Classical Mechanics Symmetry and Conservation Laws

Dipan Kumar Ghosh UM-DAE Centre for Excellence in Basic Sciences Kalina, Mumbai 400085 September 7, 2016

1 Concept of Symmetry

If the property of a system does not change under some defined operations, the system is said to have a symmetry with respect to so defined operations. Thought he universe is both homogeneous and isotropic, there are local inhomogeneity and lack of isotropy. Nevertheless, a closed system (i.e. a system which does not interact with particles or fields outside of the system) may exhibit space homogeneity and/or isotropy.

1.1 Homogeneity of Space: Homogeneity of space implies that the physical laws are invariant under space translation. This means that if we change the position of every particle (and ﬁeld) by the same constant vector ~, the Lagrangian of our system would remain the same. Mathematically, it implies 0 that when ~ri → ~ri = ~ri + ~, the change in the Lagrangian of our system, X ∂L X ∂L δL = · ~ = ~ · (1) ∂~r ∂~r i i i i Here the summation index i is over all the particles of the system. Symmetry under arbitrary translation implies δL = 0, which results in

X ∂L = 0 (2) ∂~r i i Using Euler-Lagrange equation, this implies

X d ∂L = 0 dt ∂~v i i

1 c D. K. Ghosh, IIT Bombay 2

For a closed system, V depends only on positions of the particles,

1 X L = m v2 − V ({~r } 2 i i i i We then get ! ! d X d X m ~v = ~p = 0 (3) dt i i dt i i i i.e. the total momentum of the system is constant. Thus homogeneity of space leads to conservation of linear momentum of the system.

1.2 Isotropy of Space: If L is independent of the orientation of the system, it results in the conservation of angular momentum of the system. Consider (see ﬁgure) the rotation of the system about the z-axis by an angle θ so that OA~ → OA~ 0. z

C δϕ

r r’ θ

It follows from the ﬁgure that OA~ 0 = OA~ + AA~ 0 where the magnitude of AA~ 0 is given by | AA~ 0 |= r sin θδϕ. Deﬁning δϕ~ to be of magnitude δϕ and directed along the axis of rotation, we can write, AA~ 0 = δϕ~ × ~r Thus as ~r → ~r0 = ~r + δ~r, δ~r = δϕ~ × ~r. We then have

~v → ~v0 = ~v + δ~v where δ~v = δϕ~ × ~v. c D. K. Ghosh, IIT Bombay 3

If the Lagrangian is invariant under rotation described agove, we have X ∂L X ∂L δL = · δ~r + · δ~r˙ ∂~r i ~ i i i i ∂r˙i X d ∂L X ∂L = · δ~r + · δ~r˙ dt ~ i ~ i i ∂r˙i i ∂r˙i X d ∂L X ∂L = − · (~r × δϕ~ ) − · (~v × δϕ~ ) (4) dt ~ i ~ i i ∂r˙i i ∂r˙i Once again, if we consider a closed system, 1 X L = m v2 − V ({~r } 2 i i i i we have ∂L = mi~r˙i = ~pi ∂~r˙i (4) can be written as X h ~ ~ i δL = − p~˙i · (~ri × δϕ) + ~pi · (~r˙i × δϕ) (5) i Writing δϕ~ =nδϕ ˆ , we can express (5) as X d δL = δϕnˆ · (~r × ~p ) (6) dt i i i Thus δL = 0 implies d X (~r × ~p ) = 0 dt i i i ~ P so that the angular momentum of the system J = i(~ri × ~pi) is constant.

1.3 Homogeneity of Time and Energy Conservation: Closed system also exhibit symmetry with respect to time translation. What this implies is that repeating an experiment under identical conditions after a lapse of some time will yield identical result. Time homogeneity is a consequence of the fact that the Lagrangian ∂L of the system has no explicit time dependence: = 0, so that L = Lq, q˙ only. ∂t The total time derivative of the Lagrangian, in such a case is given by dL X ∂L X ∂L = q¨ + q˙ dt ∂q˙ i ∂q i i i i i X ∂L X d ∂L = q¨ + q˙ ∂q˙ i dt ∂q˙ i i i i i X d ∂L = q˙ dt ∂q˙ i i i c D. K. Ghosh, IIT Bombay 4

We deﬁne the Hamiltonian of the system as

X ∂L H = q˙ − L (7) ∂q˙ i i i dH so that the preceding equation gives us = 0. For a closed system, the Hamiltonian dt also gives the total energy of the system (see below), so that the time homogeneity results in conservation of energy. Let us consider a closed system for which the kinetic energy is expressible as a bilinear product of generalized velocities and the potential energy is a function of the generalized coordinates alone, 1 X T = C q˙ q˙ 2 i,j i j i,j where Ci,j = Cj,i could be functions of generalized coordinates but not of velocities or of time. Thus we have

L = T ({q˙i}) − +V ({qi}) We then have ! X ∂L X ∂ 1 X q˙k = q˙k Ci,jq˙iq˙j ∂q˙k ∂q˙k 2 k k i,j " # X 1 X 1 X = q˙ C δ q˙ + C q˙ δ k 2 i,j i,k j 2 i,j i j,k k i,j i,j ! 1 X X X = C q˙ + C q˙ q˙ 2 k,j j i,k i k k j i X = Cj,kq˙jq˙k j,k = 2T where in the last but one step, we have used Ci,k = Ck,i and have changed the dummy summation index in the second term from i to j. Using (7) for the deﬁnition of the Hamiltonian, we then have,

X ∂L H = q˙ − L ∂q˙ i i i = 2T − L = 2T − (T − V ) = T + V which is the total energy. Thus for a closed system, homogeneity of time leads to a conservation of the total energy. c D. K. Ghosh, IIT Bombay 5

2 Invariance under Galilean Transformation

Let S and S0 be two inertial frames. Though the values of kinematical quantities will be measured diﬀerently in two frames, by deﬁnition of inertial frames, Newton’s laws will be valid in both the frames, i.e.

d2~r d2~r 0 m i = m i (= F ) i dt2 i dt2 i which gives d2(~r − ~r 0) i i = 0 dt2 which can be integrated to yield

0 ~ri − ~ri = ~u0(t − t0) ≡ ~u0t

, where u0 is a constant velocity, the last relation follows if the two frames are made to coincide at t = t0. The two frames are thus related by Galilean transformation:

0 ~ri = ~ri − ~u0t

How does the Lagrangian transform? If the potential energy depends only on the relative coordinates of two particles then

0 0 ~r2 − ~r1 = ~r2 − ~r1 so that in S0, 1 L0 = mv02 − V 2 1 = m(~v − ~u )2 − V 2 0 1 1 = mv2 − V − m~v · ~u + mu2 2 0 2 0 d 1 = L + mu2t − m~u · ~r (8) dt 2 0 0 dF ≡ L + (9) dt where the function F is defined through (8) and (9). Note that both L and L0 satisfy Euler-Lagrange equation. This is an example of gauge symmetry of the Lagrangian. If L satisfies the Euler-Lagrange equation and F is any differentiable function of (q, t), then dF L + also satisfies the Euler-Lagrange equation. dt This is readily proved by observing that the extra term in the Euler-Lagrange equation c D. K. Ghosh, IIT Bombay 6

dF due to is dt d ∂ dF ∂ dF − = dt ∂q˙i dt ∂qi dt " ! !# d ∂ X ∂F ∂F ∂ X ∂F ∂F q˙ + − q˙ + dt ∂q˙ ∂q j ∂t ∂q ∂q j ∂t i j j i j j (10)

Note that d ∂ ∂F ∂ d ∂F ∂ ∂F = = dt ∂q˙i ∂t ∂t dt ∂q˙i ∂t ∂qi since the explicit diﬀerentiation can always be carried out at the end. Further, " !# ! d ∂ X ∂F ∂ X ∂F X ∂ ∂F q˙ = q˙ = q˙ dt ∂q˙ ∂q j ∂q ∂q j ∂q ∂q j i j j i j j j j i

sinceq ˙j is independent of the coordinates and the independent coordinates qi and qj are interchangeable. Using the above two relations, the right hand side of (10) becomes

X ∂2F ∂2F X ∂2F ∂2F q˙ + − q˙ − = 0 ∂q ∂q j ∂t∂q ∂q ∂q j ∂q ∂t j j i i j i j i which proves the assertion. The reason for the existence of the gauge symmetry can be traced back to the minimisation of the action integral. Let,

dF (q, t) L0(q, q,˙ t) = L(q, q,˙ t) + dt The corresponding action integral would then be given by

Z tf S0[q(t)] = L0dt ti Z tf dF = L + dt ti dt tf = S + F (q, t) |ti = S + constant

since δq(ti) = δq(tf ) = 0 for the variational function, adding a constant to the action integral is inconsequential to the determination of t. Example: Consider the Lagrangian for a charged particle in electromagnetic ﬁeld. We know that c D. K. Ghosh, IIT Bombay 7 the following gauge transformations of the scalar and vector potentials leave the ﬁelds unchanged:

A~0 = A~ + ∇F (~r, t) ∂ ϕ0 = ϕ − F ∂t We have seen earlier that the electromagnetic forces can be included in the Lagrangian formulation if we deﬁne a velocity dependent potential, eϕ − e~v · A~, where e is the charge of the particle. Thus 1 L = mv2 − eϕ + e~v · A~ 2 dF (~r, t) Suppose we add to the Lagrangian above a term equal to e The new Lagrangian dt is then given by 1 dF L0 = mv2 − eϕ + e~v · A~ + e 2 dt 1 ∂F h i = mv2 − e ϕ − + ~v · e(A~ + e∇F ) 2 ∂t where we have used dF ∂F ∂F ∂F ∂F = v + v + v + dt ∂x x ∂y y ∂z z ∂t ∂F = ∇F · ~v + ∂t which gives 1 L0 = mv2 − eϕ0 + e~v · A~0 2

3 Symmetry Operations and Noether’s Theorem

3.1 Coordinate Transformation vrs. Symmetry Operations Suppose we have a system with Lagrangian L(q, q,˙ t). If we make a coordinate transformation such that q(t) → q0(t). (When we write q, it actually stands for the set of coordinates

{qk} which are required to describe the system. ) This is simply a way to relabel the system, the form of Lagrangian would change to L0(q0, q˙0, t) though the physical properties of the system would not change. 1 1 Example : Suppose the Lagrangian is given by L = mq˙2 − kq2. If we change the 2 2 coordinate system such that q → q0 = q + a, the new Lagrangian would be 1 1 L0 = mq˙02 − k(q0 − a)2 (11) 2 2 c D. K. Ghosh, IIT Bombay 8 which is essentially saying that L0(q0, q˙0, t) = L(q(q0, q˙0, t), q˙(q0, q˙0, t), t) (12) Notice that L0’s functional dependence on q0, q˙0, t is not the same as the functional dependence of L on q, q,˙ t. What we have done is to simply relabel the system coordinates. This could be done by either a passive manner (which is to leave the system as it is and shift the coordinate system or by an active transformation in which we keep the coordinate frame the same but moving the entire system (particles and ﬁelds) relative to the original coordinates such that a point in the system which had a coordinate q now has the coordinate q0. This is illustrated in the ﬁgure below. Note that import of both ways of doing is identical– irrespective of whether we choose active or passive transformation, the physical result is the same.

Passive

Active

Example: (1) Suppose we decided that the coordinate x alone will become x0 = x+a. In the passive transformation, we would have shifted the origin along the x-axis to lett keeping axes parallel so that a point which was at x now is at x + a. In the active transformation, we would not change the coordinate axis but will move system to the right by an amount a. (2) Suppose we rotated the coordinate axes by an angle θ so that the new coordinates of a point which was earlier (x, y, z) is x0, y0, z0 where x0 = x cos θ − y sin θ y0 = x sin θ + y cos θ z0 = z (13) In the passive case, we rotate the coordinate axes about the z axis by an angle θ so that a point which was at x, y, z now has the coordinates x0, y0, z0 given by the relationship above. c D. K. Ghosh, IIT Bombay 9

On the other hand, we could arrive at the same values for the new coordinates if we did an active transformation, in which we did not touch the coordinate axis but rotated the entire physical system by an angle −θ about the z− axis. These two examples (translation and rotation) are what are known as continuous transformation as in these there is a parameter (a in case of translation and the angle θ in case of rotation) which could be varies continuously. [An example of discrete transformation is reﬂection transformation in which x → x0 = −x]. In the following discussion we would be interested in continuous transformation alone where the coordinates depend on a continuously varying parameter s, so that we express the transformation as follows:

q → q0 = Q(s, t) (14) with the identity transformation being the situation when s = 0, i.e. Q(0, t) = q(t). Let us look at some very special types of coordinate transformation. Instead of the Lagrangian of the new situation being given by (12) suppose we demand that

L0(q0, q˙0t) = L(q0, q˙0, t) (15) i.e. instead of the relation given by (11), if we had somehow got into a situation where 1 1 L0 = mq˙02 − kq02 (16) 2 2 was valid, we would have had what is known as a Symmetry Transformation of the system. Of course, under the transformations discussed in examples (1) and (2) we cannot get (16). However, this would have happened if we had considered a reﬂection transformation where q0 = −q. (This would not have been a continuous transformation, but that is another matter!). As an example of continuous transformation consider a mass connected to a spring such that the spring force is radially inward and the Lagrangian for which is 1 1 L(x, y, z, x,˙ y,˙ z,˙ t) = m(x ˙ 2 +y ˙2 +z ˙2) − k(x2 + y2 + z2) 2 2 If we do a rotation transformation (13) on the coordinates, we would have

L0(x0, y0, z0) = L(x(x0, y0, z0), y((x0, y0, z0), z(x0, y0, z0)) 1 = m (x ˙ 0cosθ +y ˙0 sin θ)2 + (−x˙ 0 sin θ +y ˙0 cos θ)2 +z ˙02 − 2 1 k (x0cosθ + y0 sin θ)2 + (−x0 sin θ + y0 cos θ)2 +z ˙02 2 1 1 = m(x ˙ 02 +y ˙02 +z ˙02) − k(x02 + y02 + z02) (17) 2 2 which is equal to L(x0, y0, z0) showing that the rotation is a continuous symmetry operation for this Lagrangian. However, of we considered a translation operation such that x0 = x − a, y0 = y, z0 = z, the c D. K. Ghosh, IIT Bombay 10 kinetic energy term would have the same form but the potential energy term would be 1 k ((x0 + a)2 + y02 + z02) which would not result ion a symmetry operation. Thus (15) is 2 a test for symmetry operation. One can have alternate but equivalent prescription for testing a symmetry operation. That is to check if L(x0, y0, z0) = L(x, y, z) (18) Let us check this prescription for the example considered above. Consider rotation transformation (13). We have 1 1 L(x0, y0, z0) = m(x ˙ 02 +y ˙02 +z ˙02) − k(x02 + y02 + z02) 2 2 Using (13) this expression can be written as 1 L(x0, y0, z0) = (x ˙ cos θ − y˙ sin θ)2 + (x ˙ sin θ +y ˙ cos θ)2 +z ˙2 − 2 1 k (x cos θ − y sin θ)2 + (x sin θ + y cos θ)2 + z2 2 1 1 = (x ˙ 2 +y ˙2 +z ˙2) − k(x2 + y2 + z2) 2 2 = L(x, y, z)

Identical conclusion can be drawn for reﬂection operation but not for translation.

3.2 Noether’s Theorem: Whenever a physical system exhibits a continuous symmetry, there is a conserved quantity (called Noether’s charge in field theory) associated with it. [Converse of this is not necessarily true as in field theory instances are known of conservation with which no apparent symmetry has been found to be associated.] Let s be a constant parameter which can be continuously varied and whose value tells us how far are we from the identity operation. Example of the parameter is θ in case of rotation, a in case of translation (reflection, not being a continuous symmetry does not have such a parameter associated with it). Let Q(s, t) is a symmetry operation such that the Lagrangian does not depend on it: d d L[Q(s, t), Q˙ (s, t), t] = L[q(t), q˙(t), t] = 0 ds ds The statement does seem confusing as Q(s, t) has explicit dependence on s, but the Lagrangian can be written in such a way that it does not have s dependence. We then have, ∂L dQ ∂L dQ˙ + = 0 ∂Q ds ∂Q˙ ds c D. K. Ghosh, IIT Bombay 11

(the dt/ds term is zero because t and s are independent). Using Euler-Lagrange equation on the ﬁrst term of above,

d ∂L dQ ∂L dQ˙ + = 0 dt ∂Q˙ ds ∂Q˙ ds which can be written as d ∂L dQ = 0 dt ∂Q˙ ds ∂L As the canonical momentum corresponding to Q is p = , we have ∂Q˙ dQ Λ = p | = constant (19) ds s=0 We have evaluated dQ/ds at s = 0 as the product being constant, we are free to evaluate it at any value of s. In general, if the Lagrangian possesses a set of n continuous symmetry operations, there would then be n conserved quantities. If the Lagrangian is a function of a set of generalized k coordinates and velocities, the conserved Noether charges are

X dQk Λj({qk}, {q˙k}, t = pk |sj =0 (20) dsj k for n parameters sj, j = 1, . . . n.

3.3 Cyclic coordinates: A coordinate is said to be cyclic (or ignorable) if L does not explicitly depend on it. In such a case, the corresponding momentum is conserved. This follows trivially from the Euler-Lagrange equation,as

∂L d ∂L = 0 =⇒ = 0 ∂q dt ∂q˙

dp which gives = 0. It also follows trivially from Noether’s theorem, as if we consider a dt continuous transformation q(t) → q(t) + s, q˙(t) → q˙(t), we have

δL = L(q + s, q,˙ t) − L(q, q,˙ t) ∂L ≈ s = 0 ∂q if L is cyclic. c D. K. Ghosh, IIT Bombay 12

3.4 Examples from single parameter family of Transformation: Example 1: Consider a particle moving in two dimensions in a central potential. The Lagrangian is given by 1 L = m(r ˙2 + r2ϕ˙ 2) − V (r) 2 2 In this case ϕ is cyclic and hence pϕ = mr ϕ˙ = constant. Note that r and ϕ depends on t in such a way that the product r2ϕ˙ is independent of time. In this case it is appropriate to consider the transformation:

r˜(s) = r;r ˜(0) = 0 ϕ˜(s) = ϕ + s;ϕ ˜(0) = ϕ

The Noether’s charge is ∂L ∂r˜ ∂L ∂ϕ˜ Λ = | + | ∂r˙ ∂s s=0 ∂ϕ˙ ∂s s=0 = (mr˙) × 0 + (mr2ϕ˙) × 1 = mr2ϕ˙

Example 2: Let us look at the same problem in the Cartesian coordinates taking a slightly diﬀerent form of the continuous symmetry. The Lagrangian is rewritten 1 L = m(x ˙ 2 +y ˙2) − V (p(x2 + y2)) 2 Consider the following continuous transformation which leaves the Lagrangian unchanged.

x˜(s) = x cos s − y sin s y˜(s) = x sin s + y cos s

We have ∂x˜ ∂y˜ = −y˜; =x ˜ ∂s ∂s Noether’s charge is given by ∂L ∂x˜ ∂L ∂y˜ Λ = | + | ∂x˙ ∂s s=0 ∂y˙ ∂s s=0 = (mx˙) × (−y) + (my˙) × x 2 = m(xy˙ − yx˙) = m(~r × ~r˙)z ≡ mr ϕ˙

Example 3: Consider the motion of a particle where the potential is a function of distance ρ and the combination aϕ+z, where (ρ, ϕ, z) are the usual cylindrical coordinates. The Lagrangian is given by 1 L = m[ρ ˙2 + ρ2ϕ˙ 2 +z ˙2] − V (ρ, aϕ, z) 2 c D. K. Ghosh, IIT Bombay 13

Deﬁne a one parameter family of transformation,

ρ˜ = ρ, ϕ˜ = ϕ + s, z˜ = z − as

Note that aϕ˜ +z ˜ = aϕ + z. The conserved Noether’s charge is given by ∂L ∂ρ˜ ∂L ∂ϕ˜ ∂L ∂z˜ Λ = | + | + | ∂ρ˙ ∂s s=0 ∂ϕ˙ ∂s s=0 ∂z˙ ∂s s=0 = mρ˙2 × 0 + mρ2ϕ˙ × 1 + mz˙ × (−a) = mρ2ϕ˙ − maz˙

Let us verify whether this quantity is indeed conserved. Using Euler-Lagrange equations, we have,

d ∂L d ∂L ∂V = (mρ2ϕ˙) = = − dt ∂ϕ˙ dt ∂ϕ ∂ϕ ∂V ∂(aϕ + z) = − ∂(aϕ + z) ∂ϕ = −aV 0 d ∂L d ∂L ∂V = (mz˙) = = − dt ∂z˙ dt ∂z ∂z ∂V ∂(aϕ + z) = − ∂(aϕ + z) ∂z = −V 0 where we have written V 0 = ∂V/∂(aϕ + z). Thus

d (mr2ϕ˙ − maz˙) = −aV 0 + aV 0 = 0 dt Example 4: Suppose L is invariant under an inﬁnitesimal translation of the system alongn ˆ. We have x˜i = xi + snˆ. Noether’s charge is

X ∂L Λ = · nˆ = P~ · nˆ ~ i ∂x˙ i P where P = i ~pi is the total linear momentum of the system. Thus continuous translational symmetry leads to conservation of linear momentum. Example 5: If L is invariant under an inﬁnitesimal rotation about an axisn ˆ. We then have

~x˜i = R(s, nˆ)~xi 2 = ~xi + snˆ × ~xi + O(s ) c D. K. Ghosh, IIT Bombay 14

The Noether’s charge in this case is the total angular momentum of the system aboutn ˆ,

X ∂L Λ = · (ˆn × ~x ) ~ i i ∂x˙ i X ∂L =n ˆ · ~x × i ~ i ∂x˙ i X =n ˆ · ~xi × pˆi =n ˆ · L~ where L~ is the total angular momentum about the axisn ˆ.

4 Additional Advanced Information* (* not included as a part of the course.)

In proving Noether’s theorem we had required that L be invariant under a continuous transformation. This is actually a more stringent requirement than is required. What we require is that the action S be invariant, which means that the Lagrangian can vary by an overall time derivative, dF L → L0 = L + dt We have seen that presence of this additional term does not change action except for an additive constant. If

qi(t) → Qi(sα) = qi(t) + sαδqi(t) be an inﬁnitesimal continuous transformation of the generalised coordinates by parameters sα such that under this transformation

dF L → L0 = L + s α dt then the quantities given by ∂L Λα = δqi − F ∂q˙α are conserved. The proof is straightforward. The change in the Lagrangian due to change in the coordinates and the velocities is ∂L ∂L d ∂L δL = δq˙i + δqi = δqi ∂q˙i ∂qi dt ∂q˙i

For each sα, the change is given by d ∂L dF sα δqi = sα dt ∂q˙i dt c D. K. Ghosh, IIT Bombay 15 which gives d ∂L δqi − F = 0 dt ∂q˙i Let us consider a consequence of this. Consider time translation, i.e. the case of symmetry

Q(s, t) = q(t + s)

In such a case δq =q ˙ since ∂q sδq = q(t + s) − q(t) = s | = sq˙ ∂t s=0 and we get, d d L(Q(s)) | = L ds s=0 dt Thus in this case F can be taken to be L itself. Thus the conserved quantity is ∂L δq − L ∂q˙ which is just the Hamiltonian H. Normally this also turns out to be the total energy (we 1 will see later that Hamiltonian is not always the total energy). If L = mq˙2 − V (q). The 2 Hamiltonian is then given by 1 1 mq˙ · q˙ − ( mq˙2 − V ) = mq˙2 + V = T + V = E 2 2 1 Consider a free particle for which L = mq˙2 which has translational symmetry. We can 2 take

qs(t) = q(t) + sa In this case δq = a and δq˙ = 0. Since q is cyclic, δL = 0. We therefore have F = 0 in this case which leads to conservation of momentum.