Contents
Part 1
Analytical Mechanics
1
Chapter 1: Generalized coordinates
3
Chapter 2: Lagrange’s equations
9
Chapter 3: Hamiltonian dynamics
31
Part 2
47
Relativity
Chapter 4: The four-dimensional space-time
49
Chapter 5: Consequences of the Lorentz transformations
55
Chapter 6: Four-vector formalism and covariant equations
63
Chapter 7: Relativistic kinematics
67
Chapter 8: Relativistic dynamics
77
Part 3
89
Electrodynamics
Chapter 9: Maxwell’s equations
91
Chapter 10: Electromagnetic field dynamics
99
Chapter 11: Maxwell’s equations with stationary sources
105
Chapter 12: Electromagnetic radiation
113
Part 4
Classical Field Theory
129
Chapter 13: Lagrangian and Hamiltonian formulations
131
Chapter 14: Symmetry transformations
139
Chapter 15: Relativistic fields
145
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PART 1
Analytical Mechanics
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Chapter 1
Generalized coordinates Problem 1.1 Four mechanical systems are studied. In all cases the number of degrees of freedom are specified, and an appropriate set of generalized coordinates is chosen. a) The first system consists of a pendulum attached to a block which in turn is attached to a spring. We assume all motion takes place in a twodimensional, vertical plane. The block is constrained to move in the horizontal direction, and the pendulum is constrained by the constant length of the rod. Starting from two degrees of freedom for each of the two objects, the two constraints reduce the number of degrees of freedom to two, one for each object. A natural choice of generalized coordinates is the horizontal displacement x of the block and the angle θ of the rod relative to the vertical direction. b) The second system consists of a pendulum attached to a vertical disk, which rotates with a fixed angular frequency. Also here we consider the motion restricted to a two-dimensional, vertical plane. There is no degree of freedom related to the rotating disk, since it has an externally determined angular frequency. The pendulum is again only constrained by the fixed length of the rod, and the number of degrees of freedom of the system is therefore one. A natural choice of generalized coordinate is the angle θ between the pendulum rod and the vertical direction. c) In the third case a rigid rod can tilt without sliding on the top of the cylinder, while the cylinder can roll on a horizontal plane. Assuming again that the motion is restricted to a two-dimensional, vertical plane, the starting point is three degrees of freedom for each object. For the cylinder this corresponds to two coordinates for its center of mass and one for its angle of rotation. For the rod there are two coordinates needed to determine the position of its center of mass, and one coordinate to determine the angle of the rod relative to the horizontal (or vertical) direction. 3
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Solution Manual for Classical Mechanics and Electrodynamics (2nd Edition)
The constraints are the following, 1) For the cylinder the vertical coordinate of the cylinder’s center is fixed, and since the cylinder is rolling, rather than sliding, the rotation coordinate is linked to the horizontal coordinate of the cylinder. This gives two constraints for the cylinder. 2) The rod is constrained to lie on the top of the cylinder, and if we assume that it is not allowed to slide on the cylinder, the only degree of freedom for the rod is to tilt on the cylinder. This means that there are two constraints also for the rod. The number of degrees of freedom is therefore: 3 + 3 − 2 − 2 = 2. A possible choice of generalized coordinates is the horizontal coordinate of the cylinder and the tilting angle of the rod. d) In the last case a rotating top moves on a horizontal floor. For a rigid body in three dimensions the number of degrees of freedom is six, three to determine the position of its center of mass and three to determine its orientation (corresponding to the three parameters, which are needed to specify a rotation in 3D). For the rotating top there is one constraint, with the vertical coordinate of its tip being fixed by the vertical coordinate of the floor. Choice of generalized coordinates: The (x, y)-coordinates of the tip of the rotating top, and the angles (θ, φ, χ) which determine its orientation (two to determine the direction of its symmetry axis, and one to determine its rotation angle around the axis). Problem 1.2 We examine an Atwood’s machine, which consists of three weights, with masses m1 = 4m, m2 = 2m and m3 = m. The ropes, with fixed lengths l1 and l2 , and the two pulleys, are treated as massless. The number of degrees of freedom is two, since the heights of two of the weights, m1 and m2 , will determine the third one, m3 . We choose the vertical coordinates y1 and y2 , shown in Fig. 1.1, as generalized coordinates. Expressed in these variables the potential energy is V = −m1 y1 + m2 (−y2 − (l1 − y1 )) + m3 ((y1 − l1 ) + (y2 − l2 )) = −m1 y1 + m2 (y1 − y2 ) + m3 (y1 + y2 ) + const. = −4my1 + 2m(y1 − y2 ) + m(y1 + y2 ) + const. = −m(y1 + y2 ) + const. ,
(1.1)
where the constant can be absorbed in the definition of the zero point of the potential energy.
5
Generalized coordinates
y1
l1-y1
m1 y2 m2
Fig. 1.1
l2-y2
m3
Atwood’s machine with two independent coordinates y1 and y2 .
The corresponding expression for the kinetic energy is 1 1 1 T = m1 ẏ12 + m2 (ẏ1 − ẏ2 )2 + m3 (ẏ1 + ẏ2 )2 2 2 2 1 = m(4ẏ12 + 2(ẏ1 − ẏ2 )2 + (ẏ1 + ẏ2 )2 ) 2 1 = m(7ẏ12 + 3ẏ22 − 2ẏ1 ẏ2 ) . 2
(1.2)
Problem 1.3 Three identical rods of mass m and length l are connected by frictionless joints, with the distance between the points of suspension being equal to the length of the rods. The rods move in the vertical plane. We will show that the system has only one degree of freedom, where the angle θ of one of the rods can be used as generalized coordinate. The Lagrangian will be found, expressed as a function of θ and θ̇. Two coordinates are needed to determine the position of each of the joints in the vertical plane. There are three constraints, which give relations between these positions, corresponding to the fixed lengths of the three rods. The number of degrees of freedom of the system is therefore 2+2−3 = 1, and the angle θ in the figure is an obvious choice for the generalized coordinate. The four endpoints of the rods define a parallelogram, as follows from the fact that the lengths of all four sides are the same. As a consequence
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Solution Manual for Classical Mechanics and Electrodynamics (2nd Edition)
two of the rods will rotate about the end points, which are fixed to the roof, while the third rod will move without rotation, since it is always parallel to the roof. The common angular velocity of the rotating rods is ω = θ̇, and the velocity of the non-rotating rod is v = lθ̇, since this is equal to the velocity of the lower endpoints of the two rotating rods. The kinetic energy of the three rods is therefore 1 1 1 5 1 (1.3) T = 2 · Iω 2 + mv 2 = ml2 θ̇2 + ml2 θ̇2 = ml2 θ̇2 . 2 2 3 2 6 With y as the vertical distance of the horizontal rod, the potential energy of the system is y (1.4) V = 2 · mg + mgy = 2mgy = −2mgl cos θ . 2 This gives for the Lagrangian L(θ, θ̇) =
5 2 2 ml θ̇ + 2mgl cos θ . 6
(1.5)
Problem 1.4 A particle with mass m moves in three-dimensional space under the influence of a constraint. The constraint is expressed by the following relation between the Cartesian coordinates of the particle, 2
2
e−(x +y ) + z = 0 .
(1.6)
a) The constraint relation can be used to express z as a function of x and y. Thus, there are two independent variables, which means that the system has two degrees of freedom. With x and y chosen as the generalized coordinates, the position vector of the particle is 2
2
r = xi + yj − e−(x +y ) k .
(1.7)
b) A virtual displacement, with x → x + δx and y → y + δy, gives the following variation in z 2
2
δz = 2e−(x +y ) (xδx + yδy) .
(1.8)
The expression for the variation of the position vector is therefore 2
2
2
2
δr = (i + 2e−(x +y ) x k)δx + (j + 2e−(x +y ) y k)δy .
(1.9)
c) The constraint force f satisfies the condition f · δr = 0 for arbitrary variations of the form (1.9). This implies that the two coefficients of the
7
Generalized coordinates
scalar product, proportional to δx and δy respectively, have to vanish separately, 2
2
fx + 2e−(x +y ) xfz = 0 ,
2
2
fy + 2e−(x +y ) yfz = 0.
(1.10)
In vector form this is 2
2
2
2
f = fz (−2e−(x +y ) xi − 2e−(x +y ) yj + k) .
(1.11)
Only the direction of f is determined by this expression since fz is an undetermined function of x and y. To fully determine the constraint force which acts on the particle one needs to know the applied forces and the velocity of the particle.
z
x
f δr
Fig. 1.2
A section of the constraint surface with directions of a virtual displacement δr of the small body and the constraint force f acting on the body. 2
d) In the y = 0 plane the constraint is described by the curve z = −e−x . A virtual displacement δr is directed tangentially to the curve, while the constraint force f is perpendicular to the curve, as illustrated in Fig 1.2. Problem 1.5 A flexible chain can move without friction on a smooth surface with the vertical heights of the endpoints denoted zA and zB . The chain has constant (linear) mass density μ. We shall use the principle of virtual work to find how zA and zB are related when the chain is at static equilibrium. A (dimensionless) parameter s, which measures length along the chain (relative to the total length), will be used as the variable. We set s = 0 at the right end of the chain and s = 1 at the left end. For a virtual translation
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Solution Manual for Classical Mechanics and Electrodynamics (2nd Edition)
δs of the chain along the surface, there is a vertical displacement δz, which varies with the position s along the chain, dz δs . ds The corresponding virtual work, integrated along the chain is 1 dz ds = gμδs(zB − zA ) . δW = g δzdm = gμδs 0 ds δz =
(1.12)
(1.13)
At equilibrium the principle of virtual work gives δW = 0, which implies zA = zB . Note that, as a simpler argument, a small displacement of the chain along the surface is equivalent to taking a small part of the chain at one end and move it to the other end. At equilibrium the change in potential energy should vanish, which would again mean that zA = zB .
Chapter 2
Lagrange’s equations Problem 2.1 A particle with mass m moves freely in a horizontal plane. The problem to be solved is to give a Lagrangian description of the motion in a rotating coordinate system, and to compare the corresponding equations of motion with the standard Newtonian description, where the rotation introduces centrifugal and Coriolis forces. a) The coordinate transformation from the rotating Cartesian reference frame, with coordinates (ξ, η), to the fixed Cartesian frame, with coordinates (x, y), is x = ξ cos ωt − η sin ωt , y = ξ sin ωt + η cos ωt ,
(2.1)
with the corresponding transformation of velocities, ẋ = (ξ˙ − ωη) cos ωt − (ωξ + η̇) sin ωt , ẏ = (ξ˙ − ωη) sin ωt + (ωξ + η̇) cos ωt .
(2.2)
The Lagrangian is identical to the kinetic energy, 1 m(ẋ2 + ẏ 2 ) 2 1 = m (ξ˙ − ωη)2 + (ωξ + η̇)2 2 1 ˙ + ω 2 (ξ 2 + η 2 ) . = m (ξ˙2 + η̇ 2 ) + 2ω(ξ η̇ − ξη) 2
L=
(2.3)
b) The partial derivatives of L with respect to the coordinates and their time derivative are ∂L ∂L = m(ξ˙ − ωη) , = m(ω η̇ + ω 2 ξ) , ∂ξ ∂ ξ˙ ∂L = m(−ω ξ̇ + ω 2 η) , ∂η 9
∂L = m(η̇ + ωξ) . ∂ η̇
(2.4)
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Solution Manual for Classical Mechanics and Electrodynamics (2nd Edition)
This gives as Lagrange’s equations, d ∂L ∂L − =0 dt ∂ ξ˙ ∂ ξ̇ d ∂L ∂L − =0 dt ∂ η̇ ∂ η̇
⇒
ξ¨ − 2ω η̇ − ω 2 ξ = 0 ,
⇒
η̈ + 2ω ξ̇ − ω 2 η = 0 .
(2.5)
Without any external force acting on the particle, Newton’s second law takes in the rotating frame the following form mr̈ = −mω × (ω × r) − 2mω × ṙ ,
(2.6)
where the angular velocity vector ω is orthogonal to the plane of rotation. The term which is linear in ω is the Coriolis force and the term which is quadratic in ω is the centrifugal force. To compare the equations in (2.5) and (2.6) we express the position vector r in (2.6) in terms of the coordinates ξ and η as r = ξi + ηj ,
(2.7)
where i and j are rotating unit vectors. In the rotating frame these are treated as fixed and the velocity and acceleration vectors are therefore given as ˙ + η̇j , ṙ = ξi
r̈ = ξ̈i + η̈j .
(2.8)
We insert these expressions for the Coriolis and centrifugal forces ω × (ω × r) = ω(ω · r) − ω 2 r = −ω 2 (ξi + ηj ) , ˙ + η̇j ) = ω(ξj ˙ − η̇i ) . ω × ṙ = ωk × (ξi
(2.9)
Inserting these in the vector equation (2.6) and extracting the components proportional to i and j , we find ξ¨ = 2ω η̇ + ω 2 ξ , η̈ = −2ω ξ̇ + ω 2 η ,
(2.10)
which are the same two equations as in (2.5). This shows the consistency between Lagrange’s equations expressed in the (rotating) coordinates ξ and η, and the standard vector equation used for Newton’s second law in a rotating reference frame.