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Introduction
These notes were written during the Fall, 2004, and Winter, 2005, terms. They are indeed lecture notes – I literally lecture from these notes. They combine material from Hand and Finch (mostly), Thornton, and Goldstein, but cover the material in a different order than any one of these texts and deviate from them widely in some places and less so in others. The reader will no doubt ask the question I asked myself many times while writing these notes: why bother? There are a large number of mechanics textbooks available all covering this very standard material, complete with worked examples and end-of-chapter problems. I can only defend myself by saying that all teachers understand their material in a slightly different way and it is very difficult to teach from someone else’s point of view – it’s like walking in shoes that are two sizes wrong. It is inevitable that every teacher will want to present some of the material in a way that differs from the available texts. These notes simply put my particular presentation down on the page for your reference. These notes are not a substitute for a proper textbook; I have not provided nearly as many examples or illustrations, and have provided no exercises. They are a supplement. I suggest you skim them in parallel while reading one of the recommended texts for the course, focusing your attention on places where these notes deviate from the texts.
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Contents
- 1 Elementary Mechanics
- 1.1 Newtonian Mechanics
- 1.1.1 The equation of motion for a single particle
- 1.1.2 Angular Motion
- 1.1.3 Energy and Work
- 1.2 Gravitation
- 1.2.1 Gravitational Force
- 1.2.2 Gravitational Potential
- 1.3 Dynamics of Systems of Particles
- 1.3.1 Newtonian Mechanical Concepts for Systems of Particles
- 1.3.2 The Virial Theorem
- 1.3.3 Collisions of Particles
- 2 Lagrangian and Hamiltonian Dynamics
- 2.1 The Lagrangian Approach to Mechanics
- 2.1.1 Degrees of Freedom, Constraints, and Generalized Coordinates
- 2.1.2 Virtual Displacement, Virtual Work, and Generalized Forces
- 2.1.3 d’Alembert’s Principle and the Generalized Equation of Motion
- 2.1.4 The Lagrangian and the Euler-Lagrange Equations
- 2.1.5 The Hamiltonian
- 2.1.6 Cyclic Coordinates and Canonical Momenta
- 2.1.7 Summary
- 2.1.8 More examples
- 2.1.9 Special Nonconservative Cases
- Noether’s Theorem 2.1.10 Symmetry Transformations, Conserved Quantities, Cyclic Coordinates and
- 2.2 Variational Calculus and Dynamics
- 2.2.1 The Variational Calculus and the Euler Equation
- 2.2.2 The Principle of Least Action and the Euler-Lagrange Equation
- 2.2.3 Imposing Constraints in Variational Dynamics
- 2.2.4 Incorporating Nonholonomic Constraints in Variational Dynamics
- 2.3 Hamiltonian Dynamics
- 2.3.1 Legendre Transformations and Hamilton’s Equations of Motion
- 2.3.2 Phase Space and Liouville’s Theorem
- 2.4 Topics in Theoretical Mechanics
- 2.4.1 Canonical Transformations and Generating Functions
- 2.4.2 Symplectic Notation
- 2.4.3 Poisson Brackets CONTENTS
- 2.4.4 Action-Angle Variables and Adiabatic Invariance
- 2.4.5 The Hamilton-Jacobi Equation
- 3 Oscillations
- 3.1 The Simple Harmonic Oscillator
- 3.1.1 Equilibria and Oscillations
- 3.1.2 Solving the Simple Harmonic Oscillator
- 3.1.3 The Damped Simple Harmonic Oscillator
- 3.1.4 The Driven Simple and Damped Harmonic Oscillator
- 3.1.5 Behavior when Driven Near Resonance
- 3.2 Coupled Simple Harmonic Oscillators
- 3.2.1 The Coupled Pendulum Example
- 3.2.2 General Method of Solution
- 3.2.3 Examples and Applications
- 3.2.4 Degeneracy
- 3.3 Waves
- 3.3.1 The Loaded String
- 3.3.2 The Continuous String
- 3.3.3 The Wave Equation
- 3.3.4 Phase Velocity, Group Velocity, and Wave Packets
- 4 Central Force Motion and Scattering
- 4.1 The Generic Central Force Problem
- 4.1.1 The Equation of Motion
- 4.1.2 Formal Implications of the Equations of Motion
- 4.2 The Special Case of Gravity – The Kepler Problem
- 4.2.1 The Shape of Solutions of the Kepler Problem
- 4.2.2 Time Dependence of the Kepler Problem Solutions
- 4.3 Scattering Cross Sections
- 4.3.1 Setting up the Problem
- 4.3.2 The Generic Cross Section
- 4.3.3 1 r Potentials
- 5 Rotating Systems
- 5.1 The Mathematical Description of Rotations
- 5.1.1 Infinitesimal Rotations
- 5.1.2 Finite Rotations
- 5.1.3 Interpretation of Rotations
- 5.1.4 Scalars, Vectors, and Tensors
- 5.1.5 Comments on Lie Algebras and Lie Groups
- 5.2 Dynamics in Rotating Coordinate Systems
- 5.2.1 Newton’s Second Law in Rotating Coordinate Systems
- 5.2.2 Applications
- 5.2.3 Lagrangian and Hamiltonian Dynamics in Rotating Coordinate Systems
- 5.3 Rotational Dynamics of Rigid Bodies
- 5.3.1 Basic Formalism
- 5.3.2 Torque-Free Motion
- 5.3.3 Motion under the Influence of External Torques CONTENTS
- 6 Special Relativity
- 6.1 Special Relativity
- 6.1.1 The Postulates
- 6.1.2 Transformation Laws
- 6.1.3 Mathematical Description of Lorentz Transformations
- 6.1.4 Physical Implications
- 6.1.5 Lagrangian and Hamiltonian Dynamics in Relativity
- A Mathematical Appendix
- A.1 Notational Conventions for Mathematical Symbols
- A.2 Coordinate Systems
- A.3 Vector and Tensor Definitions and Algebraic Identities
- A.4 Vector Calculus
- A.5 Taylor Expansion
- A.6 Calculus of Variations
- A.7 Legendre Transformations
- B Summary of Physical Results
- B.1 Elementary Mechanics
- B.2 Lagrangian and Hamiltonian Dynamics
- B.3 Oscillations
- B.4 Central Forces and Dynamics of Scattering
- B.5 Rotating Systems
- B.6 Special Relativity
Chapter 1
Elementary Mechanics
This chapter reviews material that was covered in your first-year mechanics course – Newtonian mechanics, elementary gravitation, and dynamics of systems of particles. None of this material should be surprising or new. Special emphasis is placed on those aspects that we will return to later in the course. If you feel less than fully comfortable with this material, please take the time to review it now, before we hit the interesting new stuff! The material in this section is largely from Thornton Chapters 2, 5, and 9. Small parts of it are covered in Hand and Finch Chapter 4, but they use the language of Lagrangian mechanics that you have not yet learned. Other references are provided in the notes.
1.1 Newtonian Mechanics
Newton’s second law is F~ = ~p˙, so we therefore have
~p˙ · ~s = 0 =⇒ ~p · ~s = α (1.5)
where α is a constant. That is, there is conservation of the component of linear momentum along the direction ~s in which there is no force.
Solving simple Newtonian mechanics problems
Try to systematically perform the following steps when solving problems:
- Sketch the problem, drawing all the forces as vectors.
- Define a coordinate system in which the motion will be convenient; in particular, try to make any constraints work out simply.
- Find the net force along each coordinate axis by breaking down the forces into their components and write down Newton’s second law component by component.
- Apply the constraints, which will produce relationships among the different equations (or will show that the motion along certain coordinates is trivial).
- Solve the equations to find the acceleration along each coordinate in terms of the known forces.
- Depending on what result is desired, one either can use the acceleration equations directly or one can integrate them to find the velocity and position as a function of time, modulo initial conditions.
- If so desired, apply initial conditions to obtain the full solution.
Example 1. (Thornton Example 2.1) A block slides without friction down a fixed, inclined plane. The angle of the incline is θ = 30◦^ from horizontal. What is the acceleration of the block?
F^ ~g = m~g is the gravitational force on the block and F~N is the normal force, which is exerted by the plane on the block to keep it in place on top of the plane.
- Coordinate system: x pointing down along the surface of the incline, y perpendicular to the surface of the incline. The constraint of the block sliding on the plane forces there to be no motion along y, hence the choice of coordinate system.
CHAPTER 1. ELEMENTARY MECHANICS
m x¨ = Fg sin θ m y¨ = FN − Fg cos θ
- Apply constraints: there is no motion along the y axis, so ¨y = 0, which gives FN = Fg cos θ. The constraint actually turns out to be unnecessary for solving for the motion of the block, but in more complicated cases the constraint will be important.
- Solve the remaining equations: Here, we simply have the x equation, which gives:
x¨ =
Fg m sin θ = g sin θ
where Fg = mg is the gravitational force
- Find velocity and position as a function of time: This is just trivial integration:
d dt x˙ = g sin θ =⇒ x˙(t) = x˙(t = 0) +
∫ (^) t
0
dt′^ g sin θ = x˙ 0 + g t sin θ d dt x = ˙x(t = 0) + g t sin θ =⇒ x(t) = x 0 +
∫ (^) t
0
dt′^
[
x˙ 0 + g t′^ sin θ
]
= x 0 + ˙x 0 t +
g t^2 sin θ
where we have taken x 0 and ˙x 0 to be the initial position and velocity, the constants of integration. Of course, the solution for y is y(t) = 0, where we have made use of the initial conditions y(t = 0) = 0 and ˙y(t = 0) = 0.
Example 1. (Thornton Example 2.3) Same as Example 1.1, but now assume the block is moving (i.e., its initial velocity is nonzero) and that it is subject to sliding friction. Determine the acceleration of the block for the angle θ = 30◦^ assuming the frictional force obeys Ff = μk FN where μk = 0.3 is the coefficient of kinetic friction.
We now have an additional frictional force Ff which points along the −x direction because the block of course wants to slide to +x. Its value is fixed to be Ff = μk FN.
CHAPTER 1. ELEMENTARY MECHANICS
i.e.,
¨x ≥ g [sin θ − μs cos θ]
It becomes impossible for the block to stay motionless when the right side becomes positive. The transition angle θ′^ is of course when the right side vanishes, when
0 = sin θ′^ − μs cos θ′
or
tan θ′^ = μs
which gives θ′^ = 21. 8 ◦.
Atwood’s machine problems
Another class of problems Newtonian mechanics problems you have no doubt seen before are Atwood’s machine problems, where an Atwood’s machine is simply a smooth, massless pulley (with zero diameter) with two masses suspended from a (weightless) rope at each end and acted on by gravity. These problems again require only Newton’s second equation. Example 1. (Thornton Example 2.9) Determine the acceleration of the two masses of a simple Atwood’s machine, with one fixed pulley and two masses m 1 and m 2.
- Sketch:
- Coordinate system: There is only vertical motion, so use the z coordinates of the two masses z 1 and z 2.
1.1. NEWTONIAN MECHANICS
- Forces along each axis: Just the z-axis, but now for two particles:
m 1 z¨ 1 = −m 1 g + T m 2 z¨ 2 = −m 2 g + T
where T is the tension in the rope. We have assumed the rope perfectly transmits force from one end to the other.
- Constraints: The rope length l cannot change, so z 1 + z 2 = −l is constant, ˙z 1 = − z˙ 2 and ¨z 1 = −z¨ 2.
- Solve: Just solve the first equation for T and insert in the second equation, making use of ¨z 1 = −z¨ 2 :
T = m 1 ( ¨z 1 + g) −z¨ 1 = −g +
m 1 m 2 ( ¨z 1 + g)
which we can then solve for ¨z 1 and T :
−¨z 2 = ¨z 1 = − m 1 − m 2 m 1 + m 2
g
T =
2 m 1 m 2 m 1 + m 2 g
It is instructive to consider two limiting cases. First, take m 1 = m 2 = m. We have in this case
−z¨ 2 = ¨z 1 = 0 T = m g
As you would expect, there is no motion of either mass and the tension in the rope is the weight of either mass – the rope must exert this force to keep either mass from falling. Second, consider m 1 m 2. We then have
−z¨ 2 = ¨z 1 = −g T = 2 m 2 g
Here, the heavier mass accelerates downward at the gravitational acceleration and the other mass accelerates upward with the same acceleration. The rope has to have sufficient tension to both counteract gravity acting on the second mass as well as to accelerate it upward at g.
Example 1.
(Thornton Example 2.9) Repeat, with the pulley suspended from an elevator that is ac- celerating with acceleration a. As usual, ignore the mass and diameter of the pulley when considering the forces in and motion of the rope.
1.1. NEWTONIAN MECHANICS
- Constraints: Again, the rope length cannot change, but the constraint is more com- plicated because the pulley can move: z 1 + z 2 = 2 zp − l. The fixed rope between the pulley and the elevator forces ¨zp = ¨ze = a, so ¨z 1 + ¨z 2 = 2 a
- Solve: Just solve the first equation for T and insert in the second equation, making use of the new constraint ¨z 1 = −¨z 2 + 2a:
T = m 1 ( ¨z 1 + g) 2 a − ¨z 1 = −g +
m 1 m 2 ( ¨z 1 + g)
which we can then solve for ¨z 1 and T :
z¨ 1 = −
m 1 − m 2 m 1 + m 2 g +
2 m 2 m 1 + m 2 a
z ¨ 2 = m 1 − m 2 m 1 + m 2
g + 2 m 1 m 1 + m 2
a
T =
2 m 1 m 2 m 1 + m 2 (g + a)
We can write the accelerations relative to the elevator (i.e., in the non-inertial, accel- erating frame) by simply calculating ¨z 1 ′ = ¨z 1 − z¨p and ¨z 2 ′ = ¨z 2 − z¨p:
z¨ 1 ′ = m 2 − m 1 m 2 + m 1
(g + a)
z¨ 2 ′ = m 1 − m 2 m 1 + m 2 (g + a)
We see that, in the reference frame of the elevator, the accelerations are equal and opposite, as they must be since the two masses are coupled by the rope. Note that we never needed to solve the third and fourth equations, though we may now do so:
R = mp( ¨zp + g) + 2T = mp(a + g) + 4 m 1 m 2 m 1 + m 2
(g + a)
[
mp + 4 m 1 m 2 m 1 + m 2
]
(g + a)
E = me( ¨ze + g) + R
=
[
me + mp + 4 m 1 m 2 m 1 + m 2
]
(g + a)
That these expressions are correct can be seen by considering limiting cases. First, consider the case m 1 = m 2 = m; we find
z¨ 1 = a z ¨ 2 = a T = m (g + a) R = [mp + 2 m] (g + a) E = [me + mp + 2 m] (g + a)
That is, the two masses stay at rest relative to each other and accelerate upward with the elevator; there is no motion of the rope connecting the two (relative to the pulley)
CHAPTER 1. ELEMENTARY MECHANICS
because the two masses balance each other. The tensions in the rope holding the pulley and the elevator cable are determined by the total mass suspended on each. Next, consider the case m 1 m 2. We have
¨z 1 = −g ¨z 2 = g + 2 a T = 0 R = mp (g + a) E = [me + mp] (g + a)
m 1 falls under the force of gravity. m 2 is pulled upward, but there is a component of the acceleration in addition to just g because the rope must unwind over the pulley fast enough to deal with the accelerating motion of the pulley. R and E no longer contain terms for m 1 and m 2 because the rope holding them just unwinds, exerting no force on the pulley. The mass combination that appears in the solutions, m 1 m 2 /(m 1 + m 2 ), is the typical form one finds for transforming continuously between these two cases m 1 = m 2 and m 1 m 2 (or vice versa), as you will learn when we look at central force motion later.
Retarding Forces
(See Thornton 2.4 for more detail, but these notes cover the important material) A next level of complexity is introduced by considering forces that are not static but rather depend on the velocity of the moving object. This is interesting not just for the physics but because it introduces a higher level of mathematical complexity. Such a force can frequently be written as a power law in the velocity:
F^ ~r = F~r(v) = −k vn^ ~v v
k is a constant that depends on the details of the problem. Note that the force is always directed opposite to the velocity of the object. For the simplest power law retarding forces, the equation of motion can be solved analyt- ically. For more complicated dependence on velocity, it may be necessary to generate the solution numerically. We will come back to the latter point. Example 1. (Thornton Example 2.4). Find the velocity and position as a function of time for a particle initially having velocity v 0 along the +x axis and experiencing a linear retarding force Fr(v) = −k v.
CHAPTER 1. ELEMENTARY MECHANICS
- Sketch:
- Coordinate system: only one dimension, so trivial. Have the initial velocity ˙z 0 be along the +z direction.
- Forces along each axis: Just the z-axis.
m ¨z = −m g − k z˙
- Constraints: none
- Solve: The differential equation for z is d dt
z˙ = −g − k m
z˙(t)
Now we have both constant and velocity-dependent terms on the right side. Again, we solve by separating variables and integrating: d z˙ g + (^) mk z˙
= −dt ∫ (^) z˙(t)
z˙ 0
dy g + (^) mk y
∫ (^) t
0
dt′
m k
∫ (^) g+ k m z˙(t) g+ (^) mk z˙ 0
du u
∫ (^) t
0
dt′
log
g + k m z˙(t)
− log
g + k m z˙ 0
k m t
k m g
z˙(t) =
k m g
exp
k m
t
z ˙(t) = − m g k
( (^) m g k
exp
k m
t
We see the phenomenon of terminal velocity: as t → ∞, the second term vanishes and we see ˙z(t) → −m g/k. One would have found this asymptotic speed by also solving the equation of motion for the speed at which the acceleration vanishes. The position as a function of time is again found easily by integrating, which yields
z(t) = z 0 − m g k
t +
m^2 g k^2
m z˙ 0 k
) [
1 − exp
k m
t
)]
1.1. NEWTONIAN MECHANICS
The third term deals with the portion of the motion during which the velocity is chang- ing, and the second term deals with the terminal velocity portion of the trajectory.
Retarding Forces and Numerical Solutions
Obviously, for more complex retarding forces, it may not be possible to solve the equation of motion analytically. However, a natural numerical solution presents itself. The equation of motion is usually of the form
d dt x˙ = Fs + F ( ˙x)
This can be written in discrete form
∆ ˙x = [Fs + F ( ˙x)] ∆t
where we have discretized time into intervals of length ∆t. If we denote the times as tn = n∆t and the velocity at time tn by ˙xn, then we have
x˙n+1 = x˙n + [Fs + F ( ˙xn)] ∆t
The above procedure can be done with as small a step size as desired to obtain as precise a solution as one desires. It can obviously also be extended to two-dimensional motion. For more details, see Thornton Examples 2.7 and 2.8.
1.1.2 Angular Motion
We derive analogues of linear momentum, force, and Newton’s second law for angular motion.
Definitions
Angular velocity of a particle as a function of time with respect to a particular origin:
~v(t) = ~ω(t) × ~r(t) (1.7)
This is an implicit definition that is justified by considering a differential displacement:
1.1. NEWTONIAN MECHANICS
Conservation of Angular Momentum
Just as we proved that linear momentum is conserved in directions along which there is no force, one can prove that angular momentum is conserved in directions along which there is no torque. The proof is identical, so we do not repeat it here.
Choice of origin
There is a caveat: angular momentum and torque depend on the choice of origin. That is, in two frames 1 and 2 whose origins differ by a constant vector ~o such that ~r 2 (t) = ~r 1 (t) + ~o, we have
~L 2 (t) = ~r 2 (t) × ~p(t) = ~r 1 (t) × ~p(t) + ~o × ~p(t) = ~L 1 (t) + ~o × ~p(t) N^ ~ 2 (t) = ~r 2 (t) × F~ (t) = ~r 1 (t) × F~ (t) + ~o × F~ (t) = N~ 1 (t) + ~o × F~ (t)
where we have used the fact that ~p and F~ are the same in the two frames (~p because it involves a time derivative; F~ via its relation to ~p by Newton’s second law). Thus, while Newton’s second law and conservation of angular momentum certainly hold regardless of choice of origin, angular momentum may be constant in one frame but not in another because a torque that vanishes in one frame may not vanish in another! In contrast, if linear momentum is conserved in one frame it is conserved in any displaced frame. Thus, angular momentum and torque are imperfect analogues to linear momentum and force. Let’s consider this in more detail. We first solve the linear equations of motion for a particle moving in a circle at fixed speed as shown in the previous figure. Choose the origin of the system to be at the center of the circle and the motion to be in the xy plane. Clearly, in this frame, the particle’s position and velocity as a function of time are
~r 1 (t) = R (ˆx cos ωt + ˆy sin ωt) ~v(t) = ω R (−xˆ sin ωt + ˆy sin ωt)
where we obtained the velocity by simple differentiation. We do not subscript ~v because it is independent of the choice of origin. The mass is fixed so the momentum is just ~p(t) = m ~v(t). The force is, by Newton’s second law,
F^ ~ (t) = d~p dt = −m ω^2 R (ˆx cos ωt + ˆy sin ωt) = −m ω^2 R rˆ 1 (t)
= −m v^2 R
ˆr 1 (t)
where ˆr 1 (t) is a unit vector pointing along ~r 1 (t). Clearly, the force is back along the line to center of the circle, has magnitude F = m v^2 /R = m ω^2 R, and is perpendicular to the velocity. The velocity, momentum, and force are independent of the choice of origin. Let’s determine the angular momentum and torque. First consider the same coordinate system with position vector ~r 1 (t). Since F~ points back along ~r 1 , it holds that the torque N^ ~ = ~r 1 × F~ vanishes. The angular momentum vector is L~ 1 = ~r 1 × ~p = mv R zˆ. Since v is fixed, L~ 1 is fixed, as one would expect in the absence of torque.
CHAPTER 1. ELEMENTARY MECHANICS
Next, consider a frame whose origin is displaced from the center of the particle orbit along the z axis, as suggested by the earlier figure. Let ~r 2 denote the position vector in this frame. In this frame, the torque N~ 2 is nonzero because ~r 2 and F~ are not colinear. We can write out the torque explicitly: N^ ~ 2 (t) = ~r 2 (t) × F~ (t) = r 2 [sin α(ˆx cos ωt + ˆy sin ωt) + ˆz cos α] × F (−xˆ cos ωt − yˆ sin ωt) = r 2 F [− sin α(ˆx × yˆ cos ωt sin ωt + ˆy × xˆ sin ωt cos ωt) − cos α(ˆz × xˆ cos ωt + ˆz × ˆy sin ωt)] = r 2 F cos α (ˆx sin ωt − yˆ cos ωt)
where, between the second and third line, terms of the form ˆx × ˆx and ˆy × yˆ were dropped because they vanish. Let’s calculate the angular momentum in this system: ~L 2 (t) = ~r 2 (t) × ~p(t) = r 2 [sin α(ˆx cos ωt + ˆy sin ωt) + ˆz cos α] × mv (−xˆ sin ωt + ˆy cos ωt) = r 2 mv
[
sin α(ˆx × yˆ cos^2 ωt − yˆ × xˆ sin^2 ωt) + cos α(−zˆ × xˆ sin ωt + ˆz × ˆy cos ωt)
]
= ˆz mv r 2 sin α + mv r 2 cos α (ˆy sin ωt − ˆx cos ωt)
So, in this frame, we have a time-varying component of L~ 2 in the plane of the orbit. This time derivative of L~ 2 is due to the nonzero torque N~ 2 present in this frame, as one can demonstrate directly by differentiating ~L 2 (t) and using F = mv^2 /R = mv^2 /(r 2 cos α) and v = R ω = r 2 ω cos α. The torque is always perpendicular to the varying component of the angular momentum, so the torque causes the varying component of the angular momentum to precess in a circle. One can of course consider even more complicated cases wherein the origin displacement includes a component in the plane of the motion. Clearly, the algebra gets more complicated but none of the physics changes.
1.1.3 Energy and Work
We present the concepts of kinetic and potential energy and work and derive the implications of Newton’s second law on the relations between them.
Work and Kinetic Energy
We define the work done on a particle by a force F~ (t) in moving it from ~r 1 = ~r(t 1 ) to ~r 2 = ~r(t 2 ) to be W 12 =
∫ (^) t 2
t 1
F^ ~ · d~r (1.9)
The integral is a line integral; the idea is to integrate up the projection of the force along the instantaneous direction of motion. We can write the expression more explicitly to make this clear:
W 12 =
∫ (^) t 2
t 1
F^ ~ (t) · d~r dt dt
∫ (^) t 2
t 1
F^ ~ (t) · ~v(t) dt
