin, as shown in Fig. (b). Thus the law that the beta particles have a pref-erential velocity opposite to their spin is violated in the mirror reflection. This is how mirror symmetry is violated in a beta decay experiment.

Mathematically, the violation of the mirror symmetry is due to the appearance of nonzero value of the pseudo scalar Q = S ⋅ v in the Lagrangian. Here S is the spin of the Cobalt-60 nucleus, and v is the velocity of the elec-tron. Such quantities vanish in experiments respecting mirror symmetry (e.g., gravitational and electromagnetic interactions). The potential of weak nuclear force how-ever is a combination of a proper scalar and a pseudo sca-lar. This kind of potential was first proposed by Marshak and Sudarshan, and Feynman and Gell-mann in 1957.

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Charge Conjugation q → − q

In nuclear physics, C, P, T all are violated, but CPT to-gether is not violated.

Exercises1. Which of the following forces would violate mirror symmetry?

(a) F=q v×B (b) F=q v×E

(c) F=q v×B+mg (d) F=E+B

(e) F=E×B

In the above, E is the electric field, B is the magnetic fields, and g is the acceleration due to gravity.

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We can derive conservation laws for continuous transformation. A general transformation is

t′� = t + ϵτ

q′ �μ = qμ + ϵζμ

The new action is

∫ L′�dt′�− ∫ Ldt = ∫ [L′�dt′�dt

− L] dt

For minimum,

ddϵ [L′�

dt′�dt

− L]ϵ=0

= 0

or

Ld

dϵ [ dt′�dt ]

ϵ=0+ [ dL′�

dϵ ]ϵ=0

= 0

Note:

∂L′�∂ϵ

=∂L′�∂t′�

dt′ �dϵ

+∂L′�∂qμ

dq′ �μ

dϵ+

∂L′�∂ ·qμ

d ·q′�μ

dϵ

=∂L′�∂t′�

τ +∂L′�∂qμ

ζμ +∂L′�∂ ·qμ

d ·q′�μ

dϵ

Using dq′�μ

dt′�=

dqμ + ϵdζμ

dt + ϵdτ,

we obtain ddϵ

dq′�μ

dt′�= ·ζμ − ·q′ �μ ·τ.

Also, ddϵ ( dt′�

dt′�) =d

dϵ(1 + ϵ ·τ) = ·τ

Therefore

L ·τ +∂L∂t

τ + ·pμζμ + pμ( ·ζμ − ·q′�μ ·τ) =dFdt

or

−H ·τ −∂H∂t

τ + ·pμζμ + pμ·ζμ =

dFdt

or

pμζμ − Hτ − F = const.

Section 4

Noether’s theorem

43

This is the Noether’s theorem.

Examples:

(1): By choosing τ = 1,ζμ = 0,F = 0, we obtain H = constant.

(2): By choosing τ = 0,ζμ = 1,F = 0, we obtain pμ = constant.

(3) Choose τ = 0,F = 0, x′� = x − ϵy, y′� = ϵx + y that yields Mz =

constant.

(4) Galilean invariance: τ = 0,F = RCM ⋅ Vr, ζμ = Vμr t. It yields a

conserved quantity as

PCM ⋅ Vrt − RCM ⋅ Vr = const

which is identically zero since RCM = PCMt.

(5) Damped linear oscillator:

L = ( 12

m ·x2 −12

k x2) exp(bt /m)

The above Lagrangian is invariant under the transform t′� = t + ϵτ and x′� = x + ϵζ with τ = 1 and ζ = − bx /2m. Note that

·x′� =·x + ϵ ·ζ1 + ϵ ·τ

= ·x (1 − ϵb

2m )Substitution of the above implies the invariance of L under the above transformation.

Therefore, the conserved quantity with F = 0 is

pζ − H = constant

where p =∂L∂ ·x

= m ·x exp(bt /m).

Hence pζ = m ·x (−bx2m ) exp(bt /m) = − bx ·x exp(bt /m).

Also H = p ·x − L = ( 12

m ·x2 +12

k x2) exp(bt /m)

Therefore the conserved quantity is

( 12

m ·x2 +12

k x2 +12

bx ·x) exp(bt /m)

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Relativistic Dynamics

A particle follows a trajectory for which the elapsed time is minimum. This action is Lorentz invariant.

S = − α∫ dτ

= − α∫ 1 − v2 /c2dt

In the nonrelativistic limit,

S = − α∫ [1 −12

v2

c2 ] dt.

Hence, when we choose α = mc2, we obtain the usual Lagrangian as L = mv2 /2. Hence, the relativistic Lagrangian is

L0 = − mc2 1 − v2 /c2

The linear momentum is

p0 =∂L∂v

=mv

1 − v2 /c2

Therefore the energy or the Hamiltonian of the free particle is

H0 = p ⋅ v − L =mc2

1 − v2 /c2

For a charged particle in an electromagnetic field

L = L0 − qAμvμ

= L0 − q(ϕ − A ⋅ v/c)

The generalized momentum is

p =∂L∂v

= p0 + qA /c = γmv + qA /c

Hence γ2 = 1 + (p − qA /c)2 /m2c2.

Hence the energy of the system is

H = p ⋅ v − L = γmc2 + qϕ = m2c4 + (pc − qA)2 + qϕ

Section 5

Relativistic Lagrangian

45

In the nonrelativistic limit,

H =(p − qA /c)2

2m+ qϕ

Similarly the Lagrangian is

L =12

mv2 − q(ϕ − A ⋅ v/c)

The equation of motion is

ddt

(mvi + qAi /c) = − ∂iϕ + (∂i Aj)vj

Using d Ai

dt=

∂Ai

∂t+

∂Ai

∂xjvj, we obtain

ddt

(mvi) = − q [∂iϕ +∂Aj

∂t ] + (v × B)i = q(E + v × B)i

which is the Lorentz equation.

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