PHYSICS COMPREHENSIVE REVIEW - USNA · 2015-04-16 · Fields Test Example Questions Guide Physics Handout Series – fields.tank with jeff and jeff page 2 CAUTION: This guide covers

Fields Test Example Questions Guide

Physics Handout Series – fields.tank with jeff and jeff page 1

PHYSICS COMPREHENSIVE

REVIEW By Larry, Jeff and Jeff

Add thin film optics

Not for circulation outside USNA



CAUTION: This guide covers 35 problems that will not appear on your Fields Test. Also under

stand that the guide makes statements that are usually true without adding qualifying remarks or

apologies. Advanced or special cases are not treated. In some cases, one or two steps are omitted.

Solve each problem as you read through this guide.

STRATEGY: The best approach is to recognize the problem and solve it. If you are unsure, find

physical reasons to eliminate as many of the five options as possible. Make an educated guess as you

pick from among the remaining options. Mark the problem so that you can return to it later at which

time you can attempt a more thorough analysis. As you work through the guide, pay special attention

to the examples for which options are found not to be viable.

DO NOT MEMORIZE: Memorizing values and facts from this guide is not recommended. Do

think about the concepts and orders of magnitude as you read this guide. Think and read just

carefully enough to be left with a hazy recollection of the experience. THINK, think, think while you

take the test.

The sample questions appear boxed. Search for “Ex.” to find the next problem.

MECHANICS AND RELATIVITY:

Newton’s Laws are best written for particle mechanics. The laws are written in forms appropriate for

inertia reference observes. A particle is a body for which every point in the body has the same

velocity. If one considers a rotating billiard ball, there is no one velocity and hence no one

acceleration for all the mass points comprising the ball. The center-of-mass theorems are the bridge

that allows one to generalize Newton’s laws to forms that apply to more general bodies.

I.) A particle maintains its state of uniform motion unless acted upon by an (external) force.

Uniform motion is motion at a constant velocity v

.



II.) The net force that acts on a particle causes that particle to accelerate. The net force is the vector

sum of all the forces that are acting on the particle, and that net force is equal to the mass of that

particle times its acceleration. (The word causes is of cosmic importance in physics.)

III.) If particle A exerts a force on particle B then particle B exerts a force of equal magnitude and

opposite direction on particle A. AB BAF F

Newton’s Third Law deals with forces and particles.

General Reciprocity (NOT Newton’s Law, an extension of Newton’s third law): If a physical

entity A exerts an influence on entity B, then B exerts a proportionate counter influence on A. For

example, an electric charge causes an electric field and an electric field exerts a force on electric

charges. A moving charge creates a magnetic field, and a magnetic field exerts forces on moving

charges. Note that the entities in these examples were not equivalent so the counter influences were

not equivalent. The counter nature is observed when a time-varying magnetic field induces a current

in a conductor. The magnetic field caused by the induced current fights (counters) the variation in

the magnetic field. The magnetic field in the interior of a perfect conductor must be constant in time

as the currents induced in such a conductor are as large as required to completely counter the

disturbing influence. Newton was only setting down laws for forces and particles, not for these

more complex situations.

Newton’s Laws: Use when asked about forces and acceleration at a single position and time.

Circular Motion : 22 ˆ ˆˆ ˆcirclev dvr dta r t r r r t

The centripetal part is always in play. The tangential part applies if the speed is changing.

Friction: Static friction acts with a magnitude from 0 to s N to keep the contacting surfaces from

slipping relative to one another. Once the surfaces are slipping, the friction force on each surface has

magnitude equal to k N directed to oppose the relative motion. Factoid: k s.

Kinematics, the description of motion, is always in play. It is used in all mechanics problems.



Use Newton’s laws to solve Here & Now problems (related to force and acceleration).

Work-Energy: Work is W = F dr

suggesting that work-energy by used when a change is

associated with a change in position.

Mechanical energy: Ei = K1 i + K2 i + … + UA i + UB i + …..

K = 2½ i iparticles i

m v = ½ McmVcm2 + ½ Icm2 = ½ McmVcm

2 + ½ mvrel 2 + ... = ....

U = ½ k (stretch)2 + {mgh; - GMm/r} + {q V; kQq/r} choose near-earth or Universal form choose appropriate form

Ei + Wnon-conserve = Ef Wnc-frintion = - k N *(distance that surfaces slip relative to one another)

A conservative is any force for which the work integral can be evaluated as the difference in the

values of a scalar function at the integration endpoints. (Can do if the work integral around any

closed path is zero; curl of the force is zero).

Use Work-Energy to solve Here to There (r) problems.

Momentum and Collisions: Impulse is 2

1

t

tF dt

suggesting that impulse-momentum methods be

tried for a change that occurs as time changes – even for the time change from before to after.

, , ,

tf

k k initial net external k k finaltim v F m v

Internal forces do not change the total momentum.

Use Impulse-Momentum to solve Now to Then (Before to After, t) problems.

(1) Collisions: Before to after or now to then problems

Rule 1: Conserve momentum vectorially (component by component).

Rule 2: Kinetic energy is unchanged for an elastic collision. Some kinetic energy is converted to

other forms if the collision is partially or totally inelastic. As the total momentum is conserved, only

the kinetic energy associated with motion relative to the center of mass is available to be lost

(converted).

2 2 2,½ ½ ½total i i i relative i CM relative

i i

K m v M V m v K K ; 2

2CMtotalP

MK



M: total mass; V: speed of the center of mass (CM); vrelative, i : speed of mass i relative to CM.

In a totally inelastic collision, the particles stick together so Krelative is lost. KCM remains as required

by conservation of momentum.

Ex. 1) The figure to the left shows two

particles with masses and velocities as

indicated. The objects are moving on a flat,

frictionless surface. When they collide, the

objects stick together. Their speed after the

collection is most nearly:

(A) 0.67vo

(B) 0.87vo

(C) vo

(D) 1.15 vo

(E) 1.73vo

Type: Before to After impulse-momentum method collision sub-type

Rule 1 is to conserve momentum vectorially so a coordinate system is adopted.

300

300

m, 2 vo

2 m, vo



The momentum of the particles before the collision are 1 03

2ˆ ˆ(2 )[ ]½p m v i j

and

2 03

2ˆ ˆ(2 ) [ ]½p m v i j

. The total mass is 3 m and the total momentum is 0

32

ˆ4 [ ]totP m v i

.

0 03

2ˆ4 [ ] 2 ˆ

3 3total

finaltotal

P mv i vv i

M m

(inelastic; stick together) D

Was kinetic energy converted to other forms as a result of the collision?

The collision was totally inelastic. Was all the kinetic energy lost?

(2) Motion in a uniform gravitational field:

Ex. 2) Ball 1 is dropped from a height h and ball 2 is dropped form a height ½ h. Which of the

following gives the ratio of the speed of ball 1 to that of ball 2 just before they impact? (Assume that

air resistance is negligible.)

(A) 2 (B) 2 ½ (C) 1 (D) 2-½

(E) ½

First, the problem is a constant acceleration problem. Write down the master equations.

x(t) = xo + vo t + ½ a t 2; v(t) = vo + a t ; v 2 = vo2 + 2 a (x – xo); vave = ½(v +vo)

300

300

m, 2 vo

2 m, vo

y

x



The balls fall from rest for a distance H. Using v 2 = vo2 + 2 a (x – xo), v

2 = 2 g H or 2v g H .

11

2 2

2

2 ½

g Hv h

v g H h = 2 B

Sanity Check. Three answers can be discarded immediately. Which three? Why?

If the answer were (A), ball 1 would have twice the average speed of ball 2. What would be the ratio

of the fall times in that case given that the first ball falls twice as far? Would you expect them to hit

at the same time if dropped at the same time?

(3) Law of Universal gravitation:

Ex. 3) Two planets of mass m and M respectively have center-to-center spacing R. At what distance

from the planet of mass M do the gravitational forces of the planets cancel each other.

(A) mM R (B) 1 m

M R (C) mM R (D) 1 m

M R (E) 1 mM

R

One need only consider points along the line joining the planets. Why? What must be true about the

directions of two vectors that sum to zero?

Prepare a sketch: A large well-drawn sketch provides the greatest benefit. Assume M > m.

The distance from M is sought. Assign the symbol d to that distance.

Thought 1: Gravitational forces are attractive. Therefore, the ‘zero – force’ point must be between

the planets and along the line joining them. Compare/contrast this situation with that in Ex. 10.

Thought 2: The force is an inverse square (with distance) law force.

2

2 2 2

( )

( )

G m G M R d m R d mMR d d d M d

R

d R - d

M m



Think about the result. At the point where their influences balance, the distance from the smaller

planet is smaller than that from the large planet. (Include sanity checks as you proceed through your

solution.) Assume m < M as you reason through the possibilities,

11

Rm mR d d R d dM M mM

E

Since mM is less than one, d > ½ R. The point is farther from the more massive planet. Reread

the question. The distance d of the point from the more massive M planet is requested. The unknown

label d is assigned to the value requested. Always review the question to ensure that you have

answered the question that was asked.

Assume m <<< M, say m = 0.01 M: Answers A, B, C and D correspond to points closer to M or to

points not between the planets. Why must the correct point be between the planets and be farther

from the more massive planet?

If you expect d to be greater than ½ R, you might revise the figure to reflect this expectation.

METHOD: Try a set of values that simplifies the problem: Midn Jasperson (2011) proposed that

one might just look at the case that m = M in which case the equilibrium point should be at the mid-

point of the line joining the masses, distance equal ½ R. Only response (E) satisfies that test.

(4) Statics and Archimedes Principle

In equilibrium, the sum of the forces acting on a rigid body is zero, and the sum of the torques (about

any axis free to choose it where you wish) acting on the body is zero.

A body is buoyed up by the weight of the fluid that it displaces. FB = fluid Vdisplaced g.

Ex. 4) A metal block is suspended in an empty tank from a scale that indicates a weight of W. The

tank is then filled with water until the block is covered. If the density of the metal is three times the

density of water, what apparent weight of the block does the scale now read?

(A) ½ W (B) 2/3 W (C) W (D) 3/2 W (E) 3 W



Examine your options. The new reading should be less than the original reading. Answer (B) could

be chosen without much more as the one third density means that the mass will be buoyed up by 1/3

of its weight if completely immersed, but one should solve the problem before choosing an answer.

Prepare drawings

Wdry = m V g = 3 w V g Wwet =m V g - w V g = (m - w) V g

* The ratio of the density of a material to that of water is called the specific gravity of the material.

Wwet = Wdry ( - w/m) = Wdry ( - /) B

(5) Simple Harmonic Motion

Equation of Motion: x(t) = A cos[ t + ] = C cos[ t ] + D sin[ t ]

Take derivatives to find expressions for v(t) and a(t).

v(t) = - A sin[ t + ]; a(t) = - 2 A cos[ t + ]

Energy Conservation: ½ m v2 + ½ k x2 = ½ m (vmax)2 = ½ k A2 = Etotal mechanical

With the convention that U = 0 when the particle is at the equilibrium position, the energy is all

kinetic as the particle passes through the equilibrium position and all potential at the turning points.

= stiffnessinertial property

gkm I

A mass hanging from a linear spring executes simple harmonic motion about its equilibrium

position. Potential energies are associated with systems of interacting entities. Potential energies are

associated with pairs of things (at least) or the system while a kinetic energy can be owned by a

mVg

T = Wwet

wVgmVg

T = Wdry



single entity. The potential energy can be shifted by an additive constant. For SHM, it is often set to

zero for a particle at the equilibrium position.

Ex. 5) A 1-kilogram particle is attached to a spring and exhibits one-dimensional simple harmonic

motion. The particle’s distance (!/* displacement) from the equilibrium position is given by the

expression: y(t) = A sin[ t + /2], where A = 1 meter and = 0.5 rad/s. If the potential energy of the

particle (system) at its equilibrium position is null (or zero), which of the following gives the total

energy of the particle (system)?

(A) 2 J (B) 1 J (C) ½ J (D) 1/8 J (E) 0 J

v(t) = - A sin[ t + ]; a(t) = - 2 A cos[ t + ]

vmax = A; amax = 2 A

½ m v2 + ½ k x2 = ½ m (vmax)2 = ½ k A2 = Etotal mechanical

The total energy is kinetic at equilibrium as the potential is set to zero at that point.

½ m (vmax)2 = ½ k A2 = Etotal mechanical

The relation v(t) = - A sin[ t + ] shows that vmax = A = 0.5 m/s. Using m = 1 kg,

Etotal = ½ m (vmax)2 = ½ (1 kg) (0.5 m/s)

2 = 1/8 J. D

Alternative: Etotal = ½ k A2 = ½ (m 2) A2 = ½ (1 kg) (0.5 rad/s)2 (1 m)2 = 1/8 J.

Write down the equation that is to be used. Substitute numerical values with units for each symbol

writing the values in the same geometric pattern that was used for the symbols. Adopt procedures

that reduce the chance of making a careless error.

(6) Equilibrium with torques

In equilibrium, the sum of the forces acting on a rigid body is zero, and the sum of the torques about

any axis ( free to choose axis where you wish) acting on the body is zero.

; sinr F r F r F r F

The displacement from the axis to the point of application of the force is r

, CCW is positive, and

is the angle that takes you from the direction of r

to the direction of F

.



Vector cross product: Extend the fingers of your right hand in along the direction of r

. Orient your

hand so that the fingers can curl toward the direction of F

. The thumb of the right hand will be in

the direction of r F

.

Ex. 6) The figure shows a uniform rod of mass 4 kilograms that is pivoted at one end and supported

by a string at the other end. If the rod is at rest (in equilibrium) the tension in the string is most nearly

(A) 20 N (B) 26 N (C) 31 N (D) 40 N (E) 62 N

Use the figure The rod does not have an assigned length. Assign a length of , 1 m or even 4 m. If

the value is not given, the answer must be independent of its value.

Choose an axis: The pivot point is a natural choice for the axis. Picture the situation. If the tension

were changed, about what point would the rod rotate?

The tension force tends to cause a CCW (+)

rotation so the associated torque about the

pivot is T = + T sin(400). To compute

torque, the weight of the boom can be

assumed to be applied at its center of mass.

The weight has a lever arm of ½ and tends

to cause a CW rotation. W = - W (½ ).

The sum of the torques must be zero so:

0 0 0

½ (4 ) (9.8 )½ ½

sin(40 ) sin(40 ) sin(40 )

NkgkgW m g

T = 30.5 N C

500

400

400½

W

T

Tr

500

axis r

F

line of action of F

r



½ (4 ) (9.8 ) 2028.3

.707 .707

Nkgkg N

N

lost calculator approximate using sine of 450; 40 < 45 so r

smaller and T larger for same = r

T

We expect an answer larger than 28 N so response (C) is chosen. The actual value is about 30.5 N.

(7) Relativity Relative speed v

Lengths are contracted along the direction of relative motion, but lengths running transverse to the

motion are unchanged. 21 ( ) ;vc

A clock in motion relative to the observer runs slowly. 21 ( )vc

tt

If a time t = 10 s elapses as measured by a moving clock, and it appears to be running slowly as

viewed by the primed observer, then the corresponding interval t is longer than 10 s.

Consider a primed observer moving at ˆv i relative to an unprimed observer with the initial

conditions that their respective axes are parallel and the origins coincided at t = t = 0.

Adopt the notations: 2 2and 1 ( ) 1v vc c .

1

[ ]] [c t c t x x x cty y

y y z z t t

Ex. 7) A stick of length L lies in the x-y plane as shown. An observer moving at 0.8 c in the x

direction measures the length of the stick. Which of the following gives the components of the

length as measured by the moving observer?

Lx

Ly

(A)

L cos 0.60 L sin

(B)

0.6 L cos 0.60 L sin

(C)

0.6 L cos L sin

(D)

0.64 L cos 0.64 L sin

(E)

0.78 L cos 0.78 L sin

L

x

y

O



The component along the direction of motion should be contracted by a factor of = 0.6 while the

transverse length should be unchanged.

Lx = 0.6 (L cos) ; Ly = Ly = L sin

Note that only answer (C) has an unchanged transverse component. C

The following relativity material is under development.

There are four vectors, and they all transform according to the same transformation rules (Assuming

the same ˆv i relative velocity and with 2and 1vc .).

0 0 1 2 2

1 1 0 3 3

[ ]

[ ]

V V V V V

V V V V V

For the position vector, x0 = ct, x1 = x, x2 = y and x3 = z. The four-momentum is: (p0, p1, p2, p3) =

(E/c, px, py, pz). Note the manner in which factors of c are used to ensure that all components have the same dimensions.

The transformations ensure that there is an invariant metric product.

2 2 2 2 2 2 2 2

0 1 2 3 0 1 2 3V V V V V V V V

Invariant interval: s2 = (c t)2 - (x)2 - (y)2 - (z)2 = (c t)2 - (x)2 - (y)2 - (z)2

This result is useful when one considers the interval between two events. If s2 = (c t)2 - (x)2 -

(y)2 - (z)2 is positive, the interval is called time-like and the two events can be causally related. If

s2 is negative the events are separated by a distance greater than that which light can travel in a

time |t|, and the interval is called space-like and the two events cannot be causally related. Points at

http://en.wikipedia.org/wiki/Light_cone

a time-like interval from the current event are in the future and past light

cones of that event and can linked causally to events inside the light cone.

Points at a space-like interval form the event are outside the light cone

and can not be causally related to the event at the apex of the cones.

Similarly: p2 = (E/c)2 - (px)

2 - (py)2 - (pz)

2 = (E/c)2 - (px)2 - (py)2 - (pz)2

When one considers the momentum of a single particle, the invariant p2 can be evaluated in the rest

frame of the particle with the result that 2 2 2

particlep m c in all frames where m is the mass of the

particle. The general expression for the moment of a particle is p = (v mc, v m v

) where v

is the

standard three velocity of the particle, m is the (rest) mass and 21 ( )vv v c

.



Using the invariant p2 = (E/c)2 - (px)

2 - (py)2 - (pz)

2 =m c2, E2 = m2c4 + c2 p p

= Eo2 + c2 p p

E2 = Eo2 + c2 (v m v)2 = Eo

2 + c2 (v Eo v/c2)2 = (v Eo)

2 E = v Eo. The kinetic energy of the

particle is the increase over the rest energy. K = E = E Eo = (v – 1) Eo.

Total energy: E = v Eo.

Kinetic energy: E = (v – 1) Eo 2 3

8

42½ ...v

cmv m (small v expansion; binomial)

Sample calculations:

In one frame, event B occurs 4 s after event A and event B is 1 km from A. In a second frame,

event B occurs 5 s after event A. a.) Could event A have caused event B? b.) What is the observed

distance d between the events in the second frame?

a.)s2 = (c t)2 - (x)2 - (y)2 - (z)2 = (3 x 108 m/s * 4 x 10-6 s)2 – (1000 m)2 = 440000 m2 > 0

The interval is time-like so event A could have caused event B.

b.) The interval is invariant sos2 = 440000 m2 = (3 x 108 m/s * 5 x 10-6 s)2 – d2 d = 1345 m.

An electron has a rest momentum (0.511 MeV/c, 0, 0, 0). a.) What is the electron’s energy when it is if

it is moving at 0.8 c relative to the observer? b.) Assuming that the electron is moving in the positive

x direction relative to the observer, what is its velocity if its energy is 1.25 * 0.511 MeV?

Using the invariant p2 = (E/c)2 - (px)

2 - (py)2 - (pz)

2 =m c2, E2 = m2c4 + c2 p p

= Eo2 + c2 p p

E2 = Eo2 + c2 (v m v)2 = Eo

2 + c2 (v Eo v/c2)2 = (v Eo)

2 E = v Eo

a.) E = v mc = v Eo = [1 – 0.82]-½ = 0.852 MeV.

b.) E = v mc = Eo [1 – 2]-½ = 1.25 MeV. 1 – 2 = 1/(1.25)2. = 0.6 so v = 0.6 c.

With the other assumptions, the particle is traveling at 1.8 x 108 m/s in the positive x direction.

Alternative solution: p = (v mc, v m v

) (0.511 MeV/c, 0, 0, 0) in the rest frame. Apply the

transformations: 0 0 1 1 1 0 2 2 3 3' [ ]; [ ], ' ; 'p p p p p p p p p p

a.) pc = [0.511 MeV/c - (0)] = c-1 Eo E = Eo ()

Attempt to solve b.) using the transformation method.



Particle Decay: Consider a particle of mass M decaying into a particle of mass m and a massless

particle. Find an expression for the magnitude of the momentum of the particle of mass m.

It is assumed that the original particle decays while at rest. Conservation of momentum requires that

m and the massless particle have oppositely directed, equal magnitude (p) momenta in the final state.

Energy conservation requires that: 2 2 4 2 2 2 20Mc m c p c p c .

2 2 4 2 2 2 4 3 22 2 24 2; 2Mc pc m c p c M c p c p cMpc m c

2 2

2 4 2 4 322

M m cM c m c Mpc p

M

?: What is the expression for the energy of a massless particle with momentum p?

ELECTROMAGNETISM AND CIRCUITS:

(8) Basic Circuits and Circuit Elements

The three basic passive elements are the capacitor, resistor and inductor. Each is characterized by its

voltage rule. VC =( 1/C) Q ; VR = I R ; VL = L

dI/dt (Note 1/C in the VC relation.)

Series Combos: 1 2

1 1 1

seriesC C C Rseires= R1 + R2 Lseries= L1 + L2

Parallel Combos: Cparallel = C1 + C2

1 2

1 1 1

parallelR R R

1 2

1 1 1

parallelL L L

Study the location C relative to those of R and L in the voltage relations. Note the inverse location

of C in the voltage relations and hence in the rules for series and parallel combinations.

The voltage across a capacitor is a continuous function of time. Why?

The current through an inductor is a continuous function of time. Why?

Kirchhoff’s voltage rule: The sum of the voltages (potential changes) around a closed path is zero.

Kirchhoff’s current rule: The algebraic sum of the currents into a node is zero.

Elements are in series if the wiring of the circuit requires that the current through the elements is

the same. There can be no branch points between them.



The voltage across a series combination is the sum of the voltages across the individual elements.

Elements are in parallel if the wiring of the circuit requires that the voltage across the elements is

the same. One lead from each goes to a common point and the other two leads from the elements

join at another common point (connection path to join of high conductivity).

The current through a parallel combination is the sum of the currents through the individual

elements.

SERIES PARALLEL

http://en.wikipedia.org/wiki/File:Parallel_circuit.svg

Elements are in parallel if the wiring of the circuit requires that the voltage across the elements is the

same. One lead from each goes to a common point and the other two leads from the elements join at

another common point (connection path to join of high conductivity).

Ex. 8) A capacitor of capacitance 125 microfarads is initially charged such that the voltage across

the plates is found to be 100 volts. If the capacitor is the connected in series to (with) a pure inductor

of inductance 0.2 Henry, what is the maximum value of the current that is observed in the inductor?

(A) 1.0 A (B) 2.5 A (C) 4.5 A (D) 5.0 A (E) 10.0 A

Prepare a sketch:



When the switch is closed, the two elements will be connected in series and parallel. The capacitor initially has a charge Q = C VC = (125 F)(100 V) = 12500 C.

KVR: Q/C + L dI

/dt = 0. Using I = - dQ

/dt,

2 22

2 2

10 compare 0

d Q d xQ x

dt LC dt

A simple harmonic oscillation at = (LC)- ½.

Q(t) = Qi cos[t + ]; I(t) = - Qi sin[t + ]

The current through an inductor is a continuous function of time. It is zero just before the switch is

closed so it is zero just after it is closed. I(t) = - Qi sin[] = 0 = 0 and Qi is just Q(t = 0) or

12500 F. The frequency is (LC)- ½ = (125 x 0.2 x 10-6 )- ½ = (25 x 10-6 )- ½ = 200 rad/s.

I(t) = - Qi sin[t + ] Imax = Qi = (12500 C) (200 rad/s) = 2,500,000 A. B

The potential drops as one moves down in both

elements. Note the pattern in which the static

charges accumulate on the inductor to cause the dI/dt . Starting at the lower left and proceeding CW,

the sum of the potential changes is Q/C – (L dI/dt) =

0.When the current is in the direction chosen as

positive for dI/dt, the charge Q decreases so I = -

dQ/dt. Better: Recall that an LC circuit is an

oscillator. 2

2

10

d QQ

dt LC

Alternative: maxmax max max

max

1 100500 2.5

0.2

dI V V AI I I Asdt L HLC

The total electric field is the sum of the electrostatic field and the circulating field S CE E E

.

+++ C 100 V

+Q

0.2 H -Q

2

2d QdI

dt dt

- - -

Switch closed at t = 0.

L

C 100 V

125 F 0.2 H



ˆ0S S cB

tE d E V E d ndA

Kirchhoff’s law states that the sum of the potential drops is zero. The potential drop between A and

B is 0B

SAE d

. The electric field in the integral is the electrostatic field only! A terminal to

terminal integration path can be adopted to avoid the regions with circulating field contributions.

C

100 V

+Q

-Q

0b c d a

S S S S Sa b c dE d E d E d E d E d

0 ( 0) ( 0) 0SQ

CE d LdIdt

ALTRNATIVE: Ex. 8) A capacitor of capacitance 125 microfarads is initially charged such that the

voltage across the plates is found to be 100 volts. If the capacitor is the connected in series to (with)

a pure inductor of inductance 0.2 Henry, what is the maximum value of the current that is observed

in the inductor? (A) 1.0 A (B) 2.5 A (C) 4.5 A (D) 5.0 A (E) 10.0 A

The energy initially stored initially in the capacitor is all stored in the inductor when the current is a

maximum. ½ L Imax2 = Qmax

2/2C. Imax = LC]-½ Qmax =(200 s-1) (1.25 x 10-2 C) = 2.50 A

Energy methods are more efficient when they work!.



Ex. 9) If the V is the potential difference between points I

and II in the diagram above and all three resistance have the

same resistance R, what is the total current between I and II?

(A) V/3R (B) 3 VR (C) 2V/3R (D) 3VR/2 (E) 3V/2R

What is not said? The lower conductor continues out of the field of view. The components illustrated

are embedded as a unit in some larger circuit.

R R

R

I II

Put the sub-assembly in a circuit and use Kirchhoff’s Laws.

Solution: Start with the voltage versus current relation: VR = I R. The voltage across a resistor is

equal to the current through it times its resistance. The first conclusion is that I = V/R for an

individual resistor with each symbol representing its value for that resistor.

1st Conclusion: Responses (B) and (D) have incorrect dimensions and so are eliminated.

2nd Conclusion: There is a current V/R in the lowest resistor. A smaller current exists in the upper

branch of two resistors. The currents are to be summed. V/R < I < 2 V/R. They make it rather easy

as only response (E) corresponds to a current greater than V/R. E

Related Techniques: When you encounter a network of passive elements, you should make series

and parallel combinations in sequence to reduce the network. The upper branch resistors are

R R

R

I II



combined in series (What is required for two elements to be in series?) to yield an effective

resistance of 2R. Then 2R and the R in the lower branch are then combined in parallel to yield a net

effective resistance of 2/3 R. If one is asked about the current, voltages, … for individual elements,

one should start with the fully collapsed network and find all the values. Next, back out one step (in

our case to R and 2R in parallel. Analyze to find all the values. Next, back out one more step and

analyze …. .

(10) Coulomb’s Law and Electrostatics

A physics major should know Coulomb’s law, the Biot-Savart law and Maxwell’s equations.

equation integral form differential form

Gauss’s Law 00

ˆV V

insideQE n dA dV

0

E

Faraday’s Law Â A

BtE d n dA

BtE

Gauss’s Law - Magnetism ˆ 0V

B n dA

0B

Ampere (Maxwell's 4th) 0 0 0 Â AEtB d J n dA

o o oEtB J

Charge Conservation (Continuity)

ˆV V

insidet

dQdtJ n dA dV

tJ

Lorentz Force Law q others q othersF q E v B

q others q othersF q E v B

/

3 3

2 2 20 00

( ) ( )ˆ ˆ

4 44( )

s

p ss s s s sp sp sp

sp spall r p s all rp s

r rr d r r d r qr r

r rr rr rE r

Coulomb for E

3

0 0 03 3 3

( )

4 4 4( )

s

s s p S s p S s s p S

p

all r p S p S p S

J r d r r r I d r r q v r r

r r r r r rB r

Biot-Savart B

Maxwell’s Equations Explicated

The Helmholtz theorem of vector calculus states that vector fields have two basic behaviors,

diverging and circulating. Maxwell has four equations.

' 4electric flux out divergence

Maxwell s ormagnetic circulation curl



The flux of a field through a surface is ˆF n dA

and in, Â

F n dA

is the net flux out of the field

out of a closed surface. In the field line picture, this is the number of field lines that start in the

enclosed volume. The net flux out of a small (infinitesimal) volume divided by the volume is the

divergence of the field.

Divergence = Flux out density

The first of Maxwell’s equations 00

ˆV V

insideQE n dA dV

states that electric charge density

is a divergence source for the electric field. Equivalently, electric charge starts electric field lines.

The notation V directs that the flux is to be that through the surface V that bounds the volume V.

The circulation integral of a field is C

F d , the integral of the tangential component with respect

to path length around a closed path. The component of the curl of the field in the (RHR) normal

direction to the area enclosed by the path is the circulation divided by the area enclosed by the path

for a small path.

curl = circulation/area or circulation density

The second of Maxwell’s equations Â A

BtE d n dA

states that the electric field

circulates in a region of space in which the magnetic field varies in time. This equation represents

the induction part of the Faraday’s flux rule. The notation A directs that the circulation path is the

path (curve) A that bounds the surface patch A.

The third of Maxwell’s equation ˆ 0V

B n dA

states that lines of the magnetic lines do not start

or stop. (They may close on themselves.) Equivalently there is no such thing as magnetic charge.

The magnetic field does not have a divergence source.

The fourth of Maxwell’s equation 0 0 0 Â AEtB d J n dA

contains the fact that the

magnetic field circulates around its source which is moving charge. 0

steady state

Â AB d J n dA .

The full equation states that the magnetic field circulates in a region of space in which the electric

field is time dependent. This behavior is the reciprocity action to that described in the second

(Faraday’s) law.



In the case of general time dependence with associated charge acceleration and time dependent

charge densities, electromagnetic radiation is generated.

charge electric field moving charge magnetic field

accelerated charge E-M radiation

Ex. 10) A charge of +Q is placed on the x-axis at x = - 1

meters, and a charge of -2Q is placed at x = + 1 meters,

as shown to the left. At what position on the x-axis will a

test charge of +q experience zero net force?

(A) (3 8) m (B) – 1/3 m (C) 0 m

(D) 1/3 m (E) (3 8) m

One should think about a problem before launching into a solution. The net force on the test charge

due to the two existing charges is to be zero. Two vectors can sum to zero only if they are anti-

parallel. A little reflection makes it clear that the solution points must lie on the (extended) line

joining the two charges. That is: The solution point is somewhere on the x axis.

In order to sum to zero, the forces must have equal magnitude and opposite direction. For points on

the x-axis, the test charge must lie to the right of both charges or to the left of both charge to meet

the opposite direction requirement. To meet the equal magnitude requirement, the test charge must

be closer to the smaller magnitude charge. That places the solution point in region I, x < - 1 m. The

particular value of x is found using the inverse square law nature of the Coulomb force.

x

y

-2Q Q

I II III

x

y

-2Q Q

1 m 1 m

q

x

y

-2Q Q

1 m 1 m



Look for equal magnitudes for points in region three. The distance from Q is | x – (-1)| and the

distance from -2Q is | x – 1|.

2 22 2

(2 )2( 1) ( 1)

( 1) ( 1)

k Q k Qx x

x x

or x2 + 6 x + 1 = 0

This equation has a root (3 8) m in region I and a root (3 8) m in region II. Our previous

arguments have identified region I (x < - 1 m) as only region in which both the equal magnitude and

opposite direction requirements can be met. Recall that quadratic equations often return one

physical root (answer) and one not-so-physical root.

Review your options. Which of the proposed answers corresponds to a point to the left of both

charges? Test them, simplest value first, to see if the equal magnitude condition is met.

You need x < -1 m; only A is possible.

(11) Gauss’s Law, Ampere’s Law and E&M Knowledge Points:

Gauss:

10 0

ˆ ( )V V

encQE n dA dV

0ˆ

V

encQE n dA EA

(a wa y)enc

o

QE

A

directed away

A : The enclosing surface through the point where the field is to be determined is to be made of

portions perpendicular to the field and parallel to it. A : net area that is perpendicular to the field.

A plane = 2 A ; A cyl = 2 r L ; A sph = 4 r2

For applications to conductors, make a sketch. (For equilibrium,) The net charge is on the surface(s).

The field at points in the conducting material is zero. A small patch on the surface is essentially

planar, and the field only pokes outside. Qenc = local A; A = A; local

o

E

away.

Ampere:

ênc encA AB d I J n dA I



0i i i encAi i

B d B I

o encIB

directed to circulate (RHR)

The encircling path through the point where the field is to be determined is to be made of portions

perpendicular to the field and parallel to it. : the net parallel length along which the field has a

constant non-zero magnitude. long st. wire = 2 r ; solenoid = ; toroid = 2 r

Applied to arbitrary paths: Note the direction of

each integration path. Give the value of

d B s for each path illustrated.

CCW: out is positive; CW: in is positive

C1: CCW; C2: CW; C3: CW C

d B s = o Ienc

1Cd B s

= o Ienc = o [5 A - 2 A] = m

2Cd B s

= o [- 1 A - 5 A] = - 7.54 T m

3Cd B s

= o [- 1 A + 2 A] = 1.26 T m

0 0 ênc RHRC A Ad I J n dA

B s

The notation C = A means that C is the curve (path) than bounds the area A. The RHR subscript

means that the normal to the area is in the direction dictated by the RHR. In the cases above, an

arrow indicating a CCW path means that the normal and hence the positive sense for currents is out

of the page. For CW paths, the positive direction is into the page.

Knowledge points:

(a) Field and potential of a point charge: 0

( )4sp

sp

qV r

r ;

3 20 0

ˆ( )

4 4sp sp

spsp sp

q r q rE r

r r

spr

: displacement from source point to field point

(b) The field of a uniform spherical shell of charge is zero inside and the same as a point charge with

qtotal located at the center outside.



(c) The field of a spherical ball of charge increases linearly with r inside and is the same as a point

charge with qtotal located at the center for points outside.

(d) The field of a long uniform line of charge is the charge per length divided by 2 times the

distance from the wire. The field is directed away from a uniform line of positive charge.

(e) much, much more …

Ex. 11) A uniform insulating sphere of radius R has a total charge Q and a uniform charge density.

The electric field at a point R/3 from the center of the sphere is given by which of the following?

(A) 2

012

Q

R (B)

208

Q

R (C)

206

Q

R (D)

204

Q

R (E)

20

3

4

Q

R

By knowledge: The field grows linearly inside a uniformly charge sphere so the field has 1/3 of the

strength at r = R or 20

13 4

Q

R . The direction is radially away by symmetry. A

By Gauss: enc

o

QE

A

directed away. For the point R/3 from the center, the Gaussian surface is a

sphere surface of radius R/3 concentric with the sphere. (The point at which the field is to be found

must lie on the Gaussian surface. The surface has the symmetry of the problem.)

Qenc = (1/3)3 Q; A = (1/3)

2 (4 R2)

(12) Faraday’s Law

Â A

BE d n dA

t

Faraday’s Flux Rule

= Emf = - Bd

dt

ˆB A

B n dA B A



Schematically:

cos

coscos cos

cos cos sin[ ]

Bd dBA

dt dt

dB dA dA B BA

dt dt dt

dB dAA B BA t

dt dt

pure induction motional motional

transformer slidewire gen. rotating generator

In some cases, one needs to multiply by the number of turns.

Ex. 12) Four meters of wire form a square that is placed perpendicular to a uniform magnetic field of

strength 0.1 Tesla. The wire is reduced in length by 4.0 centimeters per second while still

maintaining its square shape. Which one of the following gives the initial induced emf across the

ends of the wire?

(A) 1 mV (B) 2 mV (C) 4 mV (D) 8 mV (E) 16 mV

The magnetic field is not varying in time, so we will use the flux rule form: Emf = - Bd

dt

.

The easy way: The side s is the perimeter divided by 4. The side at time zero is s(0) = 1 m. At one

second, s(1 sec) = 0.99 m. Using B B A , B(0 s) = (0.1 T) (1 m2) and B(1 s) = (0.1 T) (.992 m2)

Emf = - Bd

dt

2 2(1 ) (0 ) (0.1 )(1 ) (0.1 )((.99)(.99) )(0.1 )[1 0.98]

1 1B Bs s T m T m

T Vs s

= 2 mV

As a function of time. Use: Area = side2 = (perimeter/4)2

B B A = B side2 = B (perimeter/4)2 = (0.1 ) (4.0 – 0.04 t )2/16 Tm2

d/dt B =(0.1 ) ( [2 (4.0 – 0.04 t) (- 0.04)]/16) Tm2s-1

At time zero, the magnitude of the emf is |(0.1) ( [8 (- 0.04)]/16)| Tm2s-1 = 2 mV. B



The final technique to compute a motional emf is as ( )v B d B v . In this problem you

have 4 meters of wire moving inward at ½ cm/s at t = 0. Bv (0.1T)(4m)(0.005 m/s) = 2 mV.

OPTIC, WAVES, THERMO AND STAT:

(13) Properties of an E&M plane wave and the Poynting Vector

EM waves are transverse which means that the electric and magnetic fields are perpendicular to the

direction of propagation of the wave. The electric and magnetic fields are mutually perpendicular.

The wave propagates in the direction of the Poynting vector, 10( )S E H E B

. The E and

B fields oscillate in phase with one another and B/E = v, the wave speed (= c in vacuum). The

Poynting vector has units W/m2 and represents the power flow of the wave.

Energy Densities: E = ½ o E2 ; B = ½

2/o

Ex. 13) A linearly polarized electromagnetic plane

wave carries energy in the positive z direction. At

some positions r

and time t, the magnetic field points

along the positive x-axis, as shown in the figure to the

right.

At that position and time , the electric field points

along the

(A) positive y-axis

(B) negative y-axis

(C) positive z-axis

(D) negative z-axis

(E) negative x-axis

EM waves are transverse which means that the electric and magnetic fields are perpendicular to the

direction of propagation of the wave. The electric and magnetic fields are mutually perpendicular.

As the wave propagates in the z direction and at that point and time has the magnetic field in the x

x

z

y

B



direction, the electric field must be in the y direction. The Poynting vector S

represents the

intensity of the wave times its propagation direction. Find the direction of S

.

10 ( )S E H E B

10

ˆ ˆˆ ˆ ˆ ˆ( ) and and ( )S E B j i k j i k

The electric field is in the negative y direction at that point and time. B

(14) Diffraction and Interference Standing waves

As a first guess, diffraction limits angular resolution to min = /d where d is the width of the aperture.

(If the aperture is circular, use min = 1.22 /D where D is the diameter of the aperture.)

The N slit pattern constructive interference condition: d sinbright = m where m is the order of

the interference.

Array of equally spaced finite slits. A array of 100 identical slits spaced by 32 m and of width a

yields a diffraction pattern in which the first three orders of the multiple slit pattern appear, but the

fourth m = 4 peaks are missing (dark). What is the width a of the individual slits?

Intensity pattern = single slit * multi-slit pattern

The single slit pattern zeros: a sin = m m 0

The multi-slit bright: d sin = m m = integer

The fourth bright is missing: d sin = 4 anda sin = m

m = 1, because this is the first missing peak.

4

32sin 8.0

4 4 4

d a d ma m

The figures below show the intensity pattern of six narrow (width << ). The blue line is the

intensity pattern of a single slit with width of ¼ of the slit to slit spacing. The next figure shows the

intensity pattern of six slits each with width of one-fourth of the slit to slit spacing. The final figure

is the pattern of 20 equally spaced slits each with width of one-fourth of the slit to slit spacing. The



individual intensity spikes are narrower. A diffraction grating has a great many slits and so produces

very narrow spikes. The single slit pattern multiplies the N slit pattern.

Six Narrow Slits + finite Single Slit Pattern of Six Finite Slits Pattern of 20 Finite Slits

(Ex. 14) An observer looks through a slit of width 5 x 10-4 meters at two lanterns a distance of 1

kilometer from the slit. The lantern emits light of wavelength 5 x 10-7 meters. The minimum

separation of the lanterns at which the observer can resolve the lantern lights is most nearly

(A) 0.01 m

(B) 0.1 m

(C) 1 m

(D) 10 m

(E) 100 m

In this case, = 5 x 10-7 meters and d = 5 x 10-4 meters so min = 10-3 radian or one milli-radian. One

should prepare a sketch, but one can guess that the answer is 10-3 of one kilometer or one meter. C

Polarization Methods:

Selective Absorption: Light can be polarized by selective absorption which is the method used in

most sheet polarizers. The material absorbs light polarized along one of the transverse directions and

passes light polarized along the pass axis (which is perpendicular to the absorption axis).

Reflection: The fraction of the light of each of the two polarizations that is reflected depends on the

angle of incidence. At Brewster’s angle, the refracted ray make a 900 angle with respect to the

reflected ray, and the reflected light is 100% polarized perpendicular to the plane of incidence. n1

sinp = n2 sin2 and p + 2 = 900 so sinp = cos2 tanp = n2/n1.



Birefringence (double refraction): Certain materials have different indices of refraction for the two

polarization so the two polarizations are refracted at different angles leading to a separation of the

incident beam into two beams of pure linear polarization (that are polarized perpendicular to one

another).

Scattering: The electric field accelerates the charges in material along its direction. The scattered

light has electric field (polarization) components in the direction of the charges acceleration that are

perpendicular to the direction of propagation.



Illustrating the polarizing of radiation due to light scattering.

The unpolarized light has an electric field that is directed randomly in all of

the directions perpendicular to the direction of propagation. The average

values of Ex2 and Ex

2 where x and y are assumed to be the two independent

transverse directions.

Polazation, polarizers: The interaction of E-M waves with matter is described by the Lorentz force

law, q others q othersF q E v B

. For a wave, |B| c-1 E so Fmagnetic (vq/c) Felectric. The electric

interaction dominates so the polarization of the wave is associated with the direction of the electric

field. That field is transverse to the direction of propagation and there are two independent transverse

directions so there are two independent polarizations for a given propagation direction. These

polarizations might be chosen to be vertical and horizontal linear polarizations or left- and right-

handed circular polarizations. Circular polarizations carry angular momentum along the direction of

propagation. Linear polarizations carry zero net angular momentum. Natural (unpolarized light) can

be assumed to be an equal mix of the two independent polarizations.

A linear polarizer absorbs the component of the polarization perpendicular to its pass axis. It does

this by adding a wave polarized perpendicular to the pass axis that is out of phase with the

incident wave and of equal amplitude.

The incident wave has an E field that makes an

angle w.r.t. the pass axis. The charges in the

polarizer move perpendicular to the pass axis and

absorb energy effectively adding a field

perpendicular to the pass axis that is out of

phase. Superposing, the perpendicular part is

eliminated on only the part along the axis passes.

The wave immediately after a polarizer has a polarization parallel to the pass axis and an average

electric field strength of Eincident cos. As the intensity of a wave is proportional to the square of its

Incident

components

added by polarizer

transmitted

pass axis



amplitude, Itransmitted = Iincident cos2. (The Law of Malus.) Natural (unpolarized) light is an equal

mix of the two polarizations. If natural light is incident on a polarizer, light of one-half the intensity

that is 100% polarized along the pass axis is transmitted.

One half the intensity of the incident intensity of natural light passes through the first polarizer. All

the light that passes is polarized along the pass axis direction. Incident polarized light obeys the Law

of Malus, and the transmitted light is 100% polarized along the pass axis of the last polarizer.

(15) Lens Problems

C FOptical axis

Optical interface plane

Ray Tracing a Positive (Converging) Mirror Virtual for object inside focal point

The parallel ray: In parallel; out through the focus

The focal ray: In (as if) through the focal point; out parallel The chief ray: In at optical axis point

in the interface plane. r = i.

Center Ray: In through the center of curvature; out through the center of curvature

Io natural ½ Io vertical ½ Io cos2 along last pass axis ½ Io cos2cos2 along last pass axis



CFOptical axis

Optical interface plane

Ray Tracing a Convex (Diverging/Negative) Mirror Virtual Image

The parallel ray: In parallel; out as if from the back focal point

The focal ray: In (as if) toward the focal point; out parallel

The chief ray: In at optical axis point in the interface plane. r = i.

Center Ray: In through the center of curvature; out through the center of curvature

You must bend rays at the plane of the optic (optical

interface plane) in order to get accurate ray trace answers.



All the ray bends are made at the optical interface plane in

order to have a predictive and somewhat quantitative plot.

(1)

(3)

(2)

Rays actually converge atThe position of a REAL image

Positive lens. Three rays: (1) in parallel, out through focus; (2) in through focus, out parallel; (3)

Chief Ray: straight through the lens center. Shows the image point if there were no second lens.

Ray Tracing Negative lenses

The parallel ray: In parallel; out as if from the front side focal point

The focal ray: In at the backside focal point; out parallel

The chief ray: Straight through the lens center



(Ex. 15) Two thin converging lenses A and B, each having a focal length of 6 centimeters. are

placed 10 centimeters apart, as shown in the figure above. If an object is placed 10 centimeters to

the left of the lens A, the final image is

(A) 30 cm to the right of lens B

(B) 30/11 cm to the right of lens B

(C) 30/10 cm to the right of lens B

(D) 30/11 cm to the left of lens B

(E) 30/10 cm to the left of lens B

A positive lens increases the convergence of the light that passes through it. Light diverges from a

real object. Real objects have positive object distances. Light converges to a real image with a

positive image distance.



The focal length of the lenses in this problem is f = 6 cm. The object is 10 cm before the left lens

and so it has an object distance + 10 cm. Using the lens equation,

Obj Img

1 1 1

d d f

Img Img Img

1 1 1 3 1 5 1 2

10 6 30 30 30d d d

Lens A would form an image 15 cm to its right if there were no second lens. That would be 5 cm to

the right of lens B. Lens B is converging so the rays should converge closer to lens B than 5 cm.

(Lens B is converging so we are focusing on responses (B) and (C).) The appearance of 5 cm and 6

cm makes the 30/11 cm value look attractive.

The light arriving at lens B is converging. That case is called a virtual object and corresponds to a negative object distance. (Light diverges from a real object and converges to form real image. These cases correspond to positive object and image distances. Converging light incident on a lens and diverging light leaving the lens corresponds to virtual objects and virtual images negative distances for the lens formula.) Rays are converging as they reach the second optic lens two has a virtual object negative object distance. The rays as converging to a point 5 cm after the lens dObj,2 = - 5 cm. Applying the formula to lens B,

Obj Img

1 1 1

d d f

Img Img Img

1 1 1 6 1 5 1 11

5 6 30 30 30d d d

The image distance is positive so the light converges to a point on the side of the lens opposite to

that from which the light was incident. right side at 30/11 cm. B

First lens. Three rays: (1) in parallel, out through focus; (2) in through focus, out parallel; (3) straight

through the lens center. Shows the image point if there were no second lens.

To trace the second lens, you generate construction rays that are directed to converge at the position

of the image that would have been formed by the first lens in the absence of the second lens. New

image.



The blue rays shows the final image after the action of the second lens. Real image: the rays actually converge to and pass through a real image point.

Second lens: Three rays: (1) in parallel directed at the image point of the first lens, out through focus; (2) in through focus directed at the image point of the first lens, out parallel; (3) straight through the lens center directed at the image point of the first lens. Shows the intermediate and final images. Rays are converging as they reach the second optic lens two has a virtual object negative object distance. The rays as converging to a point 5 cm after the lens dobj,2 = - 5 cm.



Virtual image: The rays appear

to diverge from an image point

that the rays did not converge to

and pass through. Rays are

diverging immediately after the

last optic so the image is virtual.

(16) Basic Wave Properties

Basic Wave Plug*: v = f The wave-speed is the frequency of the wave times its wavelength.

THE BASIC WAVE PLUG (BWP) is the most important wave property relation.

Wave Speed: v = /

Tension stiffness

mass length inertial density

Traveling Wave: y(x,t) = A sin[ k x - t + ] k = 2; T = 2

2 radians per cyle k *(cycle distance ) = 2 * (cycle time T) = 2

Standing Wave: y(x,t) = A cos[t + ] sin[ k x + ]

node to node spacing: ½ node to anti-node spacing: ¼

Wave types: longitudinal and transverse

There are two transverse directions so transverse waves have polarizations

Waves reflect with a sign change when they reflect from a region with lower wave speed.

Waves reflect with a sign change when they reflect from a fixed end.

Waves reflect with no sign change when they reflect from a region with higher wave speed.

Waves reflect with no sign change when they reflect from a free end.

(Ex. 16) Two identical sinusoidal waves travel in opposite directions in a wire 15 meters long and

produce a standing wave in the wire. The traveling waves have a speed of 12 meters per second and

the standing wave has 6 nodes, including those at the two ends. Which of the following gives the

wavelength and frequency of the standing wave?

Wavelength

Frequency

(A)

3 m 2 Hz

(B)

3 m 4 Hz

(C)

6 m 2 Hz



(D)

6 m 3 Hz

(E)

12 m 2 Hz

Prepare a sketch! The four interior nodes divide the wire into five equal length segments, node-to-

node spacings. Thus, 5 (/2) = 15 m or = 30 m/5 = 6 m. From the BWP, f = v/

= 12/6 s -1 = 2 Hz. C

(17) Kinetic Theory of Gases

Gas molecules have an average translational kinetic energy of 3/2 k T (three-halves times

Boltzmann’s constant times the absolute temperature). The pressure arises due to the many

molecular collisions per second with the walls so it increases for higher speeds v (v increases if

temperature increases) and for higher density. The molecules have a cross sectional area for

interaction and so sweep out a volume v t in time t. The mean time between collisions is found

as the time for a molecule to sweep out the average volume per molecule.

v tcollision free = V/N so tfree = V/Nv

The mean free path free is the distance traveled to sweep out V/N. free = V/N.

Ex. 17) Cubical tanks X and Y have the same volumes and

share a common wall. There is 1 gram of helium in tank X

and 2 grams of helium in tank Y, and both samples are held

at the same temperature. Which of the following is the same

for both samples?

(A) the number of molecular collisions per second on the

common wall

(B) the average speed of the molecules

(C) the pressure exerted by the helium

(D) the density of the helium



(E) the mean free path of the molecules

The number of collisions per second, pressure and density for sample Y are double the X values. The

mean free path for sample Y is half that for sample X because there are twice as many things to hit in

the same volume. The average speed depends on the temperature and molecular mass. It is identical

for samples X and Y. B

(18) Ideal Gas Law

isothermal: process at constant temperature

isobaric: process at constant pressure

iso-entropic: process at constant entropy

adiabatic: process in which there is no heat transfer; insulated from the environment

isochoric, isometric, iso-volumetric -------- process at constant volume

The Ideal Gas Law: P V = n R T P V = N k T

number of moles: n

ideal gas constant R = 8.31 (J/ mol-K)

number of molecules: N

Boltzmann’s constant: k = 1.38 x 10-23 J/K

LaChatlier’s Principle: A stable system acts to counter any change in its parameters. If a the

volume of a sample of gas is reduced, its pressure increases to fight additional decreases in volume.

(Ex. 18) If one mole of an ideal gas doubles its volume as it undergoes an isothermal expansion, it

pressure is

(A) quadrupled

(B) doubled

(C) unchanged

(D) halved

(E) quartered

The process is isothermal (same temperature). P1 V1 = n R T = P2 V2



12 1 1

2½VP P PV

D

Use ratios whenever possible.

(19) Thermodynamics and efficiency

Entropy Change: revQS

T

Ideal Heat Engine Efficiency: high low

high

T Te

T

Use absolute temperatures. 0 0C = 273 K

Ex 19.) A power plant takes in steam at 527 0C to power turbines and then exhausts the steam at

127 0C. In any given time, it consumes 100 megawatts of heat energy from the steam. The maximum

output power of the plane is

(A) 10 MW (B) 20 MW (C) 50 MW (D) 75 MW (E) 100 MW

The problem statement is a little fuzzy with regard to its use of the terms power and energy!

Using 0 0C = 273 K, Thigh = 800 K and Tlow = 400 K. The ideal thermal efficiency for a reversible

engine operating between these temperatures is 50% so the maximum possible mechanical/electrical

power out is 50 MW. C

QUANTUM MECHANICS AND ATOMIC PHYSICS:

Particles have associated wave properties. As a crude characterization, particles propagate like

waves and interact as (point) particles. It may be helpful to assume that each particle has a guide

wave that feels out the space to generate the probability distribution for the particle’s location.

Waves such as optical radiation also display particle-like character. When they interact, energy

transferred into and out of the wave in quanta (photons) each with energy hv (and momentum hv/c.)

34 34; ; 6.626 10 ; 1.0546 10h p k E hv h Js Js

The information that can be known about a system is encoded in a wavefunction (x,t) that satisfies

the Schrödinger equation. The probability to find the particle between x and x + dx is

(x,t)(x,t) dx (Born interpretation). Additional information can be extracted from the wavefunction

using the operator for the dynamical quantity of interest. The operator for x is x. The operator for px



is - i x. Dynamic quantities in Hamiltonian mechanics are functions of position, momentum and

time. The quantum operator for the quantity is formed by keeping the coordinates and the time while

replacing each momentum pq with –i/q. px - i x; p- i ; …..

Q(x,y,z, px,py,pz,t) Q̂ (x,y,z, –i x, –i y, –i z, t); K 2

2

2m

; ( ) ( )V r V r

Lz = xpy – ypx ˆ ( ) ( )z y xL x i y i

Shrödinger’s equation identifies two operators for energy, the Hamiltonian and i t. In other

words, the Hamiltonian represents the total energy (kinetic + potential), and it is the time-

development operator for wavefunction (multiplied by –i/).

22ˆ ( , ) ( ) ( , ) ( , )

2H r t V r r t i r t

m t

(Use separation.)

22( , ) ( ) ( ); ( ) ( ) ( ); ( ) ( )

2 n n n n n nr t u r T t V r u r E u r E T t i T tm t

The values En are the energy eigenvalues, the total energies, a set of real values that are

characteristic of the problem and the associated temporal dependence is: Tn(t) = nE

n i ti te e .

A general solution has the form, ( , ) ( )n nall states n

ni tr t a u r e .

( ,0) ( ) ( , ) ( )n n n nall states n all states n

ni tr a u r r t a u r e

Operators operate on wavefunctions to return other (or perhaps the same) wavefunctions.

Operators operate on their eigenfunctions to return a scalar multiplying the same wavefunction.

Eigenvalue equation: px u(x) = –i x u(x) = (momentum eigenvalue) u(x) = p u(x)

–i x u(x) = p u(x) u(x) = A ( )i xe p

A e i (kx - t) is a momentum eigenfunction with eigenvalue xp k

H n(x,t) = En n (x,t)



Operators for physical observables are Hermitian (or self-adjoint) an, as such, they have real

eigenvalues and expectation values.

Wavefunction Behavior: 22

2( , ) ( ) ( , )n

mr t E V r r t

In a classically allowed region, the

kinetic energy K, E – V, is greater than zero so the net curvature of has a sign opposite to that of .

The function value oscillates back and forth across zero. In the case that V > E or K < 0 (which is

forbidden classically; regions for which V > E are classically forbidden), the curvature and function

have the same sign behavior like growing and decaying exponentials.

All the functions oscillate in the classically allowed region and decay in the forbidden regions. The

decay is more rapid for larger energy deficits. Each higher state has an additional node.

Notice that the spatial rate of oscillation increases as the kinetic energy increases (or the ‘local

wavelength decreases).

2 22 2

2(2 )

2 2kKm m

The n = 36 QHO wavefunction is plotted for t = 0. Note the more rapid oscillation around x = 0 where the

kinetic energy is greatest.

Free Particle States for piecewise constant potentials are of the form ( )i kx te where

2 ( )mk E V x . At any abrupt change in V(x), the wave is partially transmitted and partially

reflected. The wave function ca tunnel through classically forbidden regions (V > E k = i and the

wave is like ex).

-10 -5 5 10-0.2

0.2

0.4



Fails to show the reflected wave!

Energy eigenstates are stationary the associated probability density * is time-independent.

2( , ) ( ) ( ) ; ( , ) ( , ) ( )

n

n n n n n nn

Ei ti tr t u r e u r e r t r t u r

Energy Level Spacing: Energy levels in an infinite well with width a have the form 2 2

222

nm a

so

the n to n + 1 spacing is 2 2

2(2 1)

2n

m a

; the level to level spacing grows with n and is larger for

smaller a (tighter confinement). The finite-well states penetrate into the forbidden region and so are

less tightly confined leading to lower energies as compared to the infinite well with the same inside

width. For the hydrogen atom problem, the electron’s range approaches infinitely wide as the energy

approaches zero from below. The level spacing approaches zero in this limit. For positive energies,

the particle is not confined and the allowed energies run continuously (zero spacing).

Transitions between levels: A quantum system absorbs or emits energy in chunks or quanta equal

to the energy difference between the initial and final state of the system.

A hydrogen atom in an n = 4 state can make a spontaneous transition to an n =2 state emitting a

photon of energy 2.55 eV.

- 13.6 eV/42 – (- 13.6 eV/22) = 2.25 eV = 486 nm a fantastic shade of blue or blue-green!

Commutation: If the operators for two quantities do not commute, then those quantities have a



minimum uncertainty product. [x,px] = i x px ½ .

General Uncertainty Relation: 2 2 12

ˆ ˆ( ) ( ) [ , ]A B iA B A B

If operators commute, they can have simultaneous eigenvalues. For the hydrogen problem, H, L2 and

Lz (plus the operator for electron spin) commute. The hydrogen atom states are labeled by nm

representing the eigenvalues of energy ( - 13.6 eV/n2), orbital angular momentum squared ([+1]2)

and z component of orbital angular momentum (m) (plus spin – up or down).

Generalized Uncertainty Principle. Experimental measurements are performed on systems in

order to extract average values and the corresponding uncertainties. In quantum systems, constraints

can arise between two uncertainties due to the nature of the corresponding operators. Given two

operators A and B, and knowledge of the commutator [A,B] = iC, then there exists a hard lower limit

on the product of uncertainties A B >= ½ |<C>|. Example: [x,px] = i x px ½ .

If two operators commute, then there is no lower-level uncertainty constraint on knowledge

of the simultaneous values of operators (the outcomes of measurements), and the expectation values

of these operators serve as a very useful label to identify and characterize a system. For the bare

Coulomb potential (simple H-atom problem), the operators H, L2, and Lz commute, so the values of

the three can be known exactly at the same time. Therefore we identify individual states by the

eigenvalues (energy, a.m. magnitude, a.m. z-comp) or their corresponding labels n,,m (quantum

numbers).

Expectation value: 3ˆ*( , ) ( , )everywherethatis non zero

Q r t Q r t d r

Mixed State: ( , ) ( )n nall states n

ni tr t a u r e more than one an 0. Here, an is the amplitude to be

found in state n and | an |2 is the probability for the system to be found in state n. As energy

eigenstates have been used above, E 2

n nall states n

a E . A precise measurement of the energy will

yield one of the energy eigenvalues. For a set of identically prepared systems, the eigenvalue En will



be found with probability | an |2. Making a precise measurement and finding the value En places the

system in a state that has that energy. If a measurement on the state:

100 100 210 200 211 210 211 211 21 1 21 1 310 310hydrogen a a a a a a

returns the energy, - 3.40 eV, the energy for the n = 2 states, then the atom is left in a state of the

form by the measurement: 210 200 211 210 211 211 21 1 21 1after b b b b immediately after the

measurement. (All the n = 2 parts remain; the parts for other n’s are removed.) The act of making a

precise measurement of a physical observable collapses the state to a state that is consistent with the

measured value. (Removes the parts that are not consistent; preserves remainder with relative phases.)

After the collapse, the state will time develop as directed by the Hamiltonian, and the state may not

be consistent with the measured eigenvalue after some time has passed. If the operator for the

eigenvalue commutes with Hamiltonian, the expectation value for the quantity will be time

independent.

ˆˆ ,d Q i Q

H Qdt t

time development of expectation values

ˆˆ ,d Q i

H Qdt

for operators Q without explicit time dependence

Time independent operators that commute with the hamiltonian have time independent expectation values.

Example with non-zero commutator: p i ˆ , ( )

d p iH p V r

dt

The time rate of change is momentum is the ‘force’. Ehrenfest’s Theorem: Quantum mechanical

expectation values obey the corresponding classical equations of motion.

Every Hermitian operator has a complete set of eigenstates (Q m = qm m) which can be used as an

basis set for representing a general solution. ( , ) ( , )m meigenstates m

r t c r t . It follows that:

2

m meigenstates m

Q c q . The value |ck|2 is the probability that the measurement returns the eigenvalue

qk for the state k. A precise measurement of Q will return an eigenvalue and place the system in a

state with that eigenvalue immediately after the measurement. The eigenvalues of the measurement

operator are the only possible outcomes of a good measurement.



(20) Ionization and the Periodic Chart

According to L. Pauling, “the power of an atom in a molecule to attract electrons to itself.”

Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons.

Electro-negativity increases from bottom to top of a column.

Electro-negativity increases from left to right across a row.

→ Atomic radius decreases → Ionization energy increases → Electronegativity increases →



The inner electrons screen the nuclear charge seen by the outer electrons. The apparent unscreened charge increases as you move across the table filling an electron shell. It is a maximum for group VIII and a minimum for the single electron outside a closed shell in group I. The effective unscreened charge increases as you move down the table to the higher Z atoms. Group I has one electron outside a closed shell and are the easiest to ionize (easiest at top after H; lowest screened charge). In group VII, the effective charge gets higher as you move down the table most electron hungry (highest electron affinity). Electron affinity increases as you move down and across from I to VII. VIII is the closed shell nobles; they wish neither to give nor receive electrons. Outer valence shell has two S plus 6 P states.



Electron Shell Filling Sequence: Valence electrons appear rightmost

Z Atom Configuration Comments 1 H (1s) 2S½ Filling the 1s shell ; single active electron 2 He (1s)2 1S0 Shell filled spherical, zero angular momentum 1S03 Li (2s) 2S½ Filling 2s; single active electron outside closed shell 4 Be (2s)2 1S0 Sub-shell filled spherical, zero angular momentum 1S05 B (He)(2s)2(3p)1 2P½ Filling the 2p sub-shell 6 C (He)(2s)2(3p)2 3P0 Maximize unpaired electrons to minimize exchange energy7 N (He)(2s)2(3p)3 4S3/2 Half filled sub-shell; max net spin J = |L – S| 8 O (He)(2s)2(3p)4 3P2 Spin decreases as electrons added pairing previous ones 9 F (He)(2s)2(3p)5 2P3/2 J = |L + S| 10 Ne (He)(2s)2(3p)6 1S0 Sub-shell filled spherical, zero angular momentum 1S0 11 Na (Ne)(3s)1 2S½ single active electron outside closed shell 12 Mg (Ne)(3s)2 1S0 Shell filled spherical, zero angular momentum 1S0 13 Al (Ne)(3s)2(3p)1 2P½ Filling the 2p sub-shell 14 Si (Ne)(3s)2(3p)2 3P0 Maximize unpaired electrons to minimize exchange energy15 P (Ne)(3s)2(3p)3 4S3/2 Half filled sub-shell; max net spin J = |L – S| 16 S (Ne)(3s)2(3p)4 3P2 Spin decreases as electrons added pairing previous ones 17 Cl (Ne)(3s)2(3p)5 2P3/2 J = |L + S| 18 Ar (Ne)(3s)2(3p)6 1S0 Sub-shell filled spherical, zero angular momentum 1S0 19 K (Ar)(4s)1 2S½ single active electron outside closed shell 20 Ca (Ar)(4s)2 1S0 Shell filled spherical, zero angular momentum 1S0 21 Sc (Ar)(4s)2(3d)1 2D3/2 Filling inner sub-shell hidden inside the 4s electrons 22 Ti (Ar)(4s)2(3d)2 3F2 The full set has similar chemical character. 23 V (Ar)(4s)2(3d)3 4F3/2 Transition metals filling d sub-shell* 24 Cr (Ar)(4s)2(3d)4 7S3 25 Mn (Ar)(4s)2(3d)5 6S5/2 26 Fe (Ar)(4s)2(3d)6 5D4 27 Co (Ar)(4s)2(3d)7 4F9/2 28 Ni (Ar)(4s)2(3d)8 2F4 29 Cu (Ar)(4s)2(3d)9 2S½ 30 Zn (Ar)(4s)2(3d)10 1S0 Sub-shell filled spherical, zero angular momentum 1S0 31 Ga (Ar)(4s)2(3d)10(4p)1 2P½ Filling the 2p sub-shell 32 Ge (Ar)(4s)2(3d)10(4p)2 3P0 Maximize unpaired electrons to minimize exchange energy33 AS (Ar)(4s)2(3d)10(4p)3 4S3/2 Half filled sub-shell; max net spin J = |L – S| 34 Se (Ar)(4s)2(3d)10(4p)4 3P2 Spin decreases as electrons added pairing previous ones 35 Br (Ar)(4s)2(3d)10(4p)5 2P3/2 J = |L + S| 36 Kr (Ar)(4s)2(3d)10(4p)6 1S0 Sub-shell filled spherical, zero angular momentum 1S0 The fine structure energies complicate the filling sequence post Kr so we stop. *The sequences of elements arising while an inner f sub-shell is filling are called rare earths.





Electron shell filling sequence. Except for (H, He), the first electron in an S shell leads to an alkali metal (strong electron donor). Two s electrons yield a metal. The nth valence shell consists of the two ns and six np states. Beginning with period n = 4, inner shell (d and f electrons of lower n) fill after the nS, but before the nP. Shells fill first with spins parallel (say up) and add spins anti-parallel only after the sub-shell is half filled. Elements with approximately half-filled d and f sub-shells exhibit magnetic properties associated with the unpaired spins. The D, F, shells fill inside the valence electrons and have little impact on bounding. The transition metals and rare earths are the large group of elements differing by the number of electrons in the partially filled D and F sub-shells.

Ex 20.) For which of the following elements is the ionization energy of a neutral atom the lowest?

(Z) is the atomic number.

(A) Oxygen (Z = 8) (B) Fluorine (Z = 9) (C) Neon (Z = 10)

(D) Sodium (Z = 11) (E) Magnesium (Z = 12)

The first period of elements includes hydrogen and helium with one and two electrons. In the next

period (3 , there are up to eight electrons in the valance shell. Oxygen and fluorine need

electrons to close the shell and so are hard to ionize. Neon is the very stable closed shell

configuration. Sodium has just one electron outside a closed shell and so is easy to ionize. For

Magnesium, the nucleus plus inner electron shells combined have an effective charge of plus 2 and

so the screened nuclear charge is larger and it binds the outer electrons more tightly than does the

sodium nucleus plus inner shell electrons (net charge +1). D

Sodium is a light alkali metal and so is an electron donor. In compounds such as NaCl, it is strongly

ionic with the sodium being positive which means its electron has transferred to the chlorine. After

hydrogen, the Group I atoms are willing electron donors and easy to ionize. Group 7 is the group of

the strongest electron acceptors. Top Left strong donor; Bottom Right (grp.7) strong acceptor.

(21) Hydrogen Atom spectrum levels, …



2 2 2

13.6 1 113.6upper lower

upper lowern

eV cE E h h E E eVn n n

20

0 2

4a

Ze

, ao = 0.0529 nm for hydrogen, reduced mass, Z nuclear charge

The radius varies as (the mass of the orbiting particle)-1 and as (the nuclear charge)-1.

H-atom Transition Table

Lyman Series (n 1) Series Limit: nupper = to nseries

Transition Series Limit Designation L L L L L L: 13.6 eV Wavelength nm 102 nm 97.2 nm 94.9 nm 93.8 nm 91.2 nmColor VUV VUV VUV VUV VUV VUV

Balmer Series (n 2)

Transition Series Limit Designation H H H H H H: 3.4 eV Wavelength nm 486.1 nm 434.1 nm 410.2 nm 397.0 nm 364.6 nmColor red blue-green violet violet violet UV

Paschen Series (n 3)

Transition Series Limit Designation P P P P B P: 1.51 eV Wavelength 1876 nm 1282 nm 1094 nm 1005 nm 955 nm 820.6 nmColor IR IR IR IR IR near IR

Brackett (4) limit 1458.03 nm; Pfund (5) limit 2278.17 nm; Humphreys (6) 3280.56 nm

Ex 21.) The Paschen series for hydrogen corresponds to transitions that end in states with quantum

number n = 3. The shortest wavelength line in the Paschen series is closest to which of the

following? (The ionization energy of hydrogen is 13.6 eV and hc = 1240 eV nm)

(A) 125 nm (B) 250 nm (C) 400 nm (D) 800 nm (E) 1800 nm

The shortest wavelength would correspond to the transition with the largest E from the highest

bound level down to the n = 3 level. You are given 13.6 eV; You must deduce En = -13.6 n-2.

2 2

1 1 13.613.6

3 9

eVcE h eV

; 9 9 1240

80013.6 13.6

hc eV nmnm

eV eV

(22) Blackbody radiation



Total power per area = Stephan Boltzmann constant times T to the fourth power = T 4

Wien’s Displacement Law: peak T = constant = 2.898 x 106 nm K (or vmax = constant * T)

Planck’s Law: 3

2

2 1( )

1hv

kT

hvI dv dv

c e

2

5

2 1( )

1hc

kT

c hI d d

e

Wien: I() is a function of T so the peak occurs

for a set value of T peakT = constant.

Stephan: 0

( ) ?I dv

change of variable

33 3

2 2

2 1 2 1

11hv u

kT

hv kT hu kTdv du

c h c e he

4 34 4

3 20 0

2( )

1u

k u duI dv T T

h c e

4 3

3 2 0

2

1u

k u du

h c e

Ex 22.) Which of the following describes the effect of doubling the absolute temperature of a

blackbody on its power output per square meter and on the wavelength where the radiation

distribution is a maximum?

(A) The output power is increased by a factor of 16 and the maximum of the distribution shifts to

twice its original wavelength.

(B) The output power is increased by a factor of 16 and the maximum of the distribution shifts to

half its original wavelength.

C) The output power is increased by a factor of 8 and the maximum of the distribution shifts to twice

its original wavelength.

(D) The output power is increased by a factor of 8 and the maximum of the distribution shifts to half


(E) The output power is increased by a factor of 2 and the maximum of the distribution shifts to four


Higher temperature suggests higher energy and hence shorter wavelength. Answers (B) and (D) are



viable! The power increase factors of 24 or 23 for (B) and (D). The fourth power should be familiar

as it is Stephan’s law, Power T 4. Wien: peakT = constant. Doubling the temperature moves the

peak to one half of the wavelength. B

hcE hv 1 eV 1240 nm

1 eV photon = 1240 nm 2 eV photon = 620 nm; 4 eV photon = 310 nm

hc = 197 eV·nm (atomic physics) = 197 MeV·fm (nuclear physics)

color (nm) v (1012 Hz) E(eV)

IR > 700 < 428 < 1.77

red 625-700 405-480 1.77-2.0

orange 585-620 484-513

yellow 585-570 513-526

green 570-505 526-594

blue 500-440 600-682

violet 440-400 682-750

UV < 400 > 750 > 3.10

ranges are approximate and arbitrary no universal agreement on the ranges

Ex 23.) Let H denote a Hermitian operator and suppose that H |= a |, where | is an

eigenvector of H. Which of the following is true of the eigenvalue a? (The symbols Re and Im

denote the real and imaginary parts, respectively.)

(A) Re(a) = Im(a)

(B) Re(a) = - Im(a)

C) Re(a) = 0

(D) Im(a) = 0

(E) H = a

SP352 Knowledge points: The eigenvalues of a Hermitian operator are real. The eigenvectors associated with

distinct eigenvalues are orthogonal. A Hermitian operator can be diagonalized by a unitary transformation. D

Ex 24.) The quantum numbers used to label the radial wave function solutions to the Shroedinger



equation for hydrogen atom are the principal quantum number n and the angular momentum

quantum number . If the principal quantum number is n = 2, which of the following gives the

possible values for the angular momentum quantum number ?

(A) 1, 0 (B) (C) 2, 1, 0 (D) (E) 3/2, 1/2

The states of the hydrogen atom are labeled by the quantum numbers n, and m plus the intrinsic

spin. The allowed values are n = 1, 2, 3, ….; = 0, 1, … , (n -1); m = -, - +1, … , 0, 1, …, ).

That is: m assumes the 2+ 1 values spaced by 1 running from - to . Note: n and are

For n = 2, can be 0 or 1 and m can be -1, 0, +1. A Each combination can be combined with spin

up or down. The value is the orbital angular momentum which assumes integer values. The total

angular momentum is the orbital plus spin, and it assumes half-integer values for an electron in hydrogen.

Ex 25.) The three operators (Lx, Ly, Lz) for the components of the angular momentum commute with

the Hamiltonian of a particular particle. Therefore, the angular momentum of the particle is

(A) equal to zero

(B) equal to the energy in magnitude

(C) always equal to Lz

(D) a unit vector

(E) a constant of the motion

The Hamiltonian is the energy and time-development operator for a quantum system. It results that

[ , ]d i

Q H Qdt

for time-independent operators. Quantities with time-independent operators

that commute with the Hamiltonian are constants of the motion. E A quantum state can be

labeled with good quantum numbers corresponding to it eigen-energy quantum number and quantum

numbers each additional member of the largest set of mutually commuting operators that includes

the Hamiltonian. For example, the members of the set (Lx, Ly, Lz) do not commute with one another.

They do commute with L2 = Lx2 + Ly

2 + Lz2. Considering that it is a given that (Lx, Ly, Lz) commute

with the Hamiltonian, then L2 will also commute with the Hamiltonian. In such a case, there would

be good quantum numbers corresponding to H, L2 and one of the components of angular momentum.



Lz has the simplest representation in spherical coordinates so it is usually chosen. For the hydrogen

atom problem, the quantum numbers n, and m are associated with eigenvalues for H, L2 and Lz.

Ex 26.) A particle of energy E is in an eigenstate of the square well potential shown above, with

wave function (x). Which of the following is a correct expression for the expectation value of x2

for this particle?

(A) 2 ( )a

ax x dx

(B) 2 ( )x x dx

(C) *( ) ( )x x x dx

(D) 2*( ) ( )a

ax x x dx

(E) 2*( ) ( )x x x dx

The correct procedure to compute an expectation value is to slap the operator for the quantity

between *(x) and (x) and then to integrate of the full range of x for which x) is non-zero. This

square well has a finite depth so the particle penetrates into the classically forbidden region |x| > a. In

this case the integrals must run over the full range (-,). E

SPECIAL TOPICS:

(27) The Pauli Matrices:



x = 0 1

1 0

; y = 0

0

i

i

; z = 1 0

0 1

Ex 27.) Given the Pauli spin matrices shown above and the operators defined by + = x + i y and

- = x - i y, which of the following is NOT correct.

(A) +1 0

0 0

(B) +

0 12

1 0

(C) -

1 02

0 1

(D) z

1 1

0 0

(E) z

0 0

1 1

- = 0 0

2 0

0 0 1 0

22 0 0 1

C is false

Method: Realize that you have no idea what the problem is about, but also realize that it just

requires some matrix multiplication. Find the explicit forms for + and -, and brute force compute

each matrix product until you find the one that is NOT correct.

Background: ** Skip to the next problem if you are in a hurry. (28) Counting Statistics

Suppose that a set of 2 x 2 matrices is sought such that any 2 x 2 matrix can be represented as a sum

of the members of that set. Clearly the set must have at least four members as 2 x 2 matrices have

four independent elements. One choice that could be made is:

1 0 0 1 0 0 0 0, , ,

0 0 0 0 1 0 0 1

This set of matrices is not very exciting. Spice it up; add the requirement that each member of the set

be Hermitian (equal to the complex conjugate of its transpose). The simplest matrix that has

imaginary elements that meets this requirement is:

0

0

i

i

An independent off-diagonal matrix needs to have elements that are equal rather than the negative of

one another. Equal off-diagonal elements and Hermitian restricts the elements to be real.

0 1

1 0

Following the equal and negative scheme for the on diagonal matrices, the remaining members are:

1 0

0 1

and 1 0

0 1



This particular set of 2 X 2 matrices is the set of Pauli matrices. They have assigned labels:

= 1 0

0 1

; x = 1 =0 1

1 0

; y = 2 =0

0

i

i

; z = 3 = 1 0

0 1

The first matrix is the identity, and the final three are the Pauli matrices. The set of the four matrices

is a basis for the collection of all 2 X 2 matrices.

1 0 0 1 0 1 0

0 1 1 0 0 0 12 2 2 2

a b ia d b c c b a d

c d ii

An arbitrary 2 x 2 matrix can be represented as a linear combination of the four matrices, but not as

linear combination of any set with fewer than four members. (A basis must be complete in the sense

that all elements (matrices) of interest can be represented as linear combinations of the members of

the set and economical in the sense that every member is necessary. If even one member of the set is

removed then there will be at least one 2 x 2 matrix that cannot be represented as a linear

combination of the remaining members.)

Why are the Pauli matrices introduced?

The Pauli matrices are introduced to act on two-row column matrices. Matrices of the forms:

a

b

, 1

0

and 0

1

. The column vectors 1

0

and 0

1

are a basis set for all thea

b

. The column

vectors 1

0

and 0

1

are to be called (spin) UP () and (spin) DOWN ().

If you are not familiar with modern physics, skip this paragraph.

In quantum mechanics, the Pauli matrices add the flexibility that allows a wavefunction to represent

the spin character of an electron. A two row column vector is appended to a function of position and

time.

( , )a

r tb

The column vector 1

0

represents spin up while 0

1

represents spin down. The spin part is to be



normalized independently so: * *

* * 1a a aa b

a a bbb b b

†

.

Special properties of the Pauli matrices:

1 1 = 2 2 = 3 3 =

The determinants are each of the three Pauli matrices is – 1. The determinant of the identity is, of

course, equal to one.

| 1| = | 2| = | 3|= 1

Actions of the Pauli Set:

x 1

0

= 0 1 1 0

1 0 0 1

; x

0

1

= 0 1 0 1

1 0 1 0

The operator x lowers the UP state to DOWN and raises the DOWN state to UP.

y 1

0

= 0 1 0

0 0 1

ii

i

; y

0

1

= 0 0 1

0 1 0

ii

i

The operator y lowers the UP state to i times the DOWN and raises the DOWN state to –i times UP.

z 1

0

= 1 0 1 1

( 1)0 1 0 0

; z 1

0

= 1 0 0 0

( 1)0 1 1 1

The operator z on the UP state to 1 times the UP and, on the DOWN state, returns – 1 times the

DOWN state. The operator z measures the up or down character of the state.

The combination of operators + = x + i y annihilates UP and raises DOWN to 2 times UP.

+ 1

0

= 0 2 1 0

0 0 0 0

; +

0

1

= 0 2 0 1

20 0 1 0

The operator + is called the raising operator.

Exercise: Find the action of - = x + i y on the states UP and DOWN. Propose a name for -.



** The Pauli matrices are multiplied by ½ when they are to be used in anger. They become the

operators for the components of the electron’s spin angular momentum.

z =½ z = ½ 1 0

0 1

; 2 =(½ )2 [ x2 + y

2 + z2] = 3/4 2

1 0

0 1

The eigenvalues of z are ½ , and the eigenvalue of 2 is 3/4 2 = s(s + 1)2

(28) Counting Statistics

We will assume that the quick and dirty answers are enough to fake our way through the problem.

Memorize: When a counting experiment yields N counts, the uncertainty in that number is N ½ (or a

fractional uncertainty of N -½ large count numbers are relatively more precise.). The value N ½ is

the expected standard deviation. If the counting experiment is repeated many times, the expected

results are assumed to follow a standard (Gaussian) distribution.

68% of the time the result is within

95% of the time the result is within 2

99.7% of the time the result is within 3

The average count number is N , and we assume

the standard deviation is N ½.

** Know: 68% within and 95% within

2( )[ ]1( )

2½ n N

p n e

https://en.wikipedia.org/wiki/Normal_distribution



Ex. 28) An experimenter measures the counting rate from a radioactive source as 10,150 counts in

100 minutes. Without changing any conditions, the experimenter counts for one minute. There is a

probability of about 15% that the number of counts recorded will be less than

(A) 50 (B) 70 (C) 90 (D) 100 (E) 110

The first line established the average count rate as 101.5 per minute. We expect 101.5 counts in one

minute on average. The standard deviation of the count number in one minute is assumed to be the

square root of that value or 10.075. There is a 68% probability that the counts falls between 101.5 –

10.075 and 101.5 + 10.075 or a 32% chance that the number of counts is outside this range. The

counts can be high or low, so there is a 16% chance that the number of counts in a minute is less

than N - = 101.5 – 10.075 = 91.42. about 15% likelihood to be less than 90. C

We expect 101.5 counts means that if the one minute count were repeated a large number of times

that various integer results would be recorded each time. The results 101 and 102 would occur about

equally, and larger and smaller values would occur at a lower frequency. The values around N

occurring about 0.6 times as often as those around N . 2( )[ ]1

( )2

½ n Np n e

, e-½ 0.6.

(29) Graphs and Basic Lab

x(t) = 0.3190 t 2 + 0.0013

Ex.29) A motion sensor is used to measure the

position x versus time t for a cart traveling down

a ramp. A spreadsheet is then used to make a

linear fit to the plot of x vs. t, as shown in the

graph to the left. The equation for the best fit line

appears on the graph. Which of the following

gives the acceleration of the cart?

m/s2 m/s2 Cm/s2 Dm/s2 Em/s2



This one is a gift! The acceleration is the second derivative of x(t) with respect to t or 0.6380.

Adding dimensions, x(t) = 0.3190 m/s 2 t 2 + 0.0013 m yielding a = 0.6380 m/s 2. D

Whenever possible, work from the definition. You are always correct to do so.

Alternative: Compare with the kinematical equation x(t) = ½ a t2 + vot + xo. ½ a = 0.3190 m/s 2

(30) Lagrangian and Hamiltonian Mechanics:

We restrict our treatment to the simple cases for which the constraints are good, the potentials are

time independent and for which there is no dissipation.

The lagrangian L is T – U, the kinetic energy minus the potential energy. The lagrangian is a natural

function of the coordinates qi, the coordinate velocities iq and time t. The natural variables are the

ones that you must use!

The Lagrange equations of motion are:

i i

d L L

dt q q

The signs and details can be recovered by considering: 2( , , ) ½ ( )L x x t m x V x .

xd L L V

m x Fdt x x x

(recovered Newton’s 2nd Law)

It is crucial to note that the time derivative is a total derivative while the others are partial

derivatives.

The momentum conjugate (associated with) a coordinate is found as: ii

Lp

q

. Using the example

2( , , ) ½ ( )L x x t m x V x , xp m x . Another familiar result is 2p m r which appears for the

planar orbit problem for which 2 2 2( , , , ) ½ [ ] ( )L r r m r r V r .

Conservation or constants of the motion:

ii

Lp

q

i i

d L L

dt q q

ii

d Lp

dt q



The momentum pi is a constant of the motion if the lagrangian does not depend on qi, the coordinate

conjugate to that momentum. Given: 2 2 2( , , , ) ½ [ ] ( )L r r m r r V r , there is no dependence of

the angular coordinate so the conjugate momentum 2p m r is a constant of the motion.

In Hamiltonian mechanics, the coordinates and the momenta are the natural variables. The

hamiltonian is defined to be H = i

i ip q - L. Here pi is the momentum conjugate to qi as defined

based on the lagrangian above. For the simple case, 2( , , ) ½ ( )L x x t m x V x , we find xp m x .

H(x, px) = 2 2(½ ( )) ½ ( )x xp x m x V x p x m x V x . We find:

2 2

2 (( , ( )) ½ )xxp

mH x p m x V Vx x

The hamiltonian must be expressed as a function of its natural variables, the coordinates and

momenta, so the second form is the one that must be used. For time-independent problems (the type

usually encountered), the hamiltonian represents the total energy of the system.

The equations of motion are: ii

Hq

p

and ii

Hp

q

. Using the example:

2

2( , ) ( )xxp

mH x p V x ,

xi

i

pHq x

p m

and xii

H Vp m x F

q x

Constants of the Motion: The relation ii

Hp

q

shows that pi is a constant of the motion of the

hamiltonian is independent of qi. (Recall that we are restricting our attention to hamiltonians that do

not depend explicitly on time.)

Ex. 30) Two masses m1 and m2 on a

horizontal straight frictionless track are

connected by a spring of spring constant k, as

shown in the figure to the left. The spring is

initially at its equilibrium length. If x1 and x2

give the displacements of the masses from their equilibrium positions, the lagrangian L for the

system is given by which of the following? The dot denotes differentiation with respect to time.



Note that the generalized coordinates chosen (see the figure) are the displacements of each mass

from equilibrium. This choice is standard choice. The lagrangian is to be the kinetic energy minus

the potential energy. The kinetic energy is the sum of the kinetic energies of the two particles, T (or

K) = 2 21 1 2 2½ ½m x m x , and the potential energy is ½ times the spring constant times (x2 – x1)

2, the

stretch of the spring squared. U = ½ k (x2 – x1)2. Note that it helps to roll the words around in

your mind as you attempt to decipher a problem.

Conclusion: 2 2 21 2 1 2 1 1 2 2 2 1( , , , ) ½ ½ ½ ( )L x x x x m x m x k x x B

Note that four of the candidates identify the kinetic energy as: 2 21 1 2 2½ ½m x m x , and three identify

the potential as ½ k (x2 – x1)2. Only two have the negative sign for L = T – U. Possibility (E)

includes the awe-inspiring reduced mass; it has the wrong sign for the potential. In addition, (E)

depends only on the relative position and relative velocity. That is: the center of mass motion has

been suppressed. For example, if q = x2 – x1, then (E) is ½ 2q + ½ k q2. Forgetting that it should be

½ 2q - ½ k q2, there is the issue of the kinetic energy.

2 21 1 2 2½ ½m x m x = 2

1 2½( )m m X + ½ 2q where X is the center of mass velocity.

Kinetic energy in terms of the center of mass and relative coordinates.

Let V

be the velocity of the center of mass of a system of particles, iv

be the velocity of particle i

and iu

be the velocity of particle i relative to the center of mass.



2 2 2or ½ ½ ½total CM relative i i i ii i

K K K m v M V m u

For a system with only two particles, the reduced mass can be used to get the special form:2 2 2 2 2 2

1 1 2 2 2 1½ ½ ½ ½ | | ½ ½ relm v m v M V v v M V v

where M = m1 + m2, 1 2

1 2

m m

m m

and 2 1relv v v

Note that for rigid bodies the motion decomposes into the center of mass motion plus rotation about the CM. 2 2 2or ½ ½ ½total CM relative i i CM

i

K K K m v M V I

(31) Nuclear Decays

Note that the inset image expands the energy scale.

Each nucleus has Z protons (aka atomic number) and N neutrons with a mass number of A =Z+N. A

particular nucleus is denoted AZ NX ; such as 14

6 8C , however the Z and N are usually omitted because

they are redundant and are known from A and the symbol used for X.

particle charge nucleon # lepton # nucleus ZXA Z A 0

photon 0 0 0 alpha 2 4 0

beta minus e- -1 0 +1



beta plus e+ +1 0 -1 electron-neutrino ve 0 0 +1

electron-antineutrino ev 0 0 -1

An alpha particle is a tightly bound system of 2 protons and 2 neutrons (A=4). The very high total

binding energy (~28 MeV; 7 MeV per nucleon) of the alpha and its very high first internal excited

state make it favorable to be treated as a unit object. (The same could be said for the proton and

neutron which in reality are quark composites.) The maximum binding energy per nucleon occurs

for 56Fe and is about 8.8 MeV. Elements to the left can in principle fuse to form 56Fe and those to the

right can fission.

Nuclei will spontaneously decay into another nucleus if the final system has lower rest energy than

the initial. There are four decay modes; gamma decay, beta decay, alpha decay, and fission. The

conserved quantities are total energy, linear and angular momentum, along with nucleon number and

lepton number. The rest-mass energy difference between the initial and final systems is called the

Q-value and appears as kinetic energy and internal excitation energy of the products.

In gamma decay, an excited nucleus decreases its level of excitation by emitting a gamma ray

(photon)The process may be denoted: A(X*) AX + 0 or just AX + The number of neutrons

and protons is unchanged. No leptons are involved so lepton number need not be considered.

(Leptons: electron, positrons and neutrinos and . -- for nuclear)

In alpha decay, a nucleus emits an alpha particle (2 protons and 2 nucleons for a total of 4

nucleons). The final state nucleus, the daughter, has Z–2 protons, N-2 neutrons and A-4 nucleons.

This can be denoted AZ X 4

2AZ Y + 4

2 or just A-4Y + . No leptons are involved so lepton number

need not be considered.

mn =1.675 x 10-27 kg = 939.57 MeV/c2;

mp =1.672 x 10-27 kg = 938.27 MeV/c2

me =9.11 x 10-31 kg = 0.511 MeV/c2

There are three fundamental processes in beta decay. For convenience they can be considered to

be (with Q-values given).

beta minus decay 0.78en p e v MeV



beta plus decay ( 1.8 )ep n e v MeV

electron capture ( 0.79 )ep e n v MeV

Beta minus decay can occur for an isolated neutron, but because the Q-values for the remaining two

processes are negative, the lower two reactions can only occur in regions where the proton is

‘interacting with an energetic background’; such as in a nucleus. Electron capture occurs when a

‘proton in the nucleus grabs an electron out of its s-state orbit’.

Note that the reactions above conserve charge, nucleon number, and lepton number.

For nuclei undergoing beta decay, the reactions would be written,

beta minus decay AZ X 1

AZ Y + 0

1 e or just 1A

Z Y + e. + ev

beta plus decay AZ X 1

AZ Y + ee

electron capture AZ X 1

AZ Y + e

Note that the reactions above conserve charge, nucleon number, and lepton number. The existence

of three-bodies in the final state means that the available energy or Q value of the reaction is not

apportioned to the particles in a set ratio, but rather the particles of each type emerge with a

distribution of energies.

In fission, the nucleus breaks into two large pieces (fission fragments) and a few neutrons with Q-

values on the order of 200 MeV. A commonly used radioactive fission source is 252Cf. The nuclei 235U and 239Pu do not themselves spontaneously fission, but fission can be induced by striking them

with a neutron. The conserved quantities are charge, total energy, linear and angular momentum,

along with nucleon number and lepton number

Ex. 31) Thorium with atomic mass 228, decays by alpha emission to a daughter nucleus which also

decays by alpha decay to radon. Which of the following is true of the decay product, radon? (The

atomic number of thorium is 90.

Atomic Mass Atomic Number

(A) 220 82

(B) 220 86

(C) 224 82

(D) 224 88



(E) 228 91

The process is a sequence of two alpha decays:

AZ X 4

2AZ Y + 4

2 84

AZ Y + 4

2+ 42

Atomic Mass: A = 228 Afinal – 4– 4

Atomic Number: Z = 90 Zfinal = 90 – 2 – 2 = 86

228Th 224Ra + 220Rn + +

Thorium Radium + alpha Radon + alpha + alpha

(32) Spontaneous Particle Decays:

As with other decays, the electric charge, energy, linear and angular momentum must be conserved.

Because there is only one initial particle, the Q-value for the reaction must be positive. The excess

energy or Q appears as kinetic energy and excitation energy of the products. Anti-particles have the

same rest energy as their corresponding particles.

Muons and taus are heavy electrons each with its associated neutrino and anti-neutrino. Conserve

electron, muon and tau lepton numbers separately! me = 0.511 MeV/c2; m = 206 me; m = 3478 me

mpc2 = 938.27 MeV * mnc

2 = 939. 57 MeV * mnc

2 - mpc

2 = 1.3 MeV > mec2

mec2

= 0.511 MeV * mc

2 = 106 MeV * mc

2 = 1775 MeV mc

2 = 0 *

mc2

= 140 MeV charged; 135 MeV neutral mc2

= 0 *

Be aware of the values marked * and that a pion is a little more massive than a muon.

Ex. 32) Which of the following decays is possible in a vacuum? (A) + - +

(- + + e- + e+

(C)0 e- + p

(D)p n + e+ + e



(E)n p + e- + e

(A) Fails. Does not conserve charge. Q-value > 0 is OK.

(B) Fails for two reasons. Does not conserve charge. Does not conserve energy because Q<0, that

is, the products have more mass than the initial particle.

(C) Fails. Conserves charge, however the products have much more massive than the pion. Electron

lepton number is not conserved.

(D) Fails. Conserves charge and electron lepton number, however, the neutron is slightly more

massive than the proton. The mass of the products is greater than the mass of the reactants not a

spontaneous decay.

(E) Possible. Conserves charge and the Q > 0. An isolated neutron is unstable and decays as shown

with a lifetime of 886 seconds. Neutrons are stabilized in the nucleus by their binding energy which

is greater than the Q of decay for an isolated neutron. E

Because the fundamental building blocks are the 6 quarks (u,d,s,c,b,t), 6 leptons (e,ve,,v,,v) and

their anti-particles, there are actually additional conservation rules. These are not as trivial to apply

because one must be told the quark content of the composite particles.

Review the quark model section of your modern physics text if you have time.

http://hyperphysics.phy-astr.gsu.edu/hbase/Particles/quark.html

Quarks and Leptons are the building blocks which build up matter, i.e., they are seen as the "elementary particles". In the present standard model, there are six "flavors" of quarks. They can successfully account for all known mesons and baryons (over 200). The most familiar baryons are the proton and neutron, which are each constructed from up and down quarks. Quarks are observed to occur only in combinations of two quarks (mesons), three quarks (baryons), and the recently discovered particles with five quarks (pentaquark).

Quark Symbol Spin ChargeBaryonNumber

S C B T Mass*

Up U 1/2 +2/3 1/3 0 0 0 0 360 MeV

Down D 1/2 -1/3 1/3 0 0 0 0 360 MeV

Charm C 1/2 +2/3 1/3 0 +1 0 0 1500 MeV

Strange S 1/2 -1/3 1/3 -1 0 0 0 540 MeV



Top T 1/2 +2/3 1/3 0 0 0 +1 174 GeV

Bottom B 1/2 -1/3 1/3 0 0 -1 0 5 GeV

Up and Down Quarks

The up and down quarks are the most common and least massive quarks, being the constituents of protons and neutrons and thus of most ordinary matter.

The fact that the free neutron decays

and nuclei decay by beta decay in processes like

is thought to be the result of a more fundamental quark process

Particles and Intrinsic Spin

Intrinsic spin S

of most objects (electron, proton, neutron, quarks, leptons) is described by the label

(quantum number) s = ½ . The length is the intrinsic spin vector is calculated as ( 1)s s with

possible z-components of sm , where sm can have the values + ½ or – ½ . These z-components

are often referred to as spin-up and spin-down, respectively.

Particles with half-integer spins are fermions, particle that obey the Pauli exclusion principle, only

one particle per each distinct quantum state. The pattern of electrons filling atomic states is an

example of the one-per-state filling by fermions. The building blocks of matter are typically

fermions. Particles with integer-valued intrinsic spin are bosons which can, and in fact prefer, to

multiply occupy states. Examples of bosons include pions, photons, gluons and gravitons. Many

bosons are associated with forces or interactions. Those with zero rest mass give rise to forces with

infinite range (1/r2) forces such as the Coulomb and gravitational forces. Quantum states with many

bosons in the same or nearby states exhibit classical behavior; an example would of electromagnetic

waves with fluxes of many, many photons per meter2-sec. Interactions exchanging massive bosons



decrease in strength exponentially with increasing separation. Interactions (forces) mediated by

massive bosons are short range forces ( r-2 e –mr). More massive shorter range. (m 10-15 m)

Mediator Spin Force gluon 1 quark-quark photon 1 charge-charge

Z, Wbosons 1 weak decay graviton 2 mass-mass

Composite Particles: A group of particles that is bound together acts like a particle of the net spin

as long as it is probed only weakly compared to its binding energy. Pairs of quarks can bind to form

mesons (bosons, ½ and ½ integer value). Triplets of quarks for fermions such as protons and

neutrons (fermion, ½ and ½ and ½ ½ integer value). Effective bosons such as helium atoms (2

protons + 2 neutrons + 2 electrons integer) can collect in large numbers in a single (ground) state,

the condensate, leading to the superfluid state of liquid helium below 2.2 K. Electrons bound in pairs

(bosons) form analogous condensates in super-conductors.

A nucleus can be modeled as a composite particle (with binding about 8 MeV per nucleon) if it is

probed weakly using say electric and magnetic fields. If you shoot a 40 MeV proton at it, the probe

will see the inner structure invalidating the composite particle model.

Condensed Matter

Ex. 33) For atoms in a simple cubic structure, the maximum percentage of the total available volume

that can be occupied by the atoms is approximately

(A)

(

(C)

(D)60%

(E)70%

The idea is to stack solid spheres in the prescribed pattern and to compute the fraction of the total

volume that is in the spheres. A cube has eight corners and a side length s. A sphere of radius ½ s

can be centered on each corner. One eighth of each sphere will be in the cube times eight corner or

one sphere of radius ½s per cube of side s. One complete sphere of radius ½ s per cube of volume s3.

(Prepare a sketch!)



3

3

4 (½ ) 1 40.52

3 3 8 3½s

fractions

C

The closest packing of hard spheres (74%) is achieved in the hexagonal close-packed and face-

centered cubic configurations. Body-centered cubic is about 68 %.

simple cublic body-centered cubic face-centered cubic

hexangonal c.p.

MISCELLANEOUS TOPICS:

(34) Astro-Knowledge

Ex. 34) Which of the following gives the distance in light years to Andromeda (M31), the spiral

galaxy nearest to the Milky Way?

(A) 2 x 100

(2 x 102

(C)2 x 104

(D)2 x 106



(E)2 x 108

The earth is 8 light minutes from the sun.

The sun is 4 light years from the nearest star Proxima Centauri.

Galaxies have lots of stars so Galaxies should be thousands of light years in dimension ( ).

During a 1965 Star Trek episode, science officer Spock remarked that it was 40,000 light years to

the galactic center (Milky Way, our galaxy). Google says that the main disk of the Milky Way is

about 105 light years in diameter. Galaxies should be separated by to 100 galaxy diameters. Look

for 10-100 x 105 D

The universe is 14 x 109 years of age and so is about 14 x 109 light years in extent.

The separation of galaxies should be pretty small compared to the extent of the universe

(< 10-3*Universe 107 lt. yr.) so response E is not attractive. Possible range: 40,000 x 10 < answer <

10-3 (14 x 109). The geometric mean: 10 3(40,000) *10 14 10 *10 = 2.37 x 106 lt. yr.

(35) Math Methods (Hardly a challenge for a 351/2 survivor.)

Ex. 35) Which of the following is the derivative with respect to x of the function x2 cos[3 x4 + 1]?

The function is the product of x2 and cos[3 x4 + 1] so by the product rule, df/dx = 2x cos[3 x4 + 1] + x2 d/dx(cos[3 x4 + 1])

Using the chain rule, d/dx(cos[3 x4 + 1]) = - sin[3 x4 + 1] (12 x3)

Combining terms: df/dx = 2x cos[3 x4 + 1] – 12 x5 sin[3 x4 + 1] C

All students should be intimately familiar with every concept and detail presented in SP351/2.



Astrophysics Annex (Jeff2)

While the MFT contains "astrophysics" questions, don't think that you cannot answer them without

any classes in the subject! Yes, last year some questions involved stars, but stars were simply a

convenient real-world blackbody for Wein's Law and the Stefan Boltzmann Relation -- actually just

thermodynamics! This section is a quick introduction to the types of "astrophysics" questions

which may be expected and some explanation of some of the exotic units which might be used in the

problem.

Angles: Astronomical measurements typically use

angles to report the apparent size or separation

between two objects (as well as to establish a

coordinate system on the sky). The formula for

angular size or separation is:

tan = s/d

where d is the distance to the object and

s is either a physical size of an object or the

distance separating two objects.

Many of these angles are smaller than a degree and

have time-based meanings as well so follow the analogy. 1 degree can be split into smaller units

(arcminutes, or ') where 60' = 1 degree. 1' is the finest angular separation an average eye can discern.

An arcminute is also split into 60 arcseconds (or "). In short: 1o = 60' = 3600".

Distance: Distance is hard to measure precisely in astronomy and in astrophysics and the exotic

units used typically hint at the technique used in the measurement. With the exception of light-year

I would expect to find any of the other units defined on the exam.

1 Astronomical Unit: The distance from scaled to the distance of the Earth to the Sun: 1.496 X 108

km



1 Parsec: The distance from the Sun to a star with an annual parallax angle of 1": 3.09 X 1013 km

1 Light X (where X= a unit of time like year, day, second) distance is from light travel time: Use

d=ct, where t is the appropriate X in seconds. 1 Parsec 3.26 LtYr

Escape Velocity: An object trapped in a gravitational potential well of a larger primary object has

a negative gravitational potential energy. In order to escape (to infinite distance) where the potential

energy is zero and the kinetic energy must be 0. If the mass of the object is m, the mass of the

primary object is M and the separation between the two is r, the escape velocity (ve) is determined

by:

2 20 ½ 0e e

GMm GMKE PE m v v

r r

Kepler's Laws: Gravity drives the universe. Planetary motions

are typically described in terms of Kepler's laws:

1) All planets move in elliptical orbits with the Sun at one focus.

2) The line that joins a planet to the sun sweeps out equal areas

of their orbit in equal times. (Closer to the Sun you move more

quickly -- angular momentum conservation; m r v )

3) For objects of planet size or smaller orbiting the Sun, if we

express the period (P) in years and the average distance

(a, semi-major axis) in astronomical units then P2 a3.

How can years squared equal astronomical units cubed? There is

a hidden scaling! Kepler didn't know Newton's laws, just what

he could observe which was periods and astronomical units (both

defined in terms of the Earth's 1 year and 1 AU). Try

approaching Kepler's Third Law by deriving it from Newton. Consider a planet (mass m) in circular

motion at distance R around the Sun (mass M, M >> m). The only force acting on the planet is

gravity and it creates a centripetal acceleration:



F= ma= GmMR2 but for circular motion, a=

v2

R so

v2= GMr . But the motion is circular so 2circumference R

Pperiodv and 2 3 24 R G M P

Okay, but that also means that 2 3

2

41

R

G M P

, so for two planets with distances R1and R2 and periods

P1 and P2, 2 3 2 3

1 22 2

1 2

4 41

R R

G M P G M P

. Cancelling,

3 31 22 2

1 2

R R

P P or 2

2 11

32RP P R

. Choosing planet 1 to

be the earth, R1 = 1 AU and P1 = 1 year, and 22

32

.(1 . .)RP yrAU

Blackbodies: Stars are blackbodies as discussed in section (22). They have a wavelength at which

they emit the greatest intensity of light (given by Wein's law) and they emit a total amount of

radiation consistent with the Stefan Boltzmann relation.

Wein's law: peak TKelvin = 2897 m or fpeak TKelvin

Stefan Boltzmann: irradiance (radiative energy flux density) = T4; = 5.67 x 10-8 J/(s m2K4)

Stellar Spectra: The composition of stars can be determined because the inner core of a star

radiates like a blackbody. The outer cooler parts of the star absorb certain colors of light due to

radiative excitation and the rest of the spectrum makes it to us minus those colors. The exact

wavelengths of light removed are determined using the techniques in section (21). Each element

has a unique set of energy levels, therefore we can tell what elements are present based on their

spectral lines.

Stellar Structure and Evolution: Stars are held together by gravity but are prevented from

collapsing by fusion in their cores. This balance of pressures keeps them stable. For the majority of

their lives, stars turn H into He by fusion and these stars are called "main sequence". It is possible to

fuse heavier elements (most stars are massive enough to generate sufficient pressures and



temperatures to fuse He into C late in their lives when they have depleted the H supply in their

cores).

Stars are in hydrostatic equilibrium which you will remember from your general physics class.

2

( )G M r rdP

dr r

In this equation, P is pressure (force/area), r is distance from the center, M(r) is the net mass within r

or the center (net mass located inside a sphere of radius r concentric with the , ρ(r) is the density at

distance r and G is the universal gravitational constant.

Distances (actually best expressed as light travel time):

Moon: When you listen to the communications of the Apollo astronauts there is a couple second lag

between question and answer. They are not thinking hard -- the moon is approximately 1 light

second or 300,000 km away and the pause is the travel time of the radio waves.

The Sun is 8 light minutes away. If the Sun suddenly stopped producing energy, it will still take 8

minutes for the Earth to know it. Incidentally the sun is about 1 light second in radius as well.

Pluto is the edge of the planets we accept in the solar system. It would take 4.5 hours for light to

reach Pluto and twice that time to have a round trip. As we explore the solar system with robots, this

time lag has to be taken into account.

The next star is approximately 4 light years away. In the Galaxy stars are typically

1 light year apart so we live in an uncrowded neighborhood. That's good, because interactions with

stars are relatively disruptive to planetary orbits... Also stars are small compared to their separations

-- a star is a light second across but a light year away from its neighbor.

Our Milky Way Galaxy is approximately 100,000 light years across (We are 26,000 light years from

the center) and contains 100 billion stars. By rough coincidence there are approximately 100 billion

galaxies in the universe so that there 10 thousand billion billion stars in the universe.

Galaxies are large compared to their separation. Our galaxy is 100,000 light years across but the

next large galaxy Andromeda is only 2 million light years away. Forget about a round trip

conversation, though.



The edge of the observable universe is still debated is between 14 and 180

billion light years across. I'd personally bet on the lower side of

that...

--------------------------------------------------------------------------------

Times

Age of universe - about 14 billion years old Age of the Sun - about 4.5 billion years old Lifetime of

the Sun as a Hydrogen burning star - about 9 billion years Time for the Moon to go around the Earth

- 27 days. Time for the Sun to rotate: 27 hours

Time for a Sun-sized pulsar to rotate: 1 millisecond Time for the Earth to go around the Sun -

365.25 days Time for Pluto to go around the Sun - 248 years. Time for the Sun to go around the

Galaxy - 200 million years Time when Andromeda collides with the Milky Way - 3 billion years

- The sun is still a hydrogen burning main sequence star!

---------------------------------------------------------------------------------------

Alternately try the "Galaxy Song" by Monty Python from the meaning of life (which can be

downloaded in mp3 from http://www.mwscomp.com/sound.html):

Just remember that you're standing on a planet that's evolving And revolving at nine hundred miles

an hour, That's orbiting at nineteen miles a second, so it's reckoned, A sun that is the source of all

our power. The sun and you and me and all the stars that we can see Are moving at a million miles a

day. In an outer spiral arm, at forty thousand miles an hour, Of the galaxy we call the 'Milky Way'.

Our galaxy itself contains a hundred billion stars. It's a hundred thousand light years side to side.

It bulges in the middle, sixteen thousand light years thick, But out by us, it's just three thousand light

years wide. We're thirty thousand light years from galactic central point. We go 'round every two

hundred million years, And our galaxy is only one of millions of billions In this amazing and

expanding universe. The universe itself keeps on expanding and expanding In all of the directions it

can whizz As fast as it can go, at the speed of light, you know, Twelve million miles a minute, and

that's the fastest speed there is.

So remember, when you're feeling very small and insecure, How amazingly unlikely is your birth,

And pray that there's intelligent life somewhere up in space, 'Cause there's bugger all down here on

Earth.



Additions:

1. Give definitions of the following terms in prose form. Use complete sentences.

Message: Gather more general knowledge about the topic. *** Be absolutely sure that you know

everything that earn the designation LAW. In particular know Maxwell’s equations.

What fraction of these question can be answered using only chapter summary material?

a.) Write down the equation that represents the Biot-Savart Law:

02

ˆ

4

Idd

r

s rB

b.) Are there magnetic charges? Write down the Maxwell equation that provides the answer to

this question.

No. in

0

qd E A

shows that electric field lines are started and stopped by electric charges. The equation

0d B A

shows that magnetic field lines do not start or stop; that is there are no isolated magnetic

charges (monopoles). The simplest magnetic field pattern is a dipole pattern.

c.) List a device that works by using an induced (not a motional) emf.

Transformer, inductor, wire loop antenna for a radio.

d.) Define the current density J

. The current density is the current per cross sectional area in a

conductor; it has the direction of the current. d

IJ nq

A v ; dJ nq

v

e.) Identify a device that operates in a manner described by the equation sinNAB t E .

This is the generator equation. N turn coil of area A spinning at in a field of strength B.



2. In the circuit shown, a current of 3 A is

directed up through the 15 V battery. The

reference potential is set to zero at the

ground symbol.

a.) Find the potential at points A, B and C. The potential is zero at the center of the lower conductor. It changes by +15V crossing the battery in

the left edge. VA = 15V. The potential drops (3A)(4) or 12V crossing the resistor to the right so VB

= 3V. Moving from B to C, the potential rises 9V across the battery in the right edge. VC = 12V.

VA = 15V. VB = 3V. VC = 12V.

The electrical power delivered to an element is the current through it times the potential drop across

the element in the direction of the current. The potential drops across a resistor so power is flowing

into the resistor in which it is dissipate (converted to thermal forms). The potential rises across the

battery (in the direction of the current through it) to it is providing power to the circuit rather than

receiving power from it. (If current is forced through a battery in the reverse direction, the battery is

being charged.)

b.) Give the power delivered by the 15 V battery. I V c = (3A)(15V) = 45 W.

c.) Compute the power dissipated in the 4 resistor. 2 2(3 ) (4 ) 36I V I R A W c



d.) Use the voltages found in part a.) to find the currents in the 3 and 6 resistors. Point B is 3 V above ground. 3V across 3 I3 = 1A down. Point C is 12 V above ground so I6 = 2A to the left.

***** The value of a current is not complete without its direction. ******

4

15V 9V3

6

A Node

C

Pick your own branch currents and give three Kirchhoff’s rules equations for this circuit that could

be solved to find the currents. All three have been chosen to be “up”. If the current in a branch is

“down”, the current found for that branch will be negative with the convention chosen.

Current Node Rule: I1 + I2 + I3 = 0

Voltage Rule Left Loop from lower left: + 15 V – 4 I1 + 3 I2 = 0

Voltage Rule Right Loop from lower left: - 3 I2 + 9 V + 6 I3 = 0

Substitute “– In” for each current that you chose to be in the opposite direction.

The easy solution follows by setting I1 = 3A.

3 A + I2 + I3 = 0

+ 15 V – 4 (3A) + 3 I2 = 0

The equation above immediately yields I2 = - 1 A. The node equations then gives I3 = - 2A. Solving

the equations in this manner is equivalent to walking the circuit.

Walk the circuit values:



4

15V 9V3

6

A Node

C

I1 I2 I3

3. Use Lenz’s law to answer the following questions.

a.) What is the direction of the induced current in the

resistor as the magnet is moved to the left?

b.) What is the direction of the induced current in the

resistor R immediately after the switch is closed?

c.) What is the direction of the induced current in the

resistor R as the current I decreases rapidly to zero?

d.) A rod is perpendicular to the magnetic field and it is

moved to the right as shown. The result is the charge

separation illustrated. What is the direction of the

magnetic field?

a.) The flux from the magnet is to the right and it decreases as the magnet is

moved to the left. The change in flux is directed to the left. To fight the

change, the induced current must cause a field to the right. By the RHR, the

current must come out of the page at the top of the coil. the current is to

the right through the resistor.

initial + change = final

b.) After the switch is closed, the current builds up to cause a B field to the left. The field goes from

zero to something to the left. The change is to the left. The induced current must cause a field to the

right. By the RHR the induced current must be into the page at the bottom side of the rod around



which the coil is wrapped. The current comes out of the page through the resistor (or a little left to

right).

c.) By the RHR, the B field circulates around the current so that it is into the loop containing the

resistor. As the current decreases, the flux into the loop decreases so the change in out of the page

through the loop. To fight the out of the page change, the induced current causes a field into the

page. The induced current must be clockwise to the right through the resistor.

d.) As a first try, we look for B perpendicular to the rod and to the velocity. Check the action for B

into and out of the page. If B is into the page, v B

is up the page so that is the direction that

positive charges are forced. Negative charges would be forced down the page. Look no further!

4.) The switch S in the circuit to the right has

been closed for a long time before it is

opened at time t = 0 s.

a. What is the current through R just before the

switch is opened?

b. What was the charge on the capacitor just

before the switch was opened?

= 50V ; R = 250

IQ

-Q

L = 10 mH ; C = 100 F

Pure LC circuit (no damping) after switch opened.

c. What was the current through the inductor just after the switch is opened?

d. What is the oscillation frequency of the circuit after the switch is opened?

e. The charge on the capacitor oscillates for t > 0. What is the maximum charge on C at any time

after the switch is opened?

f. Give an expression for Q(t), the charge on the capacitor for t > 0. a.) After a long time, dI/dt is zero so the voltage across the inductor is zero. All the voltage is across

the resistor so I = /R = 0.20 A to the right through R and down through L. As the capacitor is in

parallel with the inductor, its voltage is zero so it is (b.) uncharged . c.) The current through the

inductor is a continuous function of time. It was I = /R just before so it is I = 0.20 A down through

L just after. d.) The oscillation angular frequency is = [LC]-½ = [(10 x 10-3) (100 x 10-6)]-½ = 1000

rad/s. The frequency is /2 or 159 Hz .

e.) All the energy stored in the inductor initially will be stored in the capacitor whenever the



capacitor has it maximum charge.

max

2 1 1 1max max max

2max½ ½ (1000 ) (0.2 ) 200,Q

CLI LC I I s A CQ

f.) Q(t) = Qmax cos[t + ] I = + dQ/dt = - Qmax sin[t + ] NOTE: I flows onto +Q.

Q(0) = 200C cos[] = 0. Try = ½ Q(t) = 200C cos[t + ½ ] Q(t) = 200C sin[t]

Q(t) = 200C sin[radst]

General method Q(t) = Qmax cos[t + ] I = dQ/dt = - Qmax sin[t + ]

Q(0) = Qmax cos[] = 0 I(0) = dQ/dt = - Qmax sin[] = 0.2 A

Solve: Qmax cos[] = 0 and - Qmax sin[] = 0.2 A with rads

5.) A conductor consists of a circular loop

of radius R and two long straight

sections as shown in the figure.

a.) Find the magnitude and direction of the magnetic field due to the current in the conductor at the

center of the loop.

The field is the sum of that due to a long straight wire at a distance R and the field due to a single

turn circular loop at its center. SUPERPOSITION: Btotal = Blsw + Bloop

32

20 0 0

2 2ˆ ˆ ˆ; at center

2 22[ ]lsw RHR loop RHR RHR

I I R IB B n n

R RR z

70 0 1 1(4 10 )(10 ) 414

2 2 2 (0.02 ) 2(0.02 )

414 into to the page

TmA

lsw loop

I IB A T

R R m m

B B T

b.) Evaluate the magnitude of the field for I = 10 A and R = 2 cm.

By the RHR, the field at the center is into the page. 414 T into the page



6.) Note the direction of each integration path. Give

the value of d B s for each path illustrated.

CCW: out is positive; CW: in is positive

C1: CCW; C2: CW; C3: CW C

d B s = o Ienc

1Cd B s

= o Ienc = o [5 A - 2 A] = m

2Cd B s

= o [- 1 A - 5 A] = - 7.54 T m

3Cd B s

= o [- 1 A + 2 A] = 1.26 T m

0 0 ênc RHRC A Ad I J n dA

B s

The notation C = A means that C is the curve (path) than bounds the area A. The RHR subscript means

that the normal to the area is in the direction dictated by the RHR. In the cases above, an arrow

indicating a CCW path means that the normal and hence the positive sense for currents is out of the

page. For CW paths, the positive direction is into the page.

7.) Motion of charges in fields:

a,b,c, d.) Give the direction in which the charge is

deflected by the magnetic field in each situation

illustrated to the right.

e.) An electron traveling at 2.00 x 106 m/s is in a

region in which the electric field is 100 V/m

directed northwest. Give the acceleration of the

electron.

{ }, ; ; onlyEBq

mmF q E v B F qv B F q E E Ea a

Parts with a magnetic field and no electric field:

a.) v B

is up the page so a positive charge is deflected up the page.

b.) v B

is into the page so a negative charge is deflected out of the page.

c.) v B

vanishes so the charge is not deflected.



d.) v B

is into the page so the positive charge is deflected into the page.

Part with an electric field and no magnetic field: electron negative charge

e.) 2

1931

131.602 109.11 10 100 1.76 10V m

m sC

kgq

ma E NW NW

2

131.76 10 ms

a SE

8.) In the long solenoid [shown in cross section] in

the figure to the right, the magnetic field varies

according to the equation:

B(t) = (2.00 t3 – 4.00 t2 + 0.800) T.

The radius of the solenoid is R = 2.5 cm and r2 =

5 cm. Calculate the magnitude and direction of the

force on an electron at P2 at time t = 2 s. When

would the force on that electron be zero?

The circle is an E line chosen to circulate in the CCW sense.

The path should be concentric with the solenoid. I missed just

a little!

Â A

BtE d n dA

The electric field circulates around Bt

. If Bt

is into the

page, E circulates in the out of the page sense (CCW).

E

E circulates CCW for Bt > 0.

ˆ 2A A enclosed

B Bt tE d n dA E r Area

If r < R, then Bt

fills all the area enclosed by the E field line path.

2ˆ 2A A

B Bt tE d n dA E r r

(r < R)

If r > R, then Bt

fills the area out to R only

2ˆ 2A A

B Bt tE d n dA E r R

(r > R)

Solving for E and adding our CCW sense convention:



2

for2

for2

CCW

r Br R

tE

R Br R

r t

For r < R, the path is completely filled with time varying B field. For r > R, the path is only filled

with time varying B field for the central circle of radius R. The magnetic field is non-zero only inside

the solenoid.

B(t) = (2.00 t3 – 4.00 t2 + 0.800) T B/t = (6.00 t2 – 8.00 t) T /s = 8.00 T /s for t = 2 s.

r2 > R so 2 2(0.025 )

(8.00 ) 0.052 2(0.05 )

NTs CCCW

R B mE

r t m

The force on an electron will be eE

or 8.01 x 10-21 N

The electric field is tangent to the concentric circular field lines in the CCW sense.

The force on an electron will be eE

or 8.01 x 10-21 N tangent in the CW sense.

B/t = (6.00 t2 – 8.00 t) T /s vanishes for t = 0 s and t = 4/3 s. E will vanish at these times.

No Equation Sheet for the actual test !!!!!! Included for REVIEW only.

Constants 12 2 20 8.85 10 C /N m 1 9 2 2

0[4 ] 8.99 10 N m /Cek 7 20 4 10 N/A

191.60 10 Ce 319.11 10 kgem 83.00 10 m/sc

Electric Fields 1 212 122

ê

q qk

rF r

0

e

q

FE

2ˆ

e

qk

rE r

2

ê

dqk

r E r

Gauss’s Law cosE EA E d E A

in0

qd E A

Electric Potential 0

B

AU q d E s

0

UV

q

B

AV d E s

V Ed

e

qV k

r e

dqV k

r 1 2

12e

q qU k

r x

VE

x

Capacitance Q

CV

eq 1 2C C C eq 1 2

1 1 1

C C C 0C C 0 A

Cd

2p aq

p E

U p E

2

2

QU

C 21

2U C V

1

2U Q V 2

0

1

2Eu E



Current and Circuits dQ

Idt

av dI nq A v d

IJ nq

A v

EJ E

VR

I

R

A

0 01 T T I V c 2

2 VI R

R

c eq 1 2R R R

eq 1 2

1 1 1

R R R

in outI I closed loop

0V /( ) 1 tq t Q e /( ) tI t eR

E

/( ) tq t Qe /( ) tQI t e

RC

RC

Magnetic Fields B q F B

v sinBF q B v B I F L B

Bd Id F s B

I A

; B

; U B

; m

rqB

v

; 02

ˆ

4

Idd

r

s rB

0 1 2

2B I IF

a

0d I B s

0

2lsw

IB

r

0

2toroid

NIB

r

0 0sol

NB I nI

20 02 2 3/2

02 [ ] 2loopz

I R IB

R z R

B d B A

0d B A

Induction/Inductance Bd

dt

E B vE sinNAB t E L

dIL

dt E

BNL

I

20N A

L

L R /1 tI eR

E

/tI eR

E

21

2U LI

2

02B

Bu

2 12

121

NM

I

1

2

dIM

dt E max cos( )Q Q t

maxsin( )

dQI Q t

dt 1

LC /2

max cosRt LdQ Q e t

1 2212dR

LC L

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PHYSICS COMPREHENSIVE REVIEW - USNA · 2015-04-16 · Fields Test Example Questions Guide Physics Handout Series – fields.tank with jeff and jeff page 2 CAUTION: This guide covers