PHY340

PHY 340: Mechanics and Vibrations

Ulrich Zurcher∗Physics Department, Cleveland State University, Cleveland, OH 44115

(Dated: October 14, 2010)

We do not have the intention to have complete lecturenotes: we have our text Thornoton and Marion for that[1]. Here, we use them twofold: (1) as a brief synopsisfor a quick overview and (2) to include material that isnot covered in the text, e.g., numerical methods.

I. MATRICES, VECTORS, AND VECTORCALCULUS

A. Vectors and Matrices

Scalars and vectors are defined how they change undera coordinate transformation (x1, x2x3) → (x′1, x

′2, x′3).

Example: x, y)→ (x′, y′) rotation with angle θ:

(x,y)=(x’,y’)

It is easy to see that

x′ = cos θ · x+ sin θ · y (1)y′ = − sin θ · x+ cos θ · y (2)

we can write this in short hand:(x′

y′

)=(

cos θ sin θ− sin θ cos θ

)·(xy

). (3)

∗Electronic address: u.zurcher@csuohio.edu

We consider linear transformations:

x′i =∑j

λijxj (4)

where the coefficients λij are the direction cosines.We have λ = λ(θ). We have the following properties:λ(θ1) · λ(θ2) = λ(θ1 + θ2), (ii) [λ(θ1) · λ(θ2)] · λ(θ3) =λ(θ1) · [λ(θ2) · λ(θ3)], (iii) e = δij is the unit matrix, (iv)there is an inverse λ−1(θ) = λ(−θ) such that λ−1 · λ =λ · λ−1 = e. These four porperties show that rotationsare a group.

The coefficients are written as a matrix: λ11 λ12 λ13

λ21 λ22 λ23

λ31 λ32 λ33

. (5)

Rotation about the z-axis:

λ =

cos θ sin θ 0− sin θ cos θ 0

0 0 1

. (6)

Matrices can be multiplied: “row times column,” orfor matrix Aij and matrix Bij , we have C = A ·B with

Cij =∑k

AikBkj . (7)

Of course, this implies that the number of columns of Ais equal to the number of rows of B.

Rotations are examples of ‘orthogonal’ matrices:∑j

λijλjk = δik, (8)

where δik is the Kronecker symbol. That is, for othrog-onal matrices, the inverse λ−1 is equal to the transposematrix λT:

λ−1 = λT. (9)

Here, (λT)ij = λji; i.e., λT is the mirror image of λ withrespect to the diagonal i = j.

Scalar quantities are invariant under a transformation:

S(x′1, x′2, x′3) = S(x1, x2, x3). (10)

Vectors transform like real coordinates, i.e.,

V ′i (x′1, x′2, x′3) =

∑j

λijVj(x1, x2, x3). (11)

This is obviously the case for the position, ~V = ~r.

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Scalar product between two vectors ~A and ~B:

~A · ~B =∑i

AiBi = AB cos( ~A, ~B). (12)

Unit vector: ~e = ~A/A. Vector product ~C = ~A× ~B, where

Ci =∑j,k

εijkAjBk. (13)

Here εijk is the Lev-Civitta symbol. We have for themagnitude:

|~C| = AB sin( ~A, ~B). (14)

B. Vector Calculus

The derivative of a vector ~A = ~A(s) is defined as ex-pected

d ~A(s)ds

= lim∆s→0

~A(s+ ∆s)− ~A(s)∆s

. (15)

Examples are velocities and accelerations.Plane Polar coordinates Cartesian coordinates areglobally orthogonal. We consider coordinate systemsthat are locally orthogonal: plane polar coordinates[cylinder coordinates] and spherical coordinates.In the plane unit vectors: er and eθ. We have the timederivatives:

˙er = θeθ (16)˙eθ = −θer. (17)

We thus have for the coordinate vector ~r = rer. Thevelocity

~v =d

dt~r = rer + r · θeθ. (18)

and the acceleration:

~a =d

dt~v = (r − rθ2)er + (rθ + 2rθ)eθ. (19)

Example: Motion along a cardioid at constant speed [TM-1-6]. The object is constrained on the caridoid:

r = k(1 + cos θ). (20)

We then have

r = −kθ sin θ (21)

r = −k[θ2 cos θ + θ sin θ]. (22)

We have for the velocity v2 = r2 + r2θ2 or

v2 = 2k2θ2[1 + cos θ] = const. (23)

Solve for the angular velocity:

θ =v√

2k2[1 + cos θ]=

v√2kr

. (24)

For the angular acceleration, we have θ =−(v/2

√2k)r/r3/2 so that

θ =v2 sin θ

4k2[1 + cos θ]. (25)

We have for the components of the acceleration

ar = r − rθ2 = −32kθ2[1 + cos θ] (26)

aθ = rθ + 2rr) = −3v2

4ksin θ

1 + cos θ. (27)

The magnitude of the acceleration follows:

a =3v2

4k cos θ/2, (28)

where we used a simple trig-identity.

For an angular displacement δ~θ, the change in the po-sition is

δ~r = δ~θ × ~r. (29)

The angular velocity is defined:

~ω =δ~θ

δt, (30)

so that

~v = ~ω × ~r. (31)

The “nabla” operator is defined:

grad = ∇ =∑i

ei∂

∂xi. (32)

The gradient of a scalar quantity φ is in the direction ofsteepest descent:

∇φ =∑i

ei∂φ

∂xi. (33)

The divergence of a vector ~A measures the strength ofthe “charge” [or source]:

∇ · ~A =∑i

∂Ai∂xi

. (34)

The curl is tells how much the vector field “winds:”

∇× ~A =∑ijk

εijk∂Ak∂xj

~ei. (35)

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C. Integration of Vectors

The fundamental theorem of calculus tells:∫ B

A

∇φ · d~s = φ(B)− φ(A). (36)

In E&M, we have the electric field ~E = −∇Φ, then∫ BA~E ·

d~s = ΦA − ΦB . The negative sign is convention.Same idea yields Gauss laws for divergence:∫

S~A · d~a =

∫V∇ · ~AdV, (37)

where S = ∂V is the surface enclosing the volume V.in E&M: we have ∇ · ~E = ρ/ε0 so that

∫S~E · d ~A =

ε−10

∫V ρdV = Qencl/ε0.

Stokes theorem applies for curl:∫C~A · d~s =

∫S

(∇× ~A) · d~a (38)

where C = ∂S is the contour enclosing the area S.In E&M: we have ∇ × ~B = µ0

~j so that∫C~B · d~s = µ0

∫S~j · d ~A = µ0Iencl.

II. NEWTON’S LAWS - SINGLE PARTICLE

For motion in a viscous medium:

A. Newton’s Laws

1. A body remains at rest or in uniform motion unlessacted upon by a force.

2. A body acted upon by a force moves in such a man-ner that the time rate of change of the momentumequals the force.

3. If two bodies exert forces on each other, these forcesare equal in magnitude and opposite in direction.

The first two laws are definitions. The third law canbe expressed: If two bodies constitute an ideal, isolatedsystem, then the accelerations of these bodies are alwaysin opposite directions, and the ratio of the magnitudes ofthe accelerations is constant. This constant ratio is theinverse ratio of the masses of the bodies:

a1

a2= −m2

m1. (39)

The third law can be written in terms of momenta:

d

dt(~p1 + ~p2) = 0, (40)

so that ~p1 + ~p2 = const.

B. Frame of reference

Galileian invariance: Newton’s law involves the deriva-tive of the velocity with respect to time, absolute veloc-ities are meaningless. Thus, If Newton’s laws are validin one reference frame, then they are also valid in a ref-erence frame in uniform motion. For the Earth [“Labframe”], the velocity is not constant. For many appli-cations, e.g., the collision of two gliders on an airtrack,deviations from Galilean invariance can be ignored.

C. Equation of Motion of a Single Particle

Retarding forces. Small particles [e.g., a raindrop] areaffected by viscous forces. Stokes law for laminar flowaround a spherical object:

Fv = 6πaηv, (41)

where a is the radius and the viscosity η has units [η] =Pa · s is the dynamic viscosity. Large objects [e..g., a car]is subject to inertial forces for turbuent flow:

Fi =12cW ρAv

2, (42)

where ρ is the density and A = πa2 is the cross section.The ratio of these forces defines the Reynolds number:Re = Fi/Fv = ρav/η. Laminar flow for Re < 1 andturbulent flow for Re > 105.

Example: Free fall with linear damping. Choose a coor-dinate system x

dv

dt= −g − kv (43)

where the constant has units of an inverse second [k] = s−1.We write the equation in the form

dv

dt+ kv = −g. (44)

The homogenous differential equation is obtained by settingthe RHS equal to zero:

dv

dt+ kv = 0. (45)

The general solution is obtained by adding the general so-lution of the homogeneous DE and one solution of the in-homogeneous equation.The solution v = const is one solution of the inhomogenousDE. Physically, this is is the terminal velocity:

vt = −gk. (46)

We use the method of separation of variables tor the generalsolution of the inhomogeneous equation:

dv

dt= −kv → dv

v= −kdt. (47)

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We integrate both sides:∫ v

v0

dv′

v′= −

∫ t

0

kdt′ (48)

where we set t0 = 0 and v0 is the velocity at time t = 0.The integrals are elementary:

lnv

v0= kt −→ v(t) = v0,he

−kt. (49)

The velocity v0,h is an “integration constant;” our solutionhas one such constant because it is a one-dimensional firstorder differential equation. The general solution thus is:

v(t) = −gk

+ v0,he−kt. (50)

If v(0) = v0 then v0,h = v0 + g/k so that

v(t) = v0e−kt + vt

(1− e−kt

). (51)

This form has a simple explanation.Example: A particle is projected upwards at speed v0 in

gravitational field and subject to inertial force. Calculatethe speed when it returns to the ground. [TM 2-12]. Sincea = dv/dt = vdv/dx, we have for upwards motion:

vdv

dx= −g − kv2 = −k[v2

t + v2], (52)

where vt =√g/k is the terminal speed. Separation of

variables yields:

vdv

v2 + vt= −dx

dx. (53)

where λ = k−1 is a characteristic length.Integration: for u = v2: du/(u+ v2

t ) = −dx/2λ∫ 0

v20

du

u+ v2t

= lnv2

t

v20 + v2

t

= −H2λ

(54)

where H is the maximum height of the projectile. Now onthe way down:

vdv

dx= g − kv2 −→ du

−u+ v2t

=dx

2λ. (55)

Integration:∫ v2f

0

du

−u+ v2t

= lnv2

t

−v2f + v2

t

=H

2λ(56)

We thus get

lnv2

0 + v2t

v2t

= lnv2

t

−v2f + v2

t

−→ vf =v0vt√v2

0 + v2t

. (57)

Interesiting limits: for v0 >> vt we have vf → vt and forv0 << vt, we have vf → v0.

D. Conservation Laws

Momentum: The total linear momentum ~p of a par-ticle is conserved when the total force on it is zero.Angular momentum: The angular momentum is de-fined with respect to the origin:

~L = ~r × ~p. (58)

the torque with to the origin:

~N = ~r × ~F . (59)

Newton’s second law for rotation follows:

d~L

dt= ~r × d~p

dt= ~N. (60)

Thus, the angular momentum of a particle subject to notorque is conserved.Energy: The kinetic energy is defined:

T =12mv2. (61)

Work from 1 to 2 is defined:

W12 =∫ 2

1

~F · d~r. (62)

Example: [TM 2-27]. A rope having the a total mass of0.4 kg and total length 4 m has 0.6 m of the rope hangingvertically down off a work bench. How much work must bedone to pull the entire rope on the bench?Solution: Use mass per unit length λ = M/L. Consider asmall section dy at the vertical component y. Work requiredto pull it up is

dW = gλdy · y = gλydy.

Integration:

W =∫ Y

0

dW =∫ Y

0

gλydy =gλY 2

2.

For conservative forces,∮C~F · d~r = 0, (63)

for a closed path C. For conservative force, the force canbe calculated from a potential:

~F = −∇U (64)

so that∫ 2

1~F · d~r = U1 − U2. Then the total energy is

defined

E = T + U. (65)

Example: Stability of a system [TM 2-42]:A solid cube of uniform density and sides b us in equilibrium

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on top of a cylinder of radius R. The planes of four sidesof the cube are parallel to the axis of the cylinder. Thecontact between cube and the cylinder is perfectly rough.Under what conditions is the equilibrium stable or unstable?

θ

θ

Solution: From the figure: we have for the height of theblock:

h(θ) =(R+

b

2

)cos θ +Rθ sin θ.

The potential energy follows U(θ) = mgh(θ):

dU

dθ= mg

[− b

2sin θ +Rθ cos θ

]d2U

dθ2= mg

[(R− b

2

)cos θ −Rθ sin θ

].

Equilibrium (stable or unstable) requires that dU/dθ = 0 orθeq = 0. It follows at equilibrium:

d2U

dθ2

∣∣∣∣eq

= mg

(R− b

2

).

We thus have the condition for stability, d2U/dθ2 > 0 for

R− b

2> 0 (stable)

and

R− b

2< 0 (unstable).

For R − b/2 = 0. Linear stability is not sufficient, and wehave to go the fourth-order in θ.

The total energy E of a particle in a conservative forcefield is a constant in time. In one dimension, the existenceof an integral of motion allows us to find the solution:

t− t0 =∫ x

x0

dx′√2[E − U(x′)]/m

. (66)

For free fall: Object at y0 = h at velocity v = v0. Totalenergy is E = mgh + mv2

0/2 so that E − U = mv20/2 +

mg(y0 − y) = mg[(y0 − y) + v20/2g]. Thus

√2gt =

∫ y

y0

dy′√(y0 − y′) + v2

0/2g=∫ (y−y0)+v20/2g

v20/2g

dz√z

= 2

√(y0 − y) +v2

0

2g+

v0√2g

.

We thus have

√2gt−

√2gv0 = 2

√(y0 − y) +

v20

2g,

We square both sides and simplifiy:

2gt2−4v0t = 4(y−y0) −→ y = y0−v0t+12gt2. (67)

That’s hopefully familiar!

E. Numerical Solutions

We consider a general second order differential equation:

d2y

dx2+ q(x)

dy

dx= r(x), (68)

where the functions r(x) and q(x) are known. Our goalis to find the numerical solution of y(x).Example: Driven Damped harmonic oscillator

d2y

dt2− γ dy

dt+ ω2y(t) = F (t), (69)

where γ is the damping constant, ω is the angular fre-quency ω2 = k/m, and F (t) is the (external) drivingforce.

We start by writing the second order differential equa-tion in terms of two coupled first-order differential equa-tions:

dy

dx= z(x) (70)

dz

dx= r(x)− q(x)z(x). (71)

Problem: Write the equation of motion for the dampedharmonic oscillator in terms of first-order equations.Thus, the generic problem is a set of N coupled firstorder differential equations:

dyi(x)dx

= fi(x, y1, y2, ..., yN ), i = 1, 2, ..., N. (72)

We consider a one-dimensional problem:

dy

dx= f(x, y). (73)

We choose a fixed step-size on the x axis: xn+1 = xn = h.Assume that we have the point (xn, yn, we need to find(xn+1, yn+1).Euler -Method The formula for the Euler method is:

yn+1 = yn + hf(xn, yn). (74)

This forumula is unsymmetrical: it advances the solutionthrough an interval h but uses the derivative only at thebeginning of that interval. That means that the step’s

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error is only one power of h smaller than the correction,i.e., O(h2).Euler-Richardson We consider Newton’s equation ofmotion:

d2y(t)dt2

=F (y, v, t)

m= a(y, v, t). (75)

The Euler method gives:

vn+1 = vn + an∆t (76)yn+1 = yn + vn∆t. (77)

We could instead use the velocity at the end of the timeinterval: Euler-Cromer method

vn+1 = vn + an∆t (78)yn+1 = yn + vn+1∆t. (79)

We write y(t + ∆t) as a Taylor series to second orderin (∆t)2:

y1(t+ ∆t) = y(t) + v(t)∆t+a(t)

2(∆t)2, (80)

where a(t) = a(y(t), v(t), t). The notation y1 implies thaty(t + ∆t) is related to y(t) by one time step. We dividethe step ∆t into half steps and write the first half stepy(t+ ∆t/2) as

y(t+∆t2

) = y(t) + v(t)∆t2

+a(t)

2(∆t)2. (81)

The second half step y2(t+ ∆t), can be written as

y2(t+ ∆t) = y

(t+

∆t2

)+ v

(t+

∆t2

)∆t2

+a(t+ ∆/2)

2

(∆t2

)2

. (82)

Insert Eq. (28) into Eq. (29):

y2(t+ ∆t) = y(t) +12

[v(t) + v

(t+

∆t2

)]∆t

+12

[a(t) + a

(t+

∆t2

)](∆t2

)2

.(83)

Now a(t+∆t/2) = a(t)+(a′(t)/2)∆t+ .... hence to order(∆t)2:

y2(t+∆t) = y(t)+12

[v(t) + v

(t+

∆t2

)]∆t+

2a(t)2

(∆t2

)2

.

(84)We can create an approximation that is accurate to

order (∆t)3 by combining Eq. (27) and Eq. (30) so thatthe terms to order (∆t)2 cancel. The combination thatworks is 2y2−y1, which gives the Euler-Richardson result:

yer(t+∆t) = 2y2(t+∆t)−y1(t+∆t) = y(t)+v(t+

∆t2

)∆t.

(85)

The same reasoning leads to an approximation for thevelocity accurate to (∆t)3 giving:

ver(t+∆t) = 2v2(t+∆t)−v1(t+∆t) = v(t)+a(t+

∆t2

)∆t.

(86)A bonus of the Euler-Richardson method is that thequantities |y2 − y1| and |v2 − v1| give an estimate forthe error in the procedure. We can use these estimatesto change the time step so that the error is always withinsome desired level of precision.

The Euler-Richardson algorithm thus follows:

an = F (yn, vn, tn)/m (87)

vmid = vn +12an∆t (88)

ymid = yn +12vn∆t (89)

amid = F (ymid, vmid, t)/m, (90)

then

vn+1 = vn + amid∆t (91)yn+1 = yn + vmid∆t (92)

Runge-Kutta Method We take a “trial” step at themidpoint:

k1 = hf(xn, yn) (93)

k2 = hf(xn +h

2, yn +

k1

2) (94)

yn+1 = yn + k2 +O(h3). (95)

As inidicated in the error term, this symmetrization can-cels out the first order error term, making the methodsecond order. Note that the mathod is called n-th orderif its error term is O(hn+1).

Even more elaborate, and better, is the fourth-orderRunge-Kutta formula:

k1 = hf(xn, yn) (96)

k2 = hf(xn +h

2, yn +

k1

2) (97)

k3 = hf(xn +h

2, yn +

k2

2) (98)

k4 = hf(xn + h, yn + k3) (99)

yn+1 = yn +k1

6+k2

3+k3

3+k4

6+O(h5). (100)

This is usually referred to as “Runge-Kutta,” or “RK”method.

III. OSCILLATIONS

A. Single Harmonic Oscillator

A system described by the coordinate x. Equilibriumposition x0 so that

F0 =dU

dx

∣∣∣∣x=x0

= 0. (101)

7

We choose a coordinate system so that the minimum isat the origin: x0 = 0: Expansion of the potential energyaround the minimum:

U(x) = U0 +12d2U

dx2

∣∣∣∣x0

x2 +O(x3). (102)

Since we use an expansion around a local minimum[rather than maximum], we have

d2U

dx2

∣∣∣∣0

= k > 0. (103)

We thus get for a linear restoring force:

F (x) = −kx. (104)

We have the equations of motion:

x+ ω20x = 0, (105)

where the angular frequency is

ω20 =

k

m. (106)

A soution is

x(t) = A cos(ω0t− α) (107)

where A is the amplitude and δ is the phase. The totalenergy is:

E = T + U =12kA2. (108)

Isotropic harmonic oscillator in two dimensions:

~F = −k~r. (109)

The equations of motions are:

x+ ω20x = 0 (110)

y + ω20y = 0 (111)

The solutions are:

x((t) = A cos(ω0t− α), y(t) = B cos(ωt− β). (112)

Plot y vs. x, closed orbits, shape depends on the phasedifference δ = α− β. In general, ωx 6= ωy. Closed orbitsif the frequencies are rational:

ωxωy

=k

l. (113)

The orbits are called Lissajous figures.

B. Damped Oscillations

Damping proportional to the velocity: f = −βv =−bx. The equations of motion then are

x+ 2βx+ ω20x = 0. (114)

The systems is autonomous: there is no external force.The equation is second-order linear homogeneous difffer-ential equation: Solution: use an ansatz in the form ofan exponential function:

x(t) = Aeλt. (115)

Inserted in the equation of motion gives the characteristicequation:

λ2 + 2βλ+ ω2 = 0. (116)

The solutions are:

λ± = −β ±√β2 − ω2

0 . (117)

The solution is:

x(t) = A+eλ+t +A−e

λ−t. (118)

We have two unknowns A± since it is a second-orderdifferential equation. Physically, this matches that wecan choose the position and the velocity at time zero:

x0 = A+ +A−, v0 = λ+A+ + λ−A−. (119)

Three cases:

1. Underdamped

We have

ω20 > β2 (120)

so that √β2 − ω2

0 = i√ω2

0 − β2 = iω1, (121)

with ω1 ∈ <. The solution is exponentially damped os-cillation:

x(t) = e−βt[A+e

iω1t +A−e−iω1t

], (122)

This is usually written as

x(t) = e−βt cos[ω1t− δ]. (123)

The envelope decays exponentially.

2. Overdamped

Now we have √β2 − ω2 = λ < β. (124)

The solutions is

x(t) = A+e−[β+λ]t +A−e

−[β−λ]t, (125)

that is the solution is exponentially decaying [no oscilla-tions - like good shock absorbers!]

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3. Critically damped

We have

β2 = ω20 (126)

so that the two eigenvalues are degenerate:

λ+ = λ−. (127)

However, we need to have two linearly independent solu-tions [because the equation of motion is a second-orderODE]. Easily confirm that the linear function is a solu-tion: x(t) = A+Bt: thus

x(t) = (A+Bt)e−βt. (128)

C. Physical Pendulum

For linear systems x+ ω20x = 0: if x1 and x2 are solu-

tions, then x(t) = x1(t) + x2(t) is also a solution. Thisimplies that the period does not depend on the ampli-tude. This is one of the characteristic properties of linearsystems.

For the physical pendulum, we have the equations ofmotion:

θ + ω20 sin θ = 0, (129)

where

ω20 =

g

l. (130)

Linear approximation: sin θ = θ.This is a conservative system, and thus can be inte-

grated exactly:

U(θ) = mgl(1− cos θ). (131)

For the kinetic energy

T =12ml2θ2. (132)

We assume that θ0. Thus the total energy is

E = mgl(1− cos θ0) = 2mgl sin2 θ0

2. (133)

We find the kinetic energy

T =12ml2θ2 = 2mgl

[sin2 θ0

2− sin2 θ

2

](134)

so that

θ = 2ω0

√sin2 θ0

2− sin2 θ

2. (135)

We find by integration

τ =2ω0

∫ θ0

0

dθ√sin2 θ0/2− sin2 θ/2

(136)

where we integrate from 0 to θ0 for a quarter period.We use the substitution:

z =sin θ/2sin θ0/2

=sin θ/2k

, (137)

where k = sin θ0/2. We get

τ =4ω0

∫ 1

0

dz√(1− z2)(1− k2z2)

. (138)

This is the elliptic integral of the first kind:

τ =4ω0F (sin θ0/2, 1). (139)

Time-dependence: the equation

ω0dt = ±12

dθ√k2 − sin2 θ

, (140)

has the solutions:

sinθ

2= k sn(ω0t) cos

θ

2= k dn(ω0t), (141)

where sn and dn are Jacobi Elliptic functions. See, e.g.F. Bowman, Introduction to Elliptic Functions with Ap-plications (Dover, NY, 1961).

D. Sinusoidal Driving Force

The net force is given by:

F = −kx− bx+ F0 cosωt. (142)

We assume that driving force is not at the natural fre-quency:

ω 6= ω0 =

√k

m. (143)

The equation of motion follows:

x+ 2βx+ ω20x = A cosωt. (144)

This is an inhomogeneous equation. The solution of thehomogeneous equation is transient, i.e., it will vanish forlong times and we only see the particular solution:

xp(t) = D cos [ωt− δ] . (145)

Inserted in to the equation of motion:A−D[(ω2

0 − ω2) cos δ + 2ωβ sin δ]

−D[(ω20 − ω2) sin δ − 2ωβ cos δ] = 0 (146)

We get for the phase:

tan δ =2ωβ

ω20 − ω2

(147)

9

and the amplitude:

D =A√

(ω0 − ω)2 + 4ω2β2. (148)

That is both the phase and the amplitude are frequencydependent. The amplitude is large when the driving fre-quency is close to the natural frequency:

limω→ω0

D(ω) =A

2ωβ(149)

E. Arbitrary Driving Force: Fourier Series

In general, we seek a “representation” of a functionf(x) defined along the x-axis:

f(x), −∞ < x <∞. (150)

As it turns out, this vector space has infinite dimension-ality - this is a difficult problem. So we first consider asomewhat simpler problem: We consider a function inthe finite interval x ∈ [−L,L] and then periodically con-tinue it for x /∈ [−L,L]. We need a set of basis ‘vectors.’Note that the sine- and cosine function satisfy an orthog-onality relation:

φn(x) =

sin(nπx/l)cos(nπx/L) , n = 1, 2, 3, ... (151)

Then, ∫ L

−Lφnφmdx = Lδnm (152)

where δnm is the Kronecker symbol. Also, sine and cosineare orthogondal. This suggests that the sine- and cosinefunctions are the basis ‘vectors:’

f(x) =a0

2+∞∑n=1

[an cos

(nπxL

)+ bn sin

(nπxL

)], (153)

where the ‘components’ are obtained by taking the innerproduct:

an =1L

∫ L

−Lf(x) cos

(nπxL

)(154)

bn =1L

∫ L

−Lf(x) sin

(nπxL

)(155)

If the function is even, f(−x) = f(x), bn = 0 and if oddf(−x) = −f(x), an = 0.

Theorem by Dirichlet: If f(x) is periodic of period 2πand if between −π and π it is single-valued, ha a finitenumber maximum and minimum values, and a finitenumber of discontinuities and if

∫ π−π |f(x)dx < ∞ the

Fourier series converges to f(x) at all points where f(x)

is continuous; at jumps the Fourier series converges tothe midpoint of the jump.Two examples:

.

π 2π 3π 4π−π x

f1(x)

The Fourier series is

f1(x) = 2∞∑n=1

(−1)nsin(nx)n

, (156)

.

π 2π 3π 4π−π x

f2(x)

The Fourier series is

f2(x) =4π

∞∑n=1

sin(nx)n

. (157)

π 2π 3π 4π−π x

f3(x)

The Fourier series is

f3(x) =4π

∞∑n=1

(−1)nsin(nx)n2

. (158)

We recall that∞∑n=1

1n2

<∞ (159)

10

i.e., it converges and the function f3 is continuous. Notehowever that the first derivative has a discontinuity

df3(x)dx

= f2(x). (160)

For the Fourier series representation, we get

df3(x)dx

=∞∑n=1

(−1)ncos(nx)

n. (161)

Parseval’s Theorem: It relates the average square ofthe function

P =1

2π

∫ π

−π|f(x)|2dx. (162)

Here, we use P since this is the power in the “time-domain.” Since the average of sin(nx) and cos(nx) is1/2, we get

P =a2

0

4+

12

∞∑n=1

[a2n + b2n

]. (163)

This yields the power spectrum:

.

Application: f(x) = x for −π < x < x. Then:

f(x) =2π∼∞n=1 (−1)n

sin(nπx)n

. (164)

Parseval’s theorem then gives:

13

=2π2

∞∑n=1

1n2. (165)

This gives the result:

∞∑n=1

1n2

=π2

6. (166)

Linear vector space L2: We have a finite Fourier sum[not series].

fN (x) =a0

2+

N∑n=1

[an sin

(nπxL

)+ bn cos

(nπxL

)].

(167)This corresponds to a function defined on a lattice

xi = −L+ iL

N, i = −N,−N + 1, ...,−1, 0, 1, 2, ..., N

(168)

We have the basis vectors:

|i〉 =

0...0i0...

, (169)

These basis vectors are orthogonal:

〈j|i〉 = δij . (170)

For the vector of the values of the function f(x):

|f〉 =

f0

f1

...fi−1

fifi+1

...

, (171)

We then write:

fi = 〈i|f〉 (172)

The vector |f〉 can be written as a linear superposition:

|f〉 =2N+1∑i=0

fi |i〉 =2N+1∑i=0

〈i|f〉 |i〉

=2N+1∑i=0

|i〉 〈i|f〉 =

(2N+1∑i=0

|i〉 〈i|

)|f〉 (173)

We thus have |f〉 = I · |f〉, where

I =2N+1∑i=0

|i〉 〈i| (174)

Of course, this is simply the (2N + 1) × (2N + 1) unitmatrix; we say that the set of basis vectors |i〉2N+1

i=0 iscomplete.

We write φn(x) = sin(πx/L), cos(nπx/L). The func-tions are orthogonal:

〈φn|φm〉 =∫φn(x)φm(x)dx = δnm. (175)

We write the Fourier series on the lattice:

〈i|f〉 = f(xi) =2N+1∑n=0

anφn(xi) =2N+1∑n=0

an 〈i|φn〉 , (176)

where

an =∫φn(x)f(x)dx = 〈φn|f〉 . (177)

11

The functions are orthogonal:

〈φn|φm〉 =∫φn(x)φm(x)dx = δnm. (178)

We thus have

|f〉 =2N+1∑i=0

|i〉 〈i|f〉

=2N+1∑i=0

|i〉2N+1∑n=0

an 〈i|φn〉 (179)

Now we exchange the order of taking the series:

|f〉 =2N+1∑n=0

an

2N+1∑i=0

|i〉 〈i|φn〉

=2N+1∑n=0

an

(2N+1∑i=0

|i〉 〈i|

)|φn〉

=2N+1∑n=0

an |φn〉 , (180)

where we used∑2N+1i=0 |i〉 〈i| = 1. We now use an =

〈φn|f〉, to find

|f〉 =2N+1∑n=0

〈φn|f〉 |φn〉

=2N+1∑n=0

|φn〉 〈φn|f〉

=

(2N+1∑n=0

|φn〉 〈φn|

)|f〉 (181)

that is; we have the completeness of the Fourier basisfunctions:

1 =2N+1∑n=0

|φn〉 〈φn| (182)

These two completeness relations are usually written inthe following fashion:(i) Since 〈φn|φm〉 = δnm, we have

δnm = 〈φn|

(2N+1∑i=0

|i〉 〈i|

)|φm〉

=2N+1∑i=0

〈φn〉 〈i|φm〉

=2L

2N+1∑i=0

[sin(πnxi

L

)sin(πmxi

L

)+ cos

(πnxiL

)cos(πmxi

L

)](183)

(ii) Since 〈i|j〉 = δij , we have

δij = 〈i|

(2N+1∑n=0

|φn〉 〈φn|

)|j〉

=2N+1∑n=0

〈i|φn〉 〈φn|j〉

=1L

2N+1∑n=0

[sin(nπxi

L

)sin(nπxj

L

)+ cos

(nπxiL

)cos(nπxj

L

)]. (184)

IV. FOURIER TRANSFORM

What happens in the limit L→ that is, when the pe-riodicity goes to infinity:

A. Response of iinear Oscillators

V. GRAVITATION

The attractive force on mass m due to mass M [New-ton’s law of gravitation]

~F = −GmMr2

er, (185)

where r is the unit vector from M to m. If the object Mis extended with density ρ = ρ(~r), we get by integration

~F = −Gm∫V

ρ(~r′)er′r′2

dv′. (186)

The gravitational field is defined:

~g =~F

m= −GM

r2er = −G

∫V

ρ(~r′)r′2

dv′. (187)

Example: Rod of mass M and length L along x-axis:−L/2 < x < L/2. Find the gravitational field at (x, 0).[TM 5-7].The gravitational field due to the mass element between[x, x+ dx]. We have for λ = M/L,

dgx = −G λ

(x2 + y2)x√

x2 + y2dx (188)

dgy = Gλ

(x2 + y2)y√

x2 + y2dx (189)

Note that gx(−x) = −gx(x) [odd function) so that

gx =∫ L/2

−L/2dgx = 0. (190)

12

We have for gy:

gy = Gλy

∫ L/2

−L/2

dx

(x2 + y2)3/2

=2Gλy

∫ L/2y

0

du

(1 + u2)3/2. (191)

In the limit y >> L/2, we have 1/(1 +u2)3/2 ' 1 so thatusing M = Lλ for the total mass,

gy =2Gλy· L

2y=GM

y2, (192)

that is, we recover the inverse square law, as it should.In general, we have

qy =2Gλy

[u√

1 + u2

]L/2y0

=GM

y2

1√1 + (L/2y)2

. (193)

A. Gravitational Potential

The gravitational potential Φ is defined

~g = −∇Φ (194)

For a point mass M , where

Φ = −GMr

(195)

For a continuous mass distribution:

Φ = −G∫V

ρ(~r′)r

dv′ (196)

Example: Rod of mass M and length L along x-axis:−L/2 < x < L/2. Find the gravitational field at (x, 0).[TM 5-7].The gravitational field due to the mass element between[x, x+ dx]. We have for λ = M/L,

Φ = −Gλ∫ L/2

−L/2

dx√x2 + y2

= −2Gλ∫ L/2y

0

du√1 + u2

= −2Gλ ln

L2y

+

√1 +

(L

2y

)2 . (197)

For y >> L/2, we have in leading order

ln

L2y

+

√1 +

(L

2y

)2 ' L

2y,

so that

Φ = 2Gλ · L2y

= −GMy

, (198)

where M = Lλ.Spherical shell: a < r < b: For a point P at the radiusR from the center of the shell. The volume element isspherical coordinates

dV = r2 sin θdθdφ. (199)

choose z-axis through the point, and integrate with re-spect to φ. This gives

Φ = −2πGρ∫ b

a

r′2dr′∫ π

0

sin θr

dθ, (200)

where

r2 = R2 + r′2 − 2Rr′ cos θ. (201)

We get for r′ = const:

2rdr = 2r′R sin θdθ. (202)

Thus

sin θr

dθ =dr

r′R(203)

so that

Φ = −2πGρR

∫ b

a

r′dr′∫ rmax

rmin

dr (204)

We have for P outside the spherical shell:

rmin = R− r′ and max = R+ r′. (205)

and for P inside the spherical shell:

rmin = r′ −R and rmax = r′ +R. (206)

We thus have∫ rmax

rmin

dr =

2r′ P ∈ shell2R P 3 shell (207)

We get for P in the shell:

Φ = −43πρG

R(a2 − b2) = −GM

R, (208)

where M = (4π/3)ρ(a3− b3) is the mass of the shell. ForP inside the shell:

Φ = 2πρG(a2 − b2) = const. (209)

That is, for a point outside the shell, the gravitationalpotential of a shell is the same as the potential from apoint mass at the center.

13

Poisson equation:We calculate the surface integral:

Φm =∫S

~n · ~gda. (210)

where

~n · ~g = −GM cos θr2

. (211)

and θ is the angle between ~n and ~g. We find

Φm = −Gm∫S

cos θr2

da. (212)

We note that cos θda/r2 = dΩ is the element of the solidangle: thus

Φm = −Gm∫

4π

dΩ = −4πGm. (213)

Note that this answer does not depend on the shape ofthe surface. if we have several masses mi inside V :∫

~n · ~gda = −4πG∑

mi. (214)

or for a mass distribution:∫S

~n · ~gda = −4πG∫V

ρdv. (215)

Gauss’ theorem shows:∫S

~n · ~gda =∫V

∇ · ~gdv (216)

it follows that

∇ · ~g = −4πGρ. (217)

Since ~g = −∇Φ.

∇2Φ = 4πGρ. (218)

where ρ is the mass density. This is called In a regionwith no mass, we have ρ = 0, and the Laplace equation:

∇2Φ = 0. (219)

TM 5-16: A uniformly solid sphere of mass M and radiusR is fixed at a distanceh above a thin infinite sheet of massdensity σ. Find the force between sheet and the sphere.We choose the sheet in the (x, y)-plane. We use small ringsof thickness [r, r + dr] in the sheet. Repalce the sphere bya point mass. Then:

Fz = GM2πσ∫ ∞

0

cos θ rdrr2 + h2

= 2πσMh

∫ ∞0

rdr

(r2 + h2)3/2.

(220)

where we used cos θ = h/√r2 + h2. Then

Fz = −2πGMσh

[1

r2 + h2

]∞0

= 2πσGM. (221)

Alternative: Gaussian “pill box” of height 2h and area Asymmetric around sheet. Then∫

~g · nda = −4πG∑i

mi (222)

so that

2gzA = 4πGσA (223)

and

gz = −2πGσ = const. (224)

TM 5-21: A point mass m is located at a distance x fromthe nearest end of a thin rod of mass M and length L alongthe axis of the rod. Find the force on the particle m. Wehave dm = (M/L)ds for the We have

F =GMm

L

∫ x+L

x

ds

s2=GMm

L

[1

L+ x− 1x

]=

GMm

x(L+ x).

(225)Find the potential:

Φ =GM

L

∫ L+x

x

ds

s=GM

L[ln(L+ x)− lnx] . (226)

We assume that x < L. This corresponds to the case whenL→∞ and M →∞ such that λ = M/L = const. Then

Φ = −Gλ lnL

x(227)

Calculate how long it will take to reach the rod. We assumethat we start from rest at x0. Then

v2

2= −Gλ

[lnL

x0− ln

L

x

]= Gλ ln

x0

x(228)

Then

v =√

2Gλ lnx0

x(229)

We find for the time to reach the coordinate 0 < x < x0:

t(x) =∫ x

x0

dx′√2Gλ ln(x0/x′)

(230)

Subsitution u = ln(x0/x′) so that x′ = x0e

−u and dx′ =x0e

udu. Thus

t(x) =x0√Gλ

∫ ln(x0/x)

0

e−u√udu (231)

14

Substitution:√u = v so that du/

√u = 2dv. Then

t(x) =2x0√Gλ

∫ √ln(x0/y)

0

e−v2dv

=2x0√Gλ

2√π

erf

([lnx0

x

]1/2)(232)

where we used the error function. We assume that x < x0

so that

lnx0

yx> 1. (233)

For large arguments z:

erf(z) = 1− erfx(z) ' 1− 1√πze−z

2, (234)

so that

t(x) =2x0√Gλ

2√π

[1− 1√π√

ln(x0/x

x

x0

]. (235)

This shows that the time to reach the rod is finite:

tfinal = (236)

TM 5-13: A planet with density ρ1 (spherical core, radiusR1) with a thick spherical cloud of dust (density ρ2, radiusR2] is discovered. What is the force on a particle of massm placed inside the dust cloud?We assume R1 < r < R2. Then the force is

F =GM(r)m

r2, (237)

where M(r) is the mass inside the radius r:

M(r) =4π3R3

1ρ1 +4π3

(r3 −R31)ρ2. (238)

We get

F =4π3Gm

[(ρ1 − ρ2)R3

1

r2+ ρ2r

]. (239)

TM 5-14: Calculate the self-energy of a sphere with massM and radius R.Build the sphere from scratch: Assume that we have builta sphere of mass m and radius r:

m(r) = M( rR

)3

. (240)

The gravitational potential at the the surface:

φ = −Gmr. (241)

Thus the gravitational energy for a shell of thickness dr:

dU = φdm = φ3M

4πR34πr2dr = −3GM

R6r4dr. (242)

Thus, the total energy is

U =∫dU = −3GM

R6

∫ R

0

r4dr = −3GM5R

. (243)

VI. SOME METHODS IN THE CALCULUS OFVARIATION

Example: Lifeguard problem.A particle moves in the (x, y)-plane from (xi, yi) to(xf , yf ). Assume that yi > 0 and yf < 0. The parti-cle travels in piecewise straight paths as shown.

y

x

(xi,yi)

(xf,yf)

ξ v1

v2

The speed of particle is v1 and v2 in the upper- [y > 0] andlower half plane [y < 0], respectively. [You can imagine,for example, that a life guard sits at the beach at (xi, yi)and needs to rescue a drowning swimmer at (xf , yf ). Hewalks on the beach at the speed v1 and swims at thespeed v2 < v1].Assume that the particle travels through the intermedi-ate point (ξ, 0), and calculate the time of travel:

t(ξ) =

√(ξ − xi)2 + y2

i

v1+

√(xf − ξ)2 + y2

f

v2. (244)

We seek the minimum: dt/dξ = 0 so that

1v1

ξ − xi√(ξ − xi)2 + y2

i

− 1v2

(xf − ξ√xf − ξ)2 + y2

f

= 0 (245)

We have

ξ − xi√(ξ − xi)2 + y2

i

= sin θi,(xf − ξ√

xf − ξ)2 + y2f

= sin θf

15

y

x

(xi,yi)

(xf,yf)

ξ v1

v2

θi

θf

so that we recover Snell’s law:

1v1

sin θ1 =1v2

sin θ2. (246)

For mechanical systems, the action is defined by

S(x, t|x0, t0) =∫ t

t0

Ldt, (247)

where

L(x, x) = T − V (248)

is the Lagrangian (function).We choose a midpoint (ξ, τ) for fixed time τ :

S(ξ) = S(x, t|ξ, τ) + S(ξ, τ |x0, t0). (249)

This is a convex function: d2S/dξ2 > 0. We find thatthis function has global minimum at the coordinate cor-responding to the equation of motion:

dSdξ

= 0 −→ ξ = x(τ). (250)

Free Particle: The action is

S(xf , tfxi, ti) =m

2(xf − xi)2

tf − ti(251)

ttfti

xi

xf

x

τ

ξ

We thus have for tf = t and ti = 0:

S(ξ) =m

2

[(xf − ξ)2

t− τ+

(ξ − xi)2

τ

](252)

We find

S(ξ) =m

2t

(t− τ)τ

[x2f

τ

t+ x2

i

t− τt−

2ξ(xfτ

t+ xi

t− τt

)+ ξ2

]=

m

2

[ξ −

(xfτ

t+ xi

t− τt

)]2

+m

2(xf − xi)2

t.

We thus have a minimum of the action for

ξmin = xfτ

t+ xi

t− ττ

= xi +xf − xi

tτ = xcl(τ). (253)

This suggests that the solution of the equation of motionminimizes the action.

Harmonic Oscillator: Without loss of generality, we setti = 0 and tf = τ . Then we have for the coordinate as afunction of time:

x(t) =xf − xi cosωτ

sinωτsinωt+ xi cosωt. (254)

and for the velocity:

v(t) =ω[xf − xi cosωτ ]

sinωτcosωt− ωxi sinωt. (255)

We get for the Lagrangrian: We thus find for the actionfor a harmonic oscillator with (angular) frequency ω:

S(xi, ti|xf , tf ) (256)

=mω

2 sinω(tf − ti)[(x2i + x2

f ) cosω(tf − ti)− 2xixf].

16

Choose an intermediate time ti < τ < tf , and assumethat the particle is at the coordinate ξ: x(τ) = ξ. Calcu-late the action:

We get

S(ξ) =mω

2 sinωτ · sinω(t− τ)[x2i cosωτ sinω(t− τ)

+x2f cosω(t− τ) sinωτ

+ξ2 sinωt− 2ξ xi sinω(t− τ) + xf sinωτ]

We thus see that S(ξ) is a parabola so that we have asingle global minimum,

dS

dξ

∣∣∣∣xmin

= 0 −→ xmin(τ) = xisinω(t− τ)

sinωt+xf

sinωτsinωt

(257)Since sinω(t− τ) = sinωt cosωτ cosωt sinωτ , we get

xmin = xi cosωτ +[xf − xi cosωt

sinωt

]sinωτ. (258)

That is, the minimum xmin(τ) coincides with the solutionof the harmonic oscillator!

A. Hamilton’s Principle - Lagrangian andHamiltonian mechanics

We assume that we can describe the different path bya parameter α:

x(t) = x(αt) = x(0, t) + αη(t) (259)

Then for the velocity:

x = x(0, t) + αdη

dt. (260)

We seek the minimum (extrema) of the action:

∂S

∂α= 0. (261)

We write

∂S

∂α=∫ (

∂L

∂x

∂x

∂α+∂L

∂x

∂x

∂α

)dt. (262)

Since

∂x

∂α= η

∂S

∂x∂α =

dη

dt, (263)

We have

∂S

∂α=∫ (

∂L

∂xη(t) +

∂L

∂x

dη

dt

)dt. (264)

Integration by parts of the second term yields:

∂S

∂α=∫ (

∂L

∂x− d

dt

∂L

∂x

)η(t) dt = 0, (265)

We have δ∫ t2t1L(x, x)dt = 0 which gives Lagrange

equations of motion:

∂L

∂x− d

dt

∂L

∂x= 0. (266)

Generalized Coordinates: We use notation qi then

L = L(qiqi). (267)

We only consider autonomous systems so that L is not afunction of time. Atwood’s machine: We have y1 = yand y2 = l − y so that y1 = y and y2 = −y. Then

T =m1 +m2

2y2

and the potential energy

U = m1gy1 +m2gy2 = m2gl + (n1 −m2)gy.

The Lagrangian is thus

L(y, y) =m1 = m2

2y2 − (m1 −m2)gy.

now for the Euler-Lagrange equations:

∂L

∂y= −(m1 −m2)g

and

∂L

∂y= (m1 +m2)y.

The Euler-Lagrange equation are thus

d

dt(m1 +m2)y + (m1 −m2)g = 0 (268)

or

y = −m1 −m2

m1 +m2g.

That is, the tension in the rope does not appear only“physical” forces derived from a potential energy. Ten-sion is derived from the constraint

y1 + y2 = l = const (269)

This a holonomic constraint. If it involves velocities[rolling] it’s called non-holonomic. Method of Lagrangianmultiplier.

Ball in a bowl: Use polar coordinates (r, θφ) definedby

x = r sin θ cosφy = r sin θ sinφz = r cos θ.

Then for the velocities:

x = rθ cos θ cosφ− rφ sin θ sinφy = rθ cos θ sinφ+ rφ sin θ cosφz = −rθ sin θ.

17

Thus the kinetic energy is

K =m

2(x2 + y2 + z2

)(270)

=m

2r2(θ2 + φ2 sin2 θ

). (271)

and the potential energy:

U = mgz = mgr cos θ (272)

The Lagrangian follows

L = L(θ, φ, θ, φ) =m

2r2(θ2 + sin2 θ φ2

)−mgr cos θ.

(273)We have

∂L

∂θ= mr2φ2 sin θ cos θ +mgr sin θ (274)

∂L

∂φ= 0 (275)

∂L

∂θ= mr2θ (276)

∂L

∂φ= mr2 sin2 θφ. (277)

We thus get from the Euler Lagrange equation

d

dt

(mr2θ

)−mr2φ2 sin θ cos θ −mgr sin θ = 0(278)

d

dt

(mr2 sin2 θφ

)= 0(279)

Thus

mr2 sin2 θφ = const (280)

This is the generalized momentum, here the angular mo-mentum.

Example: Two point masses m1 and m2 6= m1 areconnected by a string of length l passing through a ahole in a horizontal table. There is no friction betweentable and the mass m1.

m1

m2

a) What is the initial velocity of m1 so that m2 remainsmotionless without friction?We have

m1v2

l − d= m2g −→ v =

√m2

m1(l − d)g ≡ v0. (281)

b) If m2 is slightly displaced in vertical direction, smalloscillations ensue. Use Lagrange’s equation to find thefrequency.

We have the Lagrangian in terms of (r, θ):

L =m1

2

(r2 + r2θ2

)+m2

2r2 +msg(l − r). (282)

We get the Lagrange’s equations:

m1r2θ = const (283)

(m1 +m2)r −m1rθ2 +m2g = 0 (284)

Initial condition at t = 0: r = l − d and v = v0. Then

θ0 =v0

l − d=√m2

m1

g

l − d. (285)

We thus have

m1r2θ2 = m1(l − d)θ2

0 = m1

√m2

m1(l − d)3g (286)

so that

rθ2 =r4θ2

r3=m2

m1

(l − dr

)3

g. (287)

Inserted into Eq. () yields:

(m1 +m2)r −m2

(l − dr

)3

g +m2g = 0. (288)

18

We now make the assumption of small oscillations:

r = (l − d) + ρρ

l − d<< 1. (289)

Thus r = ρ and

r−3 = (l−d)−3

(1 +

ρ

l − d

)−3

' (l−d)−3

(1− 3

ρ

l − d

).

(290)Then the equation of motion for the radius becomes:

ρ+3m2g

(m1 +m2)(l − d)ρ = 0, (291)

so that ρ oscillates around ρ = 0 with frequency

ω2 =3m2g

(m1 +m2)(l − d). (292)

B. Canonical Equations of Motion - HamiltonDynamics

Lagrange equations of motion are second order in time.In mathematics [see, eq, Boas] we never integrate secondorder differential equations. We convert to coupled firstorder differential equations.

We define generalized momentum:

pi =∂L

∂qi. (293)

The Hamiltonian is defined vi a a Legendre transform toeliminate the velocity:

H = H(pi, qi) =∑i

pi · qi − L. (294)

The equations of motion then follow:

qi =∂H

∂pi, pi = −∂H

∂qi. (295)

VII. CENTRAL FORCE MOTION

A. Reduced Mass

In one-dimension: we have mass m1 at x1 and massm2 at coordinate x2. We assume a central force, i.e.,

F (x1, x2) = F (x1 − x2). (296)

The equations of motion then follow

m1x1 = F (x1 − x2) (297)m2x2 = −F (x1 − x2) (298)

Then

m1x1 +m2x2 = 0 (299)

so that

m1x1 +m2x2 = (m1 +m2)vcm = const. (300)

That is the center of mass (CM) moves with constantvelocity:

xcm =m1x1 +m2x2

m1 +m2= vcmt+ xcm,0 (301)

Barycentric coordinates: xi = x′i + xcm with

x′1 =m2

m1 +m2(x1 − x2) (302)

x′2 =m1

m1 +m2(x2 − x1). (303)

Since x′i = xi, we get

m1m2

m1 +m2

d2

dt2(x1 − x2) = F (x1 − x2). (304)

This define the reduced mass µ = m1m2/(m1 +m2) or

1µ

=1m1

+1m2

. (305)

If m1 >> m2 [e.g., sun and a planet] then

M ' m1 µ ' m2. (306)

This is correct in two and three dimensions as well.

B. Conservation Theorems

For central potential

~F (~r) = F (r)er. (307)

Thus the torque is zero:

τ = ~r × ~F = 0 (308)

and the angular momentum is constant

~L = ~r × ~p = const. (309)

This imlies that the motion is in the plane perpendicularto ~L: polar coordinates (r, θ). The Lagrangian is

L = L(r, θ, r, θ) =12m(r2 + r2θ2)− U(r). (310)

Since ∂L/∂θ = 0, we have conservation of angular mo-mentum:

pθ =∂L

∂θ= mr2θ = l = const. (311)

We have the area dA = r ·rdθ/2 so that the areal velocityis

dA

dt=

12r2θ =

l

2m. (312)

19

That’s Kepler’s second law.The conservation of energy yields: E = T + U , or

E =m

2r2 +

12l2

mr2+ U(r). (313)

This defines the “effective” potential:

Ueff = U(r) +12l2

mr2(314)

The second term is the “centrifugal” potential.

C. Equations of Motion

We get the equation of motion for the radius:

r =

√2m

[E − U(r)]− l2

m2r2. (315)

Since dθ = (θ/r)dr, we find the trajectory:

θ(r) =∫

±l/r2√2m[E − U ]− l2/2mr2

dr. (316)

The equations of motion along the radius. From

d

dt

∂L

∂r− ∂L

∂r= 0, (317)

we get

m(r − rθ2) =∂U

∂r= F (r). (318)

We write

u =1r

(319)

then

du

dθ= − 1

r2

r

θ= −m

lr, (320)

where we used θ = l/mr2. For the second derivative:

d2u

dθ2= −m

2

l2r2r. (321)

We then have

r = − l2

m2u2 d

2u

dθ2(322)

rθ2 =l2

m2u3 (323)

We thus have

d2u

dθ2+ u = −m

l21u2F (1/u). (324)

For the inverse-square law

F (r) = − k

r2(325)

We get

d2u

dθ2+ u = +

km

l2≡ 1α

(326)

With solution

u(θ) = A cos θ +1α

(327)

We thus have r−1 = α−1(1 + ε cos θ with A = ε/α, or

r =α

1 + ε cos θ. (328)

We have r = 0 for θ = 0 [or θ = π]. Then for the energy:

E =12

l2

mα2(1 + ε)2 − k (1 + ε)2

α

2

= (329)

So that

ε =

√1 +

2El2

mk2. (330)

For 0 < ε < 1, we get an ellipse with axis:

a =α

1− ε2, b =

α√1− ε2

. (331)

We get for the perhelion and aphelion radii:

rmin = a(1− ε) =α

1ε, rmax = a(1 + ε) =

α

1 + ε(332)

Since dA/dt = l/2m, we have for the period

T =2mlA =

2ml

(333)

The area of an ellipse is A = πab so that

T =2lπ

l

2|E|· √

2m|E|. (334)

Since α = l2/mk, we get

T 2 =4π2m

ka3, (335)

which is Kepler’s third law. We have k = Gm1m2 andm−1 = m−1

1 +m−12 so that

T 2 =4π2a3

G(m1 +m2). (336)

Planet hopping: We assume that the planets have cir-cular orbits r1 and r2 > r1. We have for circular motionmv2/r = k/r2 so that T = mv2/2 = +k/2r. The total

20

energy thus follows E = T + U = +k/2r − k/r = −k/2rThis is also correct for elliptical trajectories. The satellitemoves at speed v1 at an orbit with radius r1:

E = − k

2r1=

12mv2

1 −k

r1, (337)

so that

v1 =√

k

mr1. (338)

We seek an elliptical trajectory between the circular or-bits:

2a = r1 + r2 (339)

We need to add a boost to satellite:

v1,t > v1. (340)

Calculate the boosted speed using conservation of energy:

Et = − k

2a= − k

r1 + r2=

12mv2

1,t −k

r1. (341)

We then get

v1,t =

√2kmr1

(r2

r1 + r2

). (342)

Thus the necessary boost is

∆v = v1t − v1 (343)

Similar near the second planet:

v2 =√

k

mr2, v2t =

√2kmr2

(r2

r1 + r2

). (344)

Perihelion Precession The orbit of Mercury showsnoticeable deviations from the predictions using Kepler’slaws; in particular, the perihelion precesses around thesun at a rate of 40 arc-seconds per century. A full de-scription requires the Schwarzschild space-time geome-try. In the lowest approximation, this can be describedin terms of an additional potential energy to Newtoniangravity V (r) = −GMm/r;

V (r) = −GMm

r+

ε

r2,

where ε is a small constant and M is the mass of the Sun,and m is the mass of mercury.Calculate the rate of precession of the planetary perihe-lion, to lowest order in ε. Assume that the orbit is nearlycircular [that is, the eccentricity is small].

The total angular momentum is L = mr2θ. Then

E =12mr2 +

L2

2mr2+ V (r).

We write

r =dr

dθ

dθ

dt= θ

dr

dθ.

Thus

dr =

√2m(E − V )− L2

r2· r

2

Ldθ

so that

θ =∫

Ldr

r2√

2m(E − V )− L2/r2+ const.

For V = V0 the orbit is an ellipse which is a closed orbitso that ∆θ = 0 as r varies from rmin to rmax. Thus

∆θ = 2∫ rmax

rmin

Ldr

r2√

2m(E − V )− L2/r2

= −2∂

∂L

∫ rmax

rmin

√2m(E − V )− L2

r2dr. (345)

Now write V = V0 +V1 and use a Taylor series expansionfor the integrand:

f(V ) = f(V0 +(∂f

∂V

)V0

V1 + ... (346)

The zeroth-order term gives ∆θ = 2π, and the first termis

∆φ =∂

∂L

∫ rmax

rmin

2mV1(r)dr√2m[E − V0(r)]− L2/r2

.

Somewhat “miracleously,” this integral can be computedexactly:

r = θdr

dθ=

L

mr2

dr

dθ

so that

1m

√2m[E − V0(r)]− L2

r2=

L

mr2

dr

dθ.

The last integral can thus be written:

∆θ =∂

∂L

[2mL

∫ π

0

r2V1(r)dθ]

Since V1 = ε/r2, we find

∆θ =∂

∂L

[2mL

∫ π

0

εdθ

]= −2πεm

L2.

21

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April 20, 2010

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22

SectionDynamics of a System of Particles

D. Center of Mass

E. Linear Momentum of the System

F. Angular Momentum of the System

G. Energy of the System

MathCad: Animal Locomotion

H. Elastic Collision of Two Particles

I. Systems with varying Mass

Rockets....sectionMotion in a Noninertial Reference Frame

J. Rotating Coordinate System

K. Centrifugal and Coriolis Forces

L. Motion Relative to the Earth

Foucault Pendulum.

VIII. DYNAMICS OF RIGID BODIES

A. Simple Planar Motion

B. Inertia Tensor

C. Angular Momentum

D. Principle Axes of Inerta

E. Moment of Inertia for Different BodyCoordinate Systems

F. Euler’s Equation for a Rigid Body

IX. COUPLED OSCILLATIONS

A. Two Coupled Harmonic Oscillators

As Homework....

[1] S. T. Thornton and J. B. Marion, Classical Dyanmics ofparticles and Systems (Thomson Brooks-Cole, Bedford,

CA).