M.Sc. (MATHEMATICS) MECHANICS

MAL-513 MECHANICS

DIRECTORATE OF DISTANCE EDUCATION GURU JAMBHESHWAR UNIVERSITY OF SCIENCE & TECHNOLOGY HISAR-125001 1

TABLE OF CONTENTS LESSON No. NAME OF THE LESSON 1 Moment of Inertia-1 2 Moment of Inertia-2 3 Generalized co-ordinates and Lagrange’s Equations 4 Hamilton’s Equations of Motion 5 Canonical Transformations 6 Attraction and Potential

Author Prof. Kuldip Singh Department of Mathematics GJUS & T Hisar 2

Lesson: 1 Moment of Inertia-1 1.1 Some definations:- Inertia:- Inertia of a body is the inability of the body to change by itself its state of rest or state of uniform motion along a straight line. Inertia of motion:- It is the inability of a body to change by itself its state of motion. Moment of Inertia:- A quantity that measures the inertia of rotational motion of body is called rotational inertia or moment of inertia of body. M.I. is rotational analogue of mass in linear motion. We shall denote it by I. Let there are n particles of masses mi, then moment of inertia of the system is

O 2 2 2 I = m1 d1 + m2d2 +...+ mndn d1 m1 n d2 2 m2 = ∑midi i=1 d3 m3 2 dn ∴ I = Σ md mn

L where di are the ⊥ distances of particles from the axis. (i) M.I. in three dimension: - Let us consider a three dimensional body of volume

V. Let OL be axis of rotation. Consider an infinitesimal small element of mass dm, then mass of small element dm = ρdv where dv = volume of infinitesimal small element and O d ρ is the density of material. Then moment of inertia of body is dm I = d d 2 ∫∫∫ m v L or I = ∫∫∫ρd2 dv v

(ii) M.I. in two dimension O d Here mass of small element dm = ρdS d and moment of inertia is I = d d 2 m ∫∫ m S or I = ∫∫ρd2 dS L S where dS = surface area of small element O λ (iii) M.I. in one dimension Consider a body (a line or curve) in one d dimension. Consider a small element of length ds ds and mass dm. Then mass of small element, L dm = ρds 2 M.I. of small element = dmd ∴ M.I. of body I = d d2 ∫ m s or I = ∫ρd2 ds s Radius of gyration:- Radius of gyration of a body about a given axis is the ⊥ distance of point P from the axis where its whole mass of body were concentrated, the body shall have the same moment of inertia as it has with the actual distribution of mass. This distance is represented by K. When K is radius of gyration, I′ = I

2 2 2 2 ⇒ MK = m(r1 + r2 +...+ rn )

mn(r 2 + r 2 +...+ r 2 ) = 1 2 n n M(r 2 + r 2 +....+ r 2 ) ⇒ MK2 = 1 2 n n

r 2 + r 2 +...+ r 2 ⇒ K = 1 2 n n 4

where n is the number of particles of the body, each of mass ‘m’ and r1, r2… rn be the perpendicular distances of these particles from axis of rotation. Where M = m × n = total mass of body. m m r3 r1 K M m P r2

Hence radius of gyration of a body about a given axis is equal to root mean square distance of the constituent particles of the body from the given axis. Example:-M.I. of a uniform rod of length ‘2a’ about an axis passing through one end and ⊥ to the rod:- O 2a

A δx B x L Let M = mass of rod of length 2a. OL = axis of rotation passing through one end A and ⊥ to rod. M ∴ Mass per unit length of rod = 2a Consider a small element of breadth δx at a distance ‘x’ from end A. M ∴ Mass of this small element = δx 2a M ∴ M.I. of small element about axis OL or AL = δx x2 2a

2a M ∴ M.I. of rod about OL = ∫ x2 dx 0 2a

2a M ⎡x3 ⎤ ∴ I = ⎢ ⎥ 2a 3 ⎣ ⎦0 5

M 8a 4 I = = Ma2 2a 3 3

4 2 ⇒ IOL = Ma 3 Example:- M.I. of a rod about an axis passing through mid-point and ⊥ to rod

Here LL′ is the axis of rotation passing L through mid-point ‘0’ of rod having 2a length 2a. 0 Consider a small element of breadth δx at a distance ‘x’ from A δx B x mid-point of rod O. M L′ ∴ Mass of this small element = δx 2a M ∴ M.I. of small element about LL′ = δx. x2 2a

M a ∴ M.I. of rod about LL′ = ∫ x 2dx 2a −a

a 3 a 2M 2 M ⎡x ⎤ ∴ ILL′ = x dx = ⎢ ⎥ 2a ∫ a 3 0 ⎣ ⎦0 M Ma 2 = a 3 = 3a 3

1 2 ⇒ ILL′ = Ma 3 Example:-M.I. of a rectangular lamina about an axis (line) passing through centre and parallel to one side

PQ D C

N G L x

A δx B 6

Let ABCD be a rectangular lamina of mass ‘M’ and NL be the line about which M.I. is to be calculated. Let AB = 2a, BC = 2b M Mass per unit area of lamina = 4ab Consider an elementary strip PQ of length (BC = 2b) and breadth δx and at a distance ‘x’ from G and parallel to AD. M ∴ Mass of elementary strip = . 2b δx 4ab M = δx 2a b2 M.I. of this strip about NL = (Mass of strip) 3 M b2 = δx. 2a 3 ∴ M.I. of rectangular lamina about NL

M b2 a = ∫δx 2a 3 −a

M b2 Mb2 ⇒ I = 2a = 2a 3 3 Example:- M.I. of rectangular L lamina about a line ⊥ to lamina and D C passing through centre : d Let GL = axis of rotation passing through G y x centre ‘G’ and ⊥ to lamina ABCD. Consider a A small element of surface area δS = δxδy B

Here ⊥ distance of small element from axis GL, d = x 2 + y2

∴ Mass of small element = ρδx δy

M.I. of this small element about GL = ρ δx δy (x2 + y2)

b a ∴ M.I. of lamina = ∫∫ρ (x2 + y2)dx dy −−b a

b a = 4ρ ∫∫(x 2 + y2) dx dy 0 0

b 3 a b 3 ⎛ x 2 ⎞ ⎛ a 2 ⎞ = 4ρ ∫∫⎜ + xy ⎟ dy = 4ρ ⎜ + ay ⎟dy 0 ⎝ 3 ⎠0 o ⎝ 3 ⎠

b ⎛ a3 ay3 ⎞ = 4ρ⎜ y + ⎟ ⎜ 3 3 ⎟ ⎝ ⎠0 4ρa 4ρab = (a 2b + b3 ) = (a2 + b2) 3 3 M ∴ I = (a2 + b2) [using mass of lamina M = 4ρ ab] 3 1.2 Moments and products of inertia about co-ordinate axes:- (I) For a particle system Q z m P(x,y,z)

O(0,0,0 y ) d P′(x,y,0) x Consider a single particle P of mass ‘m’ having co-ordinates (x, y, z) Here d = ⊥ distance of particle P of mass m from z-axis

= PQ = OP′ = x 2 + y2

Therefore, M.I. of particle of mass ‘m’ about z-axis = md2 = m (x2 + y2) 8

∴ M.I. of system of particles about z-axis 2 2 2 Ioz = Σmd = Σm(x + y ) 2 2 And Standard notation for M.I about z-axis is C, i.e., C = Σm (x + y ) = Ioz Similarly, we can obtain M. I. about x and y-axis which are denoted as under: 2 2 About x-axis, A = Σ(y + z ) = Iox 2 2 About y-axis, B = Σm (z + x ) = Ioy Product of Inertia The quantities D = Σ myz E = Σ mzx and F = Σ mxy are called products of inertia w.r.t. pair of axes (oy, oz), (oz, ox) and (ox, oy) respectively. For a continuous body: The M.I. about z-axis, x-axis and y-axis are defined as under C = ∫∫∫ρ(x 2 + y2 ) dx dy dz V A = ∫∫∫ρ (y2 + z2)dx dy dz V B = ∫∫∫ρ (z2 + x2)dx dy dz V Similarly, the products of inertia w.r.t. pair of axes (oy, oz), (oz, ox) and (ox, oy) respectively are as under D = = ∫∫∫ρ yz dv ,E = ∫∫∫ρ zx dV and F = = ∫∫∫ρ xy dV V V V For laminas in xy plane, we put z = 0, then A = ∫∫ρ y2 dxdy S B = ∫∫ρ x2 dx dy S C = ∫∫ρ (x2 + y2)dxdy S D = E = 0, F = ∫∫ρ xy dx dy S 9

1.3 M.I. of a body about a line (an axis) whose direction cosines are <λ, μ, ν> :- Let aˆ is a unit vector in axis OL z whose direction cosines are <λ, μ, ν>. <λ, μ, ν> N Then d ρ P(x, y, z) aˆ = λˆi + μˆj + νkˆ ...(1) a ρ θ r Let P(x, y, z) be any point(particle) of mass of the body. 0 y ρ Then its position vector r is given by ρ OP = r = xˆi + yˆj+ zkˆ …(2)

⊥ distance of P from OL, x ρ d = PN = OP sinθ = | r ×aˆ | …(3)

⇒ d = (xˆi + yˆj+ zkˆ)×(λˆi + μˆj+ νkˆ)

= |(νy − μz) ˆi + (λz − νx)ˆj+ (μx − λy)kˆ |

= (νy − μz)2 + (λz − νx)2 + (μx − λy)2

⇒ d = λ2 (y2 + z2 ) + μ 2 (z2 + x 2 ) + ν2 (x 2 + y2 ) − 2μνyz − 2λνxy − 2λμxy

Therefore, M.I. of body about an axis whose direction cosine are λ, μ, ν is

2 2 2 2 2 2 2 2 2 IOL = Σm{λ (y + z ) + μ (z + x ) + ν (x + y ) − 2μνyz − 2λνxz − 2λμxy} 2 2 2 ⇒ IOL = Aλ + Bμ + Cν − 2μνD − 2λνE − 2λμF 1.4 Kinetic Energy (K.E.) of a body rotating about O:-

Let axis of rotation be OL through O, then angular velocity about OL is ρ w = w aˆ 1 ρ ρ Then K.E., T = Σ m (v. v) 2 1 ρ = Σm | v |2 2 1 ρ ρ ρ ρ ρ ⇒ T = = Σm | aˆ × r |2 w 2 [Θ v = w × r = waˆ × r 2 10

1 2 2 = w Σ md [using equation (3)] 2

1 2 ⇒ T = w IOL 2 This is the required expression for kinetic energy in terms of moment of inertia 1.5 Parallel axis theorem Statement:- For a body of mass ‘M’, we have C = C′ + Md2 where C′ = M. I of body about a line GL through C.G.( centre of mass) and parallel to z-axis C = M.I. of body about z-axis (i.e. a line parallel to GL) and at a distance ‘d’ from GL. z z′ L (x′,y′,z′) Proof : P(x,y,z) d G (x, y, z) N y′ x′

o y (0,0,0)

Let M = Mass of body and P is any point whose co-ordinates w.r.t. oxyz are (x,y,z),

G is the centre of mass whose co-ordinates w.r.t. oxyz are ( (x, y, z) ). Let us introduce a new co-ordinate system Gx′y′z′ through G and Co-ordinates of P ρ w.r.t. this system are (x′,y′,z′). Let rG be the position vector of G and ri position vector of mass mi w.r.t. oxyz system. Now by definition of centre of mass of body, ρ Σm r r = i i G M 11

when centre of mass cocides with origin at G w.r.t new co-ordinate system G x′y′z′, σ we have rG = 0 . Therefor

Σmi ri = 0 ⇒ Σ mi ri = 0 M Σm x' Σmy' Σmz' ⇒ = 0, = 0, = 0 M M M where ρ r'= x'ˆi + y'ˆj+ z'kˆ ρ and r = xˆi + yˆj+ zkˆ So, we have Σm x′ = Σmy′ = Σmz′ = 0 …(1)

Now d2 = (GN)2 = (OG)2 − (ON)2 = x 2 + y2 + z 2 − z 2

= x 2 + y2 …(2) Co-ordinates of P w.r.t. (ox, oy, oz) axes is (x, y, z) Co-ordinates of P w.r.t. (Gx′, Gy′, Gz′) axes is (x′, y′, z′) Then x = x + x', y = y + y', z = z + z′ Thus, M.I. about z-axis is C = Σm(x2 + y2)

2 2 = Σm[( x + x') + (y + y') ]

⇒ C = Σm [ x 2 + x'2 +2x x'+y2 + y'2 +2y y']

2 2 2 2 = Σm(x′ + y′ ) + Σm( x +Σy ) + 2x mx′ + 2y Σm y′

C = Σm (x′2 + y′2) + (x 2 + y2 ) Σm + 0 [from (1)]

⇒ C = C′ + Md2 [using (2) and Σm = M, total mass] Similarly, M.I. about x and y-axis are given by A = A′ + Md2 B = B′ + Md2 where d is perpendicular distance of P from x and y-axis 12

For Product of Inertia Here Product of Inertia w.r.t. pair (ox,oy) is F = Σmxy = Σm( x + x')(y + y')

= Σm( x y + x'y + x y'+x'y')

= Σm x′ y′ + x y Σm +Σx Σm y'+y m x′

= F′ + M x y + 0 [using (1)]

⇒ F = F′ + M xy Similarly, for products of Inertia w.r.t. pair (oy, oz) and (oz, ox) respectively, we have D = D′ + M y z and E = E′ + M z x . 1.6 Perpendicular axis theorem (For Two dimensional bodies or mass distribution) Statement:- The M.I. of a plane mass distribution (lamina) w.r.t. any normal axis is equal to sum of the moments of inertia about any two ⊥ axis in the plane of mass distribution (lamina) and passing through the intersection of the normal with the lamina. Proof : z Let ox, oy are the axes in the plane of lamina and oz be the normal axis, i.e., xy is the plane of lamina. o y Let C is the M.I. about ⊥ axis, i.e., oz axis Here to prove C = A + B By definition, M.I. of plane lamina about z-axis, x C = ∫∫ρ (x2 + y2) dS [for a continuous body] S = ∫∫ρ x2 dS + ∫∫ρ y2 dS S S ⇒ C = B + A For mass distribution, C = Σm (x2 + y2) = Σm x2 + Σmy2 13

⇒ C = B + A For two dimensional body, D = E = 0 and F = Σmxy. Converse of perpendicular axis theorem : Given C = A + B To prove it is a plane lamina. Proof:- Here A = Σm(y2 + z2) B = Σm(z2 + x2), C = ΣM(x2 + y2) Now given C = A + B ⇒ Σ(x2 + y2) = Σm (y2 + z2) + Σm(z2 + x2) 2 2 2 = Σm(y + 2z + x ) ⇒ Σmx2 + Σmy2 = Σmy2 + 2Σmz2 + Σmx2 ⇒ 2Σmz2 = 0 ⇒ Σmz2 = 0 for all distribution of mass. For a single particle of mass ‘m’, mz2 = 0 ⇒ z = 0 as m ≠ 0 ⇒ It is a plane mass distribution or it is a plane lamina. 1.7 Angular momentum of a rigid body about a fixed point and about a fixed axis:- The turning effect of a particle about the axis of rotation is called angular Momentum.

Let O be the fixed point and OL be an axis passing z ω L through the fixed point. ρ w = angular velocity about OL ρ ρ P(m) r = position vector of P(x, y, z) r O y (fixed ρ ˆ ˆ ˆ point) ⇒ r = OP = x i + y j+ z k x ρ ρ ρ Also linear velocity of P, v = w × r …(1) The angular momentum of body about O is ρ ρ ρ ρ ρ ρ H = Σ(r × mv) = Σ[r × m(w × r)] …(2) 14

→ ρ ρ ρ H = Σm [ r × (w × r)] ρρ ρ ρ ρ ρ = Σm [(r.r)w − (r.w)r]

[Θ AX(B × C) = (A. C) B − (A . B)C] ρ ρ ρ ρ = Σm r 2w − (r.w)r] ρ ρ ρ ρ ρ ⇒ H = (Σmr 2 )w − Σm(r.w)r …(3) ρ ˆ ˆ ˆ If H = h1 i + h 2 j+ h3 k ρ ˆ ˆ ˆ w = w1 i + w 2 j+ w 3 k …(4) ρ ρ Then r.w = w1x + w 2 y + w 3z ∴ from (3), ˆ ˆ ˆ 2 ˆ ˆ ˆ h1 i +−h 2 j+ h3 k = (Σm r )(w1i + w 2 j+ w 3 k) ˆ ˆ ˆ Σm(w1x + w2y + w3z) (x i + y j+ zk)

Equating coefficients of ˆi on both sides, 2 2 2 h1 = Σm (x + y + z ) w1 − Σm(w1x + w2y + w3z)x 2 2 2 2 = Σm(y + z )w1 + Σm x w1 − Σmw1 x − Σm(w2y + w3z)x 2 2 = Σm(y + z ) w1 − (Σm xy) w2 − (Σm xz)w3

∴ h1 = Aw1 − Fw2 − Ew3 Similarly,

h2 = Bw2 − Dw3 − Fw1

h3 = Cw3 − Ew1 − Dw2 …(5)

⎡h1 ⎤ ⎡ A − F − E⎤ ⎡w1 ⎤ ⎢h ⎥ ⎢ F B D⎥ ⎢w ⎥ ⎢ 2 ⎥ = ⎢− − ⎥ ⎢ 2 ⎥ ⎣⎢h3 ⎦⎥ ⎣⎢− E − D C ⎦⎥ ⎣⎢w 3 ⎦⎥ Inertia matrix (symmetric 3 × 3 matrix) 1.8 Principal axis and their determination ρ ρ Definition :- If the axis of rotation w is parallel to the angular momentum H , then the axis is known as principal axis. 15

ρ ρ If w =| w |aˆ = w aˆ ρ ρ ρ Hˆ =| H |aˆ ⇒ H = n w , where n is a constant ⇒ H = nw 1.8.1 Theorem:- Prove that in general, there are three principal axes through a point of rigid body. Proof : For principal axis, ρ ρ H = n w ⇒ H = nw …(1) ρ ρ Let H = Haˆ , w = w aˆ where aˆ is a unit vector along principal axis of body through O. ρ By definition of H , ρ ρ ρ H = Σ(r × mv) ρ ρ ρ ρ ρ ⇒ H = (Σmr2 ) w − Σm(r.w)r ρ ρ Using H = n w , we get ρ ρ ρ ρ ρ n w = (Σmr 2 )w − Σm(r.w)r ρ Using w = w aˆ , ρ ρ n w aˆ = Σmr2 waˆ − Σm(r.w aˆ)r Cancelling w on both sides & rearranging, ρ ρ (Σmr2 − n)aˆ = Σm(r.aˆ)r …(2) ρ Let r = x ˆi + yˆj+ zkˆ, aˆ = λˆi + μˆj+ νkˆ …(3) where <λ, μ, ν> are direction cosine of principal axis.

⇒ (Σmr2 − n) (λ ˆi + μˆj+ ν kˆ) = Σm[λx + μy + νz) (x ˆi + yˆj+ zkˆ ) ]

Equating coefficients of ˆ i on both sides, ⇒ [Σm(x2 + y2 + z2)−n] λ = Σm(λx2 + μxy + νxz) ⇒ [Σm(y2 + z2)−n]λ = Σm[μxy + νxz] [canceling Σm λx2 on both sides] ⇒ (A − n)λ − Fμ − Eν = 0 Similarly (B −n)μ − Dν − Fλ = 0 …(4) 16

(C −n)ν − Eλ − Dμ = 0 or (A −n)λ − Fμ − Eν = 0 −Fλ + (B−n)μ − D ν = 0 …(5) −Eλ − Dμ + (C −n)ν = 0 Equation (5) has a non-zero solution only if A − n − F − E − F B − n − D = 0 …(6) − E − D C − n

This determinental equation is a cubic in n and it is called characteristic equation of symmetric inertia matrix. This characterstic equation has three roots n1, n2, n3 (say), so n1, n2, n3 are real. Corresponding to n = (n1, n2, n3) (solving equation (5) or (6) for <λ, μ, ν>). Let the values of (λ, μ, ν) be

(λ1, μ1, ν1) → n = n1

(λ2, μ2, ν2) → n = n2

(λ3, μ3, ν3) → n = n3

These three sets of value determine three principal axes aˆ1,aˆ 2 , aˆ 3 given by ˆ ˆ ˆ aˆ p = λpi + μ p j+ νp k where p = 1, 2, 3. 1.8.2 Theorem :- Three principal axes through a point of a rigid body are mutually orthogonal.

Proof : Let the three principal axes corresponding to roots n1, n2, n3 of characteristic equation A − n − F − E − F B − n − D = 0 − E − D C − n be aˆ1, aˆ 2 , aˆ 3 .

Let nˆ 1,nˆ 2 , nˆ 3 are all different.

Then from equation, ρ ρ (Σm r2−n) aˆ = Σm(r.aˆ)r We have 2 ρ ρ ρ (Σmr − n1) aˆ1 = Σm(r.a1)r …(1) 2 ρ ρ (Σmr − n2) aˆ 2 = Σm(r.aˆ 2 )r …(2) 2 ρ ρ (Σmr − n3) aˆ 3 = Σm(r.aˆ 3 )r …(3)

Multiply scalarly equation (1) by aˆ 2 and equation (2) with aˆ1 and then substracts, we get

(n1 − n2) aˆ1.aˆ 2 = 0

⇒ aˆ1.aˆ 2 = 0 as n1 ≠ n2

Similarly aˆ 2 , aˆ 3 = 0 and aˆ 3.aˆ1 = 0

⇒ aˆ1, aˆ 2 ,aˆ 3 are mutually orthogonal.

Remarks :- (i) If n1≠ n2 ≠ n3 then there are exactly three mutually ⊥ axis through 0.

(ii) If n2 = n3 (i.e. two characteristic aˆ1 roots are equal). There is one principal axis corresponding to n1 through 0. 0 Then every line through 0 & ⊥ to this

aˆ1 is a principal axis. Infinite set of principal axis with the condition that

aˆ1 is fixed. (iii) If n = n = n , then 1 2 3 aˆ 3 ˆ Any three mutually ⊥ axes a 2 through O(centre of sphere) are aˆ1 principal axes. Z

1.9 Moments and products of Inertia about principal axes and hence to find angular momentum of body. aˆ 3 ρ P(X,Y,Z) Let aˆ1,aˆ 2 , aˆ 3 are the principal axes. r O Y Let us take co-ordinates axes along the aˆ 2

aˆ1 principal axes. X ρ r = OP = Xaˆ1 + Yaˆ 2 + Zaˆ 3 ∴ r2 = X2 + Y2 + Z2 From equation ρ ρ (Σmr2−n) aˆ = Σm(r.aˆ)r We have 2 ρ ρ (Σmr − n1) aˆ1 = Σm(r.aˆ 2 )r …(1) 2 ρ ρ (Σmr − n2) aˆ 2 = Σm(r.aˆ 2 )r …(2) 2 ρ ρ (Σmr − n3) aˆ 3 = Σm(r.aˆ 3 )r …(3) From (1), 2 (Σmr − n1)]aˆ1 = Σm[(Xaˆ1 + Yaˆ 2 + Zaˆ 3 ).aˆ1[Xaˆ1 + Yaˆ 2 + Zaˆ 3

= Σm×()Xaˆ1 + Yaˆ 2 + Zaˆ 3

Equating coefficients of aˆ1 aˆ 2 , aˆ 3 ,

2 2 2 2 Σm(X + Y + Z ) − n1 = ΣmX 0 = ΣmXY 0 = ΣmXZ …(4) 2 2 or n1 = Σm(Y + Z ) = A* & F* = 0, E* = 0 Similarly from (2) & (3), we get

n2 = B*, D* = 0, F* = 0

& n3 = C*, E* = 0, D* = 0 19

where A*, B*, C* are M.I. and D*, E*, F* are product of Inertia about principal axes. Inertia matrix for principal axes through O is ⎛n 0 0 ⎞ ⎛A* 0 0 ⎞ ⎜ 1 ⎟ ⎜ ⎟ ⎜ 0 n 2 0 ⎟ = ⎜ 0 B* 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 0 0 n3 ⎠ ⎝ 0 0 C*⎠

Definition: Three mutually ⊥ lines through any point of a body which are such that the product of inertia about them vanishes are known as principal axes. ρ Expression for angular momentum ( H ) Here D* = E* = F* = 0, then from equation,

h1 = Aw1 − Fw2 − Ew3 we have

h1 = A*w1 − F*w2 − E* w3

⇒ h1 = A*w1 [Θ F* = E* = 0]

Similarly h2 = B* w2, h3 = C* w3 ρ ∴ H = h1 aˆ1 + h 2 aˆ 2 + h3 aˆ 3

= A* w1 aˆ1 + B* w 2aˆ 2 + C*w 3aˆ 3 where (w1, w2, w3) are components of angular velocity about ( aˆ1,aˆ 2 ,aˆ 3 ). A*, B*, C* are also called principal moments of inertia. 1.10 Momental Ellipsoid:- We

know that M.I., IOL of a body

about the line whose d.c.’s are

<λ, μ, ν> is

2 2 2 IOL = I = Aλ + Bμ + Cν − 2Dμν

− 2Eνλ − 2Fλμ …(1) 20

ρ Let P(x,y, z) be any point on OL and OP = R, then R = R(λˆi + μˆj+ νkˆ) = xˆi + yˆj+ z kˆ x y z ⇒ λ = , μ = , ν = …(2) R R R Now let P moves in such a way that IR2 remains constant, then from (1), (2), we get Ax2 + By2 + Cz2 − 2Dyz − 2Ezx − 2Fxy = IR2 = constant Since coefficients of x2, y2, z2 i.e. A, B, C all are positive, this equation represents an ellipsoid known as momental ellipsoid. Example:- A uniform solid rectangular block is of mass ‘m’ and dimension 2a × 2b × 2c. Find the equation of the momental ellipsoid for a corner ‘O’ of the block, referred to the edges through O as co-ordinates axes and hence determine M.I. about OO′ where O′ is the point diagonally opposite to O. Solution : z

(0, 0, 2c)

0′(2a,2b,2c)

2c 2b (0,2b,0) 2a 0(0,0,0) y

(2a,0,0) x

Taking x, y, z axes along the edges of lengths 2a, 2b, 2c, we obtain A = ∫∫∫ρ(y2 + z2 )dv V

2a2b2c 2 2 = ∫∫∫ρ (y + z )dz dy dx 0 0 0

2a2b 3 2c ⎛ 2 z ⎞ = ρ⎜ y z + ⎟ dy dx ∫∫ ⎜ 3 ⎟ 0 0 ⎝ ⎠0

2a2b ⎛ 1 ⎞ = ρ ∫∫⎜ y2 2c + 8c3 ⎟ dy dx 0 0 ⎝ 3 ⎠ 21

2a2b ⎛ 4 ⎞ = ρ. 2c ∫∫⎜ y2 + c2 ⎟ dy dx 0 0 ⎝ 3 ⎠

2b 2a⎛ y3 4 ⎞ = ρ. 2c ⎜ + c2 y⎟ dx ∫⎜ 3 3 ⎟ 0 ⎝ ⎠0

ρ2c 2a = ∫(8b3 + 4c2 2b) dx 3 0

2c 2a = ρ 8b ∫ (b2 + c2 )dx 3 0

16bc 2 2 4 2 2 = ρ (b + c )2a = (8abc ρ) (b + c ) 3 3 4M ⇒ A = (b2 + c2) 3 4M 4M Similarly B = (c2 + a 2 ), C = (a 2 + b2 ) 3 3

2a2b2c D = ∫∫∫ρyz dV = ρ ∫ ∫ ∫ yz dz dy dx v 0 0 0

2c 2a2b ⎡z2 ⎤ = π y⎢ ⎥ dy dx ∫∫ 2 0 0 ⎣ ⎦0

ρ 2a2b = ∫∫y(4c2 ) dy dx 2 0 0

2a2b 2a 2 2b 2 2 ⎡ y ⎤ = 2c ρ ydy dx = 2c ρ ⎢ ⎥ dx ∫∫ ∫ 2 0 0 0 ⎣ ⎦0

2a 2a = c2 ρ ∫∫4b2 dx = 4b2c2 ρ dx 0 0 ⇒ D = 4b2c2ρ. 2a = (8abc ρ)bc = M bc Similarly E = Mca, F = Mab Using these in standard equation of momental ellipsoid, we get 22

4M [(b2 + c2)x2 + (c2 + a2) y2 + (a2 + b2)z2] 3 − 2M[bcyz + cazx + ab xy] = IR2 …(1) which is required equation of momental ellipsoid. To find M.I. about OO′ :- using x = 2a, y = 2b, z = 2c as O′(2a, 2b, 2c) and R2 = 4(a2 + b2 + c2) from (1), 4M [(b2 + c2 )4a 2 + (c2 + a 2 )4b2 + (a 2 + b2 )4c2 ]−8M(b2c2 + c2a 2 + a 2b2 ) 3 IOO′ = 4(a 2 + b2 + c2 )

8M ⎡2(2a 2b2 + 2b2c2 + 2c2a 2 ) − 3(a 2c2 + a 2b2 + b2c2 )⎤ ⇒ IOO′ = ⎢ 2 2 2 ⎥ 3 ⎣ 4(a + b + c ) ⎦ 2M (b2c2 + c2a 2 + a 2b2 ) ⇒ IOO′ = 3 (a 2 + b2 + c2 )

Lesson-2 Moment of Inertia-2

2.1 Equimomental Systems:- Two systems are said to be equimomental if they have equal M.I. about every line in space. 2.1.1 Theorem:- The necessary and sufficient condition for two systems to be equimomental are : (i) They have same total mass. (ii) They have same centroid. (iii) They have same principal axes. Proof:- Part A : The condition (i) to (iii) are sufficient. Here we assume that if (i) to (iii) hold, we shall prove that two systems are equimomental. Let M be the total mass of each system. II system I system z

λ′(<λ, μ, ν>) G y M x M h

λ(<λ, μ, ν>)

Let G be the common centroid of both the system. Let A*, B*, C* be the principal M.I. about principal axes through G for both the systems. Let λ be any line in space with d.c. <λ, μ, ν>. We draw a line λ′ similar to λ passing through G. Let h = ⊥ distance of G fromλ. M.I. about λ′ for both the system is 2 2 2 Iλ′ = A*λ + B*μ + C*ν [Θ Product of inertia about principal axes i.e. D* = E* = F* = 0] 24

So by parallel axes theorem, the M.I. of both the system about λ is 2 Iλ = Iλ′ + Mh 2 2 2 2 ⇒ Iλ = A*λ + B*μ + C*ν + Mh Hence both the system have same M.I. about any line of space. So they are equimomental. Part B:- The conditions are necessary. Here we assume that the two systems are equimomental and derive condition (i) to (iii). Let M1 and M2 be the total masses of the two systems respectively and G1 & G2 are their centroid respectively.

Condition (i) λ h M1

G1 G2

Since the systems are equimomental i.e. they have same M.I., ‘I’ (say) about line

G1G2(in particular). Let λ be the line in space which is parallel to G1 G2 at a 2 distance h. Then by parallel axes theorem, M.I. of Ist system about λ = I + M1h and 2 M.I. of IInd system about λ = I + M2h . Since the two systems are equimomental, therefore we have, 2 2 I + M1h = I + M2h

⇒ M1 = M2 = M (say) This implies that both the systems have same total mass.

Condition (ii) H1 H2

G1 G2 M

M 25

Let G1H1 and G2H2 be two parallel lines each being ⊥ to G1 G2. Let I* be the M.I. of either system about a line G1H1 and ⊥ to G1G2 (through G1) Using parallel axes theorem, 2 M.I. of Ist system about G2H2 = I* + M (G1G2) 2 M.I. of IInd system about G2H2 = I* − M (G1G2) As the systems are equimomental, therefore 2 2 I* + M (G1G2) = I* M(G1G2) 2 ⇒ (G1G2) = 0 as M ≠ 0

⇒ G1 = G2 = G (say) ⇒ Both the systems have same centroid. Condition (iii):- Since the two systems are equimomental, they have the same M.I. about every line through their common centroid. Hence they have same principal axes and principal moments of inertia. 2.2 Coplanar distribution:- 2.2.1Theorem:- (i) Show that for a two dimensional mass distribution (lamina), one of the principal axes at O is inclined at an angle θ to the x-axis through O 2F such that tan 2θ = B − A where A, B, F have their usual meanings. (ii) Show that maximum and minimum values of M.I. at O are attained along principal axes. OR Theorem: - For a 2-D mass distribution (lamina), the value of maximum and minimum M.I. about lines passing through a point O are attained through principal axes at O. y′ Proof :- y (x′,y′) x′ x′ P(x,y) y′ N L Let us consider an arbitrary particle of θmass m at P whose coordinates w. r.t. axes x 0(0, 0) 26

through O are (x,y), then for mass distribution, we have M.I. about x-axis i.e. A = Σmy2 M.I. about y-axis i.e. B = Σmx2 …(1) and Product of inertia F = Σmxy We take another set of ⊥ axes ox′, oy′ such that ox′ is inclined at an angle θ with x- axis. Then equation of line ox′ is given by y = x tanθ ⇒ y cos θ − x sin θ = 0 …(2) π Changing θ to θ + , equation of oy′ is 2 − y sin θ − x cos θ = 0 ⇒ y sin θ + x cos θ = 0 …(3) Let P(x′, y′) be co-ordinates of P relative to new system of axes ox′, oy′, then PL = y′ = length of ⊥ from P on ox′ ycosθ − xsinθ = cos2 θ + sin 2 θ = y cos θ − x sin θ …(4) Similarly x′ = PN = length of ⊥ from P on oy′ ysinθ + x cosθ = cos2 θ + sin 2 θ = y sinθ + x cos θ …(5) Therefore, M.I. of mass distribution (lamina) about ox′ is 2 2 Iox′ = Σmy′ = Σm(y cos θ − x sin θ) 2 2 2 2 = Σ m(y cos θ + x sin θ − 2xy sin θ cos θ) = cos2θ Σ my2 + sin2θ Σ mx2 − 2 sinθ cosθ Σmxy = A cos2θ + B sin2θ − F sin2θ …(6) Similarly M.I. of mass distribution (lamina) about oy′is given by 27

2 ⎛ π ⎞ 2 ⎛ π ⎞ ⎛ π ⎞ Ioy′ = A cos ⎜ + θ⎟ + B sin ⎜ + θ⎟ − F sin 2⎜ + θ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ = A sin2θ + B cos2θ + F sin2θ …(7) Product of inertia w.r.t pair of axes (ox′, oy′),

Ix′y′ = Σmx′y′ = Σm(y sinθ + x cosθ) (y cosθ − x sinθ) 2 2 ⇒ Ix′y′ = sinθ cosθ Σmy − sinθ cosθ Σ mx − sin2θ Σmxy + cos2θ Σmxy = A sinθ cosθ − B sinθ cosθ + (cos2θ − sin2θ)F sin 2θ = (A−B) + F cos2θ …(8) 2 The axes ox′, oy′ will be principal axes if

Ix′y′ = 0 Using equation (8), 1 (A−B) sin2θ + F cos2θ = 0 2 2F ⇒ tan2θ = B − A 1 2F ⇒ θ = tan −1 …(9) 2 B − A This determines the direction of principal axes relative to co-ordinates axes. We shall now show that maximum/minimum (extreme) values of Iox′, Ioy′ are obtained when θ is determined from (9),

We rewrite, Iox′ and Ioy′ as 1 1 Iox′ = (A + B) − [(B−A) cos 2θ + 2 F sin 2θ] 2 2 1 1 Ioy′ = (A +B) + [(B−A) cos 2θ + 2F sin 2θ] 2 2

For maximum and minimum value of Iox′, Ioy′, 28

d d (I ) = 0 and (I ) = 0 dθ ox' dθ oy' d i.e. [(B−A) cos 2θ + 2F sin 2θ] = 0 dθ ⇒ −(B −A)2 sin 2θ + 4F cos 2θ = 0 2F ⇒ tan 2θ = …(11) B − A d Similarly [(B−A) cos 2θ + 2F sin 2θ] = 0 dθ ⇒ −(B−A) 2 sin 2θ + 4F cos 2θ = 0 2F ⇒ tan 2θ = B − A

So extreme values of Iox′ and Ioy′ are attained for θ given by equation (11) already obtained in (9). Therefore, the greatest and least values of M.I. for mass distribution (lamina) through O are obtained along the principal axes. The extreme values are obtained as under sin 2θ 2F We have, tan 2θ = = cos2θ B − A sin 2θ cos2θ 1 ⇒ = = 2F B − A 4F2 + (B − A)2

2F ⇒ sin 2θ = 4F2 + (B − A)2

B − A & cos 2θ = 4F2 + (B − A)2

Now writing 1 1 Iox′ = (A + B) − [(B−A) cos 2θ + 2 F sin 2θ] 2 2

Using values of cos 2θ and sin 2θ , we obtain the extreme values of Iox′ and Ioy′ as under 29

1 1 ⎡ (B − A)(B − A) 4F2 ⎤ Iox′ = (A + B) − ⎢ + ⎥ 2 2 2 2 2 2 ⎣⎢ 4F + (B − A) 4F + (B − A) ⎦⎥

1 1 ⎡ (B − A)2 + 4F2 ⎤ = (A + B) − ⎢ ⎥ 2 2 2 2 ⎣⎢ 4F + (B − A) ⎦⎥ 1 1 = (A + B) − [ 4F2 + (B − A)2 ] 2 2

1 1 2 2 Similarly Ioy′ = (A + B) + [ 4F + (B − A) ] 2 2 Example 1:- A square of side ‘a’ has particles of masses m, 2m, 3m, 4m at its vertices. Show that the principal M. I. at centre of the square are 2ma2, 3ma2, 5ma2. Also find the directions of principal axes. y Solution : y′ ⎛ a a ⎞ ⎛ − a a ⎞ R⎜ , ⎟ S⎜ , ⎟ ⎝ 2 2 ⎠ ⎝ 2 2 ⎠ 4m 3m x′

θ 0 x

m 2m ⎛ − a − a ⎞ ⎛ a − a ⎞ P⎜ , ⎟ Q⎜ , ⎟ ⎝ 2 2 ⎠ ⎝ 2 2 ⎠

Taking origin O at the centre of square and axes as shown in the figure, we have A = M.I. of system of particles about x-axis.

4 2 2 2 2 2 ⎛ − a ⎞ ⎛ − a ⎞ ⎛ a ⎞ ⎛ a ⎞ = ∑mi yi = m⎜ ⎟ + 2m⎜ ⎟ + 3m⎜ ⎟ + 4m⎜ ⎟ i=1 ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ 5 ⇒ A = ma2 …(1) 2

2 2 2 2 2 ⎛ − a ⎞ ⎛ − a ⎞ ⎛ a ⎞ ⎛ a ⎞ B = Σmixi = m⎜ ⎟ + 2m⎜ ⎟ + 3m⎜ ⎟ + 4m⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ 30

5 ⇒ B = ma2 …(2) 2 ∴ C = B + A = 5ma2 For a two-dimensional mass distribution, D = E = 0 and ⎛ − a ⎞⎛ − a ⎞ ⎛ a ⎞⎛ − a ⎞ ⎛ a ⎞⎛ a ⎞ ⎛ − a ⎞⎛ a ⎞ F = Σmixiyi = m⎜ ⎟⎜ ⎟ + 2m⎜ ⎟⎜ ⎟ + 3m⎜ ⎟⎜ ⎟ + 4m⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠

2 2 2 2 ma ⎛ 2ma ⎞ 3ma 4ma 2 3 2 ⇒ F = + ⎜ ⎟ + − = ma − ma 4 ⎝ 4 ⎠ 4 4 2 −1 ⇒ F = ma2 2 Let ox′, oy′ be the principal axes at O s. t. ∠x′ox = θ. 2 2 Then, we have Iox′ = A cos θ − 2F sinθ cosθ + B sin θ 2 2 Ioy′ = A sin θ + 2F sinθ cosθ + B cos θ 1 and Ix′y′ = (A −B) sin2θ + F cos2θ 2

Since ox′ & oy′ are principal axes, therefore Ix′y′ = 0 1 ⇒ (A−B) sin 2θ + F cos 2θ = 0 …(3) 2 2F ⇒ tan 2θ = B − A 5 (3) ⇒ cos 2θ = 0 [Θ A = B = ma2] 2 π π ⇒ 2θ = ⇒ θ = 2 4 ⇒ Diagonals OR & OS are principal axes.

5 1 ⎛ − ma 2 ⎞ 1 5 1 2 ⎛ ⎞ ⎜ ⎟⎛ ⎞ 2 ⎛ ⎞ Therefore, IOR = ma ⎜ ⎟ − 2⎜ ⎟⎜ ⎟ + ma ⎜ ⎟ [using equation (I)] 2 ⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠ 2 ⎝ 2 ⎠ 2 ⇒ IOR = 3ma

2 ⎡ 5 2 ⎛ 1 ⎞ 1 2 ⎛ 1 ⎞ 5 2 ⎛ 1 ⎞⎤ and Ios = 2ma ⎢Θ Ioy' = ma ⎜ ⎟ − 2. ma ⎜ ⎟ + ma ⎜ ⎟⎥ ⎣ 2 ⎝ 2 ⎠ 2 ⎝ 2 ⎠ 2 ⎝ 2 ⎠⎦ 31

M.I. about z-axis is C = B + A 2 2 ⇒ C = IOR+ IOS = 3ma + 2ma ⇒ C = 5ma2. Example 2 :- Show that a uniform rod of mass ‘M’ is equimomental to three particles situated one at each end of the rod and one at its middle point, the masses M M 2M of the particle being , and respectively. 6 6 3 Solution:- Let AB = 2a is the length of rod having mass ‘M’. L

A G B

2M M M 3 6 6

Let m, M−2m, m are the masses at A, G, B respectively. This system of particles has same centroid and same total mass M. This system of particles has the same M.I. (i.e. each zero) about AB, passing through common centroid ‘G’. Therefore, systems are equimomental. To find m:- we take M.I. of two systems (one system is rod of mass ‘M’) and other system consists of particles. Ma 2 M.I. of rod about GL = 3 M.I. of particles about GL = ma2 + 0 + ma2 = 2ma2 As systems are equimomental, Ma 2 ∴ 2ma2 = 3 M ⇒ m = 6 M 2M & M − 2m = M− = 3 3 M 2M M So masses of particles at A, G, B are , , respectively. 6 3 6 32

Example 3 :- Find equimomental system for a uniform triangular lamina. Solution:- Let M = Mass of Δ lamina. A Let ⊥ distance of A from BC = h B′ C′ δx i.e. AD = h First find M.I. of Δ lamina ABC about BC. x 1 M = ah σ, where σ = surface density of lamina B D C 2 a M ⇒ σ = (density = Mass/area) ⎛ ah ⎞ ⎜ ⎟ ⎝ 2 ⎠ B'C' h − x Now = BC h a(h − x) ⇒ B′C′ = = length of strip h a(h − x) Area of strip B′C′ = δx h M a(h − x)δx Mass of strip = . ⎛ ah ⎞ h ⎜ ⎟ ⎝ 2 ⎠ 2M = (h − x)δx h 2 2M ∴ M.I. of strip B′C′ about BC = (h−x)x2 δx h 2 ∴ M.I. of Δ lamina ABC about BC

h 2M = (h − x) x2 dx ∫ 2 0 h h 2M ⎡hx3 x 4 ⎤ = ⎢ − ⎥ 2 3 4 h ⎣ ⎦0

2M ⎛ − h 4 h 4 ⎞ 2M h 4 = ⎜ + ⎟ = 2 ⎜ ⎟ 2 h ⎝ 4 3 ⎠ h 12 1 ⇒ I = Mh2 …(1) 6 33

Now we apply this result to general case of finding M.I. about any line λ in the plane of lamina.

D′ λ h1 D h λ′ 2 A h3 I B G G

Let h1, h2, h3 are length of ⊥ drawn from corners (or points) A, B, C respectively of

ΔABC. s. t. h1 < h2 < h3 We extend BC to meet a point ‘D’ on line λ′. We draw a line λ′ through A and parallel to λ.

Distances of C and B from λ′ are h3 − h1, h2 − h1

Let M1 is the mass of ΔACD and M2 is the mass of ΔABD

This ⇒ M = M1− M2 1 σ AD(h − h ) M 3 1 and 1 = 2 M 1 2 σ. AD(h − h ) 2 2 1 M h − h ⇒ 1 = 3 1 M2 h 2 − h1 M M M − M = M ⇒ 1 = 2 = 1 2 h3 − h1 h 2 − h1 h3 − h 2

M(h3 − h1) M(h 2 − h1) ⇒ M1 = , M2 = …(2) h3 − h 2 h3 − h 2

We denote Iλ as the M. I. of ΔABC about λ and Iλ′ as the M.I. of ΔABC about λ′ and IG as the M.I. of ΔABC about a line parallel to λ or λ′ through centre of mass (G) of ΔABC. So then

Iλ′ = M.I. of ΔACD − M.I. of ΔABD 34

1 2 1 2 = M1(h3 − h1) − M2(h2 − h) 6 6

M ⎡(h − h )3 − (h − h )3 ⎤ = ⎢ 3 1 2 1 ⎥ [using equation (2)] 6 ⎣ h3 − h 2 ⎦

M ⎡(h3 − h 2 )⎤ 2 2 = ⎢ ⎥ [(h3 − h1) + (h2 − h1) + (h3 − h1) (h2 − h1)] 6 ⎣(h3 − h 2 )⎦ [Θ a3 − b3 = (a −b) (a2 + b2 + ab)] M = []h 2 + h 2 − 2h h + h 2 + h 2 − 2h h + h h − h h − h h + h 2 6 3 1 1 3 2 1 1 2 2 3 1 2 1 3 1 M = [3h 2 + h 2 + h 2 + h h − 3h h − 3h h ] …(3) 6 1 2 3 2 3 3 1 1 2 (h + h + h ) Now ⊥ distance of G from λ = 1 2 3 …(4) 3

⎛ h1 + h 2 + h3 ⎞ and ⊥ distance of G from λ′ = ⎜ − h1 ⎟ …(5) ⎝ 3 ⎠ Using parallel axes theorem,

M 2 Iλ = IG + (h1 + h2 + h3) …(6) 9

M 2 and Iλ′ = IG + (h2 + h3 − 2h1) …(7) 9

M 2 2 Iλ = (3 h + h + h + h h − 3h h − 3h h + 4h − 2c + h h ) 6 1 2 3 2 3 1 3 1 2 1 3

M 2 (7) ⇒ IG = Iλ′ − (h2 + h3 − 2h1) …(8) 9 Put equation (8) in (6),

M 2 M 2 Iλ = Iλ′ + (h1 + h2 + h3) − (h2 + h3 − 2h1) 9 9

M 2 2 2 = [3h + h + h + h2 h3 − 3h3h1 − 3h1h2] 6 1 2 3

M 2 M 2 − (h2 + h3 −2h1) + (h1 + h2 + h3) 9 9 35

M 2 2 2 M 2 2 2 = [3h + h + h + h h − 3h h3 − 3h1h2] + [ h + h + h + 2h1h2 6 1 2 3 2 3 1 9 1 2 3

2 2 2 + 2h2h3 + 2h3h1 − h 2 − h3 − 4h1 − 2h 2h3 + 4h1h3 +4h1h2]

M 2 2 2 ⇒ Iλ = [h + h + h + h1 h2 + h2 h3 + h1 h3] 6 1 2 3 ⎡ 2 2 2 ⎤ M ⎛ h1 + h 2 ⎞ ⎛ h 2 + h3 ⎞ ⎛ h3 + h1 ⎞ ⇒ Iλ = ⎢⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎥ 3 ⎣⎢⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎦⎥ M = M. I. of mass placed at mid-point of A and B + 3 M M.I. of M.I. of mass placed at mid-point of B and C + 3 M M.I. of mass placed at mid-point of C and A. 3 M i.e. which is same as M.I. of equal particles of masses at the mid-points of sides 3 of ΔABC. Example 4:- Find equimomental system for a uniform solid cuboid. OR Show that a uniform solid cuboid of mass ‘M’ is equimomental with M M (i) Masses at the mid-points of its edges & at its centre. 24 2 M 2M (ii) Masses at its corners & at its centroid. 24 3 z Solution:- β7 α5(0,2a,2a) β8 (2a,2a,2a) β12 α7 (2a,0,2a)α6 β11 Z (a,a,a) β6 β3 G β 9 Y β 0 X 10 α2(0,2a,0) Let length of edge of cuboid = 2a β1 β2 y α1 β 5 x (2a,0,0) β4 Coordinates of mid-point of edges of cuboid are

β1 = (a, 0, 0), β2 = (0, a, 0), β3 = (0, 0, a), β4 = (2a, a, 0), β5 = (a, 2a, 0) 36

β6 = (0, 2a, a), β7 = )0, a, 2a), β8 = (a, 0, 2a), β9 = (2a, 0, a), β10 = (0, 2a, a)

β11 = (a, 2a, 2a), β12 = (2a, a, 2a) Let G be centroid & ρ is the density of cuboid, then M = ρV = ρ(2a)3 = 8ρa3, …(1) Now we find M.I. and Product of Inertia of cuboid about co-ordinates axes. Therefore, A = M.I. of cuboid about x-axis

2a2a2a 2 2 2 2 = ∫∫∫ρ(y + z )dv = ρ ∫ ∫ ∫( y + z )dx dy dz v 0 0 0 8 8 = a 2 (8ρa3 ) = Ma2 [using (1)] 3 3 8 Similarly B = M. I. of cuboid about y-axis = Ma2 3 8 C = M.I. of cuboid about z-axis = Ma2 3 Now D = product of inertia of cuboid w.r.t. pair (oy, oz)

2a2a2a ⇒ D = ∫∫∫ρ yz dx dy dz = (da3 ρ) a2 0 0 0 ⇒ D = Ma2 Similarly E= F = Ma2 M (i) Now consider a system of particles in which 12 particles each of mass are 24 situated at mid-point of edges. M i.e. at βi (i = 1 to 12) and a particle of mass at G. 2 ⎛ M ⎞ M Total mass of this system = 12⎜ ⎟ + ⎝ 24 ⎠ 2 M M = + = M 2 2

⇒ The two systems have same mass. Also the centroid of these particles at βi and G is the point G itself which is centroid of cuboid. ⇒ the two systems have same centroid. 37

Let A′ = M.I. of system of particles at (βi, G) about x-axis. M = Σm(y2 + z2) + (2a2) 2 M = [0 + a2 + a2 + a2 + 4a2 + 5a2 + 5a2 + 4a2 + a2 + 5a2 + 8a2 + 5a2] 24 M + (2a2) 2 M 64 8 ⇒ A′ = (40a 2 ) + Ma 2 = Ma 2 = Ma2 24 24 3 Similarly B′ = M.I. of system of particle about y-axis = Σm(z2 + x2) 8 ⇒ B′ = Ma2 3 8 Similarly C′ = Ma2 3 Now D′ = M.I. of system of particles w.r.t. (oy, oz) = Σmyz M M = [0 + 0 + 0 + 0a2 + 0 + 2a2 + 2a2 + 0 + 0 + 2a2 + 4a2 + 2a2]+ (a 2 ) 24 2 M M M M ⇒ D′ = (12a 2 ) + a 2 = a 2 + a 2 = Ma 2 24 2 2 2 Similarly E′ = F′ = Ma2 ⇒ Both the systems have same M.I. and product of inertia referred to co-ordinate axes through O. Using parallel axes theorem, both systems (i.e. cuboid & particles) have identical moments and products of inertia referred to parallel axes through common centroid G. So both the systems have same principal axes and principal M.I. Therefore both the systems are equimomental.

(ii) Now let A′′ = M.I. of system of particles at (αi & G) about x-axis M 2 = (0 + 4a2 + 4a2 + 4a2 + 8a2 + 4a2 + 8a2) + M (2a2) 24 3 38

M 4 4 4 = (32 Ma2) + Ma2 = Ma2 + Ma2 24 3 3 3 8 ⇒ A′′ = Ma2 3 8 Similarly B′′ = C′′ = Ma2 3 Also D′′ = P.I. of system of particles w.r.t. (oy, oz) axes M 2M = [0 + 0+ 0 + 0 + 4a2 + 0 + 4a2] + (a2) 24 3 8 2M 1 2 = Ma 2 + a 2 = Ma 2 + Ma 2 24 3 3 3 ⇒ D′′ = Ma2 Similarly E′′ = F′′ = Ma2 ⇒ Both the systems have same M.I. and product of inertia referred to co-ordinate axes through O. Using parallel axes theorem, both systems (i.e. cuboid & particles) have identical moments and products of inertia referred to parallel axes through common centroid G. So both the systems have same principal axes and principal M.I. Therefore both the systems are equimomental. Self Assessment Questions 1. Find Principal direction at one corner of a rectangular lamina of dimension 2a and 2b. 2. Find equimomental system for a parallelogram or parallelogram is equimomental with particles of masses M/6 at mid-points of sides of ||gm & M at the intersection of diagonals. 3

Lesson-3 Generalized co-ordinates and Lagrange’s Equations 3.1 Some definations 3.1.1 Generalized Co-ordinates A dynamical system is a system which consists of particles. It may also include rigid bodies. A Rigid body is that body in which distance between two points remains invariant. Considering a system of N particles of masses m1, m2,….mN or mi (1 ≤ i ≤ N). Let (x, y, z) be the co-ordinates of any particle of the system referred to rectangular axes. Let position of each particle is specified by n independent variables q1, q2,….qn at time t. That is

x = x(q1, q2,… qn;t)

y = y(q1, q2,… qn; t)

z = z(q1, q2,…qn; t)

The independent variables qj are called as “generalized co-ordinates” of the system. Here we use ‘⋅’ to denote total differentiation w.r.t. time. dq dq ∴ q& = 1 , q& = j etc. (j = 1,2, 3…..n) 1 dt j dt dq The n quantities q& = j are called “generalized velocities”. j dt

3.1.2 Holomonic system :- If the n generalized co-ordinate (q1, q2,…qn) of a given dynamical system are such that we can change only one of them say q1 to (q1 + δq1) without making any changes in the remaining (n−1) co-ordinates, the system is said to be Holonomic otherwise it is said to be “Non-Holonomic” system. 40

3.1.3 Result :- Let system consisits of N particles of masses mi (1 ≤ i ≤ N) & qj (j = 1 to n) are generalized coordinates. ρ Let ri be the position vector of particle of mass mi at time t. Then ρ ρ ri = ri (q1, q2,…, qn; t) …(1) ρ ρ dr Then r&= i i dt ρ ρ ρ ρ ρ& ∂ri dq1 ∂ri dq2 ∂ri dqn ∂ri ⇒ ri = + +...+ + ∂q1 dt ∂q2 dt ∂qn dt ∂t ρ ρ ρ ρ ρ& ∂ri ∂ri ∂ri ∂ri ⇒ ri = q&1 + q&2 +...+ q&n + ∂q1 ∂q2 ∂qn ∂t

We regard q&1, q&2...q&n , t as independent variables. So, ρ ρ ∂r& ∂r i = i ∂q&j ∂q j 3.1.4 Virtual displacement:- Suppose the particles of a dynamical system undergo a small instantaneous displacement independent of time, consistent with the constraint of the system and such that all internal and external forces remain unchanged in magnitude & direction during the displacement. 3.1.5 Virtual Work & Generalised forces:- Consider a dynamical system consisting of N particles of masses mi (1 ≤ i ≤ N). Let mi is the mass of ith particle ρ with position vector ri at time t, it undergo a virtual displacement to position ρ ρ ri + δri . ρ Let Fi = External forces acting on mi ρ Fi ' = Internal forces acting on mi ρ Therefore, virtual work done on mi during the displacement δri is ρ ρ ρ (Fi + Fi′).δri ∴ Total work done on all particles of system is, N ρ ρ ρ δW = ∑(Fi + Fi′).δri i=1 41

N ρ ρ N ρ ρ = ∑ Fi. δri + ∑ Fi′. δri i=1 i=1 where δW is called virtual work function. If internal foces do not work in virtual displacement, N ρ ρ then ∑Fi′. δri = 0 i=1 N ρ ρ so δW = ∑Fi . δri i=1 ρ ρ Let Xi, Yi, Zi are the components of Fi and δxi, δyi, δzi are the components of δri ρ ρ i.e. Fi = (Xi ,Yi ,Zi ) and δri = (δxi ,δyi ,δzi )

N Then δW = ∑(Xiδxi + Yi δyi + Ziδzi ) . If the system is Holonomic ,i.e., the co- i=1 ordinate qj changes to qj + δqj without making any change in other (n−1) co- ordinate. Let this virtual displacement take effect & suppose the corresponding work done on N ρ ρ the dynamical system to be Qj δqj , then Qj δqj = ∑Fi .δri i=1

Now, if we make similar variations in each of generalized co-ordinate qj, then n N ρ ρ δW = ∑∑Q j δq j = Fi .δri j==1 i 1

Here Qj are known as Generalised forces and δqj are known as generalised virtual displacements. 3.2 Constraints of Motion:- When the motion of a system is restricted in some way, constraints are said to have been introduced.

Constraints

Holonomic Non-Holonomic Co-ordinates are related by Co-ordinates through ρ ρ ρ equations f(r1, r2...rn ,t ) = 0 inequalities 42

Holonomic

Scleronomic Rheonomic

1. Time independent i.e. Time dependent i.e. derivative derivative w.r.t t is w.r.t. t is non-zero. zero

2. Independent of Constraints depends explicitly velocities (x&, y&,z&) on time

Example of Holonomic constraints ρ ρ 2 1. (ri − rj ) = constant ρ ρ 2. f(r1,..., rn , t ) = 0 Example of non-Holonomic constraints Motion of particle on the surface of sphere. Constraints of motion is (r2 − a2) ≥ 0 where a is radius of sphere.

3.3 Lagrange’s equations for a Holonomic dynamical system:- Lagrange’s equations for a Holonomic dynamical system specified by n-generalised co- ordinates qj (j = 1, 2, 3….. n) are

d ⎛ ∂T ⎞ ∂T ⎜ ⎟ − = Q , ⎜ & ⎟ j dt ⎝ ∂q j ⎠ ∂q j where T = K.E. of system at time t and Qj = generalized forces. ρ Consider a dynamical system consisting of N particles. Let mi, ri be the mass, position vector of ith particle at time t and undergoes a virtual displacement to ρ ρ position ri + δri . ρ Let Fi = External fore acting on mi ρ Fi′ = Internal force acting on mi

Then equation of motion of ith particle of mass mi is 43

ρ ρ ρ Fi + Fi′ = mi r&i …(1) The total K.E. of the system is,

N 1 ρ2 T = ∑mi r&i …(2) 2 i=1 ρ ρ d ⎡ ∂r ⎤ ⎛ ∂ n ∂ ⎞⎛ ∂r ⎞ Now i = ⎜ + q& ⎟⎜ i ⎟ …(3) ⎢ ⎥ ⎜ ∑ k ⎟⎜ ⎟ dt ⎣⎢∂q j ⎦⎥ ⎝ ∂t k=1 ∂qk ⎠⎝ ∂q j ⎠ ρ ρ ρ ρ ⎡ dr ρ ∂r n ∂r d ∂ n ∂r ⎤ i & i & i & i ⎢Θ = ri = + ∑qk ⇒ = + ∑qk ⎥ ⎣ dt ∂t k=1 ∂qk dt ∂t k=1 ∂qk ⎦ ρ ρ ρ d ⎛ ∂r ⎞ ∂ ⎛ ∂r ⎞ n ∂ ⎛ ∂r ⎞ (3) ⇒ ⎜ i ⎟ = ⎜ i ⎟ + q& ⎜ i ⎟ ⎜ ⎟ ⎜ ⎟ ∑ k ⎜ ⎟ dt ⎝ ∂q j ⎠ ∂t ⎝ ∂q j ⎠ k=1 ∂qk ⎝ ∂q j ⎠ ρ ρ ∂ ∂r n ∂ ⎛ ∂r ⎞ ⎛ i ⎞ & i = ⎜ ⎟ + ∑qk ⎜ ⎟ ∂q j ⎝ ∂t ⎠ k=1 ∂q j ⎝ ∂qk ⎠

ρ n ρ ∂ ⎛ ∂ri ⎞ ∂ ⎡ ∂ri ⎤ = ⎜ ⎟ + ⎢∑q&k ⎥ [Θ q&k are independent of qj] ∂q j ⎝ ∂t ⎠ ∂q j ⎣k=1 ∂qk ⎦

∂ ⎡ ∂ n ∂ ⎤ ρ = ⎢ + ∑q&k ⎥ ri ∂q j ⎣∂t k=1 ∂qk ⎦ ρ ∂ dr ∂ ρ ⎡ i ⎤ & = ⎢ ⎥ = (ri ) ∂q j ⎣ dt ⎦ ∂q j ρ ρ d ⎛ ∂r ⎞ ∂r& ⇒ ⎜ i ⎟ = i …(4) ⎜ ⎟ dt ⎝ ∂q j ⎠ ∂q j Also we know that ρ ρ ∂r& ∂r i = i …(5) ∂q&j ∂q j Consider ρ ρ ρ d ⎡ρ ∂r ⎤ ρ ∂r ρ d ⎛ ∂r ⎞ r i = r& i + r& ⎜ i ⎟ ⎢ i ⎥ i i ⎜ ⎟ dt ⎣⎢ ∂q j ⎦⎥ ∂q j dt ⎝ ∂q j ⎠ 44

ρ ρ ρ ρ ∂ri ∂r&i = r&i + r&i [using (4)] ∂q j ∂q j ρ ρ ρ ρ ∂r d ⎡ρ ∂r&⎤ ⎛ ρ ∂r&⎞ ⇒ r& i = ⎢r& i ⎥ − ⎜ r& i ⎟ [using (5)] i i & ⎜ i ⎟ ∂q j dt ⎣⎢ ∂q j ⎦⎥ ⎝ ∂q j ⎠

1 d ⎡ ∂ ρ ⎤ 1 ⎡ ∂ ρ2 ⎤ = ⎢ (r&)2 ⎥ − ⎢ (r& )⎥ & i i 2 dt ⎣⎢∂q j ⎦⎥ 2 ⎣⎢∂q j ⎦⎥

Multiplying both sides by mi & taking summation over i = 1 to N.

N ρ N ρ ∂r d ⎡ ∂ ⎛ 1 ρ2 ⎞⎤ ∂ ⎛ 1 ρ2 ⎞ ⇒ m r& i = ⎢ ⎜ m r& ⎟⎥ − ⎜ Σm r& ⎟ ∑∑i i & ⎜ i i ⎟ i i i==1 ∂q j dt ⎣⎢∂q j ⎝ 2 i 1 ⎠⎦⎥ ∂q j ⎝ 2 ⎠ ρ N ρ ∂r d ⎛ ∂T ⎞ ∂T ⇒ (F + F′) i = ⎜ ⎟ − [using (1) & (2)] …(6) ∑ i i ⎜ ⎟ i=1 ∂q j dt ⎝ ∂q j ⎠ ∂q j Also we have the relation, n N ρ ρ N ρ ρ ρ δW = ∑∑∑Q j δq j = δq j = Fi∂ri = [Fi + Fi′]δri …(7) j===1 i 1 i 1

Since the system is Holonomic, we regard all generalized co-ordinates except qj as constant. Then, (7) gives N ρ ρ ρ Qj δqj = ∑(Fi + Fi′)δri …(8) i=1 N ρ ρ ρ δri ⇒ Qj = ∑(Fi + Fi′) i=1 δq j

N ρ ρ ρ ∂ri ⇒ Q j = ∑ (Fi + Fi′) …(9) i=1 ∂q j from (6) & (9), we get

d ⎛ ∂T ⎞ ∂T ⎜ ⎟ − = Q , j = 1, 2,…n ⎜ & ⎟ j dt ⎝ ∂q j ⎠ ∂q j This is a system of n equations known as Lagrange equations.

3.4 Example:- Planetary Motion :-

(m) P(r,θ) P → Planet μm S → Sun(fixed)

r 2 θ S(fixed) Initial line Let (r, θ) be the polar co-ordinates of P w.r.t. S at time t. μm Under the action of inverse square law of attraction, force = r2 radial velocity = r& transverse velocity = rθ& Here (r, θ) are the generalized co-ordinates of the system and K.E. is 1 T = m(r&2 + r 2θ&2 ) [Θ θ2 = r&2 + r 2 θ&2 ] 2 where r, θ, r&, θ& are independent. As the system is Holomonic, the virtual work function is given by ⎛ − μm ⎞ δW = ⎜ ⎟δr + 0 [Θ δW = ΣQ j δq j = Q1 δq1 + Q2 δq2 = Qr δr + Q2 δθ] ⎝ r 2 ⎠ − μm ⇒ Qr = r 2

Qθ = 0 ∂T ∂ ⎡1 ⎤ Now = m(r&2 + r 2θ&2 ) ∂r ∂r ⎣⎢2 ⎦⎥ ∂T ⇒ = mrθ&2 ∂r ∂T and = 0 ∂θ ∂T ∂T = mr&, = mr2 θ& ∂r& ∂θ& Therefore Lagrange’s equations are 46

d ⎛ ∂T ⎞ ∂T ⎜ ⎟ − = Qr …(1) dt ⎝ ∂r&⎠ ∂r d ⎛ ∂T ⎞ ∂T and ⎜ ⎟ − = Qθ …(2) dt ⎝ ∂θ&⎠ ∂θ d − μm (1) ⇒ (mr&) = mrθ&2 = …(3) dt r 2 − μm and m&r&− mrθ&2 = r 2 μm (3) ⇒ m&r&− mrθ&2 = − r 2 − μ ⇒ &r&− rθ&2 = r 2 d and (4) ⇒ (r 2 θ&) = 0 dt 3.5 Lagrange equation for a conservative system of forces Suppose that the forces are conservative & the system is specified by the generalized co-ordinate qj (j = 1, 2,….n). So we can find a potential function

V(q1, q2,…, qn) ∂V ∂V ∂V such that δW = −δV, where δV = δq1 + δq2 +...+ δqn ∂q1 ∂q2 ∂qn

n ⎛ ∂V ⎞ ⇒ δW = − ⎜ ⎟δq ∑⎜ ⎟ j j=1 ⎝ ∂q j ⎠

n n ⎛ ∂V ⎞ ⇒ Qδq = − ⎜ ⎟δq ∑∑j j ⎜ ⎟ j j==1 j 1 ⎝ ∂q j ⎠ − ∂V ⇒ Qj = ∂q j Therefor, Lagrange’s equation for a conservative holonomic dynamical system becomes 47

d ⎛ ∂T ⎞ ∂T ∂V ⎜ ⎟ − = − ⎜ & ⎟ dt ⎝ ∂q j ⎠ ∂q j ∂q j

d ⎛ ∂T ⎞ ∂ or ⎜ ⎟ − (T − V) = 0 ⎜ & ⎟ dt ⎝ ∂q j ⎠ ∂q j Let L = T − V where L = Lagrange’s function or L = K. E. − P. E

d ⎛ ∂T ⎞ ∂L ⇒ ⎜ ⎟ − = 0 ⎜ & ⎟ dt ⎝ ∂q j ⎠ ∂q j

Since V does not depend upon q&1,q&2 ,...,q&n ∂V ⇒ = 0 ∂q&j

d ⎛ ∂L ⎞ ∂L ⇒ ⎜ ⎟ − = 0 ⎜ & ⎟ dt ⎝ ∂q j ⎠ ∂q j 3.6 Generalised components of momentum and impulse

Let qj (j = 1, 2,…n) be generalized co-ordinate at time t for a Holonomic dynamical system. Let T = T(q1, q2,…., qn, q&1, q&2....q&n ,t) . Then, the n quantities pj is defined by ∂T pj = are called generalized components of momentum. ∂q&j We know that Lagrange equation is

d ⎛ ∂T ⎞ ∂T ⎜ ⎟ − = 0 ⎜ & ⎟ dt ⎝ ∂q j ⎠ ∂q j d ∂T ⇒ (p j ) − = 0 dt ∂q j

1 1 ρ 1 Now T = m v2 = m r&2 = m(x&2 + y&2 + z&2 ) 2 2 2 ∂T Then px = = mx& ∂x& 48

Similarly py = m y&, pz = mz&

For generalized forces Qj (j = 1, 2,….n) for dynamical system, the n quantities Jj defined by ⎡ τ ⎤ Lt ⎢ Q j dt⎥ = J j (finite) when limit exists are called generalised Q →∞ ∫ j ⎢0 ⎥ τ→0 ⎣ ⎦ impulses.

n Since δW = ∑Qj δq j j=1

τ n ⎡ τ ⎤ ∴ ∫δW dt = ∑ δq j ⎢∫ Q j dt⎥ 0 j=1 ⎣⎢0 ⎦⎥

τ n ⎡ τ ⎤ ⎢ ⎥ ∴ LtδW dt = δq j Lt Q j dt Q →∞ ∫ ∑ ⎢Q →∞ ∫ ⎥ j 0 j=1 j 0 τ→0 ⎣⎢τ→0 ⎦⎥

n ⇒ δU = ∑Jj δq j j=1 where δU is called impulsive virtual work function. 3.7 Lagrange’s equation for Impulsive forces It states that generalized momentum increment is equal to generalized impulsive force associated with each generalized co-ordinate. Derivation:- We know that Lagrange’s equation for Holonomic system are

d ⎛ ∂T ⎞ ∂T ⎜ ⎟ − = Q ⎜ & ⎟ j dt ⎝ ∂q j ⎠ ∂q j d ∂T ⇒ (p j ) − = Q j …(1) dt ∂q j

Integrating this equation from t = 0 to t = τ we get

τ ∂T τ (pj)t=τ − (pj)t=0 = ∫∫dt + Q j dt 0 ∂q j 0

Let Qj→∞, τ→0 in such a way that 49

LimQ j dt = J j (finite) (j = 1, 2,…., n) Qj→∞ ∫ τ→0 0

Further as the co-ordinate qj do not change suddenly,

τ ∂T Lt dt = 0 τ→0 ∫ 0 ∂q j

Writing Δpj = Lt [(pj)t=τ − (pj)t=0], τ→0 We thus obtain Lagrange’s equation in impulsive form

Δpj = Jj, j = 1, 2,….n 3.8 Kinetic energy as a quadratic function of velocities

Let at time t, the position vector of ith particle of mass mi of a Holonomic ρ system is ri , then K.E. is

N 1 ρ2 T = ∑mi r&i …(1) 2 i=1 where N is number of particles. Suppose the system to be Holonomic & specified ρ ρ by n generalized co-ordinates qj , then ri = ri (q1, q2,…., qn, t) ρ ρ ρ ρ ρ ρ dri ∂ri ∂ri ∂ri ∂ri ⇒ r&i = = q&1 + q&2 +...+ q&n + …(2) dt ∂q1 ∂q2 ∂qn ∂t From (1) & (2),

ρ ρ ρ ρ 2 1 N ⎛ ∂r ∂r ∂r ∂r ⎞ & i & i & i i T = ∑mi ⎜q1 + q2 +...+ qn + ⎟ 2 i=1 ⎝ ∂q1 ∂q2 ∂qn ∂t ⎠

ρ ρ ρ 2 1 N ⎛ ∂r ∂r ∂r ⎞ & i & i & i = ∑mi ⎜q1 + q2 +...+ qn ⎟ 2 i=1 ⎝ ∂q1 ∂q2 ∂qn ⎠

ρ 2 ρ ρ ρ 1 N ⎛ ∂r ⎞ N ∂r ⎛ ∂r ∂r ⎞ i i & i & i + ∑∑mi ⎜ ⎟ + mi ⎜q1 +...+ qn ⎟ 2 i==1 ⎝ ∂t ⎠ i 1 ∂t ⎝ ∂q1 ∂qn ⎠ 1 ⇒ T = [(a q&2 + a q&2 +...+ a q&2 + 2a q&q& +....) 2 11 1 22 2 nn n 12 1 2

+ 2(a1 q&1 + a 2q&2 +....+ a nq&n ) + a] …(3) 50

ρ ρ N ⎛ ∂r ⎞⎛ ∂r ⎞ ⎜ i ⎟⎜ i ⎟ where ars = asr = ∑ mi ⎜ ⎟⎜ ρ ⎟ , s ≥ r i=1 ⎝ ∂qr ⎠⎝ rqs ⎠ ρ ρ N ⎛ ∂r ⎞⎛ ∂r ⎞ ⎜ i ⎟ i ar = ∑mi ⎜ ⎟⎜ ⎟ i=1 ⎝ ∂qr ⎠⎝ ∂t ⎠

N ρ 2 ⎛ ∂ri ⎞ a = ∑mi ⎜ ⎟ i=1 ⎝ ∂t ⎠ equation (3) shows that T is a quadratic function of the generalized velocities. ρ ρ Special Case :- When time t is explicitly absent, then ri = ri (q1, q2,….qn) ρ ρ ρ ρ ρ dri ∂ri ∂ri ∂ri ⇒ r&i = = q&1 + q&2 +...+ q&n dt ∂q1 ∂q2 ∂qn ρ ∂r and i = 0 ∂t From (3), we get 1 T = [a q&2 + a q&2 +...+ a q&2 + 2a q&q& +....] 2 11 1 22 2 nn n 12 1 2 1 n n = ∑∑a rs q&r q&s 2 s==1 r 1 Thus the K.E. assumes the form of a Homogeneous quadratic function of the generalized velocitiesq&1,q&2...q&n . In this case, using Euler’s theorem for Homogeneous functions ∂T ∂T ∂T q&1 + q&2 +...+ q&n = 2T ∂q&1 ∂q&2 ∂q&n

⇒ q&1p1 + q&2p2 +...+ q&n pn = 2T

3.9 Donkin’s Theorem :-Let a function F(u1, u2,…., un) have explicit dependence on n independent variables u1, u2…un. Let the function F be transformed to another function G = G(v1, v2… vn) expressed in terms of a new set of n independent variables v1, v2,… vn where these new variables are connected to the old variables by a given set of relation 51

∂F Vi = , i = 1, 2…..n …(1) ∂ui & the form of G is given by

n G(v1, v2…. vn) = ∑ui vi − F(u1, u2….un) …(2) i=1 then the variables u1, u2… un satisfy the dual transformation ∂G ui = …(3) ∂vi

n & F(u1, u2… un) = ∑ui vi − G(v1, v2….vn) i=1

This transformation between function F & G and the variables ui & vi is called Legendre’s dual transformation.

Proof : Since G is given by

n G(v1, v2,…., vn) = ∑uk vk − F(u1, u2,…., un) k=1

∂G ∂ ⎡ n ⎤ Then = ⎢∑u k vk − F(u1,u 2...u n )⎥ ∂vi ∂vi ⎣k=1 ⎦

n n n ∂u k ∂Vk ∂F ∂u k = ∑∑∑vk + u k − k==1 ∂vi i 1 ∂Vi k =1 ∂u k dvi

n n ∂G ∂vk ∂H ∂u k ⇒ = ∑∑∑vk + u k δki − ∂vi k==1 ∂vi k 1 ∂u k ∂vi

n n ∂u k ∂F ∂u k = ∑∑vk + ui − k==1 ∂vi k 1 ∂u k ∂vi

n n ∂u k ∂F ∂F ∂u k ⎡ ∂F ⎤ = ∑∑+ ui − ⎢Θ (1) ⇒ vk = ⎥ k==1 ∂vi ∂u k k 1 ∂u k ∂vi ⎣ ∂u k ⎦

= ui ∂G ⇒ = ui ∂vi 52

3.10 Extension of Legender’s dual transformation

Further suppose that there is a another set of m independent variables α1,

α2…αn present in both F & G.

⇒ F = F(u1, u2….un, α1, α2…. αm)

G = G(v1, v2… vn, α1, α2….αm) then there should be some extra condition for Legendre’s dual transformation to be satisfied. These conditions are ∂F − ∂G = , j = 1, 2….m L.H.S. ∂α j ∂α j

Consider G = G(v1, v2…vn, α1, α2… αm)

n = ∑ui vi − F(u1, u2…un, α1, α2… αm) R.H.S. i=1 From L.H.S. n ∂G m ∂G δG = ∑∑δvi + δα j …(1) i==1 ∂vi j 1 ∂α j From R.H.S., n n n ∂F m ∂F δG = ∑∑∑ui δvi + vi δui − δui − ∑ δαj …(2) i===1 i 1 i 1 ∂ui j =1 ∂α j Equating (1) & (2), n ∂G m ∂G n n n ∂F m ∂F ∑∑δvi + δα j = ∑∑∑∑ui δui + vi δui − δui − δα j i==1 ∂vi j 1 ∂α j i ====1 i 1 i 1 ∂ui j 1 ∂α j ∂F ⇒ vi = are satisfied provided ∂ui ∂G ui = ∂vi ∂G − ∂F and = ∂α j ∂α j

Lesson-4 Hamilton’s Equations of Motion 4.1 Introduction So far we have discussed about Lagrangian formulation and its application. In this lesson, we assume the formal development of mechanics turning our attention to an alternative statement of the structure of the theory known as Hamilton’s

. formulation. In Lagrangian formulation, the independent variables are qi and q,i whereas in Hamiltonian formulation, the independent variables are the generalized coordinates qi and the generalized momenta pi 4.2 Energy equation for conservative fields Prove that for a dynamical system T + V = constant where T = K.E. V = P.E. or ordinary potential

Proof : Here V = V(q1, q2,…, qn)

T = T(q1, q2 …. Qn, q&1,q&2...q&n )

L = T − V = L(q1, q2…qn, q&1,q&2 ,....,q&n ) If Lagrangian function L of the system does not explicitly depend upon time t, then ∂L = 0 ∂t i.e. L = L(qj, q&j ) for j = 1, 2,…n The total time derivative of L is dL n ∂L n ∂L = ∑∑q&j + &q&j …(I) dt j==1 ∂q j j 1 ∂q&j We know that the Lagrange’s equation is given by

d ⎡ ∂L ⎤ ∂L ⎢ ⎥ − = 0 …(II) & dt ⎣⎢∂q j ⎦⎥ ∂q j

dL n ⎡ d ⎛ ∂L ⎞⎤ n ∂L (I) ⇒ = q&⎢ ⎜ ⎟⎥ + &q& [using (ii)] dt ∑ j dt ⎜ ∂q& ⎟ ∑ ∂q& j j=1 ⎣⎢ ⎝ j ⎠⎦⎥ j=1 j 54

n d ⎡ ∂L ⎤ = ⎢q& ⎥ ∑ j & j=1 dt ⎣⎢ ∂q j ⎦⎥

d ⎡ n ∂L ⎤ ⇒ ⎢ q& − L⎥ = 0 …(1) ∑ j & dt ⎣⎢ j=1 ∂q j ⎦⎥

dH n ∂L ⇒ = 0 where H = ∑q&j − L dt j=1 ∂q&j is a function called Hamiltonian

n H = ∑q&j p j − L …(A) j=1 ∂L [Θ = p j = generalized component of momentum] ∂q&j Integrating (1), n ∂L ∑q&j − L = constant …(2) j=1 ∂q&j

n ∂L n ∂T Now ∑∑q&j = q&j j==1 ∂q&j j 1 ∂q&j

n N ρ N ρ 1 ⎡ ∂ ⎧ 2 ⎫⎤ ⎡ 1 2 ⎤ = q&⎢ ⎨ m r& ⎬⎥ Θ T = m r& ∑∑j & i i ⎢ ∑ i i ⎥ 2 j==1 ⎣⎢∂q j ⎩ i 1 ⎭⎦⎥ ⎣ 2 i=1 ⎦ ρ n ⎡ N ρ⎛ ∂r&⎞⎤ = q&⎢ m r&⎜ i ⎟⎥ ∑∑j i i ⎜ ∂q& ⎟ j==1 ⎣⎢ i 1 ⎝ j ⎠⎦⎥ ρ ρ ρ n ⎡ N ρ⎛ ∂r ⎞⎤ ⎡ ∂r& ∂r ⎤ = q&⎢ m r&⎜ i ⎟⎥ ⎢Θ i = i ⎥ ∑∑j i i ⎜ ∂q ⎟ ∂q& ∂q j==1 ⎣⎢ i 1 ⎝ j ⎠⎦⎥ ⎣⎢ j j ⎦⎥

N ρ ⎡ n ρ ⎤ N ρ ρ ∂ri = ∑∑mi r&i ⎢ q&j ⎥ = ∑mi r&i r&i i==1 ⎣⎢ j 1 ∂q j ⎦⎥ i=1 = 2T …(3)

from (2) & (3), 2T − L = constant. ⇒ 2T − (T−V) = constant [Θ L = T−V] ⇒ T + V = constant. Also from (A), H = T + V = constant. ∴ Total energy T + V =H, when time t is explicitly absent. 4.3 Generalised potential

For conservative forces, Potential function V = V(q1, q2…qn), therefore δW = −δV

⎛ ∂V ⎞ ⎛ ∂V ⎞ = − ⎜ δx ⎟ = − ⎜ ⎟δq ∑⎜ i ⎟ ∑⎜ ⎟ j ⎝ ∂xi ⎠ ⎝ ∂q j ⎠

Also δW = ΣQj δqj where Qj are generalized forces.

⎛ ∂V ⎞ ⇒ Qδq = − ⎜ ⎟δq ∑∑j j ⎜ ⎟ j ⎝ ∂q j ⎠ ∂V ⇒ Qj = − ∂q j

4.4 Cyclic or Ignorable co-ordinates Lagrangian L = T − V If Lagrangian does not contain a co-ordinate explicitly, then that co-ordinate is called Ignorable or cyclic co-ordinate.

Let L = L(q1, q2…qn, q&1,q&2 ,...q&n ,t)

Let qk is absent in L, then ∂L = 0 ∂qk

Lagrange’s equation (equation of motion) corresponding to qk becomes d ⎛ ∂L ⎞ ∂L ⎜ ⎟ − 0 = 0 ⇒ = constant = pk dt ⎝ ∂q&k ⎠ ∂q&k 56

4.5 Hamiltonian and Hamiltonian variables:- In Lagrangian formulation, independent variable are generalized co-ordinates and time. Also generalised velocities appear explicitly in the formulation.

∴ L(qk, q&k , t,)

Like this Lagrangian L(qj, q&j , t), a new function is Hamiltonian H which is function of generalized co-ordinates, generalized momenta and time ,i.e., ∂L H(qj, pj, t), where pj = ∂q&j Also we have shown that

H = ∑p j q&j − L j

This quantity is also known as Hamiltonian. The independent variables q1, q2,…qn , p1, p2…pn, t are known as Hamiltonian variables. 4.6 Hamilton’s Canonical equations of motion Lagrange’s equations of motion are

d ⎛ ∂L ⎞ ∂L ⎜ ⎟ − = 0 , j = 1, 2,….n ⎜ & ⎟ dt ⎝ ∂q j ⎠ ∂q j

Now H = H(qj, pj, t) …(1)

n H = ∑p j q&j − L(q j,q&j,t) …(2) j=1 The differential of H from (1), ∂H ∂H ∂H dH = ∑∑dq j + dp j + dt …(3) ∂q j ∂p j ∂t from (2) ⇒ n ∂L ∂L − ∂L dH = ∑∑[p j dq&j + q&jdp j ]− dq j − ∑ dq&j dt j=1 ∂q j ∂q&j ∂t

n ∂L n n ∂L ⎡ ∂L ⎤ ⇒ dH = dq& + q& dp − d qj ⎢Θ p = ⎥ …(4) ∑∑∑& j j j j & j===1 ∂q j j 1 j 1 ∂q j ⎣⎢ ∂q j ⎦⎥ 57

d ∂L From Lagrange’s equation, (p j ) = dt ∂q j ∂L ⇒ p&j = …(5) ∂q j Using (4) & (5), we get n ∂L dH = ∑q&j dp j − Σp&j dq j − dt …(6) j=1 ∂t Comparing equation (3) & (6), we get ∂H − ∂H = q&j, p&j = …(7) ∂p j ∂q j ∂H − ∂L and = where j = 1, 2,…n …(8) ∂t ∂t The equation (7) is called Hamiltonian’s canonical equations of motion or Hamilton’s equations. Result:- To show that if a given co-ordinate is cyclic in Lagrangian L, then it will also be cyclic in Hamiltonian H. ∂L If L is not containing qk ,i.e., qk is cyclic, then = 0 ∂qk then p&k = 0 ⇒ pk = constant

From equation (1), H(qj, pj, t)

⇒ H(q1, q2,….., qk−1, qk+1,…., qn, p1, p2,…. pk−1, pk+1….pn, t) If H is not containing t ,i.e.,

H = H(qj, pj) dH ∂H ∂H then = ∑ q&j + ∑ p&j dt ∂q j ∂p j Using equation (7) or Hamilton’s equation

dH ∂H ∂H ∂H ⎛ ∂H ⎞ = − ⎜ ⎟ = 0 ∑∑⎜ ⎟ dt ∂q j ∂p j ∂p j ⎝ ∂q j ⎠ 58

dH ⇒ = 0 ⇒ H = constant. dt If the equation of transformation are not depending explicitly on time & if P.E. is velocity independent, then H = E (total energy) Which can also be seen from the expression as given under ρ ρ ri = ri (q1, q2,…., qn)

P.E., V = V(q1, q2,…., qn) N ρ 1 2 K.E., T = ∑mi r&i 2 i=1 ρ n ρ ∂ri Now r&i = ∑ q&j j=1 ∂q j

ρ 2 1 N ⎛ n ∂r ⎞ ⇒ T = m ⎜ i q& ⎟ ∑∑i ⎜ j ⎟ 2 i==1 ⎝ j 1 ∂q j ⎠

= (quadratic function of q&1,q&2 ,....,q&n ) Therefore by Euler’s theorem for Homogeneous function, we have ∂T ∑q&j = 2T ∂q&j ∂L ∂T H = ∑p j q&j − L = ∑q&j − L = ∑q&j − L = 2T − L ∂q&j ∂q&j

⇒ H = 2T − (T − V) = T + V = E ⇒ H = E Example:- Write the Hamiltonian & Hamilton’s equation of motion for a particle in central force field (planetary motion). Solution : Let (r, θ) be the polar co-ordinates of a particle of mass ‘m’ at any instant of time t. Now L = T−V(r) where V(r) = P.E. 1 ⇒ L = m[r&2 + r 2 θ&2 ] − V(r) …(1) P(r,θ) 2 (m) As qj = r, θ r& θ O X 59

& q&j = r&,θ

pj = pr, pθ

∂L ∂L 2 Now pr = = mr&, p = = mr θ& …(2) ∂r& θ ∂θ& 1 H = p q& − L = p r&+ p θ&− m(r&2 + r 2θ&2 ) + V(r) ∑ j j r θ 2 1 1 = m r&2 + mr2θ&2 − m(r&2 ) − mr2θ&2 r + V(r) 2 2 1 = m(r&2 + r 2θ&2 ) + V(r) …(3) 2 ⇒ H = T + V 1 from (2), r&= p m r

& 1 θ = pθ mr2 ⎡ 2 2 ⎤ 1 ⎛ pr ⎞ 2 ⎛ pθ ⎞ Then H = m⎢⎜ ⎟ + r ⎜ 2 ⎟ ⎥ + V(r) 2 ⎣⎢⎝ m ⎠ ⎝ mr ⎠ ⎦⎥

2 1 ⎡ 2 pθ ⎤ ⇒ H = ⎢pr + 2 ⎥ + V(r), which is required Hamiltonian 2m ⎣ r ⎦ Hamilton’s equations of motion are, ∂H − ∂H q&j = , p&j = ∂p j ∂q j

The two equations for q&j are

∂H pr r&= = = q&r ∂pr m ∂H p & θ & Similarly θ = = 2 = qθ ∂pθ mr and two equations for p&j are, 60

2 − ∂H pθ ∂V(r) p&r = = − ∂r mr3 ∂r − ∂H and p& = = 0 ⇒ pθ = constant θ ∂θ 4.7 Routh’s equations Routh proposed for taking some of Lagrangian variables and some of Hamiltonian variables. The Routh variables are the quantities

t, qj, qα, q&j , pα j = 1,2,….k α = k + 1, k + 2, ….n k is arbitrary fixed number less than n. Routh’s procedure involves cyclic and non- cyclic co-ordinates.

Suppose co-ordinates q1, q2,…., qk (k < n) are cyclic (or Ignorable). Then we want to find a function R, called Routhian function such that it does not contain generalized velocities corresponding to cyclic co-ordinates.

L = L(q1, q2….qn, q&1,q&2 ,...,q&n ,t )

If q1, q2….qk are cyclic, then

L(qk+1,…., qn, q&1,q&2 ,....q&n ,t) so that n ∂L n ∂L ∂L dL = ∑∑dq j + dq&j + dt j==k+1 ∂q j j 1 ∂q&j ∂t

⎛ k ∂L ⎞ n ∂L n ∂L ∂L ⇒ ⎜d L − dq& ⎟ = dq + dq& + dt …(1) ⎜ ∑ & j ⎟ ∑∑j & j ⎝ j=1 ∂q j ⎠ j=+k+1 ∂q j k 1 ∂q j ∂t

Routhian function R, in which velocities q&1, q&2....q&k corresponding to ignorable co- ordinate q1, q2,…., qk are eliminated, can be written as

R = R(qk+1, qk+2,…., qn, q&k+1,....q&n ,t) so that 61

n ∂R n ∂R ∂R dR = ∑∑dq j + dq&j + dt …(2) j=+k+1 ∂q j j=k 1 ∂q&j ∂t Further we can also define Routhian function as

k R = L − ∑q&j p j j=1

k k We want to remove ∑qj or ∑q&j from L to get R. j=1 j=1

k k dR = dL − ∑∑q&j dp j − p j dq&j j==1 j 1

k ∂L k = dL − ∑∑dq&j − q&j dpj [using (1)] j==1 ∂q&j j 1

n ∂L n ∂L ∂L k ⇒ dR = ∑∑dq j + dq&j + dt − ∑q&j dp j …(3) j=+k+1 ∂q j j==k 1 ∂q j ∂t j 1 Comparing (2) & (3) by equating the coefficients of varied quantities as they are independent, we get ∂L ∂R ∂L ∂R = , = …(4) ∂q j ∂q j ∂q&j ∂q&j ∂L ∂R = , j = k + 1, k + 2,….n ∂t ∂t Put (4) in Lagrangian’s equations,

n ⎡ d ⎛ ∂L ⎞ ∂L ⎤ ⎢ ⎜ ⎟ − ⎥ = 0 j = 1, 2,….n ∑ dt ⎜ ∂q& ⎟ ∂q j=1 ⎣⎢ ⎝ j ⎠ j ⎦⎥ we get,

n ⎡ d ⎛ ∂R ⎞ ∂R ⎤ ⎢ ⎜ ⎟ − ⎥ = 0 ∑ dt ⎜ ∂q& ⎟ ∂q j=k+1⎣⎢ ⎝ j ⎠ j ⎦⎥

d ⎛ ∂R ⎞ ∂R or ⎜ ⎟ − = 0 , j = k + 1,….n ⎜ & ⎟ dt ⎝ ∂q j ⎠ ∂q j 62

These are (n−k) 2nd order equations known as Routh’s equations. 4.8 Generalised potential When the system is not conservative. Let U is Generalised potential, say it depends on generalised velocities ( q&j ) i.e. we consider the case when in place of ordinary potential V (qj, t), there exits a generalised point U(qj, t, q&j ) in terms of which the generalised forces Qj are defined by d Qj = , j = 1,2,…n ⎡ ∂U ⎤ ∂U dt⎢ ⎥ − & ⎣⎢∂q j ⎦⎥ ∂q j [Θ L = T−V for conservative system, L = T−U for non-conservative system] Here U is called generalised potential or velocity dependent potential.

d ⎛ ∂T ⎞ ∂T d ⎛ ∂U ⎞ ∂U Here Lagrange equations are ⎜ ⎟ − = Q = ⎜ ⎟ − ⎜ & ⎟ j ⎜ & ⎟ dt ⎝ ∂q j ⎠ ∂q j dt ⎝ ∂q j ⎠ ∂q j

d ⎡ ∂ ⎤ ∂ ⇒ ⎢ (T − U)⎥ − (T − U) = 0 & dt ⎣⎢∂q j ⎦⎥ ∂q j

d ⎛ ∂L ⎞ ∂L ⇒ ⎜ ⎟ − = 0 [Θ L = T − U for non-conservative system] ⎜ & ⎟ dt ⎝ ∂q j ⎠ ∂q j

4.9 Poisson’s Bracket Let A and B are two arbitrary function of a set of canonical variables (or conjugate variables) q1, q2,…., qn, p1, p2….pn , then Poisson’s Bracket of A & B is defined as

⎛ ∂A ∂B ∂A ∂B ⎞ [A, B] = ⎜ − ⎟ q,p ∑⎜ ⎟ j ⎝ ∂q j ∂p j ∂p j ∂q j ⎠ If F is a dynamical variable, i.e.,

F = F(qj, pj, t), then 63

dF dF ⎛ ∂F ∂F ⎞ ∂F = (q ,p ,t) = ⎜ q& + p& ⎟ + …(1) j j ∑⎜ j j ⎟ dt dt j ⎝ ∂q j ∂p j ⎠ ∂t Using Hamilton’s canonical equations, ∂H − ∂H q&j = , p&j = ∂p j ∂q j

dF ⎛ ∂F ∂H ∂F ∂H ⎞ ∂F ∴ from (1), = ⎜ − ⎟ + ∑⎜ ⎟ dt ⎝ ∂q j ∂p j ∂p j ∂q j ⎠ ∂t dF ∂F ⇒ = [F, H] + dt q,p ∂t If F is not depending explicitly on t, then ∂F = 0, ∂t

dF ⎛ ∂F ∂H ∂F ∂H ⎞ so = ⎜ − ⎟ ∑⎜ ⎟ dt ⎝ ∂q j ∂p j ∂p j ∂q j ⎠

= [F, H]q,p 4.9 Properties

I. [X, Y]q,p = −[Y, X]q,p II. [X, X] = 0 III. [X, Y+Z] = [X, Y] + [X, Z] IV. [X, YZ] = Y[X, Z] + Z[X, Y]

⎛ ∂X ∂Y ∂X ∂Y ⎞ Solution :- I. By definition [X, Y] = ⎜ − ⎟ q,p ∑⎜ ⎟ j ⎝ ∂q j ∂p j ∂p j ∂q j ⎠

⎛ ∂Y ∂X ∂Y ∂X ⎞ [Y, X] = ⎜ − ⎟ q, p ∑⎜ ⎟ j ⎝ ∂q j ∂p j ∂p j ∂q j ⎠

⎛ ∂X ∂Y ∂X ∂Y ⎞ = − ⎜ − ⎟ ∑⎜ ⎟ j ⎝ ∂q j ∂p j ∂p j ∂q j ⎠

⇒ [Y, X]q,p = −[X, Y] 64

⎛ ∂X ∂X ∂X ∂X ⎞ II. [X, X] = ⎜ − ⎟ = 0 q,p ∑⎜ ⎟ j ⎝ ∂q j ∂p j ∂p j ∂q j ⎠

⎛ ∂X ∂C ∂X ∂C ⎞ Also [X, C] = ⎜ − ⎟ = 0 q,p ∑⎜ ⎟ j ⎝ ∂q j ∂p j ∂p j ∂q j ⎠

⎛ ∂X ∂(Y + Z) ∂X ∂(Y + Z) ⎞ III. [X, Y + Z] = ⎜ − ⎟ q,p ∑⎜ ⎟ j ⎝ ∂q j ∂p j ∂p j ∂q j ⎠

⎡ ∂X ⎛ ∂Y ∂Z ⎞ ∂X ⎛ ∂Y ∂Z ⎞⎤ = ⎢ ⎜ + ⎟ − ⎜ + ⎟⎥ ∑ ∂q ⎜ ∂p ∂p ⎟ ∂p ⎜ ∂q ∂q ⎟ j ⎣⎢ j ⎝ j j ⎠ j ⎝ j j ⎠⎦⎥

⎛ ∂X ∂Y ∂X ∂Y ⎞ ⎛ ∂X ∂Z ∂X ∂Z ⎞ ⇒ [X, Y + Z] = ⎜ − ⎟ + ⎜ − ⎟ q,p ∑⎜ ⎟ ∑⎜ ⎟ j ⎝ ∂q j ∂p j ∂p j ∂q j ⎠ j ⎝ ∂q j ∂p j ∂p j ∂q j ⎠ = [X, Y] + [X, Z]

⎛ ∂X ∂(YZ) ∂X ∂(YZ) ⎞ IV. [X, YZ] = ⎜ − ⎟ q,p ∑⎜ ⎟ j ⎝ ∂q j ∂p j ∂p j ∂q j ⎠

⎡ ∂X ⎛ ∂Z ∂Y ⎞ ∂X ⎛ ∂Y ∂Z ⎞⎤ = ⎜Y + Z ⎟ − ⎜ Z + Y ⎟ ⎢ ⎜ ⎟ ⎜ ⎟⎥ ⎣⎢∂q j ⎝ ∂p j ∂p j ⎠ ∂p j ⎝ ∂q j ∂q j ⎠⎦⎥

⎡ ⎛ ∂X ∂Z ∂X ∂Z ⎞⎤ ⎡ ⎛ ∂X ∂Y ∂X ∂Y ⎞⎤ = 4 ⎢ ⎜ − ⎟⎥ + Z⎢ ⎜ − ⎟⎥ ∑⎜ ∂q ∂p ∂p ∂q ⎟ ∑⎜ ∂q ∂p ∂p ∂q ⎟ ⎣⎢ j ⎝ j j j j ⎠⎦⎥ ⎣⎢ j ⎝ j j j j ⎠⎦⎥ = Y[X, Z] + Z[X, Y] Also

(i) [qi, qj]q,p = 0

(ii) [pi, pj]q,p = 0 ⎧1, i = j (iii) [qi, pj]q,p = δij = ⎨ ⎩0, i ≠ j Solution:-

⎡ ∂qi ∂q j ∂qi ∂q j ⎤ (i) [qi, qj]q,p = ∑⎢ − ⎥ …(1) k ⎣∂qk ∂pk ∂pk ∂qk ⎦ 65

Because qi or qj is not function of pk ∂q ∂q ⇒ i = 0, j = 0 ∂pk ∂pk

(1) ⇒ [qi, qj]q,p = 0.

⎡ ∂pi ∂p j ∂pi ∂p j ⎤ (ii) [pi, pj]q,p = ∑⎢ − ⎥ k ⎣∂qk ∂pk ∂pk ∂qk ⎦

As pi, pj is not a function of qk ∂p ∂p ∴ i = 0, j = 0 ∂qk ∂qk

⇒ [pi, pj]q,p= 0

⎛ ∂qi ∂p j ∂qi ∂p j ⎞ (iii) Now [qi, pj]q,p = ∑⎜ − ⎟ k ⎝ ∂qk ∂pk ∂pk ∂qk ⎠

⎛ ∂qi ∂p j ⎞ ∂qi ∂p j = ∑⎜ − 0⎟ = ∑ k ⎝ ∂qk ∂pk ⎠ k ∂qk ∂pk

= ∑δik δ jk = ∑δij = δij k k ⎧1, i = j ⇒ [qi, pj]q,p = δij = ⎨ ⎩0, i ≠ j Some other properties:- If [φ, ψ] be the Poisson Bracket of φ & ψ, then ∂ ⎡∂φ ⎤ ⎡ ∂ψ⎤ (1) [φ, ψ] = ,ψ + φ, ∂t ⎣⎢ ∂t ⎦⎥ ⎣⎢ ∂t ⎦⎥ d ⎡dφ ⎤ ⎡ dψ⎤ (2) [φ, ψ] = ,ψ + φ, dt ⎣⎢ dt ⎦⎥ ⎣⎢ dt ⎦⎥

∂ ∂ ⎡ ⎛ ∂φ ∂ψ ∂φ ∂ψ ⎞⎤ ⎜ ⎟ Solution:- (1) [φ, ψ] = ⎢∑⎜ − ⎟⎥ ∂t ∂t ⎣⎢ i ⎝ ∂qi ∂pi ∂p i ∂qi ⎠⎦⎥

∂ ⎡ ∂φ ∂ψ ∂φ ∂ψ ⎤ = ∑ ⎢ − ⎥ i ∂t ⎣∂qi ∂pi ∂pi ∂qi ⎦ 66

⎡ ∂ ⎛ ∂φ ⎞⎛ ∂ψ ⎞ ∂φ ∂ ⎛ ∂ψ ⎞⎤ = ∑⎢ ⎜ ⎟⎜ ⎟+ ⎜ ⎟⎥ i ⎣⎢∂t ⎝ ∂qi ⎠⎝ ∂pi ⎠ ∂qi ∂t ⎝ ∂pi ⎠⎦

⎡ ∂ ⎛ ∂φ ⎞ ∂ψ ∂φ ∂ ⎛ ∂ψ ⎞⎤ − ⎢ ⎜ ⎟ + ⎜ ⎟⎥ ⎣∂t ⎝ ∂pi ⎠ ∂qi ∂pi ∂t ⎝ ∂qi ⎠⎦

⎡ ∂ ∂φ ⎛ ∂ψ ⎞ ∂ψ ∂ ⎛ ∂φ ⎞⎤ ∂φ ∂ ∑⎢ ⎜ ⎟ − ⎜ ⎟⎥ + i ⎣∂qi ∂t ⎝ ∂pi ⎠ ∂qi ∂pi ⎝ ∂t ⎠⎦ ∂qi ∂pi

⎡ ∂ ⎛ ∂φ ⎞ ∂ψ ∂φ ∂ ⎛ ∂ψ ⎞ ∂ ⎛ ∂ψ ⎞⎛ ∂φ ⎞ ∂ψ ∂ ⎛ ∂φ ⎞⎤ = ∑⎢ ⎜ ⎟ + ⎜ ⎟ − ⎜ ⎟⎜ ⎟ − ⎜ ⎟⎥ i ⎣∂qi ⎝ ∂t ⎠ ∂pi ∂qi ∂pi ⎝ ∂t ⎠ ∂qi ⎝ ∂t ⎠⎝ ∂pi ⎠ ∂qi ∂pi ⎝ ∂t ⎠⎦

⎡ ∂ ⎛ ∂φ ⎞ ∂ψ ∂ψ ∂ ⎛ ∂φ ⎞⎤ = ∑⎢ ⎜ ⎟ − ⎜ ⎟⎥ i ⎣∂qi ⎝ ∂t ⎠ ∂pi ∂qi ∂pi ⎝ ∂t ⎠⎦

⎡ ∂φ ∂ ⎛ ∂ψ ⎞ ∂φ ∂ ⎛ ∂ψ ⎞⎤ + ∑⎢ ⎜ ⎟ − ⎜ ⎟⎥ i ⎣∂qi ∂pi ⎝ ∂t ⎠ ∂pi ∂qi ⎝ ∂t ⎠⎦ ∂ ⎡∂φ ⎤ ⎡ ∂ψ⎤ ⇒ [φ, ψ] = , ψ + φ, . ∂t ⎣⎢ ∂t ⎦⎥ ⎣⎢ ∂t ⎦⎥ (2) Similarly, we can prove d ⎡dφ ⎤ ⎡ dψ⎤ [φ, ψ] = ,ψ + φ, dt ⎣⎢ dt ⎦⎥ ⎣⎢ dt ⎦⎥ 4.9.1 Hamilton’s equations of motion in Poisson’s Bracket:- If H → Hamiltonian ∂H then [q, H]q,p = = q& ∂p − ∂H [p, H]q,p = = p& ∂q From Hamilton’s equations, − ∂H ∂H = p&, = q& ∂q ∂p

⎡∂q j ∂H ∂q j ∂H ⎤ [qj, H] = ∑⎢ − ⎥ i ⎣∂qi ∂pi ∂pi ∂qi ⎦ 67

∂H ⎡ ∂q j ⎤ [qj, H]= ∑δ ji ⎢Θ = 0⎥ i ∂pi ⎣ ∂pi ⎦ ∂H = ∂p j

Also [pj, H] = p&j

But [pj, H] = 0

⇒ pj = 0 ⇒ pj = constant 4.10 Jacobi’s Identity on Poisson Brackets (Poisson’s Identity):- If X, Y, Z are function of q & p only, then [X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y]] = 0 Proof : [X, [Y, Z]] + [Y, [Z, X]] = [X, [Y, Z]] − [Y, [X, Z]]

⎡ ⎛ ∂Y ∂Z ∂Y ∂Z ⎞⎤ = ⎢X, ⎜ − ⎟⎥ ∑⎜ q p p q ⎟ ⎣⎢ j ⎝ ∂ j ∂ j ∂ j ∂ j ⎠⎦⎥

⎡ ⎛ ∂X ∂Z ∂X ∂Z ⎞⎤ ⎢Y, ⎜ − ⎟⎥ …(1) ∑⎜ q p p q ⎟ ⎣⎢ j ⎝ ∂ j ∂ j ∂ j ∂ j ⎠⎦⎥ ∂Y ∂Z ∂Y ∂Z Let ∑∑= E, = F jj∂q j ∂p j ∂p j ∂q j ∂X ∂Z ∂X ∂Z ∑ = G, ∑ = H j ∂q j ∂p j j ∂p j ∂q j

∴ (1) ⇒ [X, [Y, Z]] − [Y, [X, Z]] = [X, E−F] − [Y, G−H] = [X, E] − [X, F] − [Y, G] + [Y, H] …(2)

∂Y ∂Z ⎛ ∂Y ⎞⎛ ∂Z ⎞ Let E = = ⎜ ⎟⎜ ⎟ ∑∑⎜ ⎟⎜∑ ⎟ jj∂q j ∂p j ⎝ ∂q j ⎠⎝ j ∂p j ⎠

∴ E = E1 E2

Similarly F = F1 F2, G = G1 G2, H = H1 H2 ∴ RHS of (2) becomes 68

[X, E] − [X, F]−[Y, G]+[Y, H] = [X, E1 E2]+[Y,H1 H2]−[X,F1F2]−[Y, G1G2]

= [X, E1] E2 +[X, E2] E1 −[X, F1] F2 − [X, F2] F1 −[Y, G1] G2

−[Y, G2] G1 + [Y, H1] H2 + [Y, H2] H1

⎡ ⎛ ∂Y ∂Z ⎞⎤ ⎡ ⎛ ∂Y ∂Z ⎞⎤ ∴ RHS of (2) is = ⎢X, ⎜ ⎟⎥ − ⎢X, ⎜ ⎟⎥ ∑⎜ q p ⎟ ∑⎜ p q ⎟ ⎣⎢ j ⎝ ∂ j ∂ j ⎠⎦⎥ ⎣⎢ j ⎝ ∂ j ∂ j ⎠⎦⎥

⎡ ⎛ ∂X ∂Z ⎞⎤ ⎡ ⎛ ∂X ∂Z ⎞⎤ − ⎢Y, ⎜ ⎟⎥ + ⎢Y, ⎜ ⎟⎥ ∑⎜ q p ⎟ ∑⎜ p q ⎟ ⎣⎢ j ⎝ ∂ j ∂ j ⎠⎦⎥ ⎣⎢ j ⎝ ∂ j ∂ j ⎠⎦⎥

Using properties [X, E1 E2] = [X, E1] E2 + [X, E2] E,

⎡ ∂Y ⎤ ∂Z ⎡ ∂Z ⎤ ∂Y = ⎢X, ∑ ⎥∑ + ⎢X,∑ ⎥∑ ⎣⎢ ∂q j ⎦⎥ ∂p j ⎣⎢ ∂p j ⎦⎥ ∂q j

⎡ ∂Y ⎤ ∂Z ⎡ ∂Z ⎤ ∂Y − ⎢X,∑ ⎥∑∑− ⎢X, ⎥∑ ⎣⎢ ∂p j ⎦⎥ ∂q j ⎣⎢ ∂q j ⎦⎥ ∂p j

⎡ ∂X ⎤ ∂Z ⎡ ∂Z ⎤ ∂X − ⎢Y, ∑ ⎥ ∑ − ⎢Y, ∑ ⎥∑ ⎣⎢ ∂q j ⎦⎥ ∂q j ⎣⎢ ∂p j ⎦⎥ ∂q j

⎡ ∂X ⎤ ∂Z ⎡ ∂Z ⎤ ∂X + ⎢Y, ∑ ⎥ ∑∑+ ⎢Y, ⎥∑ ⎣⎢ ∂p j ⎦⎥ ∂q j ⎣⎢ ∂q j ⎦⎥ ∂p j

⎪⎧− ∂Z ⎛ ⎡ ∂X ⎤ ⎡ ∂Y ⎤⎞ ∂Z ⎛ ⎡ ∂X ⎤⎡ ∂Y ⎤⎞⎪⎫ = ⎨ ⎜ ⎢ ,Y⎥ + ⎢X, ⎥⎟ + ⎜ ⎢ ,Y⎥⎢X, ⎥⎟⎬ ∑ ∂q ⎜ ∂p ∂p ⎟ ∂p ⎜ ∂q ∂q ⎟ j ⎩⎪ j ⎝ ⎣⎢ j ⎦⎥ ⎣⎢ j ⎦⎥⎠ j ⎝ ⎣⎢ j ⎦⎥⎣⎢ j ⎦⎥⎠⎭⎪

⎪⎧ ∂Y ⎡ ∂Z ⎤ ∂Y ⎡ ∂Z ⎤ ∂X ⎡ ∂Z ⎤ ∂X ⎡ ∂Z ⎤⎪⎫ + ⎨ ⎢X, ⎥ − ⎢X, ⎥ − ⎢Y, ⎥ + ⎢Y, ⎥⎬ ∑ q p p q q p p q j ⎩⎪∂ j ⎣⎢ ∂ j ⎦⎥ ∂ j ⎣⎢ ∂ j ⎦⎥ ∂ j ⎣⎢ ∂ j ⎦⎥ ∂ j ⎣⎢ ∂ j ⎦⎥⎭⎪ …(3) Using the identity, ∂ ⎡∂X ⎤ ⎡ ∂Y ⎤ [X, Y] = , Y + X, ∂t ⎣⎢ ∂t ⎦⎥ ⎣⎢ ∂t ⎦⎥ Then, we find that R.H.S. of equation (3) reduces to 69

⎪⎧ ∂Z ∂ ∂Z ∂[X,Y]⎪⎫ = ∑⎨− [X,Y]+ ⎬ j ⎩⎪ ∂q j ∂p j ∂p j ∂q j ⎭⎪ + 0 (All other terms are cancelled)

⎪⎧ ∂Z ∂[X,Y] ∂Z ∂[X,Y]⎪⎫ = − ∑⎨ − ⎬ j ⎩⎪∂q j ∂p j ∂p j ∂q j ⎭⎪ = −[Z, [X, Y]] or [X, [Y, Z]] + [Y, [Z, X]] = − [Z, [X, Y]] ⇒ [X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y]] = 0

Particular Case Let Z = H, then [X, [Y, H]] + [Y, [H, X]] + [Y, [X, Y]] = 0 Suppose X & Y both are constants of motion, then [X, H] = 0, [Y, H] = 0 Then Jacobi’s identity gives [H, [X, Y]] = 0 ⇒ [X, Y] is also a constant of Motion. Hence poisson’s Bracket of two constants of Motion is itself a constant of Motion. 4.11 Poisson’s Theorem The total time rate of evolution of any dynamical variable F(p, q, t) is given by dF ∂F = + [F, H] dt ∂t

dF ∂F ⎡ ∂F ∂F ⎤ Solution : (p,q,t) = + ∑ ⎢ q&j + p&j ⎥ dt ∂t j ⎣⎢∂q j ∂p j ⎦⎥

∂F ⎡ ∂F ∂H ∂F ∂H ⎤ = + ∑⎢ − ⎥ ∂t j ⎣⎢∂q j ∂p j ∂p j ∂q j ⎦⎥ 70

dF ∂F = + [F, H] dt ∂t dF If F is constant of motion, then = 0. dt Then by Poisson’s theorem, ∂F + [F, H] = 0 ∂t ∂F Further if F does not contain time explicitly, then = 0 ∂t ⇒ [F, H] = 0 This is the requirement condition for a dynamical variable to be a constant of motion. 4.12 Jacobi-Poisson Theorem :- (or Poisson’s Second theorem) If u and v are any two constants of motion of any given Holonomic dynamical system, then their Poisson bracket [u, v] is also a constant of motion. d ∂ Proof:- We consider [u,v] = [u, v] + [[u, v], H] …(1) dt ∂t using the following results, ∂ ⎡∂u ⎤ ⎡ ∂v⎤ [u,v] = , v + u, …(2) ∂t ⎣⎢ ∂t ⎦⎥ ⎣⎢ ∂t ⎦⎥ [u, [v, w]] + [v, [w, u]] + [w, [u, v]] = 0 …(3)

d ⎡∂u ⎤ ⎡ ∂v⎤ ∴ (1) ⇒ [u,v] = ,v + u, +[[u,v],H] …(4) dt ⎣⎢ ∂t ⎦⎥ ⎣⎢ ∂t ⎦⎥ Put w = H in (3), we get [H, [u, v]] = −[u, [v, H]] − [v, [H, u]] ⇒ −[[v, H], u] − [[H, u], v] = [[u, v], H] …(5) from (4) & (5), we get

d ⎡∂u ⎤ ⎡ ∂v⎤ [u,v] = ,v + u, − [[v, H], u] − [[H, u], v] dt ⎣⎢ ∂t ⎦⎥ ⎣⎢ ∂t ⎦⎥ 71

⎡∂u ⎤ ⎡ ∂v⎤ = ,v + u, + [u, (v, H]] + [[v, H], v] ⎣⎢ ∂t ⎦⎥ ⎣⎢ ∂t ⎦⎥

⎡∂u ⎤ ⎡ ∂v ⎤ = +[u,H],v + u, +[v,H] ⎣⎢ ∂t ⎦⎥ ⎣⎢ ∂t ⎦⎥

d ⎡du ⎤ ⎡ dv⎤ ⇒ [u,v] = ,v + u, …(6) dt ⎣⎢ dt ⎦⎥ ⎣⎢ dt ⎦⎥ du dv Because and both are zero as u & v were constants of motion. dt dt d ∴ (6) ⇒ [u, v] = 0 dt ⇒ The Poisson bracket [u, v] is also a constant of motion.

4.13 Derivation of Hamilton’s Principle from Lagrange’s equation:- We know that Lagrange’s equations are

d ⎛ ∂L ⎞ ∂L ⎜ ⎟ − = 0 …(1) ⎜ & ⎟ dt ⎝ ∂q j ⎠ ∂q j

t2 t2 & Now δ ∫ L dt = ∫δL(q j q j ) dt t1 t1

t t 2 2 ⎡ ⎛ ∂L ∂L ⎞⎤ ⇒ δ L dt = ⎢ ⎜ δq + δq& ⎟⎥ dt ∫∫∑⎜ ∂q j ∂q& j ⎟ t1 t1 ⎣⎢ j ⎝ j j ⎠⎦⎥

t t 2 ⎡ ∂L ⎤ 2 ∂L = ⎢ δq ⎥dt + δq& dt ∫∫∑ ∂q j ∑ ∂q& j t1 ⎣⎢ j ⎦⎥ t1 j

t t2 t 2 ∂K ∂L 2 d ⎛ ∂L ⎞ = δq dt+ δq − ⎜ ⎟δq dt …(2) ∫∑∑j & j ∫∑ ⎜ & ⎟ j t ∂q j j ∂q j t j dt ∂q j 1 t1 1 ⎝ ⎠ Since, there is no coordinate variation at the end points.

⇒ δq j = δq j = 0 t1 t2 72

t t 2 2 ⎡ ∂L − d ⎛ ∂L ⎞⎤ So (2) ⇒ δ L dt = ⎢ ⎜ ⎟⎥ δqj dt ∫∫∑ ∂q dt ⎜ ∂q& ⎟ t1 t1 ⎣⎢ j j ⎝ j ⎠⎦⎥

t2 t2 ⇒ δ ∫∫L dt = 0 δq j dt [Using (1)] t1 t1 = 0 4.14 Derivation of Lagrange’s equations from Hamilton’s principle

t2 We are given δ ∫ Ldt = 0 t1

As δqj are arbitrary & independent of each other, So its coefficients should be zero separately. So we have

⎡ ∂L d ⎛ ∂L ⎞⎤ ⎢ − ⎜ ⎟⎥ = 0 ∑ ∂q dt ⎜ ∂q& ⎟ j ⎣⎢ j ⎝ j ⎠⎦⎥

d ⎛ ∂L ⎞ ∂L ⇒ ⎜ ⎟ − = 0 for j = 1, 2,….n ⎜ & ⎟ dt ⎝ ∂q j ⎠ ∂q j 4.15 Principle of Least action

Action of a dynamical system over an interval t1 < t < t2 is

t2 A = ∫ 2T dt t1 where T = K.E. This principle states that the variation of action along the actual path between given time interval is least, i.e.,

t2 δ ∫ 2T dt = 0 …(1) t1 Now we know that T + V = E (constant) V = P.E. and L = T−V

By Hamilton’s principle, we have

t2 t2 ∫δL dt = 0 or ∫δ(T − V)dt = 0 t1 t1

t2 ⇒ ∫δ (T − E + T) dt = 0 t1

t2 ⇒ ∫[δ(2T) − δE] dt = 0 t1

t2 ⇒ ∫δ(2T) dt = 0 [using E = constant ∴ δE = 0] t1

t2 ⇒ δ ∫ 2T dt = 0 t1 4.16 Distinction between Hamilton’s Principle and Principle of least action:-

Hamilton’s principle δS = 0 is applicable when the time interval (t2 − t1) in passing from one configuration to the other is prescribed whereas the principle of least action i.e. δA = 0 is applicable when the total energy of system in passing from one configuration to other is prescribed and the time interval is in no way restricted. This is the essential distinction between two principles. 4.17 Poincare−Cartan Integral Invariant :- We derive formula for δW in the general case when the initial and terminal instant of time, just like initial & terminal co-ordinates are not fixed but are functions of a parameter α.

t2 & W(α) = ∫ L[t1 qj(t, α), q j (t, α)]dt t1 let t1 = t1(α), t2 = t2(α)

q&j = q&j (α) at t = t1

2 2 q j = q j (α) at t = t2 74

t t 2 2 ⎡ ∂L ∂L ⎤ Now δW = δ L dt = L2 δt2 − L1 δt1 + ⎢ δq + δq ⎥ dt ∫ ∫∑ ∂q j ∂q& j t1 t1 j ⎣⎢ j j ⎦⎥ Integrating by parts Then δW = L δt + p2[δq ] − L δt − p′[δq ] 2 2 ∑ j j t=t2 1 1 ∑ j j t=t1 jj

t 2 ⎡ ∂L d ⎛ ∂L ⎞⎤ + ⎢ − ⎜ ⎟⎥δq dt …(1) ∫∑ ∂q dt ⎜ ∂q& ⎟ j t1 j ⎣⎢ j ⎝ j ⎠⎦⎥

Now qj = qj(t, α)

⎡∂q j (t,α)⎤ & ∴ [δq j ]t=t = ⎢ ⎥ δα 1 ∂α ⎣ ⎦ t=t1

⎡ ∂ ⎤ & [δq ] = q (t,α) δα …(2) j t=t2 ⎢∂α j ⎥ ⎣ ⎦ t=t2 On the other hand, for the variation of terminal co-ordinates

2 2 q j = q j q[t(α), α]

2 2 ⎡∂q j ⎤ ∴ δq j = q&j δt 2 + ⎢ (t,α)⎥ δα ∂α ⎣ ⎦ t=t2

⇒ δq2 = [δq ] + q&2 δt [using (2)] j j t=t2 j 2

⇒ [δq ] = δq2 − q&2 δt …(3) j t=t2 j j 2 Similarly [δq ] = δq′ − q&′ δt …(4) j t=t1 j j 1 Put (3) & (4) in (1), we get

2 2 2 δW = L2 δt2 + ∑p j (δq j − q&j δt 2 ) − L1 δt1 j

t 2 ⎡ ∂L d ⎛ ∂L ⎞⎤ − p′ (δq′ − q&′δt ) + ⎢ − ⎜ ⎟⎥ δqjdt ∑ j j j 1 ∫∑ ∂q dt ⎜ ∂q& ⎟ j t1 j ⎣⎢ j ⎝ j ⎠⎦⎥

2 t ⎡ n ⎤ 2 ⎡ ∂L d ⎛ ∂L ⎞⎤ ⇒ δW = ⎢ p δq − Hδt⎥ + ⎢ − ⎜ ⎟⎥ δqj dt …(5) ∑ j j ∫∑ ∂q dt ⎜ ∂q& ⎟ ⎣⎢ j=1 ⎦⎥1 t1 j ⎣⎢ j ⎝ j ⎠⎦⎥ 75

2 n ⎡ ⎤ 2 2 where ⎢∑pj δq j − Hδt⎥ = ∑ p j δq j − H2 δt 2 − ∑p′j δq′j + H1 δt1 ⎣⎢ j=1 ⎦⎥1 j j

Now we know that − H = L − ∑p j q&j j

∴ −H1 = L1 − ∑p′j q&′j j

2 2 and −H2 = L2 − ∑p j q&j j

In the special case for any α, the path is extremum, the integral on R.H.S. of equation (5) is equal to zero and formula for variation of W takes the form

2 ⎡ n ⎤ δW = ⎢∑p j δq j − Hδt⎥ …(6) ⎣⎢ j=1 ⎦⎥1 Integrating, we get

⎡ n ⎤ W = ∫ ⎢∑pj δq j − Hδt⎥ dt ⎣⎢ j=1 ⎦⎥ which is known as Poincare Cartan Integral Invariant. 4.18 Whittaker’s Equations :- We consider a generalised conservative system i.e. an arbitrary system for which the function H is not explicitly dependent on time. For it, we have

H(qj, pj) = E0 (constant) …(1) where j = 1, 2,….n

(2n − dimensional phase space in which qj, pj are coordinates) Then basic integral invariant I will becomes

I = ∫()∑ p j δq j − Hδt

n ⇒ I = ∫∑p j δq j [ Θ for a conservative system, δt = 0] …(2) j=1 solving (1) for one of the momenta, for example p1.

p1 = − k1(q1, q2,…., qn, p2…pn, E0) …(3) 76

Put the expression obtained in (2) in place of p1, we get

⎡ n ⎤ I = ∫ ⎢∑pj δq j + p1 δq1 ⎥ ⎣⎢ j=2 ⎦⎥

⎡ n ⎤ = ∫ ⎢∑p j δq j − k1 δq1 ⎥ …(4) ⎣⎢ j=2 ⎦⎥ But this integral invariant (4) again has the form of Poincare – Cartan integral if it is assumed that the basic co-ordinate and momenta are quantities qj & pj (j = 2, 3...n)

& variable q1 plays the role of time variable (and in place of H, we have function k1). Therefore the motion of a generalised conservative system should satisfy the following Hamilton’s system of differential equations (2n − 2).

dq j ∂k1 − ∂k1 dp j q&j = = ; = j = 2, 3…. …(5) dt ∂p j ∂q j dq1 The equations (5) were obtaioned by Whittaker and are known as Whittaker’s equations. 4.19 Jacobi’s equations :-

Integrating the system of Whittaker’s equations, we find qj & pj (j = 2,

3,….n) as functions of variables q1 and (2n −2) arbitrary constant C1, C2…. C2n−2. Moreover, the integrals of Whittaker’s equations will contain an arbitrary constant

E0 ,i.e., they will be of the form

qj = ϕj (q1, E0, C1, C2…. C2n−2)

pj = ψj(q1, E0, C1, C2…. C2n−2) (j = 2, 3,….n) …(6) Putting expression (6) in (3), we find

p1 = ψ1 (q1, E0, C1, C2….. C2n−2) …(7) From equation dq ∂H dq 1 = ⇒ dt = 1 dt ∂p1 ⎛ ∂H ⎞ ⎜ ⎟ ⎝ ∂p1 ⎠ 77

dq1 ⇒ t = + C2n−1 …(8) ∫ ⎛ ∂H ⎞ ⎜ ⎟ ⎝ ∂p1 ⎠

⎛ ∂H ⎞ where all the variables in partial derivative ⎜ ⎟ are expressed in terms of q1 with ⎝ ∂p1 ⎠ the help of (6) & (7). The Hamiltonian system of Whittaker’s equations (5) may be replaced by an equivalent system of equations of the Lagrangian type:

d ⎛ ∂P ⎞ ∂P ⎜ ⎟ − = 0 , j = 2, 3,….n …(9) ⎜ ′ ⎟ dq1 ⎝ ∂q j ⎠ ∂q j These are (n −1) second order equations where

dq j q′j = dq1 and the function P(analogous to Lagrangian function) is connected with the function K1 (analogous of Hamiltonian function) by the equation

p = p(q1, q2….qn, q′2 ,...q′n )

n P = ∑p j q′j − K1 …(10) j=2

dq2 The momenta pj must be replaced by their expression in temrs of q'2 = ,…… dq1

dqn q′n = dq1 which may be obtained from first (n−1) equation (5). From (3) & (10), n 1 n P = ∑p j q′j + p1 = ∑pi q&i j=2 q&1 i=1 1 ⇒ P = (L + H) …(11) q&1 For conservative system, L = T − V, H = T + V 78

⇒ L + H = 2T 2T Then P = …(12) q&1 1 n & K.E., T = ∑a ik q&iq&k 2 i,k=1

2 = q&1 G(q1,q2.....qn ,q′2...q′n ) …(13) where 1 n G(q1, q2….qn, q′2...q′n ...q′n ) = ∑aik q′iq′k …(14) 2 i,k=1 From (1) & (13), we find H = E

2 T = G q&1

T H − V q& = = 1 G G

E − V ∴ q& = …(15) 1 G

2 2T 2q&1 G and P = = = 2Gq&1 q&1 q&1

E − V ⇒ P = 2G [from (15)] G

= 2 G(E − V) …(16) Differential equation (9) in which function P is of the form (16) and which belong to ordinary conservative system are called Jacobi’s equations. 4.20 Theorem of Lee-Hwa-Chung

n If I′ = ∫∑[(qAi k, pk, t) δqi + Bi(t, qk, pk)δpi] i=1 is a universal relative integral invariant, then I′ = c I, where c is a constant and I1 is Poincare integral. 79

For n = 1, I′ = ∫(Aδq + Bδp)

⇒ I′ = c ∫ pδq = cI1 ⎡ n ⎤ ⎢I1 = ∫∑[pi δqi ]⎥ ⎣ i=1 ⎦ and I1 = ∫ pδq −Hδt from Poincare Cartan integral.

Lesson-5 Canonical Transformations 5.1 Introduction:-The Hamiltonian formulation if applied in a straightforward way, usually does not decrease the difficulty of solving any given problem in mechanics; we get almost the same differential equations as are provided by the Lagrangian procedure. The advantages of the Hamiltanian formation lie not in its use as a calculation tool, but rather in the deeper insight it affords into the formal structure of mechanics. We first derive a specific procedure for tranforming one set of variables into some other set which may be more suitable. In dealing with a given dynamical system defined physically, we are free to choose whatever generalised coordinates we like. The general dynamical theory is invariant under transformations qi→Qi, by which we understand a set of n variable expressing one set of n generalised co-ordinates qi in term of another set Qi. Invariant means that any general statement in dynamical theory is true no matter which system of coordinates is used. In Hamiltonian formulation, we can make a transformation of the independent coordinates qi , pi to a new set Qi, Pi with equations of transformation

Qi = Qi (q,p,t), Pi = Pi (q,p,t) Here we will be taking transformations which in the new coordinates Q, P are canonical. 5.2 Point transformation

Qj = Qj(qj, t) Transformation of configuration space is known as point transformation. 5.3 Canonical transformation old variables → new set of variable

qj, pj → Qj, Pj

Here Qj = Qj (qj, pj, t)

Pj = Pj(qj, pj, t) …(1) If the transformation are such that the Hamilton’s canonical equations 81

∂H − ∂H q&j = , p j = ∂p j ∂q j Preserve their form in the new variables i.e.

& ∂K & − ∂K Q j = , Pj = ∂Pj ∂Q j K being Hamiltonian in the new variable, then transformations are said to be canonical transformation. Also if

H = ∑p j q&j − L in old variable, then in new variable, & K = ∑PjQ j − L' where L′ = new Lagrangian

d ⎛ ∂L' ⎞ ∂L' Now ⎜ ⎟ − = 0 ⎜ & ⎟ dt ⎝ ∂Q j ⎠ ∂Q j i.e. Lagrange’s equations are covariant w.r.t. point transformation & if we define Pj as ∂L'(Q ,Q& ) P = j j j & ∂Q j ∂K(Q P ,t) & j j Q j = ∂Pj

− ∂K(Q ,P ,t) & j j and Pj = ∂Q j

t2 Hamilton’s principle in old variable, δ ∫ Ldt = 0 t1

t2 & δ ∫[]∑p j q j − H(q,p,t) dt = 0 …(2) t1 and in new variable, 82

t2 & δ ∫[]∑Pj Q j − k(Q,P,t) dt = 0 …(3) t1

t2 & & δ ∫{}()∑p jq j − H − ()∑Pj Q j − k dt = 0 …(4) t1 Let F = F(q, p, t)

t 2 d ∴ δ F(q,p,t) = δ[F(q,p,t)]t2 ∫ dt t1 t1 = δF

t2 ∂F ∂F = δq j + δp j ∂q j ∂p j t1

t1 t2 ∂F ∂F = δq j + δp j = 0 ∂q j ∂p j t1 t1

[Since the variation in qj and pj at end point vanish]

t2 ⎧ & dF⎫ ∴ (4) ⇒ δ ⎨()∑p j q&j − H − ()∑Pj Q j − K ⎬dt = 0 ∫ ⎩ dt ⎭ t1 dF ⇒ ()p q& − H − ()P Q& − K = …(5) ∑ j j ∑ j j dt

In (5), F is considered to be function of (4n + 1) variables i.e. qj, pj, Qj, Pj, t. But two sets of variables are connected by 2n transformation equation (1) & thus out of 4n variables besides t, only 2n are independent.

Thus F can be fuinction of F1(q, Q, t), F2 (q, P, t), F3(p, Q, t), or F4(p, P, t) So transformation relation can be derived by the knowledge of function F. Therefore it is known as Generating Function.

Let F1 = F1(q, Q, t) dF p q& − H = P Q& − k + 1 (q,Q,t) …(6) ∑∑j j j j dt 83

dF ⎛ ∂F ∂F ⎞ ∂F and 1 (p,Q,t) = ⎜ 1 q& + 1 Q& ⎟ + 1 …(7) ∑⎜ j j ⎟ dt ⎝ ∂q j ∂Q j ⎠ ∂t ∴ from (6) & (7), we get

& ∂F1 ∂F1 & ∂F1 ∑∑∑p j q&j − H = ∑Pj Q j − k + q&j + Q j + j jj∂q j ∂Q j ∂t

⎡ ⎤ ⎡ ⎤ ∂F1 ∂F1 & ∂F1 or ∑⎢ − p j ⎥q&j + ∑⎢Pj + ⎥Q j + H − K + = 0 …(8) j ⎣⎢∂q j ⎦⎥ j ⎣⎢ ∂Q j ⎦⎥ ∂t

Since qj and Qj are to be considered as independent variables, equation (8) holds if the coefficients of qj and Qj separately vanish i.e.

∂F1 (q, Q, t) = pj …(9) ∂q j

− ∂F1 (q,Q,t) and Pj = …(10) ∂Q j ∂F (q,Q,t) and K = H + 1 …(11) ∂t equation (11) gives relation between old and new Hamiltonian. Solving (9), we an find Qj = Qj(qj, pj, t) which when put in (10) gives

Pj = Pj(qj, pj, t) …(12) equation (12) are desired canonical transformation. 5.4 Hamilton–Jacobi Equation :-If the new Hamiltonian vanish, i.e., K = 0, then

Qj = αj (constant)

Pj = βj (constant) ∂F Also H + 1 = K = 0 ∂t

∂F1 ⇒ H(qj, pj, t) + = 0 ∂t

⎛ ∂F ⎞ ∂F Using (11), H⎜q , 1 , t⎟ + 1 = 0 ⎜ j ⎟ ⎝ ∂q j ⎠ ∂t 84

This partial differential equation together with equation (9) is known as Hamilton- Jacobi Equation. Generating function is also called characteristic function

Let F1 is replaced by S, then ∂S ⎛ ∂S ∂S ∂S ⎞ + H⎜q1,q 2 ,...,qn , , ,..., ,t⎟ = 0 , …(9) ∂t ⎝ ∂q1 ∂q 2 ∂q n ⎠ The solution S to above equation is called Hamilton principle function or characteristic function. Equation (9) is first order non-linear partial differential equation in (n + 1) independent variables (t, q1, q2,…,qn) and one dependent variables S. So Therefore, there will be (n + 1) arbitrary constants out of which one would be simply an additive constant and remaining n arbitrary constants may appear as arguments of S so that complete solution has the form S = S(q, t, α) + A …(10) where αi = α1, α2,…, αn are n constants and A is additive constant. Jacobi proved a theorem known as Jacobi’c theorem that the system would volve in such a way that the derivatives of S w.r.t. α’s remain constant in time and we write ∂S = βi (constant) (i = 1, 2, …., n) ∂αi

αi = Ist integrals of motion

βi = IInd integrals of motion 5.5 Statement of Jacobi’s Theorem

If S(t, qi, αi) is some complete integral fo Hamilton Jacobi equation (9) then the final equations of motion of a holonomic system with the given function. H may be written in the form ∂S ∂S = pi and = βi ∂q i ∂αi where βi, αi are arbitrary constants Proof: Given the complete integral for S given by (10), we wish to prove 85

∂S = βi ∂αi Consider

d ⎛ ∂S ⎞ ∂ ⎛ ∂S ⎞ ∂ ⎛ ∂S ⎞ & ⎜ ⎟ = ⎜ ⎟ + ∑ ⎜ ⎟q j dt ⎝ ∂αi ⎠ ∂t ⎝ ∂αi ⎠ ∂q j ⎝ ∂αi ⎠

∂ ⎛ ∂S ⎞ ∂ ⎛ ∂S ⎞ = + ⎜ ⎟q& ⎜ ⎟ ⎜ ⎟ j ∂αi ⎝ ∂t ⎠ ∂αi ⎝ ∂q j ⎠

∂ ⎡ ⎛ ∂S ⎞⎤ ∂ 2S = ⎢− H⎜ t, q , = p ⎟⎥ + q& [using (9)] ∂α ⎜ j ∂q j ⎟ ∂α ∂q j i ⎣⎢ ⎝ j ⎠⎦⎥ i j

d ⎛ ∂S ⎞ − ∂H ∂q ∂H ∂ 2S ∂ 2S ⎜ ⎟ j & ⎜ ⎟ = − + q j dt ⎝ ∂αi ⎠ ∂q j ∂αi ⎛ ∂S ⎞ ∂αi∂q j ∂αi∂q j ∂⎜ ⎟ ⎜ ⎟ ⎝ ∂q j ⎠

∂q j Since qj’s and αi’s are independent, we get = 0 ∂αi

d ⎛ ∂S ⎞ ⎡ ∂H ⎤ ∂ 2S ⎜ ⎟ & ⎜ ⎟ = ⎢− + q j ⎥ = 0, Using Hamilton’s equation of motion dt ⎝ ∂αi ⎠ ⎣⎢ ∂p j ⎦⎥ ∂αi∂q j

⎡ ∂H ⎤ ⎢q&j = ⎥ ⎣⎢ ∂p j ⎦⎥

d ⎛ ∂S ⎞ ⇒ ⎜ ⎟ = 0 dt ⎝ ∂αi ⎠ ∂S ⇒ = constant = βi (i = 1, 2,…, n) ∂αi

Remark :Consider total time derivative of S(qj, αj, t) dS ∂S ∂S ∂S = ∑∑q&j + α&j + dt ∂q j ∂α j ∂t 86

∂S ∂S But we know that αj = 0 since αj are constant. Also = p j & H + = 0 gives ∂q j ∂t

∂S = −H , we have ∂t dS = p q& − H = L dt ∑ j j ⇒ S = ∫ Ldt + constant The expression differs from Hamilton’s principle in a constant show that this time integral is ofindefinite form. Thus the same integral when indefinite form shapes the Hamilton’s principle. 5.6 Method of sepration of variables:- ∂H For a generalised conservative system = 0 . Then Hamilton’s Jacobi equation ∂t has the form

∂S ⎛ ∂S ⎞ ⎜ ⎟ + H⎜qi , ⎟ = 0 ∂t ⎝ ∂qi ⎠ If Hamiltonian does not explicitly contain time, one can linearly decouple time from rest of variables in S and we write

S (q1, q2,…….,qn, t) = S1(t) + V1 (q1, q2, ……qn) and its complete solution is of the form

S = −Et + V(q1, q2,…., qn, α1,…, αn-1, E) [S = function of t + function of q

S = −Et + V(q1, q2,…, qn, α1,…., αn−1, E)

Example:- Write Hamiltonian for one-dimensional harmonic oscillator of mass m & solve Hamilton-Jacobi equation for the same. Solution :- Let q be the position co-ordinates of harmonic oscillator then q& is its velocity and 1 K.E., T = m q&2 2 87

1 P.E., V = kq2 [k is some constant] 2 Lagrangian 1 1 L = T−V = mq&2 − k q2 2 2 The momentum is ∂L p = = mq& ∂q&j p ⇒ q&= m Then, the Hamiltonian is

H = ∑pi q&i − L

⎛ 1 2 1 2 ⎞ = p q&− ⎜ mq& − kq ⎟ ⎝ 2 2 ⎠

2 p 1 p 1 2 = p − m + k q m α m2 2 1 p2 1 H = + kq2 2 m 2 Also for a conservative system,

2 2 1 p 1 2 1 ⎡p 2 ⎤ Total Energy = P.E. + K.E. = m 2 + kq = ⎢ + kq ⎥ = Hamiltonian 2 m 2 2 ⎣ m ⎦ p2 kq2 ⇒ H (q, p) = + …(1) 2m 2 ∂S Replacing p by in H, ∂q

2 ⎛ ∂S ⎞ 1 ⎛ ∂S ⎞ kq2 ∂S ⎛ ∂S ⎞ H⎜q, ⎟ = ⎜ ⎟ + . Then from + H⎜qi , ⎟ = 0, we get ⎝ ∂q ⎠ 2m ⎝ ∂q ⎠ 2 ∂t ⎝ ∂qi ⎠

2 ∂S 1 ⎛ ∂S ⎞ kq2 + ⎜ ⎟ + = 0 …(2) ∂t 2m ⎝ ∂q ⎠ 2 88

We separate variables

∂S dS1, ∂S dV S(q, t) = V(q) + S1(t) ⇒ = & = ∂t dt ∂q dq Putting in (2), we get

2 2 1 ⎡dV⎤ kq dS1 (2) ⇒ ⎢ ⎥ + + = 0 2m ⎣ dq ⎦ 2 dt

2 2 dS1 −1 ⎛ dV ⎞ kq ⇒ = ⎜ ⎟ − dt 2m ⎝ dq ⎠ 2 L.H.S. is function of t only and R.H.S. is function of q only, But it is possible only when each side is equal to a constant (−E) (say) dS Let 1 = −E …(i) dt

2 1 ⎛ dV ⎞ 1 2 & ⎜ ⎟ + kq = E …(ii) 2m ⎝ dq ⎠ 2

(i) ⇒ S1 = −Et + constant

2 ⎛ dV ⎞ ⎛ 1 2 ⎞ (ii) ⇒ ⎜ ⎟ = 2m⎜E − kq ⎟ ⎝ dq ⎠ ⎝ 2 ⎠

dV ⎛ 1 ⎞ = 2m⎜E − kq 2 ⎟ dq ⎝ 2 ⎠

1 ⎛ 1 ⎞ 2 V(q) = 2m⎜E − kq 2 ⎟ dq + cons tan t ∫ ⎝ 2 ⎠ Therefore, complete integral is

S(q, t) = S1(t) + V(q)

1 ⎛ 1 ⎞ 2 S(q, t) = −Et + 2m⎜E − Kq 2 ⎟ dq + constan t ∫ ⎝ 2 ⎠ Further by Jacobi’s theorem

Here α1 = E 89

∂S ∂S = β1 ⇒ = β1 ∂α1 ∂E

∂S 2m dq β = ⇒ β = −t + 1 ∂E 1 2 ∫ 1 ⎛ 1 ⎞ 2 ⎜E − Kq 2 ⎟ ⎝ 2 ⎠

m 2 dq β = −t + , 1 2 K ∫ 1 ⎛ 2E ⎞ 2 ⎜ − q 2 ⎟ ⎝ K ⎠

m dq ⎡ 1 −1 x ⎤ = −t + ⎢ dx = sin ⎥ ∫ 2 ∫ 2 2 a K ⎛ E ⎞ ⎣ a − x ⎦ ⎜ ⎟ 2 ⎜ ⎟ − ()q ⎝ K ⎠

m ⎛ K ⎞ −1⎜ ⎟ = − t + sin ⎜q ⎟ K ⎝ 2E ⎠ K ⎛ K ⎞ t + β = sin −1 ⎜ q⎟ ()1 ⎜ ⎟ m ⎝ 2E ⎠ k ⎡ K ⎤ q = sin⎢ ()t + β1 ⎥ 2E ⎣ m ⎦ 2E ⎡ K ⎤ q(E, t) = sin⎢ ()t + β1 ⎥ K ⎣ m ⎦ The constants β1, E can be found from initial conditions The momentum is given by ∂S ∂()S (t) + V(q) p = = 1 ∂q ∂q ∂V = ∂q

1 ⎛ 1 ⎞ 2 =2m⎜E − Kq 2 ⎟ ⎝ 2 ⎠

2m 1 = ()2E − Kq 2 2 2 90

= m()2E − Kq 2 , which can be expressed as a function of t if we put q = q(t). Example:- Central force problem in Polar co-ordinates (r, θ) ρ Here K.E, r

. 2 1 ⎡ &2 ⎤ T = m⎢r + ()rθ ⎥ P(r, θ) 2 ⎣ ⎦ θ P.E., V(r) S Then L = T − V 1 L = m[r&2 + r 2θ&2 ]− V(r) 2

∂L & pr = [q1 = r,q& = r&, q2 = θ, q& = θ ,pj = pr, pθ] ∂r& 1 2

p r pr = mr& ⇒ r&= m

∂L 2 p θ pθ = = mr θ& ⇒ θ&= ∂θ& mr 2 Therefore, Hamiltonian is & H = ∑piqi − L 1 ⎛ ⎞ & & . &2 2 &2 = pr r + pθθ − m⎜r + r θ ⎟ + V(r) 2 ⎝ ⎠

2 2 2 p p mp mr p θ& = p r + p Q − r − + V(r) r& m θ& mr 2 2mr 2 2m2r 2 p2 p2 p2 p2 =r + θ − r − θ + V(r) m mr 2 2m 2mr 2

2 2 1 ⎡pr pθ ⎤ =⎢ + 2 ⎥ + V(r) 2 ⎣ m mr ⎦

2 1 ⎡ 2 pθ ⎤ =⎢pr + 2 ⎥ + V(r) 2m ⎣ mr ⎦ H − J equation is 91

∂S + H = 0 ∂r

2 2 ∂S 1 ⎡⎛ ∂S ⎞ 1 ⎛ ∂S ⎞ ⎤ + ⎢⎜ ⎟ + 2 ⎜ ⎟ ⎥ ∂t 2m ⎣⎢⎝ ∂r ⎠ r ⎝ ∂θ ⎠ ⎦⎥ + V(r) = 0 Using the Method of separation of variable, we have

S = S1(t) + R(r) + Φ (θ) ⎡ 2 2 ⎤ dS1 −1 ⎛ dR ⎞ 1 ⎛ dΦ ⎞ ⇒ = ⎢⎜ ⎟ + 2 ⎜ ⎟ ⎥ − V(r) dt 2m ⎣⎢⎝ dr ⎠ r ⎝ dθ ⎠ ⎦⎥ L.H.S. is function of t and R.H.S. is function of r & θ but not of t, therefore it is not possible only where each is equal to constant = −E (say)

dS1 ⇒ = −E ⇒S1 (t) = − Et + constant dt

2 2 −1 ⎡⎛ dR ⎞ 1 ⎛ dΦ ⎞ ⎤ and ⎢⎜ ⎟ + 2 ⎜ ⎟ ⎥ − V(r) = −E 2m ⎣⎢⎝ dr ⎠ r ⎝ dθ ⎠ ⎦⎥

2 2 − r 2 ⎛ dR ⎞ 1 ⎛ dΦ ⎞ ⇒ ⎜ ⎟ − r 2V(r) + r 2E = ⎜ ⎟ 2m ⎝ dr ⎠ 2m ⎝ dθ ⎠ L.H.S is function of r only and R.H.S is function of θ only h 2 So each = constant = ()say 2m dΦ = h ⇒ Φ()θ = hθ + constant dθ Then r equation gives

2 − r 2 ⎛ dR ⎞ h 2 ⎜ ⎟ − r 2V(r) + Er 2 = 2m ⎝ dr ⎠ 2m

2 ⎛ dR ⎞ ⎡ h 2 ⎤ ⎜ ⎟ = ⎢2m()E − V − 2 ⎥ ⎝ dr ⎠ ⎣ r ⎦ dR 1 = [2m()E − V − h 2r −2 ]2 dr 92

R = ∫ 2m()E − V − h 2r −2 dr + constant Therefore, complete solution is

S = S1 + R + Ф S = −Et + hθ + ∫ 2m()E − V − h 2r −2 dr + constant , is required solution ∂S Now, = constant ∂E mdr ⇒ − t + = β1 ()say ∫ 2 h 2m()E − V − r 2 The other equation is ∂S = constant ∂h 1 ()− 2hr −2 dr ⇒ Φ + = β ()say 2 ∫ 1 2 []2m()E − V − h 2r −2 2 Example:-When a particle of mass m moves in a force field of potential V. Write the Hamiltonian. Solution:- Here K.E. is 1 T = m(x&2 + y&2 + z&2 ) 2

p p p ⇒ x&= x , y&= y ,z&= z and P.E. is V (x,y,z) m m m ⇒ H = T + V 1 H = (p2 + p2 + p2 )+ V()x, y,z 2m x y z Example :- A particle of mass m moves in a force whose of potential in spherical coordinates V is -μcosθ/r2 . Write Hamiltonian in spherical coordinate (r, θ, φ). Also find solution of H.J. equation. 1 Solution:- L = m[]r&2 + r 2θ&2 + r 2 sin 2 θφ&2 − V()r,θ,φ 2 93

∂L pr = = mr& ∂r&

∂L 2 2 2 2 pθ = = mr φ& ,p = mr sin θφ& ∂θ& φ& Hamiltonian is given by

1 ⎛ p 2 p 2 ⎞ μcosθ H = ⎜p 2 + θ + φ ⎟ − ⎜ r 2 2 2 ⎟ 2 2m ⎝ r r sin θ ⎠ r ∂S ∂S ∂S Writing pr = , p = , p = ∂r θ ∂θ φ ∂φ Required Hamilton Jacobi equation is

2 2 2 ∂S 1 ⎡⎛ ∂S ⎞ 1 ⎛ ∂S ⎞ 1 ⎛ ∂S ⎞ ⎤ μcosθ + + + ⎜ ⎟ − = 0 …(1) ⎢⎜ ⎟ 2 ⎜ ⎟ 2 2 ⎜ ⎟ ⎥ 2 ∂t 2m ⎣⎢⎝ ∂r ⎠ r ⎝ ∂θ ⎠ r sin θ ⎝ ∂φ ⎠ ⎦⎥ r

Let S (t, r, θ, φ) = S1(t) + S2(r) + S3(θ) + S4(φ) in (1)

2 2 2 ∂S 1 ⎡⎛ dS ⎞ 1 ⎛ dS ⎞ 1 ⎛ ds ⎞ μcosθ⎤ 1 = − 2 + 3 + ⎜ 4 ⎟ + ⎢⎜ ⎟ 2 ⎜ ⎟ 2 2 ⎜ ⎟ 2 ⎥ ∂t 2m ⎣⎢⎝ dr ⎠ r ⎝ dθ ⎠ r sin θ ⎝ dφ ⎠ r ⎦⎥ L.H.S. is function of t only, R.H.S. is function of r, θ, φ and not of t, it is possible only when each is constant (= − E) (say) dS ⇒ 1 = −E ⇒ S = −Et + cons tan t dt 1

2 2 1 ⎪⎧⎛ dS ⎞ 1 ⎛ dS ⎞ 1 ⎛ dS ⎞⎪⎫ μcosθ & 2 + 3 + ⎜ 4 ⎟ − = E ⎨⎜ ⎟ 2 ⎜ ⎟ 2 2 ⎜ ⎟⎬ 2 2m ⎩⎪⎝ dr ⎠ r ⎝ dθ ⎠ r sin θ ⎝ dφ ⎠⎭⎪ r Multiplying 2mr2 and rearranging terms, we get

2 2 2 ⎛ dS ⎞ ⎛ dS ⎞ 1 ⎛ dS ⎞ r 2 2 − 2mr 2 E = − 3 − ⎜ 4 ⎟ + 2mμcosθ ⎜ ⎟ ⎜ ⎟ 2 ⎜ ⎟ ⎝ dr ⎠ ⎝ dθ ⎠ sin θ ⎝ dφ ⎠ L.H.S. is function of r only and R.H.S. is function of θ and φ. It is possible only if each is equal to constant.

2 2 ⎛ dS2 ⎞ 2 ⇒ r ⎜ ⎟ − 2mEr = β1 …(2) ⎝ dr ⎠ 94

2 2 ⎛ dS ⎞ 1 ⎛ dS ⎞ & − 3 − ⎜ 4 ⎟ + 2mμcosθ = β …(3) ⎜ ⎟ 2 ⎜ ⎟ 1 ⎝ dθ ⎠ sin θ ⎝ dφ ⎠

β1 ⇒ S2 = 2mE + dr + constan t ∫ r 2

2 2 ⎛ dS4 ⎞ 2 2 2 ⎛ dS3 ⎞ ⇒ ⎜ ⎟ = 2mμcosθsin θ − β1 sin θ − sin θ⎜ ⎟ ⎝ dφ ⎠ ⎝ dθ ⎠ L.H.S. is function of φ whereas R.H.S. is function of θ and it is possible when each is equal to constant.

2 ⎛ dS4 ⎞ dS4 ⇒ ⎜ ⎟ = β2 ⇒ = β 2 …(4) ⎝ dφ ⎠ dφ

dS4 and pφ = …(5) dφ

⇒ S4 = pφ φ + constant

2 pφ = β 2 [from (4) & (5)] and S = 2mμcosθ − p 2 cosec2φ − β dθ + constant [using (5)] 3 ∫ φ 1 The complete solution is β S = − Et + 2mE + 1 dr + 2mμcosθ − β − −p 2 cosec2θ dθ ∫∫r 2 1 φ

+ φ pφ + constant

5.7 Lagrange’s Brackets

Lagrange’s bracket of (u, v) w.r.t. the basis (qj, pj) is defined as

⎡∂q j ∂p j ∂p j ∂q j ⎤ {u, v}q,p or (u, v)q,p = ∑⎢ . − ⎥ j ⎣ ∂u ∂v ∂u ∂v ⎦ Properties: (1) (u, v) = −(v, u)

(2) (qi, qj) = 0

(3) (pi, pj) = 0

(4) (qi, pj) = δij 95

⎛ ∂q j ∂p j ∂p j ∂q j ⎞ ⎛ ∂p j ∂q j ∂q j ∂p j ⎞ (u, v) = ∑⎜ − ⎟ = − ∑⎜ − ⎟ = -(v, u) j ⎝ ∂u ∂v ∂u ∂v ⎠ j ⎝ ∂u ∂v ∂u ∂v ⎠ ⎛ ⎞ ⎜ ∂qk ∂pk' ∂qk' ∂pk ⎟ (2) (qi, qj) = ∑⎜ . − ⎟ [ Since q’s and p’s are independent k ⎝ ∂qi ∂q j ∂q j ∂qi ⎠

∂p k' ∂p k = 0. ⇒ = o and 0 ] ∂q j ∂q i

(3) Similarly we can prove that

{pi, pj} = 0 ⎛ ⎞ ∂qk ∂pk ∂qk ∂pk ∂q k ∂p k (4) {qi, pj} = ⎜ . − ⎟ = . = δ δ = δ ∑⎜ ⎟ ∑ ∂q ∂p ∑ ki kj ij k ⎝ ∂qi ∂p j ∂p j ∂qi ⎠ k i j k

5.8 Invariance of Poisson’s Bracket under Canonical transformation:- Poisson’s bracket is

⎛ ∂u ∂v ∂u ∂v ⎞ (u, v) = ⎜ − ⎟ q, p ∑⎜ ⎟ j ⎝ ∂q j ∂p j ∂p j ∂q j ⎠ The transformation of co- ordinates in a 2n – dimensional phase space is called canonical if the transformation carries any Hamiltonian into a new hamiltonian system

To show :- [F,G]q , p = [F, G]Q,P Poisson’s brackets is

⎛ ∂F ∂G ∂F ∂G ⎞ [F, G]q,p = ∑⎜ − ⎟ i ⎝ ∂qi ∂pi ∂pi ∂qi ⎠ If q, p are functions of Q & P then q = q (Q, P) & p = p (Q, P) and F & G will also function of (q, p), we have, G = G(Qk, Pk), we have 96

⎧ ∂F ⎡ ∂G ∂Q k ∂G ∂p k ⎤ ⎫ ⎪ ⎢ ⋅ + ⋅ ⎥ ⎪ ⎪∂q i ⎣∂Q k ∂Pi ∂Pk ∂Pi ⎦ ⎪ [F, G]q, p = ∑ ⎨ ⎬ i ⎪ ∂F ⎡ ∂G ∂Q k ∂G ∂Pk ⎤⎪ ⎪− ⎢ ⋅ + ⋅ ⎥⎪ ⎩ ∂pi ⎣∂Q k ∂q i ∂Pk ∂q i ⎦⎭

⎧ ∂G ⎡ ∂F ∂Q k ∂F ∂Q k ⎤⎫ ⎪ ⎢ ⋅ − ⋅ ⎥⎪ ⎪∂Q k ⎣∂q i ∂pi ∂pi ∂q i ⎦⎪ = ∑ ⎨ ⎬ i, k ⎪ ∂G ⎡ ∂F ∂Pk ∂F ∂Pk ⎤ ⎪ ⎪+ ⎢ ⋅ − ⋅ ⎥ ⎪ ⎩ ∂Pk ⎣∂q i ∂pi ∂pi ∂q i ⎦ ⎭ ⎧ ∂G ∂G ⎫ = F,Q + F,P …(1) ∑ ⎨ []q q,p []k q.p ⎬ i, k ⎩∂Q k ∂Pk ⎭

To find [F, Qk]q, p & [F, Pk]q, p

Replacing F by Qi in (1) ∂G ∂G [Q , G] = Q ,Q + Q P i q, p ∑ []i k q,p []i, k q,p i,k ∂Q k ∂PR ∂G = 0 + ∑ δik k ∂Pk ∂G [Qi, G]q,p = ∂Pi ∂G ⇒ [G, Qi]q, p = − ∂Pi ∂F and [F, Qk]q, p = − …(2) ∂PR

Replacing F by Pi in (1) ∂G ∂G [Pi, G] = − ⇒ [G, Pi] = ∂φi ∂q i ∂F and [F, Pk] = …(3) ∂q k Put these values from (2) and (3) in (1), we get:-

⎛ ∂G ∂F ∂G ∂F ⎞ [F, G]q,p = ∑⎜− + ⋅ ⎟ ⋅ i,k ⎝ ∂φK ∂PK ∂PK ∂φK ⎠ 97

= [F, G]Q,P 5.9 Poincare integral Invariant:- Under Canonical transformation, the integral J= dq dp …(1) ∫∫∑ i i S Where S is any 2 − D (surface) phase space remains Invariant Proof :- The position of a point on any 2− D surface is specified completely by two parameters, e.g. u, v

qi = qi ()u,v ⎫ Then ⎬ …(2) pi = pi ()u,v ⎭ In order to transform integral (1) into new variables (u, v), we take the relation

∂()q i ,pi dqi dpi = du dv …(3) ∂()u, v

∂qi ∂pi ∂()q ,p where i i = ∂u ∂u as the Jacobian. ∂()u, v ∂qi ∂pi ∂v ∂v Let Canonical transformation be

Qk = Qk(q, p, t), Pk = Pk(q, p, t) …(4)

∂()φK PK then dQk dPk = du dv …(5) ∂()u,v if J is invariant under canonical transformation (4), then we can write dq dp = dQ dP ∫∫∑ i i ∫∫∑ K K S i S K ∂()q p ∂(Q ,P ) or ∫∫∑ i i du dv = ∫∫∑ K K dudv S i ∂()u, v S K ∂()u, v Because the surface S is arbitrary the expressions are equal only if the integrands are identicals, i.e., 98

∂()q ,p ∂()Q ,P ∑ i i = ∑ K K i ∂()u, v K ∂()u, v …(6) In order to prove it, we would transform(q,p) basis to (Q, P) basesthrough the generating function F2(q, p, t), With this form of generating function, we have

∂F2 ∂F2 pi = & Q K = ∂q i ∂Pk

2 2 ∂pi ∂ ⎛ ∂F2 ⎞ ⎛ ∂ F2 ∂q k ∂ F2 ∂Pk ⎞ = ⎜ ⎟ = ∑⎜ ⋅ + ⎟ ∂u ∂u ⎝ ∂q i ⎠ K ⎝ ∂q i .∂q k ∂u ∂q i .∂p k ∂u ⎠

2 2 ∂pi ∂ ⎛ ∂F2 ⎞ ⎛ ∂ F2 ∂q k ∂ F2 ∂Pk ⎞ and = ⎜ ⎟ = ∑⎜ ⋅ + ⎟ ∂v ∂v ⎝ ∂q i ⎠ K ⎝ ∂q i .∂q k ∂v ∂q i .∂p k ∂v ⎠

∂q i ∂pi ∂()q ,p Now, ∑ i i = ∑ ∂u ∂u i ∂()u, v i ∂q i ∂pi ∂v ∂v

∂q ∂ 2 F ∂q ∂ 2 F ∂P i 2 k + 2 k ∂u ∂q ∂q ∂u ∂q ∂P ∂u = i k i k ∑ ∂q ∂ 2 F ∂q ∂ 2 F ∂P i,k i k + 2 k ∂v ∂q i ∂q k ∂v ∂q i ∂Pk ∂v

2 2 ∂q i ∂ F2 ∂q k ∂q i ∂ F2 ∂Pk

∂u ∂q i ∂q k ∂u ∂u ∂q i ∂Pk ∂u = ∑∑2 + 2 ik,k ∂q i ∂ F2 ∂q k i, ∂q i ∂ F2 ∂Pk

∂v ∂q i ∂q k ∂v ∂v ∂q i ∂Pk ∂v

∂qi ∂qk ∂ 2F L. H. S of (6) = ∂u ∂u ∑ ∂q ∂q i,k ∂qi∂qi i k ∂v ∂v

∂q i ∂Pk ∂ 2 F + ∂u ∂u …(7) ∑ ∂q ∂P i,k ∂q i ∂Pk i k ∂v ∂v We see that first term on R.H.S. is antisymmetric expression under interchange of i and k, its value will be zero, 99

i.e.,

∂qi ∂q k ∂q k ∂qi ∂ 2F ∂ 2F ∂u ∂u = ∂u ∂u ∑ ∂q ∂q ∑ ∂q ∂q i,k ∂qi∂q k i k k,i ∂q k ∂qi k i ∂v ∂v ∂v ∂v

∂qi ∂q k ∂ 2F = − ∂u ∂u ∑ ∂q ∂q k,i ∂q k ∂qi i k ∂v ∂v

∂qi ∂q k ∂ 2F or ∂u ∂u = 0 …(8) ∑ ∂q ∂q i,k ∂qi∂q k i k ∂v ∂v Similarly replacing q by P we have from (8)

∂Pi ∂Pk ∂ 2F ∂u ∂u = 0 ∑ ∂P ∂P i,k ∂Pi∂Pk i k ∂v ∂v Now equation (7) can be written as

∂Pi ∂Pk ∂()q p ∂ 2F i i = ∂u ∂u ∑ ∑ ∂P ∂P i ∂()u,v i,k ∂Pi∂Pk i k ∂v ∂v

∂qi ∂Pk ∂ 2F + ∂u ∂u ∑ ∂q ∂P i,k ∂qi∂Pk i k ∂v ∂v

∂ 2 F ∂P ∂ 2 F ∂q ∂P 2 i + 2 i k ∂P ∂P ∂u ∂P ∂q ∂u ∂u = k i k i ∑ ∂ 2 F ∂P ∂ 2 F ∂q ∂P i,k i + 2 i k ∂Pk ∂Pi ∂v ∂Pk ∂q i ∂v ∂v

∂ ⎛ ∂F ⎞ ∂P ⎜ 2 ⎟ k ∂u ⎜ ∂P ⎟ ∂u =∑ ⎝ k ⎠ k ∂ ⎛ ∂F2 ⎞ ∂Pk ⎜ ⎟ ∂v ⎝ ∂Pk ⎠ ∂v 100

∂F2 Put = Q k ∂Pk

∂Q k ∂Pk ∂(Q ,P ) =∑ ∂u ∂u = ∑ k k = R.H.S of (6). k ∂Q k ∂Pk k ∂()u, v ∂v ∂v Which proves that integral is invariant under canonical transformation. 5.10 Lagrange’s bracket is invariant under Canonical transformation:- The Lagrange’s bracket of u & v is defined as

⎛ ∂qi ∂pi ∂qi ∂pi ⎞ {}u, v q,p = ∑⎜ − ⎟ i ⎝ ∂u ∂v ∂v ∂u ⎠

∂qi ∂qi = ∑ ∂u ∂v i ∂pi ∂pi ∂u ∂v ∂()q p Since ∑ i i is invariant under Canonical transformation. i ∂()u,v So Lagrange’s bracket is also invariant under canonical transformation

101

Lesson-6 Attraction and Potential 6.1. Attraction of a uniform straight rod at an external point:- AB be a rod ∠APB = α − β P X α θ p β Y

δx A Q′ Q x B M

Let m be the mass per unit length of a uniform rod AB. It is required to find the components of attraction of the rod AB at an external point P. MP = p Consider an element QQ´ of the rod where MQ = x QQ´ = dx ∠MPQ = θ , In Δ MPQ , MQ x tanθ = = ⇒ x = p tanθ …(*) MP p MP p cosθ = = PQ pQ p ⇒ PQ = ⇒ PQ = psecθ …(**) cosθ Mass of element QQ´ of rod = m dx = mp sec2θ dθ …(using*) mass mpsec2 dθ The attraction at P of the element QQ´ is = = along PQ ()dis tan ce 2 ()PQ 2 Therefore, Force of attraction at P of the element QQ´ is 102

mpsec2 θdθ = …[using(**)] p2 sec2 θ m = dθ along PQ p …(1) Let ∠MPA = α and ∠MPB = β

α m f = ∫ dθ β p let X and Y be the components of attraction of the rod parallel & ⊥r to rod, then α m X = ∫ sin θdθ β p

α m & Y = ∫ cosθdθ β p m m X = []− cosθ α = []cosβ − cosα p β p

m ⎡ α − β α − β⎤ = 2sin sin …(2) p ⎣⎢ 2 2 ⎦⎥ m m and Y = []sin θ α = []sin α − sin β p β p

m ⎡ α + β α − β⎤ = 2cos sin …(3) p ⎣⎢ 2 2 ⎦⎥ Resultant force of Attraction R is given by

R = X 2 + Y 2 2m α − β R = sin […using 2&3] p 2 2m APB = sin ∠ p 2 X −1 Resultant R makes angle tan Y 103

1 ⎡ −1⎛ X ⎞ ⎡ −1⎛ α + β ⎞⎤⎤ or ()α + β with PM ⎢Θ tan ⎜ ⎟ = ⎢tan ⎜ tan ⎟⎥⎥ 2 ⎣ ⎝ Y ⎠ ⎣ ⎝ 2 ⎠⎦⎦ i.e. it acts along bisector of angle ∠APB. m m ⎡ p p ⎤ Also X = − Θ cosβ = ,cosα = & u sin g(2) PB PA ⎣⎢ PB PA ⎦⎥ Cor :- If the rod is infinitely long, the angle APB is two right angles & Resultant 2m attraction = ⊥r to the rod. p 6.2 Potential of uniform rod :- By definition, the potential at P is given by m V = dx ∫ PQ α mpsec2 θ V = ∫ dθ β psecθ α = ∫ msecθdθ β

α ⎡ ⎛ π θ ⎞⎤ = m ⎢logtan⎜ + ⎟⎥ ⎣ ⎝ 4 2 ⎠⎦ β ⎡ ⎛ π α ⎞ ⎛ π β ⎞⎤ = m ⎢logtan⎜ + ⎟ − log tan⎜ + ⎟⎥ ⎣ ⎝ 4 2 ⎠ ⎝ 4 2 ⎠⎦ ⎡ ⎛ α π ⎞⎤ tan⎜ + ⎟ ⎢ 2 4 ⎥ = m log ⎢ ⎝ ⎠⎥ ⎢ ⎛ β π ⎞ ⎥ ⎢ tan⎜ + ⎟ ⎥ ⎣ ⎝ 2 4 ⎠ ⎦ 6.3 Potential at a point P on the axis of a Uniform circular disc or plate:- We consider a uniform circular disc of radius ‘a’ & P is a pt. on the axis of disc & the pt. P is at a distance r from the centre 0, i.e., OP = r, OQ = x, PQ =

r 2 + x 2 let us divide the disc into a number of concentric rings & let one such ring has radius ‘x’ & width dx Then, Mass of ring is = 2Πx dxρ 104

where ρ density of material of disc ρ = mass/Area 2πx dxρ Therefore, Potential at P due to this ring is given by ,dv = r 2 + x 2 P

2 2 θ r + x r

OP = r

Q′ Q O

Hence, the potential at P due to the whole disc is given by

a x dx V = 2Π ρ ∫ 2 2 0 x + r

a 1 2πρ − V = ∫ 2x()x 2 + r 2 2 dx 2 0

V = 2Π ρ [ a 2 + r 2 − r] Let Mass of disc = M = Πa2 ρ M ⇒ Π ρ = a 2 2m Then V = []a 2 + r 2 − r is requiredpotential at any pt P which lies on the axis of a 2 disc. 6.4 Attraction at any point on the axis of Uniform circular disc :- Here radius of disc = a

OP = r, PQ = x 2 + r 2 OQ = x 105

We consider two element of masses dm at the two opposite position Q and Q´as shown. Now element dm at Q causes attraction on unit mass at P in the direction

PQ. Similarly other mass dm at Q´ causes attraction on same unit mass at P in the direction P Q´ and the force of attraction is same in magnitude. These two attraction forces when resolved into two direction one along the axes PO and other at right angle PO. Components along PO are additive and component along perpendicular to PO canceling each other Mass of ring = 2Πx dx ρ Attraction at P due to ring along PO is given by ρ ( dm)cosθ df = ∑ ()PQ 2 ρ cosQ.2πx dxρ r.2πxdx ρ r df = = along PO [Θ cosθ = inΔOPQ ] ()PQ 2 ()PQ 3 PQ

2πρ.rx dx = 3 ()r 2 + x 2 2 Therefore, the resultant attraction at P due to the whole disc along PO is given by

a 3 ρ − f = πρr∫ ()2x ()r 2 + x 2 2 dx 0

1 a ⎡ 2 2 − ⎤ = πρr⎢− 2()x + r 2 ⎥ ⎣ ⎦ 0

⎡1 1 ⎤ = 2πρr⎢ − ⎥ along PO ⎣r a 2 + r 2 ⎦ Let M = mass of disc of radius a =Πa2 M Π ρ = ρ a 2

ρ 2M ⎡ r ⎤ f = 2 ⎢1− ⎥ a ⎣ a 2 + r 2 ⎦ 106

2M = []1− cosα a 2 Where α is the angle which any radius of disc subtends at P Particular cases :- π 1. If radius of disc becomes infinite, then α = 2 and ρ 2M ⎡ π⎤ f = 1− cos a 2 ⎣⎢ 2⎦⎥ 2M = = constant [here, it is independent of position of P] a 2 2. When P is at a very large distance from the disc, then α→0 ρ 2M Therefore, f = ()1− cos0 a 2 = 0 6.5 Potential of a thin spherical shell :- We consider a thin spherical shell of radius ‘a’ & surface density ‘ρ’ let P be a point at a distance ‘r’ from the center O of the shell. We consider a slice BBĆĆ in the form of ring with two planes close to each other and perpendicular to OP. Area of ring (slice) BBĆĆ =2ΠBD×BB´

where Radius of ring, BD = a sinθ B′ width of ring, BB´= a dθ B x Therefor,Mass of slice (ring) is αθ θ α = 2Π a sinθ adθ ρ O D r P = 2Πa2 ρ sinθ dθ Hence, Potential at P due to slice (ring) is C C′ 2πa 2ρsin θdθ dV = …(1) x 107

Now, from ΔBOP, BP2 = OP2 + OB2 − 2OP.OB cosθ x2 = r2 + a2 − 2ar cosθ differentiating 2x dx = 2ar sinɵ dɵ x dx = sin θdθ ar 2πa 2ρx dx Putting in (1), we get dV = x.ar 2πaρdx = (2) r Therefore, Potential for the whole spherical shell is obtained by integrating equation (2), we have 2πaρ V = dx ∫ r 2πaρ = dx r ∫ Now, we consider the following cases :- Case(i) The point P is outside the shell. In this case, the limit of integration extends from x = (r − a) to x = (r + a)

2πaρ r+a Hence V = ∫ dx r r=a

4πa 2ρ V = r Here, Mass of spherical shall = 4Πa2 ρ M Then V = r Case(ii) When P is on the spherical shell, then limits are from x = 0 to x = 2a (here r = a) 108

2πaρ 2a Then V = ∫ dx a 0

4πa 2ρ M = = a a Case(iii) When P is inside the spherical shell, limit are from x = (a − r) to (a + r) M V = 4Πa ρ = a

6.6 Attraction of a spherical shell Let us consider a slice BB´C´C at point P , the attraction due to this slice is ρ 2πa 2ρsin θdθ df = along PB x 2 The resultant attraction directed along PO is given by ρ 2πa 2ρsin θdθ df = cosα x 2 xdx We know that sinθ dθ = ar PD r − a cosθ In ΔBDP, cosα = = . PB x

ρ 2πa 2ρxdx ⎛ r − a cosθ ⎞ df = ⎜ ⎟ ar.x 2 ⎝ x ⎠ We know that x2 = a2 + r2 − 2arcosθ x2 − a2 + r2 = 2r2 − 2arcosθ x 2 − a 2 + r 2 = r − a cosθ 2r ρ 2πa 2ρxdx ()x 2 − a 2 + r 2 Then, d f = ar.x 2 2r, x

πaρ ⎛ x 2 − a 2 + r 2 ⎞ = ⎜ ⎟dx 2 ⎜ 2 ⎟ r ⎝ x ⎠ 109

aπρ ⎛ r 2 + a 2 ⎞ = ⎜1+ ⎟dx 2 ⎜ 2 ⎟ r ⎝ x ⎠ Hence the attraction for the whole spherical shell is obtained by integration

ρ πaρ ⎡ r 2 − a 2 ⎤ Therefore, f = 1+ dx 2 ∫ ⎢ 2 ⎥ r ⎣ x ⎦ Now we consider the following cases depending upon the position of P Case(i) When point P is inside the shell, the limits of integration are x = (r − a) to (r + a)

→ πaρ r+a ⎛ r 2 − a 2 ⎞ f = ⎜1+ ⎟dx 2 ∫ ⎜ 2 ⎟ r r−a ⎝ x ⎠

r+a → πaρ ⎡ ⎛ −1⎞⎤ f = x r 2 a 2 2 ⎢ + ()− ⎜ ⎟⎥ r ⎣ ⎝ x ⎠⎦ r−a

4πa 2ρ M = = r 2 r 2 Case(ii) When pt. P is on the shell, the limit of integration are x = 0 to 2a

ρ πaρ 2a⎛ r 2 − a 2 ⎞ f = ⎜1+ ⎟dx 2 ∫⎜ 2 ⎟ r 0 ⎝ x ⎠ Here integration is not possible (due to second term is becoming indeterminant), because when P is on the shell, then r = a; x = 0 Hence to evaluate the integral, we consider that pt. P is situated not on the surface but very near to the surface Let r = a +δ, where δ is very small

ρ aπρ ⎡2a+δ 2a+δ ⎧()a + δ 2 − a 2 ⎫ ⎤ Then attraction is f = ⎢ dx + ⎨ ⎬dx⎥ r 2 ∫ ∫ x 2 ⎣⎢ δ δ ⎩ ⎭ ⎦⎥

πaρ ⎡ 2a+δ 2aδ ⎤ = 2a + dx 2 ⎢ ∫ 2 ⎥ r ⎣ δ x ⎦ 110

2a+δ πaρ ⎡ ⎛ −1⎞ ⎤ = 2 ⎢2a + 2aδ⎜ ⎟ ⎥ r ⎣⎢ ⎝ x ⎠δ ⎦⎥

πaρ ⎡ 2aδ 2aδ⎤ = 2a − + r 2 ⎣⎢ 2a + δ δ ⎦⎥

2πa 2δ ⎡ δ ⎤ = 2 − as δ→o, then r = a r 2 ⎣⎢ 2a + δ⎦⎥

4πa 2ρ M = = a 2 a 2 Case (iii) Poin. P is inside the shell, limits are x = a − r to a + r

ρ πaρ a+r ⎡ r 2 + a 2 ⎤ f = 1+ dx 2 ∫ ⎢ 2 ⎥ r a−r ⎣ x ⎦

a+r πaρ ⎡ ⎛ 1 ⎞⎤ = x r 2 a 2 = 0 2 ⎢ − ()− ⎜ ⎟⎥ r ⎣ ⎝ x ⎠⎦ a−r

So, there is no resultant attraction inside the shell. 6.7 Potential of a Uniform solid sphere:- A uniform solid sphere may be supposed to be made up of a number of thin uniform concentric spherical shells. The masses of spherical shells may be supposed to be concentric at centre O. Case I :- At an external point

P 0 r

Therefore the potential due to all such shells at an external point P is given by m m V = 1 + 2 +… r r where m1, m2 … etc are the masses of shells. 1 M V = (m + m + ...) = r 1 2 r 111

where M is the mass of solid sphere. Case II:- The point P is on the sphere. In case I, put r = a M V = , where a = radius of sphere a a Case III:- At an internal point. Here point P is x P considered to be external to solid sphere of 0 radius r & internal to the shell of internal radius r, external radius = a.

Let V1 = potential due to solid sphere of radius r

V2 = potential due to thick shell ofinternal radius r and external radius a mass of sphereof radius r Then V1 = r 3 4 πr ρ 4 2 = =πr ρ 3 r 3

To calculate V2 We consider a thin concentration shell of radius ‘x’ & thicknes dx. The potential at P due to thin spherical shell under consideration is given by 4πx 2dxρ = 4πx dx ρ x Hence for the thick shell of radius r & a, the potential is given by

a V2 = 4πρ ∫ xdx r ⎛ a 2 − r 2 ⎞ ⎜ ⎟ 2 2 = 4πρ⎜ ⎟ = 2πρ(a − r ) ⎝ 2 ⎠ Therefore, the potential at P due to given solid sphere.

2 2 2 V = V1 + V2 = πρ (3a − r ) 3 4 where M = Mass of given solid sphere = πa3ρ 3 112

3M ⇒ πρ = 4πa3 2 3M M Hence V = . (3a 2 − r 2 ) = (3a2 − r2) 3 4πa3 2a 3

6.8 Attraction for a uniform solid sphere Case I : At an external point ρ m m F = 1 + 2 + …. r 2 r 2 x r ρ M 0 P F = 2 , M = m1 + m2 +…. r a M = Mass of sphere and m1 , m2…. Are masses of concentric spherical shells Case II: At a point on the sphere, Here we put r = a in above result ρ M We get F = a 2 Case III: At a point inside the sphere. The point P is external to the solid sphere of radius r and it is internal to thick spherical shell of radii r and a. And we know that attraction (forces of attraction) at an internal point in case of spherical shell is zero. Hence the resultant attraction at P is only due to solid sphere of radius r and is given by ρ mass of sphere of radius r F = r 2 4 πr3ρ 4 = =πrρ 3 r 2 3 4 3M If M = πa3ρ ⇒ πρ = 3 4a3 ρ Mr Then F = a3

113

6.9 Self attracting systems:- To find the work done by the mutual attractive forces of the particles of a self-attracting system while the particles are brought from an infinite distance to the positions, they occupy in the given system. System consists of particles of masses m1, m…. at A1, A2….. etc. in the given system A.

We first being m1 from infinity to the position m1 A1. Then the work done in this process is zero. A& Since there is no particle in the system to 1 m2 exact attraction on it next m2 is brought from & A2 infinity to its position A2. Then the work done m3 & on it by m1 is = potential of m1 at A2 × m2 A3

m1 m1m2 = m2 = r12 r12 where r12 = distances between m1 & m2 (r12 = r21)

Then these two particles m1 and m2 attracts the third particle.

Work done on m3 by m1 & m2 is m m m m = 1 3 + 2 3 r13 r23

m1m4 m2m4 m3m4 when m4 is brought from infinity to its position A4 = + + r14 r24 r34 Hence the total work done in collecting all the particles from rest at infinity to their positions in the sysem A.

m1m2 ⎛ m1m3 m2m3 ⎞ = + ⎜ + +⎟ …….. r12 ⎝ r13 r23 ⎠ m m = ∑ s t , where summation extends to every pair of particles. rst

m2 m3 Let V1 = + +… r12 r13

= potential at A1 of m2, m3 ….. s one

V2 = potential at A2 of m1, m3, m4…. 114

m m = 1 + 3 +… r21 r23

m1 m2 m4 V3 = + + m13 r23 r43

msmt 1 ∑ = [V1 m1 + V2 m2 +…+ V3m3] rst 2 1 Total work done = Σmv 2 This represents the work done by mutual attraction of the system of particles. If the system from a cont. body, then work done will be 1 = v dm 2 ∫ Conversely (if particle is scattered) the work done by the mutual attraction forces of the system as its particles are scattered at infinite distance from confinguration A, −1 −1 then work done = mv = v dm 2 ∑ 2 ∫ We can find the work done as the body changes from one configuration A to another configuration B. The work done in changing of its from A to state at infinity + work done in collecting particles in a state at infinity to configuration B −1 1 = ∫∫V dm + V' dm' 2 AB2 A → ∞ → B Example. A self attracting sphere of uniform density ρ & radius ‘a’ changes to one of uniform density & radius ‘b’. Show that the work done by its mutual attractive forces is given by 3 ⎛ 1 1 ⎞ M2 ⎜ − ⎟ 5 ⎝ b a ⎠ where M is mass of sphere. Solution: Here the work done by mutual attractive forces of the system. As the particle which constitute the sphere of radius ‘a’ are scattered to infinite distance 115

−1 w1 = v dm 2 ∫ We consider a point within the system at a distance x. The potential at this point within the spheres 2 V = πρ(3a 2 − x 2 ) 3 Let us now consider at this point, a spherical shell of radius x & thickness dx, then dm = 4π x2 dx ρ 2 V dm = πρ (3a2 − x2) 4πx2ρ dx 3 8 = π2ρ2x2 (3a2 − x2) dx 3

8 a ∫∫v(3adm = π2ρ2 x 2 2 − x2) dx 3 0

3 5 a 8 2 2 ⎡ 2 x x ⎤ = π ρ ⎢3a − ⎥ 3 3 5 ⎣ ⎦0

5 5 8 2 2 ⎡ 5 a ⎤ 8 2 2 4a = π ρ ⎢a − ⎥ = π ρ 3 ⎣ 5 ⎦ 3 5 32 = π2ρ2a5 15 M = Mass of sphere of radius a 4 3m = πa 3 ρ ⇒ρ = 3 4πa3 6 m2 v dm = ∫ 5 a −1 − 3 m2 ⇒ W1 = v dm = 2 ∫ 5 a

Similarly if W2 is work done in bringing the particle ∞ to the second configuration (a sphere of radius b) 116

1 3 m2 Then W2 = v' dm'= 2 ∫ 5 b Total work done is given by

3 2 ⎛ 1 1 ⎞ W = W1 + W2 = m ⎜ − ⎟ 5 ⎝ b a ⎠ 6.10 Laplace’s equation for potential Let V be the potential of the system of attracting particles at a point P(x, y, z) not in contact with the particles so that m V = …(1) ∑ r where m is the mass of particle at P (a,b,c) 0 r = distance of P from the P0, P P0 and r2 = (x −a)2 + (y −b)2 + (z−c)2 …(2) ∂V m ∂r m (r − a) (1) ⇒ = − = − ∂x ∑ r2 ∂x ∑ r2 r2 ⎡ ∂r ∂r x − a ⎤ Θ (2) ⇒ 2r = 2(x − a) ⇒ = ⎣⎢ ∂x ∂x r ⎦⎥ ∂V m(y − b) and = − ∂y ∑ r3 ∂V m(z − c) = − ∂z ∑ r3 ∂ 2V ∂ ∴ = [ −Σm (x − a) r−3] ∂x 2 ∂x ∂r = −Σm(x − a) (−3 r−4) ∂x − Σm r−3 (1) (x − a)2 m = Σm3 − − 5 ∑ 3 r r ∂ 2V m(y − b)2 m and = − 2 ∑ 5 ∑ 3 ∂y r r 117

∂2V m(z − c)2 m = 3 − ss ∂z2 ∑ r5 ∑ r3 ∂ 2V ∂ 2V ∂ 2V ⇒ + + = 0 ∂x 2 ∂y2 ∂z2 which is Laplace equation. V → potential dv = small volume elemnt ∴ dm = ρ dv ρdv So V = ∫ r ∂V ⎛ −1⎞ ∂r = ⎜ ⎟ ρ dv ∂x ∫⎝ r 2 ⎠ ∂x 6.11 Poisson’s equation for potential Let the point P(x, y, z) be in contact (inside) the attracting matter. We describe a R sphere of small radius R & centre (a, b, c) (a,b,c) contains the point P. ρ = density of material (sphere) Since the sphere we describe is very small, therefore we cosndier the matter inside this sphere is of uniform density ρ. So potential at P may be due to (i) the matter inside the sphere (ii) the matter outside the sphere.

V1 = contribution towards potential at P by the matter outside the sphere

V2 = contribution towards potential at P by the matter inside the sphere. Since the point P is not in contact with the matter outside the sphere. Therefore by 2 Laplace equation ∇ V1 = 0.

Here V2 = potential at P(x, y, z) inside the sphere of radius R.

2 2 2 V2 = πρ (3R − r ) 3 118

where r2 = (x − a)2 + (y − b)2 + (z − c)2 ∂V 2 ⎛ ∂r ⎞ 2 r(x − a) 2 = πρ⎜− 2r ⎟ = πρ(−2) ∂x 3 ⎝ ∂x ⎠ 3 r − 4 = πρ(x − a) 3

∂ 2V − 4 ∴ 2 = π ρ ∂x 2 3

∂ 2V − 4 ∂ 2V − 4 Similarly 2 = πρ, = πρ ∂y2 3 ∂z2 3

∂ 2V ∂ 2V ∂ 2V ∴ 2 + 2 + 2 = −4πρ ∂x 2 ∂y2 ∂z2

2 ⇒ ∇ V2 = −4πρ

Since total potential V = V1 + V2 2 2 2 ∴ ∇ V = ∇ V1 + ∇ V2 ∂ 2V ∂ 2V ∂ 2V ⇒ + + = ∇2V = −4πρ ∂x 2 ∂y2 ∂z2 This equationis known as Poisson’s equation 6.12 Equipotential Surfaces The potential V of a given attracting system is a function of coordinates x,y,z. The equation V(x, y, z) = constant represents a surface over which the potential is constant. Such surfaces are known equipotential surfaces. Condition that a family of given surfaces is a possible family of equipotential surfaces in a free space. To find the condition that the equation f(x, y, z) = constant may represent the family of equipotential surface. If the potential V is constant whenever f(x, y, z) is constant, then there must be a functional relation between V and f(x, y, z) say 119

V = φ{f(x, y, z)} V = φ(f) ∂V ∂f = φ′(f) ∂x ∂x

2 ∂ 2V ⎛ ∂f ⎞ ∂ 2f = φ′′(f) ⎜ ⎟ + φ′(f) ∂x 2 ⎝ ∂x ⎠ ∂x 2

2 ∂ 2V ⎛ ∂f ⎞ ∂ 2f = φ′′(f) ⎜ ⎟ + φ′(f) 2 ⎜ ⎟ 2 ∂y ⎝ ∂y ⎠ ∂y

2 ∂ 2V ⎛ ∂f ⎞ ∂ 2f = φ′′(f) ⎜ ⎟ + φ′(f) ∂z2 ⎝ ∂z ⎠ ∂z2 Adding

2 2 2 ⎧ 2 2 2 ⎫ 2 ⎛ ∂ f ∂ f ∂ f ⎞ ⎪⎛ ∂f ⎞ ⎛ ∂f ⎞ ⎛ ∂f ⎞ ⎪ ∇ V = φ′(f) ⎜ + + ⎟ + φ′′(f) ⎨⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎬ ⎜ 2 2 2 ⎟ ∂x ⎜ ∂y ⎟ ⎜ ∂y ⎟ ⎝ ∂x ∂y ∂z ⎠ ⎩⎪⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎭⎪ But in free space, ∇2V = 0

∂ 2f ∂ 2f ∂ 2f + + ∂x 2 ∂y2 ∂z2 − φ''(f ) ⇒ = = a function of f 2 2 2 ⎛ ∂f ⎞ ⎛ ∂f ⎞ ⎛ ∂f ⎞ φ'(f ) ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎝ ∂x ⎠ ⎝ ∂y ⎠ ⎝ ∂z ⎠ = ψ(f) (say) …(1) This is the necessary condition and when it is satisfied, the potential V can be expressed in terms of f(x, y, z). φ''(f ) Then V = φ(f), where + ψ(f) = 0 φ'(f ) Integrating, log φ′(f) = log A − ∫ ψ (f)df ⎛ φ'(f ) ⎞ ⇒ log⎜ ⎟ = − ψ (f) df ⎝ A ⎠ ∫ ⇒ φ′(f) =Ae−∫ψ(f)df Again integrating, 120

V = φ(f) = A ∫e∫ ψ(f )df df + B …(2) which is required expression in terms of f(x, y, z) for V. Example :- Show that a family of right circular cones with a common axis & vertex is a possible family of equipotential surfaces & find the potential function. Solution : Taking axis of z for common axis. The equation of family of cones is x 2 + y2 f(x, y, z) = = constant …(3) z2 ∂f 2x ∂f 2 = , = ∂x z2 ∂x 2 z2 ∂f 2y ∂ 2f 2 = , = ∂y z2 ∂y2 z2 ∂f = (x2 + y2) (−2) (z−3) ∂z ∂ 2f = 6(x2+ y2) z−4 ∂z2 Therefore condition (1) becomes ∂ 2f ∂ 2f ∂ 2f + + − φ''(f ) ∂x 2 ∂y2 ∂z2 = 2 2 2 φ'(f ) ⎛ ∂f ⎞ ⎛ ∂f ⎞ ⎛ ∂f ⎞ ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎝ ∂x ⎠ ⎝ ∂y ⎠ ⎝ ∂z ⎠ 2 2 6(x 2 + y2 ) 2 + 2 + 4 = z z z 4x 2 4y2 4(x 2 + y2 )2 + + z4 z4 z6 2z2 + 2z2 + 6(x 2 + y2 ) z6 = z4 4z2 (x 2 + y2 ) + 4(x 2 y2 )2

4z2 + 6(x 2 + y2 ) = [4z2 (x 2 + y2 ) + (x 2 + y2 )2 ] 121

⎡ 3(x 2 + y2 )⎤ 2z⎢2 + ⎥ z2 = ⎣ ⎦ 2 ⎡x 2 + y2 ⎛ x 2 + y2 ⎞ ⎤ 4z4 ⎢ + ⎜ ⎟ ⎥ ⎢ z2 ⎜ z2 ⎟ ⎥ ⎣ ⎝ ⎠ ⎦ 2 + 3f 2 + 3f = = = function of f 2(f + f 2 ) 2f (f +1) − φ''(f ) 2 + 3f ⇒ = [function of f → 0, condition (1) is satisfied] φ'(f ) 2f (1+ f ) φ''(f ) 2 + 3f ⇒ + = 0 φ'(f ) 2f (1+ f ) φ''(f ) 1 1 ⇒ + + = 0 φ'(f ) f 2(1+ f ) Integrating, 1 log φ′(f) + log f + log (1 + f) = log C 2 C ⇒ log φ′(f) = log f 1+ f C ⇒ φ′(f) = f 1+ f dφ C ⇒ = df f 1+ f C ⇒ d φ = df + C′ ∫ ∫ f 1+ f Put f = tan2θ ⇒ df = 2 tanθ sec2θ dθ 2tanθsec2 θdθ ∴ φ = C ∫ + C′ tan 2 θ. 1+ tan 2 θ

2 sec2 θ = C . dθ + C' ∫ tanθ secθ 122

secθ = 2C dθ + C′ ∫ tanθ = 2C ∫ cosecθ dθ + C′ ∴ V = φ(f) = 2C log (cosec θ − cot θ) + C′ ⎛ θ ⎞ or V = φ(f) = 2C log ⎜ tan⎟ + C′ is the required potential function. So V is ⎝ 2 ⎠ constant when θ is constant.

6.13 Variation in attraction in crossing a surface on which there exist a thin layer of attracting matter.

Let P1 and P2 be two points on the nˆ opposite side of surface. σ → surface density of small circular disc of the P2 surface between P1 and P2. For P1 potential at surface, V1 = V2 ∂V ∂V 2 − 1 = −4πσ ∂n ∂n To find the attraction of matter, when the potential is given at all points of space, then Poisson’s equation ∇2V = −4πρ gives the volume density of matter −1 ∴ ρ = ∇2V 4π

If potential is given by different functions V1, V2 on opposite side of a surface S, then surface density σ is given by

−1 ⎡∂V2 ∂V1 ⎤ σ = ⎢ − ⎥ 4π ⎣ ∂n ∂n ⎦ Example. The potential outside a certain cylindrical boundary is zero, inside it is V = x3 − 3xy2 − 9x2 + 3ay2. Find the distribution of matter. 123

Solution : Since V2 = outside potential and V1 = Inside potential

Here V2 = 0 We find the boundary. Since the potential is continuous across the boundary & zero outside the boundary. The boundary may be given by x3 − 3xy2 − ax2 + 3ay2 = 0 or (x − a) (x2 − 3y2) = 0

⇒ (x − a) (x + 3 y) (x − 3 y) = 0 y AB is equation of line x = a OB is equation of line x + 3y = 0 nˆ A 60° OA is equation of line x − 3y = 0 30° P ∗P 30° The section is an equilateral ΔOAB of height ‘a’. a M x 0 30° ∗ ∂V 1 = 3x2 − 3y2 − 2ax ∂x B ∂V 1 = − 6xy + 6ay ∂y

∂ 2V 1 = 6x − 2a ∂x 2

∂ 2V 1 = −6x + 6a ∂y2

∂ 2V 1 = 0 ∂z2 so that inside the region,

−1 2 ρ = ∇ V1 4π −1 − a = [4a] ⇒ ρ = 4π π and outside, ρ = 0 since V2 = 0 At P on AB (x = a), 124

−1 ⎡∂V2 ∂V1 ⎤ σ = ⎢ − ⎥ 4π ⎣ ∂x ∂x ⎦ x=a −1 = [ 0 − 3x2 + 3y2 + 2ax] 4π

2 −1 2 2 3 ⎡a 2 ⎤ = [3y − a ] = ⎢ − y ⎥ …(1) 4π 4π ⎣ 3 ⎦ In ΔOAM, OA2 = OM2 + AM2 1 ⇒ OA2 = a2 + (OA)2 4 3 4 ⇒ (OA)2 = a2 ⇒ (OA)2 = a 2 4 3 4 1 ⇒ (2 MA)2 = a2 ⇒ (MA)2 = a2 …(2) 3 3 ∴ from (1) & (2), we have 3 σ = [MA2 − MP2] 4π 3 = (MA + MP) (MA − MP) 4π 3 = (PB) (AP) 4π At P on OA (x = 3y)

1 ⎡∂V1 ⎤ σ = ⎢ ⎥ 4π ⎣ ∂n ⎦

1 ⎡ ∂V1 ∂V1 ⎤ = ⎢− sin 30° + cos 30° ⎥ 4π ⎣ ∂x ∂y ⎦ x= 3 y

1 ⎡−1 ∂V 3 ∂V ⎤ = ⎢ 1 + 1 ⎥ 4π 2 ∂x 2 ∂y ⎣ ⎦ x= 3 y

1 ⎡− 3 2 3 2 ⎤ = ⎢ x y + ax − 3 3xy + 3 3ay⎥ 4π ⎣ 2 2 ⎦ x= 3y 125

2 1 ⎡− 3 2 3 x 2 x ⎤ = ⎢ x + + ax − 3x + 3 3 a. ⎥ 4π ⎣ 2 2 3 3 ⎦ 1 ⇒ σ = x(a − x). π 6.14 Harmonic functions Any solution of Laplace equation ∇2 V = 0 in x, y, z is called Harmonic function or spherical harmonic. ∂2V ∂ 2V ∂ 2V i.e. + + = 0 ∂x 2 ∂y2 ∂z2 If V is a Harmonic function fo degree n, then ∂ p ∂q ∂ t is a harmonic function of degree n − p − q − t. ∂x p ∂yq ∂z t

Now ∇2V = 0 [Laplace equation] p times w.r.t. x q times w.r.t. y t times w.r.t. z

q q t 2 ⎡ ∂ ∂ ∂ ⎤ So ∇ ⎢ p q t ⎥ = 0 ⎣∂x ∂y ∂z ⎦ 6.15 Surface and solid Harmonics :- In spherical polar coordinates (r, θ, φ) Laplace equation is ∂ ⎛ ∂V ⎞ 1 ∂ ⎛ ∂V ⎞ 1 ∂ 2V ⎜r 2 ⎟ + ⎜sinθ ⎟ + = 0 …(1) ∂r ⎝ ∂r ⎠ sinθ ∂θ ⎝ ∂θ ⎠ sin 2 θ ∂φ2

n Let V = r Sn where Sn is independent of r or Sn(θ, φ).

∂ ⎡ 2 ∂V⎤ ∂ ⎡ 2 ∂ n ⎤ r = r (r Sn ) ∂r ⎣⎢ ∂r ⎦⎥ ∂r ⎣⎢ ∂r ⎦⎥

∂ 2 n−1 = [r Sn n r ] ∂r

∂ n+1 ∂ n+1 = [Sn n r ] = nSn (r ) ∂r ∂r 126

n = n (n + 1) r Sn

2 n 1 ∂ ⎛ n ∂Sn ⎞ 1 n ∂ Sn (1) ⇒ n (n + 1) r Sn + ⎜sinθ.r ⎟ + r = 0 sinθ ∂θ ⎝ ∂θ ⎠ sin 2 θ ∂θ2

n n 2 n r ∂ ⎛ ∂Sn ⎞ r ∂ Sn ⇒ n(n + 1) r Sn + ⎜sinθ ⎟ + = 0 sinθ ∂θ ⎝ ∂θ ⎠ sin 2 θ ∂φ2

2 1 ∂ ⎛ ∂Sn ⎞ 1 ∂ Sn ⇒ n(n +1) Sn + ⎜sinθ ⎟ + = 0 sinθ ∂θ ⎝ ∂θ ⎠ sin 2 θ ∂φ2

2 2 ∂Sn ∂ Sn 1 ∂ Sn ⇒ n(n +1) Sn + cot θ + + = 0 …(2) ∂θ ∂θ2 sin 2 θ ∂φ2

If cos θ = μ

2 ∂ ⎡ 2 ∂Sn ⎤ 1 ∂ Sn N(n +1) Sn + ⎢(1− μ ) ⎥ + 2 2 = 0 …(3) ∂μ ⎣ ∂μ ⎦ 1− μ ∂φ

A solution Sn of equation (2) is called a Laplace function or a surface harmonic of order n. Since n(n+1) remains unchanged when we write −(n +1) for n. So there are n −n−1 two solutios of (1) of which Sn is a factor namely r Sn and r Sn. These are known as solid Harmonic of degree n & −(n +1) respectively. Remark: U 1. If U is a Harmonic function of degree n, then is also Harmonic r 2n+1 function. n U = r Sn

n U r Sn Sn −(n+1) so that = = = Sn r r 2n+1 r 2n+1 r n+1 which is Harmonic. xyz Let xyz → 3rd degree is a solutionof Laplace equation, then is also Harmonic. r7 2. If U is a Harmonic function of degree −(n +1), then Ur2n+1 is also a Harmonic function, are may write −n−1 U = r Sn 127

2n+1 2n+1 −n−1 n so that r U = r r Sn = r Sn which is Harmonic. 6.16 Surface density in terms of surface Harmonics The potential at any point P due to a number of particles situated on the surface of sphere of radius ‘a’ can be ut in the form ∞ r n V = U , when r < a …(1) 1 ∑ n+1 n n=0 a a n V = U when r > a …(2) 2 ∑ n+1 n r where Un denotes the sum of a number of surface Harmonics (for each particle) & therefore itself a surface harmonic. We assume (1) & (2) to represent potential of a certain distribution of mass & want to find it (density) on the surface.

Here U1 is Harmonic 2 2 ⇒ ∇ V1 = 0, ∇ V2 = 0 Here on the surface of sphere, density is given by

⎡∂V2 ∂V1 ⎤ −4πσ = ⎢ − ⎥ ⎣ ∂r ∂r ⎦ r=a

1 ⎡∂V1 ∂V2 ⎤ ⇒ σ = ⎢ − ⎥ 4π ⎣ ∂r ∂r ⎦ r=a

1 ⎡ U n r n−1 U a n (n +1)⎤ = ⎢ n + n ⎥ 4π ∑∑n+1 n+2 ⎣ a r ⎦ r=a

1 ⎡ U n a n−1 U a n (n +1)⎤ = n n ⎢∑∑n+1 + n+2 ⎥ 4π ⎣ a a ⎦

1 ⎡ n (n +1)⎤ = U U ∑∑n 2 + n 2 4π ⎣⎢ a a ⎦⎥ (2n +1)U ⇒ σ = n …(3) ∑ 2 4πa If potential is given by (1) & (2), then surface density is given by (3).