Classical Virtual & d’Alembert’s Principle

Dipan Kumar Ghosh UM-DAE Centre for Excellence in Basic Sciences Kalina, Mumbai 400098 August 15, 2016

1 Constraints

Motion of a system of particles is often constrained, either geometrically or kinematically. Constraints reduce the number of degrees of freedom of a given body. Consider the of a single particle in . For free, unconstrained motion, it has three degrees of freedom which are usually expressed by three coordinates such as x, y, z or r, θ, ϕ etc. If, however, the particle is constrained to move on the surface of a sphere, we must have (taking Cartesian coordinates),

x2 + y2 + z2 = R2 which reduces the number of degrees of freedom by one. Consider two connected by a rigid rod, like a dumbbell. Two particles have 6 degrees of freedom, Since the distance between the two bodies remains constant, we have the constraint

2 2 2 2 (x1 − x2) + (y1 − y2) + (z1 − z2) = d which reduces the degree of freedom to 5. These are examples of geometric or holonomic constraints whoch are expressible as algebraic equations involving the coordinates. There are other constraints which restrict the motion of bodies . Some of these are expressible as differential equations which constrain the coordinates and components of . These are called kinematic constraints. Non-integrable kinematic constraints which cannot be reduced to holonomic constraints are called non-holonomic constraints. Thus, if m is the dimension of the configuration space (i.e., the number of generalised coordinates), holonomic constraints are expressible as equations of the form

i f (t, q1, q2, . . . , qm) = 0, 1 ≤ i ≤ k

1 c D. K. Ghosh, IIT Bombay 2 where k is the number of constraints. Holonomic constraints are called scleronomic if they do not explicitly depend on . Time dependent constraints are called rheonomic. Kinematic constraints are expressed as equations in the phase space

i f (t, q1, q2, . . . , qm;q ˙1, q˙2,..., q˙m) = 0, 1 ≤ i ≤ k

Both the constraints are classified as rheonomic if they explicitly depend on time. Sometimes a constraint may appear to be kinematic but may be in reality holonomic. For instance, a constraint of the type Ax˙ + B = 0 may actually be holonomic if there exists a function f such that A = ∂f/∂x and B = ∂f/∂t. We then have,

∂f df ∂f df = + ∂x dt ∂t which gives a holonomic constraint f = constant.

Example 1:

Consider two masses connected by pulleys, as shown. In general two particles have

6 degrees of freedom. However, m1 can only move along the x direction and m2 along the z direction. Thus y1 = z1 = 0 and x2 = y2 = 0. We are left with two degrees of freedom. However, if m1 moves along the x direction by a distance d, m2 would have to move along z direction by 2d, i.e., we get another holonomic constraint, z2 − 2x1 = 0 which reduces the degree of freedom further by one. The problem is essentially a one dimensional problem. m 1 P 1 z

x

P 2

m 2

   Non-holonomic constraint could be in the form of differential equations or algebraic inequalities. For instance, a particle constrained to move inside a sphere of radius R satisfies x2 + y2 + z2 < R2. Consider a disk rolling on a horizontal surface, on x-y plane. c D. K. Ghosh, IIT Bombay 3

z

y

P φ

θ x If the disk is rolling without slipping, keeping its plane vertical, we need four coordi- nates to describe the position of the disk. These are the x and y coordinates of the centre of the disk, the angle ϕ by which a fixed point on the rim of the disk has rotated about the axis of and the angle θ that the axis of the disk makes with the x axis. If R is the radius of the disk, the of the disk is given by

v = Rω = Rϕ˙ (1)

Since the disk remains vertical, the components of the velocity of the centre of the disk are given by dx = v sin θ (2) dt dy = −v cos θ (3) dt Using eqs. (1) to (3) we get dx dϕ = R sin θ dt dt dy dϕ = −R cos θ dt dt the minus sign is due to the sense of rotation being opposite to the positive angle. Com- bining these two we get the following pair of differential equations:

dx = R sin θdϕ dy = −R cos θdϕ

These equations cannot be further reduced and we cannot connect x, y, θ and ϕ by an algebraic equality, showing that the constraint is non-holonomic. c D. K. Ghosh, IIT Bombay 4

Tackling problems with non-holonomic constraints is more difficult and no general pre- scription can be provided for their solution. Constraints introduce two new elements into problem solving. Since the generalised coordinates are no longer independent, the are not independent either. Constraints arise from between elements whose nature is unknown. These forces are known only by the effect they have on motion of the system.

2 Virtual

A real displacement of particles constituting a system happens over a finite time. Such a displacement of, say, the i-th particle is, in general, function of all the generalised coordinates as well as of time. If the position of the i-th particle is written as

~ri = ~ri(q1, q2, . . . , qs)

The total differential of the position vector is then written as

s X ∂~ri ∂~ri d~r = dq + dt (4) i ∂q j ∂t j=1 j

A virtual or an imagined displacement is instantaneous and is consistent with the con- straints on the system. The displacements, whether real or virtual, result in admissible geometrical configuration of the system. In case of a virtual displacement, we have,

s X ∂~ri d~r = δq (5) i ∂q j j=1 j where we have use δq to indicate a virtual displacement while reserving dq for real dis- placement. For a real displacement, the forces of constraints may change. The velocity is given by s d~ri X ∂~ri ∂~ri ~r˙ = = q˙ + (6) i dt ∂q j ∂t j=1 j so that we have, ~ s ∂r˙i X ∂~ri ∂~ri = δ = (7) ∂q˙ ∂q jk ∂q k j=1 j k (Note the structure of the above equation - as if the dots cancel!)

2.1 Principle of Virtual Work

Consider a system of N particles under time dependent holonomic constraints. If q1, . . . , qs be a set of , the virtual displacement of the i-th particle is given c D. K. Ghosh, IIT Bombay 5 by eqn. (5). Suppose the system, under the action of applied forces as well as those of constraints, is in equilibrium. The total acting on each particle is then zero,

~ Fi = 0; (i = 1,...,N) ~ We then have, for the virtual work done by Fi in the displacement δ~ri is ~ Fi · δ~ri = 0 so that the total work done is

X ~ δW = Fi · δ~ri = 0 i The total force acting on any particle can be split into two: an applied part and a part due to the constraints, ~ ~ a ~ c Fi = Fi + Fi then we have

X ~ a X ~ c δW = Fi · δ~ri + Fi · δ~ri = 0 (8) i i The forces of constraints (e.g. normal reaction, tension, constraints etc.) do not do any work. This is general true of scleronomic holonomic constraints and this statement is central hypothesis in the principle of virtual work. Two examples illustrate the hypothesis. Consider two types of displacements consistent with the constraints on a rigid body. A displacement along the line joining two particles does not do any net work because the reactions are equal and opposite. In order to be consistent with rigid body constraints, for a pair of particles j and k, we must have δrk = δrj. The work done is fkj · δrj + fjk · δrk = (fkj + fjk) · δrj = 0 where we have used δrj = δrk. Thus there is no work done. The second type of displacement is along the arc of a circle normal to the line joining the particles.As the forces of constraints are normal to the direction of displacement, the work done is once again zero.

Consider the pulley arrangement in Example 1. When m1 moves by an amount δx to the right, m2 moves by 2δx downwards in order to keep the sum of the lengths of the two ropes constant. The only applied forces are the and . Thus by the principle of virtual work, we have,

−µmgδx + mg2δx = 0 which shows that for static equilibrium m2 = µm1/2. Example 2: Two frictionless blocks of m each are connected by a massless rigid rod. The system is constrained to move in the vertical plane. c D. K. Ghosh, IIT Bombay 6

mg

R

δx 1 mg

x θ δ 2 F    N

If the block on the vertical track undergoes a virtual displacement δx1 and that on the horizontal plane has a virtual displacement δx2, we have δx1 sin θ = δx2 cos θ which is the constraint which keeps the rod length constant. The gravity does work on the mass on the vertical track while the applied force F2 is responsible for work on the horizontally moving block, mgδx1 + F2δx2 = 0. Thus we have

δx1 F2 = −mg = −mg cot θ δx2 Example 3: Consider an Atwood’s in equilibrium. In this case we have the constraint y1 + y2 = l = constant.    

m m 1 2

Thus we have δy1 = −δy2. The applied forces are m1g and m2g, both acting down- wards. The virtual work done is δW = m1gδy1 + m2gδy2 = (m1 − m2)gδy1 = 0, which gives the condition for equilibrium to be m1 = m2.

2.2 Generalized Force We recall that in the presence of holonomic constraints, all the components of generalised coordinates can be made independent. If the system has s degrees of freedom (s ≤ 3N), the position vector of the i-th particle (i = 1,N) can be written as c D. K. Ghosh, IIT Bombay 7

~ri = ~ri(q1, q2, . . . , qs) The velocity of the i-th particle is given by

s d~ri X ∂~ri ∂~ri ~r˙ = = q˙ + (9) i dt ∂q j ∂t j=1 j

{q˙i} are known as the generalised velocities. (If the constraints are scleronomic, we ∂~r would have i = 0. ) ∂t Let us return to the expression for virtual work and express it in terms of the gener- alised coordinates.

X ~ a δW = Fi · δ~ri i s X X ∂~ri = F~ a · δq i ∂q α i α=1 α s X ≡ Qαδqα (10) α=1 where we have defineed a generalised force corresponding to the coordinate qα by the relation X ∂~ri Q = F~ a · (11) α i ∂q i α Note that the generalised force depends only on applied forces and not on forces of con- straints. In equilibrium, we have X δW = Qαδqα = 0 α

Since the generalised coordinates are independent, we have Qα = 0. However, this does P ~ a not apply that applied forces vanish. The condition for vanishing of i Fi · δ~ri = 0 is applicable only to static situations. d’Alembert extended this to include general motion of the system.

3 d’Alembert’s Principle d’Alembert’s principle, developed from an idea originally due to Bernoulli, is to use the fact that according to Newton’s law, the force applied on a particle results in a rate of change of its , ~ Fi = p~˙i c D. K. Ghosh, IIT Bombay 8

p~˙i is known as the inertial force or pseudo force acting on the particle. One can then think of bringing the body to equilibrium by applying a pseudo force −p~˙i on the i-th particle of the system (1 ≤ i ≤ N),

~ Fi − p~˙i = 0 ~ Note that Fi contains both applied and constraint forces acting on the i-th particle. ~ Once again, we can split Fi into two parts and write the above equation for virtual displacements as X ~ a ~ c ~ (Fi + Fi − p˙i) · δ~ri = 0 i Since the force of constraints do not do any work, we get

X ~ a ~ (Fi − p˙i) · δ~ri = 0 (12) i

Equation (12) is the statement of d’Alembert’s principle. It may be noted that the only ~ a force appearing in this equation is the applied force. Further, Fi refers to the force on the i-th particle and the sum in (12) is over all particles and not over the independent ~ a ~ generalised coordinates. Consequently, (12) does not imply Fi − p˙i = 0. An interesting consequence follows if the displacement in (12) happens to be real displace- ˙ ment instead of virtual ones. In that case the displacement can be written as d~ri = ~ridt If ~ the force is conservative and is derivable from a potential V , i.e. Fi = −∇iV , can rewrite (12) as follows:

X ~ a ~ X ~ ~ (Fi − p˙i)d~ri = [−∇iV − mr¨i]r˙idt i i X d 1 = [−∇ V · d~r − ( mr˙2)dt] i i dt 2 i i X = d( Ti + V ) = 0 i

3.1 Example 4: Consider a mass resting on a frictionless incline. The mass would slide down with an when released. A horizontal acceleration is applied on the mass to keep the mass from . We need to find the acceleration. The problem is c D. K. Ghosh, IIT Bombay 9

reasonably simple and is readily solved by Newton’s laws. N 90− θ a N = mg cos θ + ma sin θ mg ma cos θ = mg sin θ

which gives a = g tan θ θ

Let us look at the problem from d’Alembert’s principle. Suppose the mass has an instantaneous (virtual) displacement δl along the incline. We then have δx = δl cos θ and δy = −δl sin θ. The only applied force is mg along the −y axis: F~ = −mgyˆ. From the principle of virtual work it follows that

Fxδx + Fyδy − maxδx − mayδy = 0

We only apply a horizontal acceleration so that we have, ay = 0 . Since Fx = 0, we get mgδy − maxδx = 0 i.e., mgδl sin θ − maδl cos θ = 0. Thus we get, a = g tan θ. Example 5: Consider the arrangement shown in the figure. The pulley is fixed on the fixed wedge. Find the acceleration of the masses when released.

l l 1 2 m m 1 2 m g 2 m g β α 1 From d’Alembert’s principle we have ~ ~ ~ ~ (F1 − p~˙1) · δl1 + (F2 − p~˙2) · δl2 = 0 (13)

Since l1 + l2 = constant, we have

δl1 = −δl2 (14a) ¨ ¨ l1 = −l2 (14b) ¨ ¨ ¨ The inertial forces arep ˙1 = m1l1 andp ˙2 = m2l2 = −m2l1 and the only applied forces are the weight of the masses. Taking the components of (13) along the incline, we have, ¨ ¨ (m1g sin α − m1l1)δl1 + (m2g sin β − m2l2)δl2 = 0 Using (14a) and (14b) we get, ¨ ¨ (m1g sin α − m1l1 − m2g sin β − m2l1)δl1 = 0 c D. K. Ghosh, IIT Bombay 10 so that ¨ m1g sin α − −m2g sin β l1 = m1 + m2 4 Euler- Lagrange Equation

We will now derive the Euler-Lagrange equation from d’Alembert’s principle. Recall eqn. (7) where we showed ~ s ∂r˙i X ∂~ri ∂~ri = δ = (15) ∂q˙ ∂q jk ∂q k j=1 j k we also had, from (6), for scleronomic constraints,

s X ∂~ri ~r˙ = q˙ (16) i ∂q j j=1 j

Using chain rule differentiation, we can write,

  s 2 d ∂~ri X ∂ ~ri dqk = dt ∂qj ∂qj∂qk dt k=1 s ! ∂ X ∂~ri = q˙k ∂qj ∂qk k=1 ∂ = ~r˙i (17) ∂qj where in the last line we have used (16) and have used the fact thatq ˙i is independent of qi. Let us return to d’alembert’s principle

N X ~ (Fi − p~˙i) · δ~ri = 0 i=1 The inertial force term may be simplified as follows:

s ! X X X ∂~ri mi~r¨i · δ~ri = mi~r¨i · δqk ∂qk i k=1 i s " !  # X d X ∂~ri X d ∂~ri = mi~r˙i · − mi~r˙i δqk dt ∂qk dt ∂qk k=1 i s " ~ ! ~ # X d X ∂r˙i X ∂r˙i = mi~r˙i · − mi~r˙i · δqk (18) dt ∂q˙k ∂qk k=1 i where in the last line we have used the dot cancelation relationship (17). c D. K. Ghosh, IIT Bombay 11

The right hand side of the above expression can be simplified and the equation can be written as follows: s " ! !# X ~ X d ∂ X 1 2 ∂ X 1 2 mir¨i · δ~ri = mir˙i − mir˙i δqk dt ∂q˙k 2 ∂qk 2 i k=1 i i s X  d  ∂T  ∂T  = − δqk (19) dt ∂q˙k ∂qk k=1 where T is the kinetic of the system of particles. Substituting (19) in d’Alembert’s equation, we have

N X  d  ∂T  ∂T  F~ a · δ~r − − δq = 0 (20) i i dt ∂q˙ ∂q k i=1 k k Ps In terms of generalized coordinates, we could write the first term as k=1 Qkδqk. Thus we have, s X   d  ∂T  ∂T  Qk − − δqk = 0 (21) dt ∂q˙k ∂qk k=1 Since the generalized coordinates are independent, we may vary each coordinate indepen- dently and get d  ∂T  ∂T − = Qk (22) dt ∂q˙k ∂qk Equation (22) is the form of Euler-Lagrange equation which is derived from d’Alembert’s principle. If, however, the external forces acting on the system are conservative, i.e., if we ~ can express Fi = −∇iV , where ∇i implies gradient taken with respect to the coordinates of the i-th particle, we have N X ∂~ri Q = F~ · k i ∂q i=1 k N X ∂~ri = −∇ V · i ∂q i=1 k ∂V = − (23) ∂qk ∂V If the potential is a function only of the position qk, we have = 0 which enables us ∂q˙k to write the generalized force as ∂V d  ∂V  Qk = − + (24) ∂qk dt ∂q˙k On bring this expression for the generalized force to the left hand side of (23) and recog- nizing that L = T − V , we recover the Euler-Lagrange equation

d  ∂L  ∂L − = 0 (25) dt ∂q˙k ∂qk c D. K. Ghosh, IIT Bombay 12

5 Velocity Dependent Potential - Lorentz Force

The expression for the generalized force in the expression (24) can be used to define a potential function for velocity dependent forces as well. Only in such a situation, the second term on the right hand side of (24) is not identically zero but would depend on the exact dependence of the force on velocity. Thus in such situations, we could derive Euler-Lagrange equation form an appropriately defined Lagrangian function. It is an important generalisation for us because Lorentz force which a charge particle experiences in an electromagnetic field is a velocity dependent force. The force on an electric charge q (not to be confused with generalized coordinate which is also usually written by the same notation), is given by

F~ = q(E~ + ~v × B~ ) (26)

Using the expressions for the electric and the magnetic fields in terms of the scalar and the vector potentials ϕ and A~ respectively, we have

∂A~ E~ = −∇ϕ − ∂t B~ = ∇ × A~ we can write the expression for the force as " # ∂A~ F~ = q −∇ϕ − + ~v × (∇ × A~) (27) ∂t

The scalar potential depends only on position and we will need to deal only with the remaining two terms. Since ~v does not depend on position, we have the following identity,

~v × (∇ × A~) = ∇(~v · A~) − (~v · ∇)A~ (28)

Proof of (28)

Proof. To prove the identity (28), we consider the x-component of its left hand side.

~ ~ ~ [~v × (∇ × A)]x = vy(∇ × A)z − vz(∇ × A)y  ∂ ∂   ∂ ∂  = v A − A − v A − A y ∂x y ∂y x z ∂z x ∂x z  ∂A ∂A   ∂A ∂A  = v y + v z − v x + v x y ∂x z ∂x y ∂y z ∂z  ∂A ∂A ∂A   ∂A ∂A ∂A  = v x + v y + v z − v x + v x + v x x ∂x y ∂x z ∂x x ∂x y ∂y z ∂z

∂A where in the last line we have added and subtracted v x . x ∂x c D. K. Ghosh, IIT Bombay 13

Since ~v or its components do not depend on position coordinates, we can write the ∂ first term as (~v · A~) and the second term as (~v · ∇)A . Adding three components of ∂x x ~v × (∇ × A~) then yields the identity (28).

We will now rewrite the second term of (28) using a smart trick. Note that the total derivative of any component of A~ can be written as follows: dA ∂A dx ∂A dy ∂A dz ∂A x = x + x + x + x dt ∂x dt ∂y dt ∂x dt ∂t ∂A ∂A ∂A ∂A = v x + v x + v x + x x ∂x y ∂y z ∂x ∂t ∂A = (~v · ∇)A + x (29) x ∂t Combining three components, we can write, dA~ ∂A~ = (~v · ∇)A~ + (30) dt ∂t ∂A~ Using (28) and substituting the expression for from (30) into (27), we get ∂t " # dA~ F~ = q −∇(ϕ − ~v · A~) − (31) dt

Now, we can write using the fact that A~ does not depend on velocity components   dAx d ∂ = (Axvx) dt dt ∂vx d  ∂  = (Axvx + Ayvy + Azvz) dt ∂vx d  ∂  = (~v · A~) dt ∂vx d  ∂  = (~v · A~ − ϕ) (32) dt ∂vx where, in the last step, we have added a term ∂ϕ/∂vx which is zero because the scalar potential does not depend on the velocity either. Thus dA~ d   = ∇ (~v · A~ − ϕ) (33) dt dt v ∂ ∂ ˆ ∂ where ∇v = ˆi + ˆj + k is gradient with respect to the velocity vector. Substi- ∂vx ∂vy ∂vz tuting this in the expression (31), we get  d   F~ = q −∇(ϕ − ~v · A~) + ∇ (ϕ − ~v · A~) (34) dt v c D. K. Ghosh, IIT Bombay 14

Thus if we define a velocity dependent potential as

U = ϕ − ~v · A~ (35) the component of Lorentz force would be given by

∂U d ∂U  Fj = − + (36) ∂qj dt ∂q˙j which is of the form (24). With this modification, the Euler-Lagrange equation is still valid for velocity dependent potential for which the Lagrangian is given by

L = T − U

.