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Methods & Mathematical Tools

I · 01

Problem Solving & Units

Problem Solving in Physics

SI units

Base quantities, names of units, and symbols:

Length — metre — m
Time — second — s
Mass — kilogram — kg
Electric current — ampere — A
Thermodynamic temperature — kelvin — K
Amount of substance — mole — mol
Luminous intensity — candela — cd

Fundamental Units! (5 of them.)

Length — m
Time — s
Mass — kg
Charge — C
Temperature — K

Units!

\text{Force:}\quad \mathrm{kg\,m\,s^{-2}}=\mathrm{N}

\text{Energy:}\quad \mathrm{kg\,m^2\,s^{-2}}=\mathrm{N\,m}=\mathrm{J}

\text{Power:}\quad \mathrm{kg\,m^2\,s^{-3}}=\mathrm{J\,s^{-1}}=\mathrm{W}

\text{Current:}\quad \mathrm{C\,s^{-1}}=\mathrm{A}

Dimensional Analysis: Checking & Fudging

“Checking” — “always.” Same units on the same side of an equation!

Vectors vs. Scalars!

Maths — tool, not be-all and end-all. Don’t formula-fit.

Sensible answers! Check!

I · 02

A Method for Solving Problems

Solving Problems with the 7 Ds & the little s.

Diagram — big! (2/3 page), and as many as you need. Include graphs.
Directions — mark them: negative/positive.
Definitions & data — put all of them on the page; include what is given.
Diagnosis → type (how): conservation principles / force laws. Side example: angular momentum.
Derivation — equations (diagnosis in symbols); as many equations as variables. Add to diagram. Check dimensions.
Determination — algebra; box the answer.
Dimensions — check and test limiting cases. Example note: $H\to h$ ; does it make sense, is it possible?
Substitutions — if necessary, do a rough calculation by hand, with units. Include error!

Example Problem 1

A body of mass $m$ slides without friction from rest at height $H$ to lower height $h$ , where it encounters a horizontal spring of strength $k$ . How much does it compress?

Initial mass on a frictionless slope and final spring compression.

Definitions

$m$ = mass of object
$k$ = spring constant
$x$ = compression of spring
$U_E$ = elastic potential energy
$H$ = initial height
$h$ = final height
$g$ = acceleration due to gravity $=9.81\,\mathrm{m\,s^{-2}}$
$U_{\mathrm{GPE}}$ = gravitational potential energy
$E$ = energy

Diagnosis

Conservation of energy: $E$ is conserved.

Derivation

U_{\mathrm{GPE}}\longrightarrow U_{\mathrm{elastic}}

E_i=E_f

mgH=mgh+\frac12kx^2

Algebra

2mg(H-h)=kx^2

\boxed{x=\sqrt{\frac{2mg(H-h)}{k}}}

Side derivation for the spring:

F=-kx

U=\int_{x_i}^{x_f}F\,dx

-\int_0^{x_f}(-kx)\,dx=\frac12kx_f^2

Dimensions

[x]=\sqrt{\frac{(MLT^{-2})L}{MT^{-2}}}=\sqrt{L^2}=L

F=-kx,\qquad MLT^{-2}=kL,\qquad [k]=MT^{-2}

Limiting cases

$H\to h$ , so $H-h\to0$ and $x\to0$ .
$m\to0$ , so $x\to0$ .
$k\to\infty$ , so $x\to0$ .
$k\to0$ , so $x\to\infty$ .

Do these fool anyone?

Substitutions

Not needed.

I · 03

Calculus

Differentiation

a)

\frac{d}{dr}\left(6r^3(\cos 2r+5r)^7\right)

=18r^2(\cos2r+5r)^7+6r^3\cdot7(\cos2r+5r)^6(-2\sin2r+5)

=18r^2(\cos2r+5r)^7+42r^3(5-2\sin2r)(\cos2r+5r)^6

=6r^2(\cos2r+5r)^6\left(3+35r-14r\sin2r\right)

b)

x=(5te^{2t+1}+2)^8

v=\frac{dx}{dt}=8(5te^{2t+1}+2)^7(5e^{2t+1}+10te^{2t+1})

=40e^{2t+1}(5te^{2t+1}+2)^7(1+2t)

a=\frac{dv}{dt}=\frac{d^2x}{dt^2}

a=40e^{2t+1}(5te^{2t+1}+2)^6\left[4(5te^{2t+1}+2)(1+t)+35e^{2t+1}(1+2t)^2\right]

Partial Derivatives

\frac{\partial}{\partial x}f(x,y),\qquad f(x,y)=5xy+ye^x

With respect to $x$ , while the other variable is treated as fixed.

Integration

\int_a^b f(x)\,dx=\int_a^b f(u)\,du

Example

I=\int_{1/2}^{1}\sqrt{1-x^2}\,dx

Substitution (chain rule): let $x=\cos u$ , so $dx/du=-\sin u$ . Right triangle: hypotenuse 2, adjacent side 1, opposite side $\sqrt3$ , angle $\pi/3$ .

I=\int_{\pi/6}^{\pi/2}\sqrt{1-\cos^2u}\,(-\sin u)\,du=\int -\sin^2u\,du

\cos2x=\cos^2x-\sin^2x=1-2\sin^2x

I=\frac12\left(\frac{\sqrt3}{4}-\frac{\pi}{3}\right)=\frac{\sqrt3}{8}-\frac{\pi}{6}

Integration by Parts

\int u\,dv=uv-\int v\,du

\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\qquad\text{(product rule)}

Example: $\int\tan^{-1}x\,dx$

\int\tan^{-1}x\,dx=\int\frac{d}{dx}(x)\tan^{-1}x\,dx

=x\tan^{-1}x-\int\frac{x}{1+x^2}\,dx

=x\tan^{-1}x-\frac12\ln(1+x^2)+C

\frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^2},\qquad \tan^{-1}x=\arctan x

Same with sine and cosine.

\frac{d}{dx}\sin^{-1}(f(x))=\frac{f'(x)}{\sqrt{1-(f(x))^2}}

I · 04

Differential Equations

Linear / non-linear.

First order — rearrange and integrate.

Second order.

Simple Harmonic Motion

x+\omega^{-2}\frac{d^2x}{dt^2}=0

Guess:

x=A\cos(\omega t)+B\sin(\omega t)

x=Ae^{i\omega t+\phi}=Ae^{i\omega t}+Be^{-i\omega t}

Springs

\frac{d^2x}{dt^2}+\frac{k}{m}x=0

Example: $v=ktx$

x=\frac{v}{kt},\qquad v=\frac{dx}{dt},\qquad \frac{dx}{dt}=ktx

\frac{dx}{x}=kt\,dt

\int_{x_0}^{x}\frac{dx}{x}=k\int_0^t t\,dt

2\ln\frac{x}{x_0}=kt^2

\frac{x}{x_0}=e^{kt^2/2}

\boxed{x=x_0e^{kt^2/2}}

I · 05

Coordinate Systems

1) Coordinate Systems

Polar (plane) — $r,\theta$ (2D)

r=\sqrt{x^2+y^2},\qquad \theta=\tan^{-1}\!\left(\frac{y}{x}\right)

x=r\cos\theta,\qquad y=r\sin\theta

Cartesian

Axes $x$ and $y$ .

Cylindrical polar

Used for axial symmetry; coordinates $r,\theta,z$ .

Spherical polar

Used for spherical symmetry; coordinates $r,\theta,\phi$ .

0\le\theta\le\pi\quad\text{or}\quad-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}

0\le\phi\le2\pi

define what is what.

Polar, cylindrical, and spherical coordinate conventions.

I · 06

Multivariable Calculus

Multivariable Integrations

\int f(x)\,dx=g(x)+c\qquad\text{(standard)}

For a rectangular area, $dA=dx\,dy$ :

\iint_A dA=\int_c^d\left(\int_a^b dx\right)dy=(b-a)(d-c)

Non-dependent limits: separate it.

In polar coordinates:

dA=r\,dr\,d\theta

Volumes

\iiint_V dV

Or $\iint_A f\,dA$ : put in a function to find values, etc.

$\iint f\,dA$ — “flux integral — keep it parallel or perpendicular.”

Sphere volume

dV=r^2\sin\theta\,dr\,d\theta\,d\phi

V=\int_0^Rr^2dr\int_0^\pi\sin\theta\,d\theta\int_0^{2\pi}d\phi

=\frac{R^3}{3}[-\cos\theta]_0^\pi\,2\pi

\boxed{V=\frac{4\pi R^3}{3}}

Area and volume elements in Cartesian, polar, and spherical coordinates.

Let’s find the volume of a sphere using Cartesian coordinates.

x^2+y^2+z^2=r^2,\qquad \iiint_V dV

OK, easier problem: area of a circle.

\iint_A dA

Circle with differential rectangle $dx\,dy$ : $x^2+y^2=r^2$ .

A=\iint dy\,dx=2\int_{-r}^{r}y\,dx

y=\sqrt{r^2-x^2}

A=2\int_{-r}^{r}\sqrt{r^2-x^2}\,dx

Let $x=r\sin\theta$ , so $dx/d\theta=r\cos\theta$ . When $x=r$ , $\theta=\pi/2$ ; when $x=-r$ , $\theta=-\pi/2$ .

I=2\int_{-\pi/2}^{\pi/2}r\cos\theta\cdot r\cos\theta\,d\theta

=2r^2\int_{-\pi/2}^{\pi/2}\cos^2\theta\,d\theta

=\frac{r^2}{2}\int_{-\pi/2}^{\pi/2}(\cos2\theta+1)\,d\theta

\cos2\theta=2\cos^2\theta-1,\qquad \cos^2\theta=\frac{\cos2\theta+1}{2}

Cartesian circle element used to derive the area.

A=\frac{r^2}{2}\left[\frac{\sin2\theta}{2}+\theta\right]_{-\pi/2}^{\pi/2}

=\frac{r^2}{2}\left(0+\frac{\pi}{2}+0+\frac{\pi}{2}\right)

\boxed{A=\pi r^2}

OK, now extrapolate!

For a horizontal slice of a sphere, let $r_c$ be the converted (cross-sectional) radius.

r^2=z^2+r_c^2

r_c^2=r^2-z^2

dV=\pi r_c^2\,dz=\pi(r^2-z^2)\,dz

V=\pi\int_{-r}^{r}(r^2-z^2)\,dz

=\pi\left[r^2z-\frac{z^3}{3}\right]_{-r}^{r}

=\pi\left(r^3-\frac{r^3}{3}+r^3-\frac{r^3}{3}\right)

\boxed{V=\frac{4\pi r^3}{3}}

HAHA!

Horizontal circular slice through a sphere.

I · 07

Laboratory Practice

Lecture 3: Labs

Lab Safety

Sensible covered shoes.
Don’t eat and drink.
Don’t mess with equipment; be safe! Heat, HV, heavy.
Notify of accidents!
Do as you’re told.

How You Do Labs Here!

Don’t fake up results! (No fudging.)
Logs: read through the lab and what you have to do.
Write in pen.
Don’t panic about low marks.
Think before you act.

Lab book

Keep a journal as you go — what you’re doing.

Aim: what to do?
Theory: what you need to graph.
Error: estimate and reduce error quantitatively; record details of decisions and how you did it.

y=ax+b

$a$ is the gradient and $b$ the $y$ -intercept.

Error

Every measurement has error. Explain the source of error.

Adding or subtracting: add absolute errors $\Delta$ .
Multiplying or dividing: add relative errors.
Powers: multiply relative errors by the power.
One-half the smallest scale division is the limit-of-reading absolute error.

z\pm\Delta z\quad\text{(with units)},\qquad \mathrm{Err}=\frac{\Delta z}{z}

A=B+C,\qquad \Delta A=\Delta B+\Delta C

A=BC,\qquad \frac{\Delta A}{A}=\frac{\Delta B}{B}+\frac{\Delta C}{C}

z=\sin A\quad\Rightarrow\quad \Delta z=\Delta A\cos A

z=f(A,B,C):\quad \Delta z=\left|\frac{\partial z}{\partial A}\right|\Delta A+\left|\frac{\partial z}{\partial B}\right|\Delta B+\left|\frac{\partial z}{\partial C}\right|\Delta C

Example: Pendulum

T=2\pi\sqrt{\frac{l}{g}}

Vary $l$ and measure $T$ .

Errors: reaction time — take many swings and divide! Parallax error. Mention all errors and minimise them.

T^2=\frac{4\pi^2l}{g}\qquad\text{(linearise)}

Systematic errors (uncertainties) → Accuracy

Equipment. Changes by the same amount for each measurement. Can’t estimate.

Random error → Precision

Moves around. Plot on a graph.

A graph

Title
Axes
Units
Error
Origin
Sensible scale
Error flags

Draw a line of best fit; do not force it through the origin. Draw two lines of worst fit to find the uncertainty in the gradient.

Discussion

How it went.
What you could improve.
Alternative ways to do it.

Method and diagram: do a whole page — clear, detailed, labelled.

Example laboratory graph with error boxes, best-fit line, and limiting gradients.

I · 08

Linear Algebra

Matrices

Adding

\begin{pmatrix}a_1&b_1&c_1\\d_1&e_1&f_1\\g_1&h_1&i_1\end{pmatrix}+\begin{pmatrix}a_2&b_2&c_2\\d_2&e_2&f_2\\g_2&h_2&i_2\end{pmatrix}=\begin{pmatrix}a_1+a_2&b_1+b_2&c_1+c_2\\d_1+d_2&e_1+e_2&f_1+f_2\\g_1+g_2&h_1+h_2&i_1+i_2\end{pmatrix}

Adding matrices must have the same dimensions.

Multiplying: dot product of each row vector by each column vector of the other matrix.

Determinant

\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc

\det\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)

Eliminate common rows and columns.

\begin{vmatrix}2&1&-1\\-3&-1&2\\-2&1&2\end{vmatrix}=2(-8+5)-1(-6-5)+(-1)(-3-4)=-3+11-7

A cross product by little bits…

Dodgy matrix because the basis vectors are not of the same field.

Example simultaneous equations:

x+z=5,\qquad x+y+z=0,\qquad y+z=5

Row reduction: you can add, subtract, and multiply rows. If row addition produces $0=5$ , the equations are inconsistent.

Mechanics

II · 01

Forces & Friction

Problem 2

Man → piano up a hill ( $\theta$ to horizontal). A pulley joins a hanging man to a piano on the incline; the coefficient of kinetic friction is $\mu_k$ .

Pulley system and free-body diagrams for the man and piano.

Definitions

$\mu_k$ = coefficient of kinetic friction
$M$ = mass of piano
$m$ = mass of man
$g$ = gravitational acceleration
$T$ = tension
$a$ = acceleration
$\theta$ = angle of hillside

Diagnosis

Forces! Free-body diagrams. Conservation of string.

Man: $T$ upward and $mg$ downward. Piano: $T$ uphill, $N$ normal to the slope, $Mg$ downward, components $Mg\cos\theta$ normal to the slope and $Mg\sin\theta$ downslope, and $\mu_kN$ downslope.

Derivation

|F|=ma=(m+M)|a|

N=Mg\cos\theta

\sum F:\quad Ma=Mg\sin\theta+\mu_kN-mg

Direction of $a$ for the man: vertically upwards.

Direction of $a$ for the piano: down and right at angle $\theta$ from the horizontal.

\boxed{|a|=g\frac{M\sin\theta+\mu_kM\cos\theta-m}{m+M}}

Dimensions

LT^{-2}=LT^{-2}\left(\frac{M+M-M}{M}\right)

Limiting cases

If $M>m$ by enough, acceleration is to the right.
If $m>M$ by enough, acceleration is negative.
What is the median point?

For the median point, $a=0$ :

M\sin\theta+\mu_kM\cos\theta-m=0

m=M\sin\theta+\mu_kM\cos\theta

M=\frac{m}{\sin\theta+\mu_k\cos\theta}

Then friction is in the opposite direction.

Substitution!

Doing both at the same time is dodgy, so let us look at the piano. Removing $m$ from the denominator seems to require modification. Be careful about conditions of consistency.

For the man:

mg=ma-T

T=m(g-a)

|a|=g\frac{M\sin\theta+\mu_kM\cos\theta-m}{M+m}

II · 02

Kinematics

Kinematic Equations!

v=\frac{dx}{dt}

a=\frac{dv}{dt}=\frac{d^2x}{dt^2}

For constant $a$ , with initial velocity $v_0=u$ :

\int_0^t a\,dt=\int_u^v dv

v-u=at

\boxed{v=u+at}

\frac{dx}{dt}=at+u

x=\int(at+u)\,dt

\boxed{x=\frac12at^2+ut+x_0}

a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}

v\,dv=a\,dx

\int_0^x a\,dx=\int_u^v v\,dv

ax=\left[\frac{v^2}{2}\right]_u^v

2ax=v^2-u^2

\boxed{v^2=u^2+2ax}

II · 03

Energy & Forces

Energy, Energy Conservation & Forces

Energy — ability to do work; that which is conserved.

Two types:

Kinetic energy — bulk: $KE=\frac12mv^2$ ; microscopic (thermal energy), which is not heat; sound and mechanical waves.
Potential energy — can become kinetic; associated with gravitational, electromagnetic, weak nuclear, and strong nuclear forces.

\mathrm{kg\,m^2\,s^{-2}}=\mathrm{N\,m}=\mathrm{J}\qquad\text{(scalar)}

Potential-energy diagrams include a negative central-force well approaching zero as $r\to\infty$ , plus square wells and harmonic wells (quantum).

For $F\propto1/r^2$ , define zero at infinity; therefore gravitational potential energy is negative:

U=-\frac{Gm_1m_2}{r}

U_{\mathrm{harmonic}}=\frac12kx^2

Force vs Energy

\vec F=-\frac{\partial U}{\partial\vec r}

U=-Gm_1m_2r^{-1},\qquad \vec F=-\frac{Gm_1m_2}{r^2}\hat r

Example: Earth’s surface

F=-\frac{GMm}{r^2}\approx588\,\mathrm{N}

In the ISS:

F=-\frac{GMm}{(r+2\times10^5)^2}

Difference in gravitational potential energy

\Delta U=-\frac{Gm_1m_2}{r_1}+\frac{Gm_1m_2}{r_2}

=Gm_1m_2\left(\frac1{r_2}-\frac1{r_1}\right)

Quite a big change.

Central-force and harmonic potential-energy curves.

Force, Energy & Work

F=-\frac{\partial U}{\partial x}

\int dU=-\int F\,dx

U=-\int F\,dx\qquad\text{(almost)}

W=\int\vec F\cdot d\vec x

How much work does $F_G$ , acting on the Moon due to the Earth, do? 0.

Energy is always conserved!

\Delta E=0,\qquad E_i=E_f\qquad\text{for an isolated system}

\Delta E_{\mathrm{system}}+\Delta E_{\mathrm{environment}}=0

Use conservation principles — conservation of energy, for example — for problem solving.

Mass on a spring

Energy labels around the hanging mass–spring system: $U_{\mathrm{GPE}}$ , $U_{\mathrm{elastic}}$ , bulk kinetic energy, and thermal kinetic energy.

Hookian Stuff

F_{\mathrm{restoring}}=-F_{\mathrm{applied}}=-k\Delta x

F=-kx

W=\int F\,dx

\boxed{W=\frac12kx^2}

Mechanical Energy & Conservative Forces

Conservative → only $U\to KE$ (bulk). $U_{\mathrm{grav}}$ , kinetic energy, and elastic potential energy are mechanical energy.

Non-Conservative Forces

Dissipative → energy goes to thermal and microscopic energy.

U=mgh,\qquad KE=\frac12mv^2

Choose a sensible reference point, and talk about it.

Hanging spring and its energy stores.

II · 04

Momentum & Newton’s Laws

More Forces — Momentum & Newton’s 2nd & 3rd Laws

Normal force — electrostatic

Usually does no work. Loop-the-loop, elevator, and bouncing-ball cases are exceptions.

Perpendicular to the surface. Equal to the perpendicular weight component to prevent slippage. A maximum normal force can be exerted.

Example: lift accelerating upwards at $a$

\sum F=ma

mg-N=-ma

N=mg+ma=m(g+a)

Friction — parallel to surface

Cold weld → static friction. Surfaces are lumpy.

F_{s,\mathrm{fric}}\le\mu_sN

Kinetic friction

Formation and breakage of bonds.

F_{k,\mathrm{fric}}=\mu_kN,\qquad \mu_k\le\mu_s

Approximation! (Only a model.) Example: box on the back of a ute. Friction opposes relative motion.

F_{\max}=\mu_sN=\mu_smg

a_{\max}=\mu_sg\qquad\text{(independent of mass)}

Normal-force example in an accelerating lift and static friction on a box.

Air resistance

F_{\mathrm{air}}=k_{\mathrm{drag}}v^2

F_{\mathrm{air}}=W=mg\qquad\text{at terminal velocity}

Tension (not elastic)

In an ideal rope, force at one end is instantly transmitted without loss, and every part feels the same tension.

The note $k_{\mathrm{ideal\ rope}}=\infty$ is accompanied by the spring relation $F=-kx$ .

Tension is the tendency to bring the rope into a state of equilibrium in a potential well.

Ideal ropes have no mass and are inextensible; $T=\text{constant}$ .

Ideal pulleys act to change direction but nothing else.

Normal and frictional forces vary depending on the applied force.

Two components of contact force give minimum energy.

Momentum

\vec p=m\vec v

Units: $\mathrm{kg\,m\,s^{-1}}$ or $\mathrm{N\,s}$ .

\vec F=m\vec a=\frac{d\vec p}{dt}

Force changes momentum. No force, no change in momentum.

Example: two carts

One cart has mass $m$ , the other mass $2m$ . They are pushed with equal force for 3 seconds on an air track.

Equal momenta:

m(2v)=(2m)v\qquad (p=p_1+p_2)

Kinetic energies:

KE_1=\frac12m(2v)^2=2mv^2

KE_2=\frac12(2m)v^2=mv^2

Therefore the lighter cart has more kinetic energy.

II · 05

Collisions

Newton’s Third Law!

Forces act on different objects.

\vec F_{AB}=-\vec F_{BA}

Gravity and the normal force on the same object are not a third-law pair. They have different origins.

\vec F=\frac{d\vec p}{dt}

If $\vec F_{AB}=-\vec F_{BA}$ , then:

\frac{d\vec p_A}{dt}=-\frac{d\vec p_B}{dt}

Therefore momentum is always conserved.

Elastic collisions

Momentum and kinetic energy are conserved. Gravitational slingshot, air molecules, Newton’s cradle, balls? Isolated system.

\sum_i m_iv_i=C

m_av_{ai}+m_bv_{bi}=m_av_{af}+m_bv_{bf}

\sum_i\frac12m_iv_i^2=\text{constant}

m_av_{ai}^2+m_bv_{bi}^2=m_av_{af}^2+m_bv_{bf}^2

Solving — prove this:

v_{af}=\left(\frac{m_a-m_b}{m_a+m_b}\right)v_{ai}+\left(\frac{2m_b}{m_a+m_b}\right)v_{bi}

v_{bf}=\left(\frac{m_b-m_a}{m_b+m_a}\right)v_{bi}+\left(\frac{2m_a}{m_a+m_b}\right)v_{ai}

If the masses are equal, they swap velocities.

If the masses are vastly different, $v_{bi}=v_{bf}$ .

v_{SF}=2v_{BI}-v_{SI}

Inelastic Collisions

Kinetic energy is lost to deformation energy, thermal energy, etc.

Momentum is still conserved.

Example analogy: a sports car coasting. As the rain increases, the speed decreases. Drill a hole to let the water out, but it will not go any faster unless the water is pumped out stationary with respect to the road.

m_av_{ai}+m_bv_{bi}=m_av_{af}+m_bv_{bf}

m_av_{ai}^2+m_bv_{bi}^2=m_av_{af}^2+m_bv_{bf}^2\qquad\text{(elastic comparison)}

Eliminate $v_{af}$ :

v_{af}=\frac{m_av_{ai}+m_bv_{bi}-m_bv_{bf}}{m_a}

v_{af}^2=\frac{(m_av_{ai}+m_bv_{bi}-m_bv_{bf})^2}{m_a^2}

m_av_{ai}^2+m_bv_{bi}^2=\frac{(m_av_{ai}+m_bv_{bi}-m_bv_{bf})^2}{m_a}+m_bv_{bf}^2

For two masses labelled $M$ and $m$ :

MV_I+mv_I=MV_F+mv_F

MV_I^2+mv_I^2=MV_F^2+mv_F^2

V_F=\frac{MV_I+mv_I-mv_F}{M}

Continuation of the algebra eliminating one final velocity from momentum and kinetic-energy equations.

(M+m)v_F^2-2v_Iv_F(m+M)+v_I(mv_I+MV_I)=0

The quadratic formula is applied to $v_F$ ; the positive root is selected.

v_F=\frac{2v_I(m+M)\pm\sqrt{4v_I^2(m+M)^2-4(M-m)v_I(mv_I+MV_I)}}{2(M-m)}

Take positive root.

Approximation note: $M\gg m$ .

II · 06

Rotational Mechanics

Rotational Kinematics

Rotation about a single fixed axis.
Rigid body: every point moves in a circular path with the centres of the circles on a common axis.
Can be approximated as a single particle for pure rotation.

Rigid-body rotation and arc-length variables.

Variables

$\theta$ : position from axis $x'$ .

\theta=\frac{s}{r}\qquad\text{(no units)}

Positive direction is arbitrary, but define it.

s=r\theta

$s$ is arc length, $r$ radius, $\theta$ angle.

Angular velocity, $\omega$

\omega=\frac{d\theta}{dt}

Direction given by right-hand screw rule; parallel to the axis of rotation.

Angular acceleration, $\alpha$

\alpha=\frac{d\omega}{dt}=\frac{d^2\theta}{dt^2}

Right-hand screw rule again.

Equations of motion (1D, derived from translation)

\omega=\omega_0+\alpha t

\theta=\theta_0+\omega_0t+\frac12\alpha t^2

\omega^2=\omega_0^2+2\alpha\theta

Rotational Dynamics: Torque

For a rigid body on a fixed axis, angular acceleration due to an applied force depends on the direction and where on the body the force is applied.

Torque — necklace, garotte…

\vec\tau=\vec r\times\vec F\qquad\text{(cross product)}

|\tau|=|r||F|\sin\theta

Torque from a force applied at a lever arm.

Rotational inertia $I$ (moment of inertia)

Angular equivalent of mass.

For a single particle:

I=mr^2

For a collection of particles:

I=\sum_n m_nr_n^2

For a solid body:

I=\iiint_V r^2\,dm

Properties of moment of inertia

I=I_0+Mh^2\qquad\text{(parallel-axis theorem)}

$h$ is the distance between parallel axes.

I_z=I_x+I_y\qquad\text{if }I_z\text{ is thin enough}

Perpendicular-axis theorem.

Example: sphere

I=\iiint_V r^2\,dm

dm=\rho\,dV=\rho r\,dr\,d\theta\,dz

I=\rho\int_0^{2\pi}d\theta\int_{-R}^{R}\left(\int_0^{\sqrt{R^2-z^2}}r^3\,dr\right)dz

Cylindrical volume element inside a sphere for the inertia integral.

Continuation of the solid-sphere moment-of-inertia integral:

I=2\pi\rho\int_{-R}^{R}\left[\frac{r^4}{4}\right]_0^{\sqrt{R^2-z^2}}dz

=2\pi\rho\int_{-R}^{R}\frac{(R^2-z^2)^2}{4}\,dz

=\frac{2\pi\rho}{4}\int_{-R}^{R}(R^4-2R^2z^2+z^4)\,dz

=\frac{\pi\rho}{2}\left[R^4z-\frac{2R^2z^3}{3}+\frac{z^5}{5}\right]_{-R}^{R}

=\frac{\pi\rho R^5}{2}\left(2-\frac43+\frac25\right)

=\frac{8\pi\rho R^5}{15}

M=\rho V=\rho\frac43\pi R^3

\boxed{I=\frac25MR^2}

Newton’s Second Law of Rotation

\tau=I\alpha

Translational comparison: $F=ma$ .

Example: Rolling without slipping

A solid sphere of mass $M$ and radius $R$ rolls down an incline of angle $\theta$ with speed $v_0$ .

v=r\omega

a=r\alpha

N=Mg\cos\theta

Mg\sin\theta-F=Ma_{\mathrm{cm}}

\tau=R\times F=RF\sin\frac{\pi}{2}=I\alpha

RF=\frac25MR^2\alpha

F=\frac25Ma_{\mathrm{cm}}

Ma_{\mathrm{cm}}=Mg\sin\theta-\frac25Ma_{\mathrm{cm}}

\boxed{a_{\mathrm{cm}}=\frac57g\sin\theta}

Solid sphere rolling without slipping down an incline.

The page begins with the final algebra of the solid-sphere inertia integral, using $M=\rho V$ , and confirms:

I=\frac25MR^2

Newton’s Second Law of Rotation

\tau=I\alpha

F=ma

Example: Rolling without slipping

a_{\mathrm{cm}}=\frac57g\sin\theta

F=\frac25Ma_{\mathrm{cm}}

F=\frac27Mg\sin\theta

v=r\omega,\qquad \frac{dv}{dt}=r\frac{d\omega}{dt},\qquad a_{\mathrm{cm}}=r\alpha

N=Mg\cos\theta

Mg\sin\theta-F=Ma_{\mathrm{cm}}

\tau=RF=I\alpha

\alpha=\frac{5F}{2MR}

a=r\alpha=\frac{5F}{2M}

a=\frac57g\sin\theta

Force and torque relations for rolling without slipping.

Rotational Mechanics, Part 2: Angular Momentum and Energy

Energy

K_{\mathrm{rotation}}=\frac12I\omega^2

K_{\mathrm{rotation+translation}}=\frac12I\omega^2+\frac12Mv_{\mathrm{cm}}^2

W=\int\tau\,d\theta

For a brake pad acting tangentially on a disk, the friction force is F=μ_kN and the torque magnitude is rμ_kN.

\boldsymbol\tau=\mathbf r\times\mathbf F,\qquad |\tau|=r\mu_kN

Rolling bodies descend a ramp without slipping; their speed depends on I=kmr².

For rolling without slipping, v_cm=rω. If I=kmr², then

K=\frac12(kmr^2)\frac{v^2}{r^2}+\frac12mv^2=\frac12mv^2(k+1)

Lower k gives a higher speed; the sphere is fastest.

I_{\mathrm{sphere}}=\frac25MR^2,\qquad I_{\mathrm{hoop}}=MR^2,\qquad I_{\mathrm{cylinder}}=\frac12MR^2

Angular momentum L

Angular momentum is the rotational analogue of linear momentum. Linear momentum is defined relative to a frame; angular momentum is defined relative to a point.

\mathbf p=m\mathbf v,\qquad \mathbf L=\mathbf r\times\mathbf p

For a rigid axially symmetric body:

\mathbf L=I\boldsymbol\omega,\qquad \mathbf L_T=\sum_i\mathbf L_i

Angular momentum and torque

\sum\boldsymbol\tau=\frac{d\mathbf L}{dt},\qquad \sum\tau_x=\left(\frac{d\mathbf L}{dt}\right)_x

The relation may be broken into components.

Steiner’s theorem

\mathbf L=\mathbf L_0+M\mathbf R_{\mathrm{cm}}\times\mathbf V_{\mathrm{cm}}

This is analogous to the parallel-axis theorem and is useful for motion that is not simply rotation about the centre of mass.

Conservation of angular momentum

\boldsymbol\tau_{\mathrm{ext}}=0\ \Longrightarrow\ \frac{d\mathbf L}{dt}=0\ \Longrightarrow\ \mathbf L_i=\mathbf L_f

Girl on a Merry-go-round

A mass

M

descends with speed

v

on a string tangent to the rim of a merry-go-round of radius

r

and mass

m

. Its straight-line momentum has angular momentum

Mvr

about the axle.

I=\frac12mr^2

L_i=0

L_f=\frac12mr^2\omega

The incoming mass has linear momentum p=Mv and angular momentum about the centre of the platform:

\mathbf p=M\mathbf v,\qquad L_{\mathrm{mass}}=|\mathbf r\times\mathbf p|=Mvr

Conservation of angular momentum gives

\frac12mr^2\omega=rMv

\omega=\frac{2Mv}{mr}

Angular momentum does not imply conventional rotation. Objects moving in a straight line have angular momentum relative to external points.

Precession

Reference note: HRK, p. 219.

Precession sketches: an upright spinning top and a tilted top whose symmetry axis sweeps around the vertical.

Precession is caused by a small lateral torque that changes the direction, but not the magnitude, of the angular momentum.

Torque geometry and angular-momentum-vector change for a tilted spinning top. Gravity acts at the centre of mass; the perpendicular lever arm is r sin θ. During a small precession angle dφ, L changes by the transverse vector dL.

\tau=rmg\sin\theta

\boldsymbol\tau=\frac{d\mathbf L}{dt}

dL=\tau\,d\theta

Precession (continued)

\omega_p=\frac{d\phi}{dt}

dL=L\sin\theta\,d\phi

d\phi=\frac{dL}{L\sin\theta}

\omega_p=\frac{d\phi}{dt}=\frac{dL}{L\sin\theta\,dt}=\frac{rmg\sin\theta}{L\sin\theta}=\frac{rmg}{L}

The precession rate is inversely proportional to L, hence inversely proportional to the spin speed of the top.

wider, faster circles.

II · 07

Gravity

Central Forces and Gravity

Gravity is an attractive central force between two masses.

Newton’s law of gravitation

\mathbf F_{\mathrm{grav}}=-\frac{Gm_1m_2}{r^2}\,\hat{\mathbf r}

For several masses, use the vector sum of the forces (or use potentials). Spherically symmetric objects behave exactly like point masses unless the observation point is inside them. All masses behave like point masses when viewed from far enough away.

Gravitational field

The gravitational field is force per unit mass. It is an intensive vector field.

\mathbf g=-\frac{Gm_1}{r^2}\hat{\mathbf r}

Field lines and flux

Field lines show the direction of the force at each point. Gravitational field lines come from infinity and point toward mass. Their density indicates relative field strength; they crowd near sharper features.

Flux measures the number of field lines passing through a surface.

\Phi_{\mathrm{grav}}=\iint_{\mathrm{surface}}\mathbf g\cdot\hat{\mathbf n}\,dA

Gravitational field lines point inward through a closed spherical surface. The local flux element depends on g·n̂ dA.

Gauss’s law

The net gravitational flux through a closed surface depends only on the mass enclosed.

\Phi_{\mathrm{grav}}=\oiint_S\mathbf g\cdot\hat{\mathbf n}\,dA=-4\pi Gm_{\mathrm{enclosed}}

Calculation hint: choose a complete spherical surface on which the field is perpendicular and has constant magnitude.

Examples using Gauss’s law

Gravitational field outside the Earth

Use spherical symmetry and a Gaussian sphere of radius r.

\oiint\mathbf g\cdot\hat{\mathbf n}\,dA=-4\pi Gm_{\mathrm{enclosed}}

g(4\pi r^2)=-4\pi GM_E

g=-\frac{GM_E}{r^2}

“WOOH!”

Inside a hollow spherical shell

m_{\mathrm{enclosed}}=0\quad\Longrightarrow\quad\Phi=0\quad\Longrightarrow\quad g=0

Inside a constant-density Earth

Let R be the Earth’s radius and r<R the radius of the Gaussian sphere.

m_{\mathrm{enc}}=\frac{\frac43\pi r^3}{\frac43\pi R^3}M=\frac{r^3}{R^3}M

g(4\pi r^2)=-4\pi G\frac{r^3}{R^3}M

g=-\frac{GM}{R^2}\frac rR=-\frac{GMr}{R^3}

Magnitude of g versus radius: linear inside a constant-density Earth and inverse-square outside.

Potential energy

Gravitational potential energy is extensive and is defined to be zero at infinity.

U=-\int_{\infty}^{r}\mathbf F\cdot\hat{\mathbf r}\,dx=-\int_{\infty}^{r}\frac{GMm}{r^2}\,dr=-\frac{GMm}{r}

Potential

Gravitational potential is potential energy per unit mass. It is an intensive quantity.

V=-\frac{GM}{r}

\mathbf g=-\nabla V

Equipotentials

Equipotential lines or surfaces have equal potential everywhere. They are perpendicular to field lines. In a conservative field, a journey that begins and ends on the same equipotential requires no net energy.

A terrain analogy for equipotentials: contour lines are constant-height surfaces, and the steepest field direction crosses them perpendicularly.

Escape velocity

Escape velocity is the speed needed to escape the gravitational field. Set the final total energy at infinity to zero and use energy conservation.

\frac12mv_e^2-\frac{GmM_E}{r}=0

v_{\mathrm{escape}}=\sqrt{\frac{2GM_E}{r}}

Kepler’s laws

All planets move in elliptical orbits with the Sun at one focus.
The line joining a planet to the Sun sweeps out equal areas in equal times.
The square of the orbital period is proportional to the cube of the orbit’s semimajor axis a.

Kepler’s laws: diagrams and circular-orbit proof

Kepler’s first two laws: the Sun S lies at a focus of an ellipse, and equal time intervals Δt sweep equal areas. The semimajor axis is a.

T^2\propto a^3

For the special case of a circular orbit, a=r and the gravitational force supplies the centripetal force.

mr\omega^2=\frac{GmM}{r^2}

\frac{4\pi^2mr}{T^2}=\frac{GmM}{r^2}

\frac{r^3}{T^2}=\frac{GM}{4\pi^2}

The equal-area law expresses conservation of angular momentum.

Proof of Gauss’s law for a spherical surface

\mathbf g=-\frac{GM}{r^2}\hat{\mathbf r}

\oiint\mathbf g\cdot\hat{\mathbf n}\,dA=-\frac{GM}{r^2}(4\pi r^2)=-4\pi GM_{\mathrm{enc}}

For a closed surface that encloses no mass, the inward and outward contributions cancel and the net flux is zero.

A flux tube crossing a closed surface twice contributes equal and opposite flux when no mass lies inside.

Divergence-theorem note: the gravitational field has zero divergence in empty space; the exception is at mass.

\iiint_V\nabla\cdot\mathbf g\,dV=\oiint_S\mathbf g\cdot\hat{\mathbf n}\,dA

II · 08

Mechanics Exam Notes

First Mech exam

Things I didn’t know before / important notes.

Be careful about:

Diagrams (size).
Directions — on every diagram.
Laying out — clear and orderly what you know.

Anatomy of ideal presentation (neat)

Three suggested solution layouts: a numbered diagram with definitions; a page with derivation and algebra; and a page highlighting dimensions, limiting cases, and substitution.

Things to Work on

Trig (lol).
Mechanics (rotational).
Calculus.

Check Dimensions.

Watch inequalities in integral calculations.

Resolve vectors perpendicularly.

No units on algebraic equations!

Consider all factors; avoid bad assumptions!

Write what you can.

Q3

I_0\omega_0=I\omega

I=\frac{I_0\omega_0}{\omega}

0=\frac{d}{dt}(I\omega)=\omega\frac{dI}{dt}+I\frac{d\omega}{dt}

I=mr^2,\qquad \frac{dI}{dt}=2mr\frac{dr}{dt}

\frac{dr}{dt}=r\omega

\omega\,2mr^2\omega+I\frac{d\omega}{dt}=0

\frac{d\omega}{dt}=-\frac{2mr^2\omega^2}{I}=-\frac{2mr^2\omega^3}{I_0\omega_0}

\int_{\omega_0}^{\omega}\frac{d\omega}{\omega^3}=\int_0^t-\frac{2mr^2}{I_0\omega_0}\,dt

\frac{1}{2}\left(\frac{1}{\omega^2}-\frac{1}{\omega_0^2}\right)=\frac{2mr^2}{I_0\omega_0}t

\omega=\omega_0\sqrt{\frac{I_0}{I_0+4mr^2\omega_0^2t}}

III

Thermal Physics & Fluids

III · 01

Thermodynamics I

Temperature

Thermometers use a thermometric property. A constant-volume gas thermometer compares the pressure of a gas at the unknown temperature with its pressure at a reference temperature.

Constant-volume gas thermometer: a gas bulb at temperature T is connected to a liquid manometer; the height difference h measures the pressure p while the gas volume is held fixed.

T_{\mathrm{ref}}=0^{\circ}\mathrm C,\qquad T_{\mathrm{gas}}=T_r\frac{p}{p_r}

The triple point of water is a specific temperature and pressure at which water, ice, and vapour are all present. It is used as the reference temperature.

T_{\mathrm{trip}}=273.16\,\mathrm K

As the reference-gas pressure is reduced toward zero, the readings of different gas thermometers converge to the same ideal-gas temperature scale.

T=T_r\lim_{p_r\to0}\frac{p}{p_r}

Why do different thermometers agree?

Zeroth law of thermodynamics: if systems A and B are each separately in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other. C may be a thermometer. Temperature is therefore well-defined and can be standardised.

T_{\mathrm C}=T_{\mathrm K}-273.15\,\mathrm K

Heat

Heat is the net flow of thermal energy, denoted Q: it is that which flows. It is measured in joules (J). Heat flow occurs between two bodies that are not in thermal equilibrium.

Three ways heat can flow

1. Conduction

Conduction through a bar of area A and length x between temperatures T_H and T_C.

H=\frac Qt,\qquad H\propto A,\qquad H\propto\frac1x,\qquad H\propto\Delta T=T_H-T_C

H\propto\frac{A\Delta T}{\Delta x},\qquad H=-kA\frac{\Delta T}{\Delta x},\qquad H=-kA\frac{dT}{dx}

2. Radiation

Stefan–Boltzmann law. For a black body the radiated intensity is proportional to the fourth power of its absolute temperature.

I=\sigma T^4,\qquad \sigma=5.67\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}

I=\varepsilon\sigma T^4,\qquad 0<\varepsilon<1

The factor ε is the emissivity.

3. Convection (in fluids)

Heated fluid becomes less dense and rises in bubbles; turbulence allows it to cool. Heated and cooled objects set up convection cells. Many cells can occur in a fluid, like the granules visible on the Sun.

Convection cells around a heated region and a cooled region.

Thermal expansion

\Delta L\propto L\Delta T,\qquad \Delta L=\alpha L\Delta T

\Delta A\simeq2\alpha A\Delta T,\qquad \Delta V\simeq3\alpha V\Delta T\quad(\alpha\Delta T\text{ small})

For a rectangle with initial area A_i=ab:

A_f=(a+\alpha a\Delta T)(b+\alpha b\Delta T)=ab+2\alpha ab\Delta T+\alpha^2ab(\Delta T)^2

\therefore\quad\Delta A\simeq2\alpha A\Delta T

Liquids do not have a linear expansion coefficient. Over a small temperature range one writes $\Delta V=\beta V\Delta T$ , where β depends on temperature. Usually β>0, but for water it can be negative. In a mercury thermometer, $\beta_{Hg}>\beta_{glass}$ .

Phases

The three phases considered are solid, liquid, and gas.

Latent heat is the heat per kilogram required to undergo a phase change.

Q=L_{\mathrm{fusion}}m\qquad(\text{solid}\to\text{liquid})

Q=-L_{\mathrm{vaporization}}m\qquad(\text{gas}\to\text{liquid})

Heat capacity

C=\frac Q{\Delta T}

C is the heat capacity of a system, measured in J K⁻¹. It is an extensive quantity.

Specific heat capacity

Specific heat capacity is heat capacity per unit mass or per mole.

\Delta H=mc\Delta T,\qquad c=\frac Cm\ \left(\mathrm{J\,K^{-1}\,kg^{-1}}\right),\qquad c=\frac Cn\ \left(\mathrm{J\,K^{-1}\,mol^{-1}}\right)

Specific heat capacity is intensive.

First law of thermodynamics

For any process between equilibrium states A and B, the quantity Q+W is the same. Q is heat added to the system; W is work done on the system.

III · 02

Thermodynamics II

First law

For a process between equilibrium states A and B, Q+W is the same: energy is conserved. Heat and work are ways of transferring energy.

A gas at pressure p pushes a piston of area A through displacement dx. With work defined as work done on the system, expansion gives dW=−p dV.

W=\int\mathbf F\cdot d\mathbf x,\qquad |F|=pA

dW=F(-dx)=-pA\,dx,\qquad dV=A\,dx

\boxed{dW=-p\,dV}

W=\int_0^W dW=-\int_{V_i}^{V_f}p\,dV

A system may do work or have work done on it.

\Delta U=U_B-U_A=Q+W,\qquad dU=dQ+dW

The internal energy U (also written E) is a state function.

Ideal gas

The ideal-gas approximation is best at low pressure, where particles are farther apart.

p\propto T,\qquad p\propto\frac1V,\qquad p\propto n

p=\frac{nRT}{V},\qquad pV=nRT

R=8.31\,\mathrm{J\,K^{-1}\,mol^{-1}},\qquad N_A=6.0223\times10^{23}\,\mathrm{mol^{-1}}

An equation of state relates the properties and amount of a substance.

Ideal-gas assumptions

Particles have negligible volume and move randomly.
The total number of particles is large.
Newtonian mechanics applies.
Particles interact only in collisions of negligible duration.
The equipartition theorem holds: on average, each quadratic part of the kinetic energy has equal energy.

\langle U\rangle=\frac12kT\quad\text{per quadratic degree of freedom}

Equipartition

E_x=E_y=E_z=\frac12kT

T is temperature and k is Boltzmann’s constant. A point mass has three translational degrees of freedom:

K_{\mathrm{av}}=\frac32kT

A three-dimensional molecule can also rotate. Each quadratic rotational term contributes $\frac12kT$ , for example

E_{\mathrm{spin},x}=\frac12I\omega_x^2=\frac12kT

Examples:

\mathrm{He}:\ f=3,\ E_T=\frac32kT;\qquad \mathrm{N_2}:\ f=5,\ E_T=\frac52kT;\qquad \mathrm{H_2O}:\ f=6,\ E_T=3kT

U=\frac f2NkT

R=N_Ak,\qquad k=1.38\times10^{-23}\,\mathrm{J\,K^{-1}}

Pressure from molecular impacts

A particle in a cubic box of side L reverses its x-momentum at the wall. Summing the impacts of N particles produces the ideal-gas pressure.

V=L^3,\qquad \frac12kT=\frac{p_x^2}{2m},\qquad p_x=\sqrt{mkT}

\Delta p_x=2p_x,\qquad \Delta t=\frac{2L}{v}=\frac{2Lm}{p_x}

F=\frac{\Delta p_x}{\Delta t}=\frac{p_x^2}{Lm}

p=\frac FA=\frac{NkT}{L^3}=\frac{NkT}{V}

This is the equation of state in particle-number form; it is equivalent to pV=nRT because R=N_Ak.

Many types of gas

For a mixture containing N_a particles of gas a, N_b particles of gas b, and so on, the total force and pressure are sums of the species contributions.

F=\sum_iF_i,\qquad p_T=\sum_i p_i

This is Dalton’s law of partial pressures.

Root-mean-square speed

v_{\mathrm{rms}}=\sqrt{\frac{\sum_i v_i^2}{N}}

\frac12kT=\frac12m v_{x,\mathrm{rms}}^2\quad\text{(one dimension)}

K_{\mathrm{total,translational}}=\frac32NkT

K_T=\frac32pV=\frac12m\sum_i v_i^2

Nm v_{\mathrm{rms}}^2=3pV

p=\frac13\frac{Nm}{V}v_{\mathrm{rms}}^2=\frac13\rho v_{\mathrm{rms}}^2

U=\frac32NkT=\frac32pV

“Row, row, row your boat.”

III · 03

Thermodynamics III

P–V diagrams

A curve on a pressure–volume diagram represents a process from one state (p₁,V₁) to another (p₂,V₂). The path matters for heat and work, even though the change in internal energy depends only on the endpoints.

\Delta E_{\mathrm{int}}=Q+W

Infinitesimally, dE_int=dQ+dW, but Q and W are not state functions.

pV=NkT

Three common P–V paths between two states: isochoric (1), isobaric (2), and a curved path (3). The work magnitude is the area under the process curve.

Isochoric: dV=0.
Isobaric: dp=0.
Isothermal: dT=0, so pV is constant.

W=-\int p\,dV\qquad\text{(minus the area under the curve, with work-on-system convention)}

Isothermal process

W_{\mathrm{isotherm}}=-\int_{V_1}^{V_2}\frac{NkT}{V}\,dV=-NkT\ln\frac{V_2}{V_1}=NkT\ln\frac{V_1}{V_2}

For comparison with a constant-pressure route between the same volumes, the difference is

W=NkT\ln\frac{V_1}{V_2}-p_1(V_2-V_1)

\Delta U_{\mathrm{int}}=\frac f2Nk\Delta T=0\qquad(\text{isothermal ideal gas})

\Delta U=Q+W=0\quad\Longrightarrow\quad Q=-W=-NkT\ln\frac{V_1}{V_2}

Specific heat capacities at constant volume

dV=0\quad\Longrightarrow\quad dW=0

dU=dW+dQ=dQ=nc_V\,dT\qquad\text{(definition)}

U=\frac f2NkT

dU=\frac f2Nk\,dT=nc_V\,dT

N_A=\frac Nn,\qquad c_V=\frac f2\frac Nn k=\frac f2N_Ak=\frac f2R

Constant pressure

dU=\frac f2Nk\,dT,\qquad dQ=nc_p\,dT

dU=dW+dQ

dW=-p\,dV=-Nk\,dT\qquad(dp=0)

Equivalently, from U=(f/2)pV:

dU=\frac f2(V\,dp+p\,dV)=\frac f2p\,dV=\frac f2Nk\,dT

\frac f2Nk\,dT=-Nk\,dT+nc_p\,dT

c_p=R\left(1+\frac f2\right)=\frac{f+2}{2}R

Ratio of specific heats

\gamma=\frac{c_p}{c_V}=\frac{f+2}{f}=1+\frac2f

Adiabatic processes

dQ=0

dU=dQ+dW=dW=-p\,dV

U=\frac f2pV=\frac f2NkT

dU=\frac f2d(pV)=\frac f2(p\,dV+V\,dp)

-p\,dV=\frac f2(p\,dV+V\,dp)

-\left(1+\frac f2\right)p\,dV=\frac f2V\,dp

-\frac{dp}{p}=\gamma\frac{dV}{V}

-\int\frac{dp}{p}=\int\gamma\frac{dV}{V}

-\ln p=\gamma\ln V+C

\ln p+\ln V^\gamma=C\quad\Longrightarrow\quad\boxed{pV^\gamma=e^C=\mathrm{constant}}

\gamma_{\mathrm{monatomic}}=\frac53,\qquad \gamma_{\mathrm{diatomic}}=\frac75,\qquad \gamma_{\mathrm{polyatomic}}=\frac43

On a P–V diagram through the same point, an adiabat is steeper than an isotherm.

Definitions

At equilibrium the macroscopic state variables are uniform and unchanging. The temperature of a non-equilibrium system is not generally defined, so such a process cannot be represented by a precise path on a P–V diagram.

A quasistatic process keeps the system arbitrarily close to equilibrium through infinitesimal changes.

Entropy and reversibility

Many processes are irreversible and therefore cannot be reversed.

A reversible process can be reversed by an infinitesimal change in the environment and is always quasistatic. A quasistatic process is not necessarily reversible.

Heat engines

A heat engine takes heat from a hot reservoir, rejects some heat to a cold reservoir, and converts the remainder to work. A refrigerator is a heat engine run in reverse.

Heat-engine and refrigerator energy flows between reservoirs T_H and T_C.

|Q_H|=|W|+|Q_C|

Heat-engine efficiency

e=\frac{|W|}{|Q_H|}=\frac{|Q_H|-|Q_C|}{|Q_H|}=1-\frac{|Q_C|}{|Q_H|}

Refrigerator coefficient of performance

K=\frac{|Q_C|}{|W|}=\frac{|Q_C|}{|Q_H|-|Q_C|}

First-law reminder: ΔU=E_int=Q+W.

Second-law statements

Kelvin–Planck form: no one can make a perfect heat engine.

Clausius form: no one can make a perfect refrigerator.

The two statements are equivalent.

Carnot: a conceptual engine made from isotherms and adiabats

Carnot cycle on a P–V diagram. A→B and C→D are isotherms at T_C and T_H; B→C and D→A are adiabats. The enclosed area is the net work magnitude.

Clockwise operation gives negative net work on the system and acts as a heat engine. Anticlockwise operation gives positive net work on the system and acts as a refrigerator.

Heat-engine sequence

A→B: place the gas on the cold reservoir and compress it isothermally.
B→C: insulate the gas and compress it adiabatically.
C→D: place the gas on the hot reservoir and allow it to expand isothermally.
D→A: insulate the gas and allow it to expand adiabatically.

Allowing expansion is what produces work from the heat engine.

e=1-\frac{|Q_C|}{|Q_H|}

pV=NkT,\qquad U=Q+W

On an ideal-gas isotherm, dU=0 and therefore Q=−W:

Q=\int p\,dV=NkT\ln\frac{V_2}{V_1}

Q_{A\to B}=NkT_C\ln\frac{V_B}{V_A}<0\equiv Q_C

Q_{C\to D}=NkT_H\ln\frac{V_D}{V_C}>0\equiv Q_H

The adiabatic legs obey

p_BV_B^\gamma=p_CV_C^\gamma,\qquad p_DV_D^\gamma=p_AV_A^\gamma

T_CV_B^{\gamma-1}=T_HV_C^{\gamma-1},\qquad T_HV_D^{\gamma-1}=T_CV_A^{\gamma-1}

\frac{V_D}{V_C}=\frac{V_A}{V_B}

\therefore\quad\boxed{e_{\mathrm{Carnot}}=1-\frac{T_C}{T_H}}

Carnot refrigerator

K=\frac{|Q_C|}{|Q_H|-|Q_C|}=\frac{T_C}{T_H-T_C}

Carnot’s theorem

The efficiency of any heat engine operating between two specified temperatures cannot exceed the efficiency of a Carnot engine operating between the same two temperatures.

Carnot violation?

Suppose a proposed engine were more efficient than a Carnot engine. Couple it to a reversed Carnot engine (a refrigerator), using the engine’s work output W to drive the refrigerator.

The contradiction construction for Carnot’s theorem: a proposed over-efficient engine drives a Carnot refrigerator between the same reservoirs.

|Q'_H|=|W|+|Q'_C|

If the proposed engine is more efficient for the same W, then |Q′_H|<|Q_H|. Considered as a whole, the work transfers cancel. The combination has the sole net effect of moving heat from the cold reservoir to the hot reservoir without external work: a perfect refrigerator, contradicting the Clausius statement of the second law.

Carnot engine but a more efficient refrigerator?

The reverse contradiction is obtained by coupling a Carnot engine to a hypothetical refrigerator more efficient than a Carnot refrigerator. The work cancels, while the combination transfers extra heat from cold to hot. This has the same forbidden net effect.

For any Carnot cycle

e=1-\frac{T_C}{T_H}=1-\frac{|Q_C|}{|Q_H|}

\frac{T_C}{T_H}=\frac{|Q_C|}{|Q_H|}

\frac{|Q_H|}{T_H}=\frac{|Q_C|}{T_C}

Because one heat transfer is negative when signs are retained:

\frac{Q_H}{T_H}+\frac{Q_C}{T_C}=0

This relation applies to a heat engine or a refrigerator when the cycle is reversible.

Approximating a reversible cycle

A Carnot cycle is reversible. A general reversible cycle may be approximated by many small Carnot cycles. Interior heat transfers cancel in pairs; as the cells become finer, the boundary approaches the original path.

A general closed P–V path tiled by small Carnot cycles. Interior contributions cancel, leaving only the boundary integral.

\sum\left(\frac{Q_H}{T_H}+\frac{Q_C}{T_C}\right)\to0

\boxed{\oint\frac{dQ}{T}=0}\qquad(\text{reversible cycle})

Entropy from the Clausius integral

\oint_{\mathrm{cycle}}\frac{dQ}{T}=0

Take two paths from state A to state B. Following path 1 from A to B and path 2 backward from B to A makes a reversible cycle:

\int_A^B\frac{dQ}{T}\bigg|_1+\int_B^A\frac{dQ}{T}\bigg|_2=0

\therefore\quad\int_A^B\frac{dQ}{T}\bigg|_1=\int_A^B\frac{dQ}{T}\bigg|_2

Two reversible paths between the same states A and B give the same integral of dQ/T.

The path-independent integral defines a state function: entropy S.

dS=\frac{dQ_{\mathrm{rev}}}{T}

\int_A^B\frac{dQ_{\mathrm{rev}}}{T}=\int_A^B dS=\Delta S

System and surroundings

Heat received by the system comes from the surroundings, so for a reversible transfer

\int_{\mathrm{surr}}\frac{dQ}{T}=-\int_{\mathrm{sys}}\frac{dQ_{\mathrm{rev}}}{T},\qquad \frac{dQ_{\mathrm{surr}}}{T}=-\frac{dQ_{\mathrm{sys}}}{T}

\Delta S_{\mathrm{surr}}=-\Delta S_{\mathrm{sys}}

\Delta S_{\mathrm{universe}}=0\qquad(\text{quasistatic reversible process})

Irreversible heat flow

dQ_{\mathrm{sys}}+dQ_{\mathrm{surr}}=0

If dQ_sys>0, heat flows into the system and T_sys<T_surr. If dQ_sys<0, heat flows out and T_sys>T_surr. In either case the entropy gained by the colder body exceeds the entropy lost by the hotter body.

\Delta S_{\mathrm{universe}}=\int_{\mathrm{sys}}\frac{dQ}{T_{\mathrm{sys}}}+\int_{\mathrm{surr}}\frac{dQ}{T_{\mathrm{surr}}}>0\qquad(\text{irreversible heat flow})

Second law of thermodynamics: entropy form

\boxed{\Delta S_{\mathrm{universe}}\ge0}

In any process between equilibrium states, the total entropy change of the universe is non-negative.

Entropy is a state function, so its change may be calculated along any convenient known reversible path between the same endpoints.

Adiabatic and isentropic processes

dQ=0\quad\Longrightarrow\quad dS=\frac{dQ_{\mathrm{rev}}}{T}=0

A reversible adiabatic process is isentropic: ΔS=0. Conversely, no entropy change identifies an adiabatic reversible path.

T–S diagram

Carnot cycle on a T–S diagram. Isotherms are horizontal, reversible adiabats are vertical, and the enclosed area represents the net heat; for a cycle it equals the work magnitude.

Q=\int T\,dS\qquad(\text{area on a }T\text{–}S\text{ diagram})

W=\int p\,dV\qquad(\text{area on a }p\text{–}V\text{ diagram, with sign set by convention})

Ideal gases versus real gases

Ideal gases are inadequate at thermodynamic extremes, where liquefaction and other phase changes occur.

Qualitative pressure–temperature phase diagram. The sublimation, fusion, and boiling lines meet at the triple point. The liquid–gas coexistence curve terminates at the critical point, beyond which liquid and gas are not distinct.

For water, the solid–liquid coexistence line slopes toward lower temperature as pressure increases. Increasing pressure can liquefy a gas when the temperature is below the critical temperature.

III · 04

Real Gases

Gas to liquid (van der Waals model)

Temperature against supplied energy at constant pressure. Temperature rises within each phase and remains constant during fusion and vaporization; the latent heats are proportional to mass.

At constant pressure, the plateaus are latent heats. The horizontal extent represents energy transferred mechanically or as heat flow.

A kink in a temperature curve shows a phase change.

Pressure–volume isotherm through a gas–liquid transition.

A phase change needs a kink.

III · 05

Fluid Mechanics

Fluids

Fluid → that which flows. Gases and liquids. Describe ideal fluids with mass and force → to describe behaviour. Fluids are approximately continuous, so we use density, pressure, and volume instead of mass and force.

\rho=\frac{M}{V},\qquad p=\frac{F}{A}

Units: $\rho$ in $\mathrm{kg\,m^{-3}}$ ; $p$ in pascals, where $1\ \mathrm{Pa}=1\ \mathrm{N\,m^{-2}}=1\ \mathrm{kg\,m^{-1}s^{-2}}$ .

1\ \mathrm{atm}=760\ \mathrm{mmHg}=14.7\ \mathrm{psi}=1.01\times10^5\ \mathrm{Pa}

Pressure is a scalar — no direction.

Gauge pressure vs. absolute pressure

Gauge pressure equals absolute pressure minus atmospheric pressure. Example: blood pressure $120\ \mathrm{mmHg}<760\ \mathrm{mmHg}$ is gauge pressure. Tyre pressure $176\ \mathrm{kPa}$ is gauge; barometric pressure $102\ \mathrm{kPa}$ is absolute.

Hydrostatic pressure

Pressure due to a static fluid.

F=W=mg=\rho Vg=\rho Ahg

p=\frac{F}{A}=\rho gh\quad\text{(gauge pressure)}

p=\rho gh+p_0

A submerged element at depth

h

in a static fluid.

Example: excavating a dam. The wall force depends on depth through $p=\rho hg$ , not on area alone; how much the wall strength must increase depends on depth.

Archimedes' Principle

A submerged object displaces its own volume in fluid. This allows volume estimation of people and can estimate fat content.

Buoyancy

A body fully or partially submerged is buoyed up by a force equal to the weight of the fluid it displaces. Air also exerts a buoyancy force.

For a floating object:

F_{\mathrm{buoy}}=W

F_B=\rho_{\mathrm{fluid}}V_{\mathrm{disp}}g=mg

Free-body picture of a floating object and its equivalent displaced-fluid volume.

Neutral buoyancy: total submergence and $\rho_{\mathrm{object}}=\rho_{\mathrm{water}}$ .

If $\rho_{\mathrm{object}}>\rho_{\mathrm{water}}$ , then $W>F_B$ and the object sinks.
If $\rho_{\mathrm{object}}<\rho_{\mathrm{water}}$ , then $F_B<W$ before full submergence and the object floats.
When partially submerged, $F_B=W$ .

m_{\mathrm{water\ displaced}}g=m_{\mathrm{body}}g

V_{\mathrm{displaced}}=\frac{\rho_{\mathrm{object}}}{\rho_{\mathrm{fluid}}}V_{\mathrm{object}}

Example: ice cubes (identical glasses)

At the same height, $W_{\mathrm{ice\ cube}}=W_{\mathrm{displaced}}$ , therefore the masses are equal.

Example: what volume of H₂ is needed to lift a 12,000 kg airship?

The hydrogen volume is approximately the displaced-air volume.

\sum F=0

\rho_{\mathrm{air}}V_Hg-Mg-\rho_HV_Hg=0

V_H(\rho_{\mathrm{air}}-\rho_H)=M

V_H=\frac{M}{\rho_{\mathrm{air}}-\rho_H}

Airship buoyancy model with upward displaced-air force and downward ship and hydrogen weights.

Experiment for later: drop a bottle with holes in it (does water flow)? No.

Real vs. Ideal Fluids (liquid/helium)

An ideal fluid is non-viscous (no internal friction) and incompressible (density is constant).
Ideal flow is laminar, non-rotational flow.
Real fluids are viscous and may be compressible; they require more energy to flow, need pressure, and can become turbulent.
Viscous fluids slow down — important in fuel oil and blood flow.
Viscosity usually decreases with increasing temperature; engine oils get runnier and work better when warm.

Once a flow is established

Two conservation principles:

Bernoulli's equation: conservation of energy; energy density is constant.
Continuity: conservation of mass; volume rate of flow is constant.

Continuity

If incompressible, what goes in comes out.

Q=Av

Units: $\mathrm{m^3\,s^{-1}}$ . Dimensions: $L^3T^{-1}$ . $Q$ is constant.

Steady incompressible flow through a pipe that narrows from section

X

to section

Y

\pi r_X^2v_X=\pi r_Y^2v_Y

A_Xv_X=A_Yv_Y

Bernoulli

p+\frac12\rho v^2+\rho gh=\text{constant}

Conservation of energy density: electrostatic potential energy; kinetic energy; gravitational potential energy. Energy comes in two sorts here, kinetic energy and potential energy.

Water flows through this pipe

Flow from a wide section

A

, through a narrow section

B

, to a raised outlet of height

h

Continuity of flow:

v_B=v_C\quad\text{(no cavitation)},\qquad Av=\text{constant}

\Delta p=-\Delta(\rho gh)

Pressure

Pressure in Bernoulli's equation is the pressure the walls and fluid experience. This is not “ram pressure” (conservation of momentum).

Bernoulli and Venturi effect

For constant total energy, low velocity means high pressure and high velocity means low pressure.

\Delta p=-\Delta\!\left(\frac12\rho v^2\right)

Please check $\propto \tfrac32kT$ or $v$ .

Fast air over a house roof gives lower pressure above; the roof may lift. Opening windows on the lee side helps equalise the pressure.

Building design → wind-tunnel effect. Windows can get sucked out of skyscrapers. Bernoulli's equation can only be applied to a continuous stream: flow vs. flux.

Poiseuille's law

$\eta$ is viscosity.

Q=\frac{8r^4}{\pi\eta}\frac{\Delta p}{\Delta L}

Waves & Optics

IV · 01

Waves & Oscillations

Lab Hints

Do the lab and the theory.
Take lots of measurements.
Use graphs.
Do not merely list sources of error: say what you did to minimise them, calculate them, and show the calculations.
Write everything down; assume nothing.

Waves and Oscillation

Oscillation is caused by a restoring force that acts to return the system to its equilibrium position. Equilibrium occurs at a minimum of the potential energy.

U=U_{\min}\quad\text{at equilibrium},\qquad F=-\frac{dU}{dx}

A stable equilibrium is a minimum of U(x). The restoring force points toward the minimum on either side.

Variables

The period T is the time for one complete oscillation. Frequency f, in hertz, is the number of cycles per second. The wavelength is λ.

T=\frac1f

Simple harmonic oscillator

F=-kx

ma=-kx\quad\Longrightarrow\quad\frac{d^2x}{dt^2}+\frac{k}{m}x=0

x=x_0\cos(\omega t+\phi)

The amplitude x₀ and phase φ are arbitrary constants fixed by the initial conditions.

\omega=\sqrt{\frac km},\qquad \omega=2\pi f

Simple harmonic motion: x, v, and a

x=x_0\cos(\omega t+\phi)

v_x=-x_0\omega\sin(\omega t+\phi)

a_x=-x_0\omega^2\cos(\omega t+\phi)

Position, velocity, and acceleration in SHM. Velocity is a quarter-cycle out of phase with position; acceleration is opposite to position.

Find the minima and maxima of x, v, and a from the phase relations.

Potential energy

-\frac{dU}{dx}=F=-kx

\int_0^U dU=k\int_0^x x\,dx

U=\frac12kx^2

U=\frac12kx_0^2\cos^2(\omega t+\phi)

Energy in SHM

E_T=U+K

E_T=\frac12kx_0^2\cos^2(\omega t+\phi)+\frac12mx_0^2\omega^2\sin^2(\omega t+\phi)

\omega^2=\frac km\quad\Longrightarrow\quad E_T=\frac12kx_0^2

Example: simple pendulum (angles below about 15°)

Simple pendulum of length l displaced by angle θ. Tension is radial and the tangential restoring component of gravity is mg sin θ.

F=-mg\sin\theta

For small angles, sin θ≈θ and x=lθ.

F\simeq-mg\theta=-\frac{mg}{l}x

k=\frac{mg}{l}=m\omega^2

\omega^2=\frac gl,\qquad \omega=\sqrt{\frac gl}

f=\frac1{2\pi}\sqrt{\frac gl},\qquad T=2\pi\sqrt{\frac lg}

Part 2: Waves

Waves transport energy or information without transporting matter.

Mechanical waves travel through an elastic medium. Particles in the medium experience no net displacement.

Types

In a transverse wave, particle displacement is perpendicular to the direction of energy transport. In a longitudinal wave, displacement and energy transport are parallel.

Transverse and longitudinal waves, and one-, two-, and three-dimensional wavefronts.

Dimensional examples: a string carries a one-dimensional wave; pond ripples have two-dimensional circular wavefronts; sound has three-dimensional spherical wavefronts.

Wave variables include the direction of particle motion, dimension, periodicity (including finite wave packets), and the shape of wavefronts.

Travelling waves

y(x,t)=y_0\sin(kx-\omega t-\phi)\qquad(\text{travelling left to right})

k=\frac{2\pi}{\lambda},\qquad \omega=2\pi f

v=\frac\lambda T=f\lambda

This v is the velocity of a fixed phase of the wave.

Waves on a stretched string

v=\sqrt{\frac F\mu}

F is the tension and μ is the linear mass density.

\frac{dK}{dt}=\frac12\mu\omega^2y_0^2v\cos^2(kx-\omega t)

\frac{dE}{dt}=\mu\omega^2y_0^2v\cos^2(kx-\omega t)

Power and intensity

P_{\mathrm{av}}=\left(\frac{dE}{dt}\right)_{\mathrm{av}}=\frac12\mu\omega^2y_0^2v\qquad(\text{string})

Intensity is average power per unit area and is used for three-dimensional waves.

I=\frac{P_{\mathrm{av}}}{A}

Example: firework

A=4\pi r^2,\qquad I=\frac{P_{\mathrm{av}}}{4\pi r^2},\qquad I\propto\frac1{r^2}

Superposition and interference

When waves collide, the resultant displacement at any point is the sum of the individual displacements.

“You’re laughing now, but wait ’til you do the tube.”

Standing waves

A standing wave is the superposition of two waves with equal amplitude and frequency travelling in opposite directions. Nodes have zero displacement at all times; antinodes have maximum displacement.

Standing-wave snapshots. Every curve passes through fixed nodes; the antinodes oscillate between the upper and lower envelopes.

Standing-wave equation

y(x,t)=\bigl(2y_0\sin(kx)\bigr)\cos(\omega t)

Placing nodes and antinodes

\text{antinodes:}\quad kx=\left(n+\frac12\right)\pi

\text{nodes:}\quad kx=n\pi

Reflections at a boundary

At a fixed boundary, a transverse wave is reflected with a phase change of π. At a free boundary, there is no phase change. At a fixed support the reflected force is the action–reaction partner.

Pulse reflection at fixed and free boundaries. The fixed-end reflection is inverted; the free-end reflection is not.

Resonance

For a resonator of length L with two fixed ends (for example a stretched string), an integer number of half-wavelengths fits in the length.

L=\frac{n\lambda}{2}

For one closed end and one open end (for example an organ pipe), only odd quarter-wavelengths occur.

L=\frac{(2n-1)\lambda}{4}

IV · 02

Geometrical Optics

Geometrical Optics 1 → Magic

Reflection, refraction, mirrors, and ray tracing.

Light is a wave, but wavelengths are so small that wave nature can be ignored (interference, diffraction). Light is a ray; rays travel in straight lines (particle model of light).

At a boundary: reflection, refraction, or absorption. With refraction (transmission), speed and therefore direction change. Wave speed depends on an inertial property and an elastic property.

Law of reflection.

\theta_i=\theta_r

If $v_r<v_i$ , the reflected wave is out of phase by $\pi$ (half a wavelength). If $v_r>v_i$ , the reflected wave has no phase change.

Transmission is refraction (bending). Frequency does not change, with $c=\nu\lambda$ .

Refraction across a boundary, with the incident and transmitted angles measured from the normal.

\frac{\sin\theta_i}{\sin\theta_t}=\frac{v_i}{v_t}=\frac{\lambda_i}{\lambda_t}=\frac{n_t}{n_i}

Refractive index (for any wave)

n_i=\frac{v_{\mathrm{vacuum}}}{v_i}

n_1\sin\theta_1=n_2\sin\theta_2\qquad\text{(Snell's law)}

Example: apparent depth

An object under water appears shallower when viewed from air.

Check HRK and prove.

d_{\mathrm{apparent}}=d_b\frac{n_{\mathrm{air}}}{n_b}=\frac{d_b}{n_b}

d_{\mathrm{apparent,glass}}=d_g\frac{n_b}{n_g}

Fermat's principle

A light ray travelling from one point to another follows a path such that the time required is a minimum or maximum compared to other paths.

\frac{dt}{dx}=0=\frac1c\frac{dL}{dx}

Fermat's principle and the law of reflection

Two-segment reflected path from heights

a

and

b

, separated horizontally by

d

, meeting the mirror at distance

x

L=\sqrt{a^2+x^2}+\sqrt{b^2+(d-x)^2}

\frac{dL}{dx}=\frac{x}{\sqrt{a^2+x^2}}-\frac{d-x}{\sqrt{b^2+(d-x)^2}}

\frac{dL}{dx}=0

\frac{x}{\sqrt{a^2+x^2}}=\frac{d-x}{\sqrt{b^2+(d-x)^2}}

The two sides are $\sin\theta_i$ and $\sin\theta_r$ , respectively.

\sin\theta_i=\sin\theta_r\;\Longrightarrow\;\theta_i=\theta_r

Geometrical optics deals with formation of images using reflection (mirrors) and refraction (lenses).
Optical instruments can be understood using ray tracing.
Let's locate images.

Definitions

Objects: things that exist and are the source, direct or indirect, of light.
Images: what you see — light on the back of your retina.
Real images have light coming from them.
Virtual images do not have light coming from them.
Magnification: ratio of image size to object size; it can be greater than or less than one.

Plane mirrors

Plane-mirror ray construction showing a virtual image behind the mirror.

Inversion is not real, as right-side up and down are not constant definitions.

Looking at yourself in a mirror: it is a virtual image.

If you are height $h$ , how long a mirror do you need to admire your entire gorgeous body at once?

L_{\mathrm{mirror}}=\frac12h

Multiple-image mirrors

Two plane mirrors meeting at an angle produce multiple virtual images of point

P

Two plane mirrors at right angles → spastic movements.

Two parallel mirrors

Parallel mirrors produce repeated images, etc., etc.

Curved mirrors

With plane mirrors, $m=i/h_o=1$ . Curved mirrors can give $m>1$ or $m<1$ , depending on curvature and the position of the object. The image can be inverted.

Curvature and focal point

Parallel paraxial rays reflected by a concave mirror meet at the focus halfway to the centre of curvature.

f=\frac{r}{2},\qquad \kappa=\frac1r\quad\text{(curvature)}

Curvature and focal plane

Off-axis parallel rays focus in the focal plane.

Ray diagrams: three useful systems

Parallel rays from the object go through the focal point.
A focal ray from the top of the object, through the focal point, comes back parallel to the axis.
A radial ray from the top of the object through the centre of curvature is reflected back on itself.

Any two will find your image.

Use rulers.

Concave-mirror construction with the object outside the centre of curvature, giving a real, inverted, reduced image.

Concave-mirror construction for an object between the focus and mirror, giving a virtual, upright, magnified image.

Calculating positions of images and objects

\frac1s+\frac1{s'}=\frac1f

For a concave mirror, $f$ is positive. For a convex mirror, $f$ is negative. If $s'>0$ , the image is on the same side as the object; if $s'<0$ , it is on the opposite side.

Magnification

m=-\frac{s'}{s}\qquad\text{(if }s\text{ and }s'\text{ are on the same side)}

Positive $m$ : image is upright.
Negative $m$ : image is inverted.
$|m|<1$ : image reduced.
$|m|>1$ : image magnified.

Reflectors on a car are convex.

Convex-mirror construction. Reflected rays diverge and their extensions meet behind the mirror at a small, upright virtual image.

Geometrical Optics 2: Thin Lenses

Lenses

n_1\sin\theta_1=n_2\sin\theta_2

Angles are measured from the normal.

Curvy surfaces

Refraction at a spherical surface between refractive indices

n_1

and

n_2

n_1\sin\theta_1=n_2\sin\theta_2

Paraxial imaging by one spherical refracting surface.

\frac{n_1}{s}+\frac{n_2}{s'}=\frac{n_2-n_1}{r}

Sign conventions

$s'$ is positive if it is on the $n_2$ side (inside): these are real images.
$s$ is positive if it is on the outside, and negative if it is on the inside (incident and transmission side).
$r$ is positive for convex and negative for concave.

Thin lenses

Consider a double-convex lens; it refracts twice. Let the two surface-image distances be $s_1'$ and $s_2'$ .

A thin double-convex lens with two radii of curvature and object/image points.

Take into account both radii of curvature. In this double-convex case, $r_1$ is positive and $r_2$ is negative.

\frac1s+\frac1{s'}=(n_2-n_1)\left(\frac1{r_1}-\frac1{r_2}\right)

\frac1f=\frac1s+\frac1{s'}\qquad\text{(thin-lens equation)}

\frac1f=(n-1)\left(\frac1{r_1}-\frac1{r_2}\right)\qquad\text{(lensmaker's equation)}

D=\frac1f\qquad\text{(diopters; }f\text{ in metres)}

For a double-concave lens: $r_1$ is negative and $r_2$ is positive.

A double-concave diverging lens.

Converging and diverging again (two focal points).

Ray tracing for lenses

A parallel ray bends at the lens and goes through the second focal point.
A focal ray through the object and first focal point emerges parallel to the axis.
A central ray goes straight through.

\frac1s+\frac1{s'}=\frac1f

Converging lens producing a magnified, inverted, real image.

Diverging lens producing a virtual, upright, reduced image.

Q. Why is the central ray not bent? Thin lens: the surfaces are parallel!

Magnification

M=-\frac{s'}s=-\frac{i}{o}

Positive means upright. If $|m|<1$ , diminished; if $|m|>1$ , magnified.

We can produce real and virtual images. Draw a ray diagram, then do physical optics. Useful check of equations.

A lens is used to form an image on a screen. If the left half is covered, what happens? The image gets fainter, but is not changed.

Eyes — Gooo

The eye uses ciliary muscles to vary the focal length of its lens and form an image on the retina.

Produces an image on the retina.
Thicker for up close; thinner for far away.
Short vs long sighted.
Short sighted sees up close; long sighted can’t see up close.
People get long-sighted with age.

Combination of lenses (optical instruments)

Line up images in cascade.

For telescopes

\ell=f_1+f_2\qquad\text{(match focal points)}

Two-lens telescope: objective and eyepiece share a focal plane and produce parallel emerging light.

m=\frac{\theta_2}{\theta_1}

\tan\theta_1=\frac{y}{f_1},\qquad \tan\theta_2=\frac{y}{f_2}

For small angles, $\sin\theta\approx\tan\theta\approx\theta$ .

m=\frac{f_1}{f_2}

Reflecting telescopes

A reflecting telescope uses a large concave primary mirror and a small secondary mirror to direct light into the viewing area.

Advantages: no chromatic aberration; they are stronger and simpler (easier to make).

With lenses, different wavelengths travel at different speeds.

IV · 03

Physical Optics

Physical optics

Light is an EM wave.

Doesn’t need a medium; it’s a coupled electric and magnetic field.

They are created by oscillating charges.

Always travels at the same speed, $c=3\times10^8\,\mathrm{m\,s^{-1}}$ , in vacuum, and slower in other stuff.

Light transports energy and momentum — transverse wave.

v=\frac{1}{\sqrt{\varepsilon\mu}}

Light & Electricity & Magnetism

Travelling $E$ and $B$ fields are perpendicular to each other and to the direction of motion (in a vacuum).

\frac{\partial^2y}{\partial x^2}=\frac{1}{v^2}\frac{\partial^2y}{\partial t^2}\qquad\text{wave equation }(a=v)

\frac{\partial^2E}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2E}{\partial t^2}

One solution:

E(x,t)=E_0\sin(kx-\omega t)=\operatorname{Re}\left(E_0e^{i(kx-\omega t-\pi/2)}\right)

B(x,t)=B_0\sin(kx-\omega t)=\operatorname{Re}\left(B_0e^{i(kx-\omega t-\pi/2)}\right)

\omega=ck,\qquad k=\frac{2\pi}{\lambda},\qquad \omega=\frac{2\pi}{T}

Intensity (irradiance)

I=\frac{c\varepsilon_0}{2}E_0^2=\frac{c}{2\mu_0}B_0^2\qquad\text{(note the square)}

What we see.

Light propagation — Huygens’s construction

Wavelets that propagate.

Successive points on a wavefront emit circular wavelets; their envelope forms the new wavefront.

Approaching apertures

A large aperture produces mostly flat emerging wavefronts; a small aperture produces more nearly spherical wavefronts.

Spreading $\to$ diffraction when small.

Refraction is bending through mediums.

Rays vs waves

Wavefronts are perpendicular to rays.

Mirrors/lens = geometric optics.

Rays emerge from a point source perpendicular to concentric wavefronts. A second sketch marks overlap as interference.

Waves interacting with each other — superposition.

y(x,t)=\sum_i y_i(x,t)

Any waveform can be built up from a set of simple waves (Fourier analysis — a useful technique).

Fourier synthesis — making nice sounds.

Waves with the same amplitude and velocity

y_1=A\sin(kx-\omega t),\qquad y_2=A\sin(kx-\omega t+\phi)

y=y_1+y_2=A\sin(kx-\omega t)+A\sin(kx-\omega t+\phi)

y=2A\cos\left(\frac{\phi}{2}\right)\sin\left(kx-\omega t+\frac{\phi}{2}\right)

Constant amplitude.

\sin A+\sin B=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)

Or go complex

re^{i\theta}=r(\cos\theta+i\sin\theta)

\sin\theta=\operatorname{Im}(e^{i\theta}),\qquad \cos\theta=\operatorname{Re}(e^{i\theta})

y_1=A\sin(kx-\omega t)=A\cos\left(\frac{\pi}{2}-kx+\omega t\right)

y_1=\operatorname{Re}\left(Ae^{i(\pi/2-kx+\omega t)}\right)

y_2=\operatorname{Re}\left(Ae^{i(\pi/2-kx+\omega t+\phi)}\right)

y_1+y_2=\operatorname{Re}\left[Ae^{i(\pi/2-kx+\omega t)}+Ae^{i(\pi/2-kx+\omega t+\phi)}\right]

y_1+y_2=\operatorname{Re}\left[(A_1e^{i\alpha_1}+A_2e^{i\alpha_2})e^{i\omega t}\right]

The bracketed amplitude is constant with respect to $t$ and can be generalised.

The real part is important; it is what we can see.

\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}

So what does it look like?

y=2A\cos\left(\frac{\phi}{2}\right)\sin\left(kx-\omega t+\frac{\phi}{2}\right)

If $\phi=0$ , $y=2A\sin(kx-\omega t)$ : constructive interference.
If $\phi=\pi$ , $y=0$ : total destructive interference.
If $\phi$ is something in between, the frequency is the same and the peaks are halfway in between.

Three sketches compare in-phase addition, total cancellation at phase

\pi

, and partial interference at an intermediate phase.

Coherence

Coherent light maintains a constant phase relation; $\phi(t)$ is constant.

For observable interference patterns, light needs to be coherent because human eyes are too slow to see otherwise.

Phase difference from path difference

Two sources are in phase, but distance is different.

Two coherent point sources emit circular wavefronts. At a marked observation point the paths differ by one wavelength, giving a

2\pi

phase difference and constructive interference.

Distance left source to spot = $4\lambda$ ; right source to spot = $3\lambda$ .

\phi=2\pi\qquad\text{when path difference}=\lambda

Areas with $n\lambda$ out of phase give constructive interference.

Twin slits — light and electrons

Monochromatic light.

A coherent source illuminates two slits separated by

d

. Rays to a distant point

P

make angle

\theta

; for

l\gg d

, the path difference is

d\sin\theta

\text{path difference}=d\sin\theta\qquad(l\gg d)

If $l\gg d$ , put a lens in the way (Bragg’s law note).

\text{constructive interference:}\quad n\lambda=d\sin\theta

\text{destructive interference:}\quad \left(n+\frac12\right)\lambda=d\sin\theta

If you decrease $d$ , $\theta$ increases (wider fringes).

If you increase $l$ , $\theta$ is the same (smaller fringes).

One slit covered: only diffraction effects.

Pairs of variables that vary inversely to each other; e.g. Heisenberg uncertainty principle:

\Delta p\,\Delta x\geq\frac{\hbar}{2}

What about the intensity?

How bright are they going to be?

I\propto E^2

E_T=E_1+E_2

E_1(t)=E_{1,0}\sin(\omega t)

E_2(t)=E_{2,0}\sin(\omega t+\phi)

Phasors: hooray!!!

A way of representing rotating or oscillating vectors.

$E(t)=E_0\sin(\omega t)$ : a rotating vector of length $E_0$ rotates with frequency $\omega$ around the origin.

A phasor of length

E_0

at angle

\omega t

; its vertical projection gives

E_1=E_0\sin(\omega t)

Let’s add vectors! On the next page…

Two equal phasors

E_0

separated by phase

\phi

add tip-to-tail to a resultant

E_T

. The half-angle construction gives a cosine rule.

Or cosine rule:

E_T=2E_0\cos\left(\frac{\phi}{2}\right)

Phase difference $\phi$ ?

\frac{\phi}{2\pi}=\frac{d\sin\theta}{\lambda}

\phi=\frac{2\pi d\sin\theta}{\lambda}

Path length over wavelength.

E_T=2E_0\sin\left(\omega t+\frac{\phi}{2}\right)\cos\left(\frac{\phi}{2}\right)

Projection.

I\propto E^2

\frac{I_\theta}{I_0}=\frac{E_\theta^2}{E_0^2}

I_\theta=I_0\frac{\left(2E_0\cos\frac{\phi}{2}\right)^2}{E_0^2}

I_\theta=4I_0\cos^2\left(\frac{\phi}{2}\right)

\boxed{I_\theta=4I_0\cos^2\left(\frac{\pi d\sin\theta}{\lambda}\right)}

Physical optics 2 — More interference & refraction

Diffraction grating

Large number of closely spaced slits, 10,000 or more, spaced a few micrometres apart.

Consider light incident on $N$ slits, width $d$ . At $\theta=0$ , $\phi=0$ .

Parallel rays from adjacent slits separated by

d

leave at angle

\theta

; their path difference is

d\sin\theta

\text{bright spot at }\theta=0:\quad I\propto N^2I_0

d\sin\theta=m\lambda

$N$ doesn’t affect position of maxima, but affects intensity.

How wide are the maxima, or the minima? Shapes of the pattern? Very.

The many-slit intensity pattern has narrow, regularly spaced principal maxima.

Phasors

For dark of two slits, equal phasors are opposite: $\phi=\pi$ .

For three slits, three equal phasors close a triangle: $\phi=2\pi/3$ .

For 4 slits

Equal phasors for four and five slits form closed polygons at destructive interference. A star construction shows the equivalent phase step

4\pi/5

for five slits.

For N slits

\phi=\frac{2\pi m}{N}

Principal maxima are narrow and evenly spaced for many slits.

First minimum

\frac{d\sin\theta}{\lambda}=\frac{\phi}{2\pi}

\phi=\frac{2\pi d\sin\theta}{\lambda}

\phi=\frac{2\pi}{N}

\sin\delta\theta=\frac{\lambda}{dN}

\sin\delta\theta\simeq\delta\theta

To get a sharper peak, increase $N$ .

\delta\theta=\frac{\lambda}{Nd}

Making a spectroscope

Use a grating to make a spectrometer.

d\sin\theta=m\lambda

$d$ is fixed; $\theta$ varies with $\lambda$ . Depending on $d$ , might get an overlap.

A grating spectrometer uses a transmission grating, or a reflection grating (like a CD).

Spectroscopy is used for atoms.

Generally more accurate than prism spectroscopy.

Angular separation of two ends of the visible spectrum

For white light and a grating of 10,000 lines/cm, $d=10^{-6}\,\mathrm{m}$ . Red $\lambda\simeq700\,\mathrm{nm}$ ; blue $\lambda\simeq400\,\mathrm{nm}$ .

\text{blue:}\quad \sin\theta=\frac{m\lambda}{d}=\frac{400\times10^{-9}}{10^{-6}}=0.4\Rightarrow\theta\simeq0.41\,\mathrm{rad}

\text{red:}\quad \sin\theta=\frac{m\lambda}{d}=\frac{700\times10^{-9}}{10^{-6}}=0.7\Rightarrow\theta\simeq0.77\,\mathrm{rad}

\Delta\theta\simeq0.36\,\mathrm{rad}

Dispersion

Ratio of angular separation to wavelength separation gives a limit to accuracy of $\lambda$ .

d\sin\theta=m\lambda,\qquad D=\frac{d\theta}{d\lambda}

\frac{d}{d\theta}(d\sin\theta)=\frac{d}{d\lambda}(m\lambda)

d\frac{d\theta}{d\lambda}\cos\theta=m

D=\frac{d\theta}{d\lambda}=\frac{m}{d\cos\theta}\qquad\text{(not dependent on }N\text{)}

For spectrometer: small $D$ [note points toward] small $m$ and large $d$ .

Transmitted waves — Refraction

\frac{n_1}{n_2}=\frac{v_2}{v_1}=\frac{\sin\theta_2}{\sin\theta_1}=\frac{\lambda_2}{\lambda_1}

When you use a prism, you get a rainbow because $n$ is dependent on $\lambda$ .

Frequency does not change (depends on source). Only $v$ and $\lambda$ change.

A ray crosses an interface from index

n_1

n_2

, with incident and refracted angles measured from the normal.

Critical angle at which everything is reflected (nothing is transmitted):

\sin\theta_c=\frac{n_2}{n_1}\qquad\text{(only for }n_2<n_1\text{)}

Perfect transmission, as with optical fibres.

Thin films — interference & diffraction

Two reflected rays from a film of thickness

t

: one reflects at the top boundary and one after traveling down and back through the film. The optical path difference is

2t

multiplied by the film’s refractive-index factor.

If $n_1<n_2$ , then $v_2<v_1$ and the reflected wave acquires $\phi=\pi$ .

If $n_3>n_2$ , the second reflection also acquires $\phi=\pi$ .

\text{path difference}=2t

For constructive interference:

m\lambda_1=2t\frac{n_2}{n_1}

\lambda_2=\lambda_1\frac{n_1}{n_2}

$n_1=n_3$ , $n_2<n_3$ : “as in a soap bubble in air.”

At a soap film in air, the top reflection undergoes a

\pi

phase change while the lower reflection does not. The second ray also travels twice through the film thickness

t

\phi=\pi,\qquad \text{phase-equivalent path difference}=\frac{\lambda}{2}

\text{path difference}=2t+\frac{\lambda}{2}=m\lambda

2t=m\lambda-\frac{\lambda}{2}=\frac{\lambda}{2}(2m-1)

t=\frac{1}{4}(2m-1)\lambda\qquad\text{for constructive interference}

But $\lambda=\lambda_1/n_2$ , so:

t=\frac{\lambda_1}{4n_2}(2m-1)

Anti-reflection coatings prevent visible light from reflecting and glaring in photos.

So you get destructive or constructive interference.

Interferometers

Michelson interferometer

Michelson interferometer: light from a source reaches a 45° beam splitter. One arm of length

d_1

ends at a fixed mirror; the perpendicular arm of length

d_2

ends at a movable mirror. The returning beams recombine and are observed with a telescope. A compensator plate matches the glass traversed by the two paths.

Beam splitter:

Internal reflection: $\Delta\phi=0$ .
External reflection: $\Delta\phi=\pi$ .

Use a compensator plate.

d_2-d_1=\text{path difference}

At $d_2-d_1=0$ : destructive.

As $d_2-d_1$ increases, the pattern changes.

\Delta(d_2-d_1)=\frac{\lambda}{4}\quad\Rightarrow\quad\text{constructive}

Can measure as small as $\lambda/4$ !

And also glass phase difference / velocity difference due to thickness.

IV · 04

Diffraction

Physical optics III — Diffraction

Diffraction occurs when waves curve around an object or aperture whose size is similar to the wavelength of the light. Waves spread out.

Huygens' principle

Every point on a wavefront acts as a source of secondary wavelets; their envelope is the new wavefront.

Huygens construction: secondary circular wavelets advance from points on a plane wavefront, and their common envelope forms the next wavefront.

When waves pass a large object there is a shadow behind it. When waves pass a small object there is no simple shadow, but an interference pattern forms where the diffracted waves meet again.

An interference pattern caused by diffraction around an object or through an aperture is a diffraction pattern.

X-ray diffraction through a crystal structure can be used to determine crystal structure and spacing. Reflection from an object comparable to the wavelength gives way to diffraction.

Example: a person behind a tree may be heard but not seen.

X-ray diffraction

\boxed{m\lambda=2d\sin\theta}\qquad\text{Bragg's law}

Single-slit diffraction

For a slit of width $D$ , pair each wavelet in the top half with one in the bottom half. At the first minimum each pair differs in path by half a wavelength and interferes destructively.

The single slit therefore has minima at:

D\sin\theta=m\lambda

\boxed{D\sin\theta=\lambda}\qquad\text{first minimum}

The diffraction pattern has a broad central maximum with successively weaker side maxima.

Fraunhofer diffraction

Fraunhofer diffraction applies when $L\gg D$ , as opposed to Fresnel diffraction. A lens can be used to produce the far-field pattern at a finite distance.

Fraunhofer single-slit geometry and its broad central diffraction maximum.

\mathrm{p.d.}=\Delta x\sin\theta

\frac{\mathrm{p.d.}}\lambda=\frac{\Delta\phi}{2\pi}

\boxed{\Delta\phi=\frac{2\pi\Delta x\sin\theta}{\lambda}}

If $\theta$ is small, or $L\gg D$ , the electric-field amplitude contributed by each equal strip of the slit is the same; call it $\Delta E_0$ .

Phasors

At the central intensity maximum, $\Delta\phi=0$ , so all $N$ equal phasors lie in the same direction and add directly.

E=N\Delta E_0

At the first minimum, the phasors wrap through one complete turn and close, so the resultant field is zero. At the second minimum they wrap through two complete turns and again close.

E=0\qquad\text{at each minimum}

Phasor sums at the central maximum and first two minima.

What about a point somewhere in between, for an arbitrary total phase difference $\phi$ ? The phasors form an arc.

\phi=\int d\phi=\frac{2\pi D\sin\theta}{\lambda},\qquad \frac\phi2=\frac{\pi D\sin\theta}{\lambda}

If the phasor arc has radius $R$ , arc length $E_m$ , and chord $E_\theta$ , then:

R=\frac{E_m}{\phi},\qquad \sin\frac\phi2=\frac{E_\theta}{2R}

E_\theta=2R\sin\frac\phi2=\frac{2E_m}{\phi}\sin\frac\phi2

\boxed{E_\theta=E_m\frac{\sin(\phi/2)}{\phi/2}}

\boxed{\frac{I_\theta}{I_m}=\left(\frac{E_\theta}{E_m}\right)^2=\left[\frac{\sin(\phi/2)}{\phi/2}\right]^2}

\phi=\frac{2\pi D\sin\theta}{\lambda}

This agrees with the minima condition: $D\sin\theta=m\lambda$ makes $\sin(\phi/2)=0$ .

Single-slit diffraction intensity: a dominant central maximum with much smaller side maxima.

The single-slit envelope is much more severe than the twin-slit modulation.

Twin-slit interference

I_\theta=4I_0\cos^2\left(\frac\phi2\right),\qquad \phi=\frac{2\pi d\sin\theta}{\lambda}

Combine interference and diffraction

\boxed{I_\theta=4I_0\cos^2\left(\frac{\phi_{\mathrm{int}}}{2}\right)\left[\frac{\sin(\phi_{\mathrm{diff}}/2)}{\phi_{\mathrm{diff}}/2}\right]^2}

\phi_{\mathrm{int}}=\frac{2\pi d\sin\theta}{\lambda},\qquad \phi_{\mathrm{diff}}=\frac{2\pi D\sin\theta}{\lambda}

The observed twin-slit fringes are modulated by the single-slit diffraction envelope: a convolution of the two effects.

Diffraction by a circular aperture

Telescopes and microscopes use lenses, so point sources such as stars produce diffraction patterns. Lens improvements cannot eliminate this limit, although computational processing can help.

A circular aperture produces an Airy disk surrounded by rings. Its first minimum is approximately:

\boxed{\sin\theta\approx1.22\frac\lambda d}

Resolution and Rayleigh's criterion

\boxed{\alpha_c=\sin^{-1}\left(1.22\frac\lambda d\right)\approx1.22\frac\lambda d}\qquad\text{for small angles}

Two images are just resolved when the maximum of one Airy pattern lies at the first minimum of the other.

Improving resolving power

\alpha\approx1.22\frac\lambda d

For telescopes, increase the aperture diameter $d$ .
For microscopes, use a smaller wavelength, for example electrons.

\lambda=\frac hp,\qquad p=\frac Ec=\frac{h\nu}{c}=\frac h\lambda

Resolving power of a diffraction grating

\delta\theta=\frac{\lambda}{Nd\cos\theta}

This is the fringe width: the angular distance from a central maximum to its next minimum.

R=\frac\lambda{\delta\lambda}\qquad\text{resolving power}

\frac{d\theta}{d\lambda}=\frac{m}{d\cos\theta}

\delta\theta=\frac{\lambda}{Nd\cos\theta}=\frac{m\,\delta\lambda}{d\cos\theta}

\boxed{R=Nm}

$N$ is the number of slits, and $m=1,2,3,\ldots$ is the diffraction order.

IV · 05

Polarisation

Physical optics IV — Polarisation

An electromagnetic wave consists of mutually perpendicular electric and magnetic transverse waves, each perpendicular to the propagation direction.

The electric and magnetic waves are perpendicular transverse waves, and both are perpendicular to the direction of propagation.

Most light is unpolarised because of the nature of its source: the electric-field direction varies among all transverse directions.

All waves from antennas, produced by oscillating charges, are polarised: there is a preferred direction for $\mathbf E$ .

How to make polarised light

Absorption
Reflection
Scattering (sunlight)
Birefringence

1) Absorption — sheets of Polaroid

Long, thin wires or long-chain molecules arranged in parallel absorb light whose electric field is parallel to the chains, and transmit light whose electric field is perpendicular to them.

A sheet of parallel conducting chains absorbs the component of the incident electric field parallel to the chains and transmits the perpendicular component.

The chains absorb electric fields parallel to themselves because the field exerts a force on their free electrons.

Intensity: what intensity of light is transmitted?

Resolve an incident field $E$ at angle $\theta$ into components $E\cos\theta$ and $E\sin\theta$ relative to the transmission axis.

E_1^2+E_2^2=E^2

For unpolarised light, the two orthogonal components are equal on average:

E_1^2=E_2^2,\qquad E^2=2E_1^2,\qquad E_1=\frac E{\sqrt2}

\frac{I_a}{I_0}=\left(\frac{E/\sqrt2}{E}\right)^2=\frac12

\frac1{2\pi}\int_0^{2\pi}\left(E\sin\theta\right)^2d\theta=\frac{E^2}{2}

\boxed{I_a=\frac{I_0}{2}}

What about two polarisers?

$\theta$ is the angle between the two axes.

E_{\parallel}=\frac{E_0}{\sqrt2}\cos\theta

\frac{I_1}{I_0}=\frac{\left(E_0\cos\theta/\sqrt2\right)^2}{E_0^2}=\frac{\cos^2\theta}{2}

$I_0$ = prepolarised-light intensity.

Two polarisers: the first selects a vertical component and the second transmits its projection on an axis at angle θ.

3 polarisers

After the first polariser the field amplitude is $E_0/\sqrt2$ . A second axis at angle $\theta$ gives $E_0\cos\theta/\sqrt2$ ; a third axis separated by $\phi$ gives $E_0\cos\theta\cos\phi/\sqrt2$ .

\boxed{\frac{I_R}{I_0}=\frac{\cos^2\theta\cos^2\phi}{2}}

Successive projections through three polariser axes.

Useful for…

Sunglasses — remove glare (which is itself polarised).
Stress studies of structures: use light, or polarised neutrons if the structure is not transparent.
LCD displays on older mobile phones.

2) Reflection

Reflected light is partly polarised. The angle of reflection equals the angle of incidence.

Brewster's angle

At Brewster's angle the reflected and refracted rays are perpendicular, and the reflected light is completely polarised.

\theta_1+\theta_2=90^\circ

n_1\sin\theta_1=n_2\sin\theta_2

\frac{n_2}{n_1}=\frac{\sin\theta_1}{\sin(90^\circ-\theta_1)}=\frac{\sin\theta_1}{\cos\theta_1}

\boxed{\frac{n_2}{n_1}=\tan\theta_1}

Brewster geometry: reflected and refracted rays meet at a right angle.

Since reflected light is polarised, which way should sunglasses be made?

Transmission axis vertically, so you block reflected light.

3) Scattering (light in the sky)

Light causes oscillation in the perpendicular plane, and so can be transmitted like that.

Polarisation by scattering from an air molecule.

Linear polarisation (E is constant)

A linearly polarised wave has its electric field confined to one fixed plane.

Birefringence (double refraction)

The speed of light depends on the refractive index, which depends on the wavelength of the light.

For birefringent materials, $n$ depends on the polarisation direction of the light.

Therefore components of the electric field travel at different speeds (with different wavelengths) in the material, e.g. calcite.

Two different waves: one wave propagates with velocity $c/n_o$ , where $n_o$ is independent of direction. The extraordinary wave has an $n_e$ that varies with direction and has a maximum value of $n_o$ parallel to the optic axis.

$n_o$ : ordinary.
$n_e$ : extraordinary.
For the perpendicular case, $n_e<n_o$ and $v_e>v_o$ .

Calcite separates an unpolarised incident beam into ordinary and extraordinary rays; the optic-axis orientation controls their relative propagation.

The result is a single beam of elliptically polarised light.

E=E_0(\omega t)+E_0(\omega t+\phi)

The components are now out of phase. Their sum gives an $E$ that rotates around an ellipse; $E$ makes one complete rotation in time $2\pi/\omega$ .

If $\phi=0$ or $n\pi$ , the result is a straight line: linearly polarised light.
If $\phi=\pi/2$ (or an equivalent multiple), the result is circularly polarised light.

Phase-shifted perpendicular field components trace an ellipse; the quarter-cycle case traces a circle.

We want a quarter-cycle phase change. How thick must the birefringent material be?

n_o=\frac{\lambda_{\rm air}}{\lambda_o},\qquad n_e=\frac{\lambda_{\rm air}}{\lambda_e}

N_o=\frac{t}{\lambda_o}=\frac{tn_o}{\lambda_{\rm air}},\qquad N_e=\frac{t}{\lambda_e}=\frac{tn_e}{\lambda_{\rm air}}

\frac{tn_o}{\lambda_{\rm air}}=\frac{tn_e}{\lambda_{\rm air}}+\frac14

t(n_o-n_e)=\frac{\lambda_{\rm air}}4

\boxed{t=\frac{\lambda_{\rm air}}{4(n_o-n_e)}}

Quarter-wave plate of thickness t, with its optic axes resolving the incident field into components.

What determines whether we get left- or right-handed circularly polarised light?

It depends which component comes out ahead.

At an intermediate angle, $n_e<n_o$ and $v_e>v_o$ . The image can rotate around the other way.

The fast extraordinary component emerges ahead of the ordinary component, fixing the handedness of the rotating resultant.

IV · 06

Waves

Beats

Two frequencies that are close enough go in and out of phase and produce beats.

y_1=A\cos\omega_1t,\qquad y_2=A\cos\omega_2t

y=y_1+y_2=A(\cos\omega_1t+\cos\omega_2t)

\boxed{y=2A\cos(\omega_{\rm ave}t)\cos(\omega_{\rm diff}t)}

\omega_{\rm ave}=\frac12(\omega_1+\omega_2),\qquad \omega_{\rm diff}=\frac12(\omega_1-\omega_2)

The beat frequency heard is $|\omega_1-\omega_2|$ in angular-frequency units: there are two loud beats per cycle of the signed cosine envelope.

Two nearby sinusoidal frequencies and their amplitude-modulated sum.

Proof

\cos(a+b)=\cos a\cos b-\sin a\sin b

\cos(a-b)=\cos a\cos b+\sin a\sin b

\cos(a+b)+\cos(a-b)=2\cos a\cos b

a+b=\omega_1t,\qquad a-b=\omega_2t

a=\frac{(\omega_1+\omega_2)t}{2},\qquad b=\frac{(\omega_1-\omega_2)t}{2}

Doppler effect

Apparent shift of frequency due to relative motion between observer and source. Let $v$ be the speed of sound.

Moving source toward a stationary observer:

T=\frac1\nu,\qquad \lambda'=\lambda-\frac{v_s}{\nu}=\lambda\left(1-\frac{v_s}{v}\right)

\lambda'\nu'=v

\boxed{\nu'=\nu\frac{v}{v-v_s}}

For a source moving away, replace $v-v_s$ by $v+v_s$ .

Moving observer toward a stationary source:

v'=v+v_o

\nu'=\frac{v'}\lambda=\nu\frac{v+v_o}{v}

For two moving bodies, with signs chosen relative to the propagation direction:

\boxed{\nu'=\nu\frac{v+v_o}{v-v_s}}

A moving source compresses wavefronts ahead and spreads them behind; a moving observer changes the rate at which wavefronts arrive.

\nu'_{\rm air}=\nu\frac{v}{v-v_s},\qquad \nu'=\nu'_{\rm air}\frac{v-v_o}{v}

\boxed{\nu'=\nu\frac{v-v_o}{v-v_s}}\qquad\text{(with respect to the chosen direction)}

Multislit interference

$\eta$ (see flute problem).

Peak width depends on $N$ .

Many equally spaced contributions add as rotating phasors; constructive alignment gives a narrow principal peak whose width decreases as N grows.

Electricity & Magnetism

V · 01

Electric Circuits

DC Circuits

What is an electric circuit? Transfer of electrical energy through a current around a loop.

Current $i$ : flow of positive charge.

A simple cell-and-load loop transfers energy electrically.

1\ \mathrm{A}=1\ \mathrm{C\,s^{-1}}\qquad\text{(charge/time)}

Voltage $V$ is an electric potential (intensive).

Ohm's law

V=iR

$R$ is resistance, a proportionality constant, measured in ohms ( $\Omega$ ).

1\ \mathrm{V}=1\ \mathrm{A\,\Omega},\qquad 1\ \Omega=1\ \mathrm{J\,s\,C^{-2}}

Resistors take electrical energy from a circuit → heat. Two resistor symbols are shown: a zig-zag and a rectangle.

This is a cell (DC source of EMF). Potential $=\varepsilon$ , constant.

Ideal cell of emf

\varepsilon

driving a resistor

R

I=\frac{\varepsilon}{R}

For two resistors

A cell

\varepsilon

with two resistors

R_1

and

R_2

Kirchhoff's laws

Charge is conserved: charge into a junction equals charge out of a junction.
The sum of potentials around a loop is zero: conservation of energy; no perpetual motion.

i_1+i_2=i_3+i_4

\sum_{\mathrm{loop}}V_i=0

U=qV

Back to the problem: resistors in series

V-V_1-V_2=0

V-i(R_1+R_2)=0

V=i(R_1+R_2)

Equivalent to a single resistor:

R_T=R_1+R_2\qquad\text{(series)}

Resistors in parallel

i_1=\frac{V}{R_1},\qquad i_2=\frac{V}{R_2},\qquad i=i_1+i_2

\frac{V}{R_T}=\frac{V}{R_1}+\frac{V}{R_2}

\frac1{R_T}=\frac1{R_1}+\frac1{R_2}\qquad\text{(parallel)}

Power in circuits is dissipated by components

Energy transfer:

U=qV

Units:

P=\mathrm{J\,s^{-1}}

V=\mathrm{J\,C^{-1}}

I=\mathrm{C\,s^{-1}}

\therefore\ P=VI

A cell of voltage

V

drives a resistor

R

, across which the voltage is

V_R

U=q(V-V_R)

\frac{dU}{dt}=\frac{dq}{dt}V_R

P=iV_R

Using $V_R=iR$ :

P=i^2R

Using $i=V_R/R$ :

P=\frac{V_R^2}{R}

With one resistor:

P=i^2R=\frac{V^2}{R}

For two equal resistors in series, $R_T=2R$ , so the power is less:

P=\frac{V^2}{2R}

You can redraw anything as long as it is the same!

For two equal resistors in parallel, each has the set voltage $V$ , so:

Two equal resistors in series have total resistance

2R

; in parallel, both branches have the same applied voltage

V

P_T=\frac{V^2}{R}+\frac{V^2}{R}=\frac{2V^2}{R}\qquad\text{(double)}

Often voltages are set, not currents.

Measurements

A voltmeter measures voltage in volts. It is placed in parallel with the element being measured.

A voltmeter is connected in parallel across a resistor.

The ideal voltmeter has $R_{\mathrm{voltmeter}}=\infty$ , so it draws no current. Real measurements are not perfectly accurate.

An ammeter measures current in amperes (amps). It is connected in series.

An ammeter is connected in series with a resistor.

The ideal ammeter has resistance $0$ , but a real one is only pretty close.

Real power supply (internal resistance)

A real source is modeled as an ideal emf

\varepsilon

in series with internal resistance

r

For an ideal supply, $r=0$ . In old batteries, $r$ is very big.

If measured with no current, $V=\varepsilon$ , since no power is being dissipated.

R_T=r+R

i=\frac{\varepsilon}{r+R}

If $r$ is similar to $R$ , lots of power is dissipated in the battery.

V_{\mathrm{eff}}=\varepsilon-V_r=\varepsilon-ir

V_{\mathrm{eff}}=\varepsilon\left(1-\frac{r}{R+r}\right)=\varepsilon\frac{R}{R+r}

A network can either be reduced by finding the effective resistance or solved by putting a current loop in every hole on a plane. Two loop currents $i_1$ and $i_2$ give shared-branch current $i_1-i_2$ .

Two-loop resistor network used for Kirchhoff equations.

V-(i_1-i_2)R_1-i_1R_3=0

-i_2R_2-i_2R_4+(i_1-i_2)R_1=0

Let's use matrices!

i_1(R_1+R_3)-i_2R_1=V

-i_1R_1+i_2(R_1+R_2+R_4)=0

\begin{bmatrix}R_1+R_3&-R_1\\-R_1&R_1+R_2+R_4\end{bmatrix}\begin{bmatrix}i_1\\i_2\end{bmatrix}=\begin{bmatrix}V\\0\end{bmatrix}

Transients — there once, then not again; pre-equilibrium

Capacitor

A capacitor stores charge against a potential-energy difference (electric field). Farads.

V_C=\frac{q}{C}

Inductor

An inductor stores energy in a magnetic field. Henries. Same use in AC.

V_L=L\frac{dI}{dt}

Equivalent combinations

C_{\mathrm{series}}=\left(\sum_i\frac1{C_i}\right)^{-1}

C_{\mathrm{parallel}}=\sum_iC_i

L_{\mathrm{series}}=\sum_iL_i

L_{\mathrm{parallel}}=\left(\sum_i\frac1{L_i}\right)^{-1}

An RC circuit

Series RC charging circuit with source

V

, resistor

R

, capacitor

C

, charge

q

, and current

i

V-iR-\frac{q}{C}=0

i=\frac{dq}{dt}

R\frac{dq}{dt}=V-\frac{q}{C}

\frac{dq}{V-q/C}=\frac{dt}{R}

\int_0^q\frac{dq}{V-q/C}=\int_0^t\frac{dt}{R}

\ln(CV-q)-\ln(CV)=-\frac{t}{RC}

\frac{CV-q}{CV}=e^{-t/(RC)}

CV-q=CVe^{-t/(RC)}

q=CV\left(1-e^{-t/(RC)}\right)

V_C=\frac{q}{C}=V\left(1-e^{-t/(RC)}\right)

\tau=RC\qquad\text{(time constant)}

The time constant says how long it takes to charge to 67%. If the time constant is small, it charges fast, and vice versa.

Capacitor charge rises exponentially toward

CV

, reaching about

0.678\,CV

t=\tau

Remove battery

q=CV\left(e^{-t/\tau}-1\right)

An RL circuit

Series RL circuit driven by a source

V

V-iR-L\frac{di}{dt}=0

\frac{di}{dt}=\frac{V-iR}{L}

\frac{di}{V/R-i}=\frac{R}{L}\,dt

\ln\!\left(\frac{V/R-I}{V/R}\right)=-\frac{t}{L/R}

\tau=\frac{L}{R}

I=\frac{V}{R}\left(1-e^{-t/\tau}\right)

Current in a driven RL circuit rises exponentially toward

V/R

Remove battery

After removing the battery, the inductor and resistor remain in a closed loop.

I=\frac{V}{R}\left(e^{-t/\tau}-1\right)

Current decays exponentially after the battery is removed.

Energy stored in capacitor

V_C=\frac{q}{C}

dU=V\,dq=\frac{q}{C}\,dq

U=\int dU=\int_0^q\frac{q'}{C}\,dq'=\frac{1}{2}\frac{q^2}{C}=\frac{1}{2}CV^2

A capacitor with charge

q

marked on one plate.

Energy stored in inductor

V=L\frac{di}{dt}

dU=V\,dq=L\frac{di}{dt}\,dq,\qquad \frac{dq}{dt}=i

dU=Li\,di

U=\frac{1}{2}Li^2

LC circuit: Oscillatory

RLC circuit $\to$ LC circuit.

Resonant LC circuit: a capacitor and an inductor connected in one loop, with the indicated current direction.

Resonant circuit.

\frac{q}{C}+L\frac{di}{dt}=0

\frac{q}{C}+L\frac{d^2q}{dt^2}=0

\frac{d^2q}{dt^2}=-\frac{q}{LC}

\omega^2=\frac{1}{LC}

\omega=\frac{1}{\sqrt{LC}}=(LC)^{-1/2}

\frac{\bigl(m+\rho_wkt-\rho_wV\bigr)g-V(\rho_wk)}{m+\rho kt}=a

Ouch!

With circuit, use Kirchhoff!

Assessments

2nd week labs — together with lab exams, approximately 40%.
Lab exams.
Week 1 exam — approximately 5–10%.
Week 2 exam.
FSE — the exams and FSE together approximately 60%.

V · 02

Electrostatics

Unit: Electrostatics

From Electrics & Statics.

Charge: conserved quantity.

Property of particles.
Two types: positive or negative.
Measured in coulombs (C).

Fundamental unit of charge: charge on an electron.

e=-1.6\times10^{-19}\,\mathrm{C}

Charges affect each other, i.e. a force exists.

Like charges repel and unlike charges attract; the note adds that positive and negative signs can be treated with mathematics.

Coulomb’s law

\vec E=\frac{q}{4\pi\varepsilon_0}\frac{1}{r^2}\hat r

$\varepsilon_0$ = permittivity of free space (deals with electrics).

|\vec F|=\frac{q^2}{4\pi\varepsilon_0}\frac{1}{d^2}

For twice the separation:

|\vec F_2|=\frac{1}{4}|\vec F_1|

But what about an intensive property respective to only one charge?

\frac{\vec F}{q}=\vec E=\frac{q}{4\pi\varepsilon_0}\frac{1}{r^2}\hat r

Hence directions: away from a positive charge and toward a negative charge.

Vector addition of electric fields

Field lines from two nearby charges combine by the principle of superposition; cancellation produces a minimum-field point between the sources.

Principle of superposition.

If a test charge is placed in a field, the initial force is tangent to the field lines.

How to draw a field line

Start at positive charge or infinity.
End at negative charge or infinity.
Field lines never cross.
Always hit things at $90^\circ$ .
Where field lines are closer, force is stronger.

F\propto\frac{1}{r^2}\qquad\text{which is like radiant intensity}

Electric-dipole field lines leave the positive charge and terminate on the nearby negative charge; the lines close together between the charges.

Conductors

Conductors allow flow of charge. So put one in an $E$ -field.

Field lines meet a conductor normally. Free charge rearranges and clusters on the surface until the internal electric field cancels.

None clustered [inside].

F=qE

Internal charge carriers cancel out the internal electric field.

Therefore no $E$ -field in a conductor normally.

\vec E_t=\sum_i\frac{1}{4\pi\varepsilon_0}\frac{q_i}{r_i^2}\hat r_i

For lots of charges? Up to five, perhaps.

Continuous distribution

$\rho$ for volume charge density.
$\sigma$ for surface charge density.
$\lambda$ for linear charge density.

Sum after Coulomb’s law $\to$ integral.

Flux

Flux $\to$ flow of stuff through other stuff.

Flux tells you how much is going straight across.

Area vector $\hat n$ points outwards of a closed area.

A field crosses an oriented surface element

dA

; its contribution is the component normal to the surface.

\Phi=\vec F\cdot\vec A

For bulk:

\Phi=\iint_A\vec F\cdot\hat n\,dA

Let’s use symmetry.

\Phi=F\iint_A\cos\theta\,dA

For gravitation

\oiint_A\vec g\cdot\hat n\,dA=-4\pi G M_{\mathrm{enc}}

For charge (electric fields): Gauss’s law

\boxed{\oiint_A\vec E\cdot\hat n\,dA=\frac{q_{\mathrm{enc}}}{\varepsilon_0}}

Useful with symmetry. So look for constant $E$ and $\theta$ , pull $E$ out, and it works well.

\oiint_A\vec E\cdot\hat n\,dA=0\qquad(\text{net flux}=0)

Not very useful [when the Gaussian surface encloses no net charge].

Point charge

A spherical Gaussian surface of radius

r

surrounds a point charge

q

;

\vec E

is radial and parallel to

\hat n

\oiint_A\vec E\cdot\hat n\,dA=\frac{q_{\mathrm{enc}}}{\varepsilon_0}

E\oiint_A dA=\frac{q}{\varepsilon_0}

E(4\pi r^2)=\frac{q}{\varepsilon_0}

\boxed{\vec E=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\hat r}\qquad\text{ta-da!}

A line charge

End caps must be on, but $\vec E\cdot d\vec A=0$ at the end caps.

Any net field comes out either end [radially through the cylindrical side].

A cylindrical Gaussian surface of radius

r

and length

l

surrounds a line charge of density

\lambda

q_{\mathrm{enc}}=\lambda l

\oiint_A\vec E\cdot\hat n\,dA=\frac{\lambda l}{\varepsilon_0}

E(2\pi rl)=\frac{\lambda l}{\varepsilon_0}

\boxed{\vec E=\frac{1}{4\pi\varepsilon_0}\frac{2\lambda}{r}\hat r}

The interior of mass / charge

A Gaussian sphere of radius

r_1

lies inside a uniformly charged sphere of radius

r_2

Inside:

\rho\frac{4\pi r^3}{3\varepsilon_0}=\oiint_A\vec E\cdot d\vec A

\vec E=\frac{\rho r}{3\varepsilon_0}\hat r

Outside:

\vec E=\frac{1}{4\pi\varepsilon_0}\frac{\rho\frac{4}{3}\pi R^3}{r^2}\hat r

\boxed{\vec E=\frac{\rho R^3}{3\varepsilon_0r^2}\hat r}

If Gaussian surface goes through conductor

A charge

q

in the cavity of a hollow conducting sphere induces

-q

on the inner surface and

+q

on the outer surface; inside the conductor,

E=0

and the Gaussian flux is zero.

E=0,\qquad \Phi=0,\qquad q_{\mathrm{enc}}=0

Induced negative charge on the inside.

\Phi=\oiint_A\vec E\cdot\hat n\,dA

Total charge on the outer surface is $+q$ .

A hollow conducting sphere has no effect on interior charge.

Dipole

\sum\vec F=0,\qquad \sum\vec\tau\neq0

\vec\tau=\vec r\times\vec F=\frac{d}{2}qE+\frac{d}{2}qE

\sum\vec\tau=qd\,\vec E

Dipole moment:

\vec p=q\vec d

\vec\tau=\vec p\times\vec E

Energy

Minimum is when the dipole and field are parallel.

U=-\vec p\cdot\vec E

U=-qdE\cos\theta

EM II — Potential Energies & Potentials

$\vec E$ is a multidimensional gradient:

\vec E=-\nabla V

U=-\int\vec F\cdot d\vec s

U=-\int q\vec E\cdot d\vec s=q\left(-\int\vec E\cdot d\vec s\right)

The quantity in parentheses is an intensive scalar quantity.

U=qV

$U$ is potential energy; $V$ is potential (sometimes called voltage).

It’s a path integral, so route doesn’t matter because [the electric field is conservative].

Example: point charge

\vec E=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\hat r

V=-\int_{\infty}^{r}\vec E\cdot d\vec s=-\int_{\infty}^{r}\frac{q}{4\pi\varepsilon_0r^2}\,dr

V=\left[\frac{q}{4\pi\varepsilon_0}\frac{1}{r}\right]_{\infty}^{r}=\frac{1}{4\pi\varepsilon_0}\frac{q}{r}

Scalar, so just add!

V=\frac{q}{4\pi\varepsilon_0d}-\frac{q}{4\pi\varepsilon_0l}

Equipotentials

Equipotential curves around a charge and around a two-charge configuration. Zero potential does not imply zero electric field because

\vec E=-\nabla V

Zero potential does not mean zero electric field, since:

\vec E=-\nabla V\qquad\text{(a gradient)}

\text{1 dimension:}\quad \vec E=-\frac{dV}{dx}\hat x

\text{3 dimensions:}\quad \vec E=-\frac{\partial V}{\partial x}\hat x-\frac{\partial V}{\partial y}\hat y-\frac{\partial V}{\partial z}\hat z

Somewhat easier to find electric potential than electric field.

Capacitors — stores charge

Consider two parallel finite plates. The sides cancel out (no electric field); use a Gaussian pillbox with bottom area $A$ .

Two oppositely charged parallel plates separated by

d

, with a Gaussian pillbox crossing one plate. The field between the plates is uniform.

\oiint_A\vec E\cdot d\vec A=\frac{q}{\varepsilon_0}

EA=\frac{q}{\varepsilon_0}

\boxed{\vec E=\frac{q}{A\varepsilon_0}\hat k}

\frac{U}{q}=V=-\int\vec E\cdot d\vec s

V=-\frac{q}{A\varepsilon_0}d

q=V\frac{A\varepsilon_0}{d}

Proportionality constant $C$ :

C=\frac{A\varepsilon_0}{d}

For two different plates

Consider common area.

q=CV

U=qV

dU=V\,dq=\frac{q}{C}\,dq

U=\int\frac{q}{C}\,dq

\boxed{U_{\max}=\frac{1}{2}\frac{q^2}{C}=\frac{1}{2}CV^2}

If we want to store ridiculously large charge (huge capacitor), we can’t change $A$ and $d$ much, so we change $\varepsilon_0$ . So we need a dielectric.

A dielectric

An applied electric field aligns molecular dipoles. Bound surface charge appears on the dielectric, reducing the electric field inside for the same free charge.

We apply an $E$ -field; the dipoles line up.

It reduces the $E$ -field inside, which reduces the potential for the same charge.

When dipoles align, it’s called polarization.

Because of internal cancelling there is just a net surface charge.

$\vec P$ : dipole moment (a net dipole moment).

Surface charge:

\sigma=\vec P\cdot\hat n\qquad\text{(sort of a flux)}

Linear dielectrics

\vec P=\chi\varepsilon_0\vec E

A polarized dielectric between plates has bound surface charge

q_b

. A Gaussian surface encloses free charge minus bound charge.

\sigma=\vec P\cdot\hat n=\chi\varepsilon_0E

q_{\mathrm{bound}}=A\chi\varepsilon_0E

E_{\mathrm{real}}A=\frac{q-q_b}{\varepsilon_0}=\frac{q}{\varepsilon_0}-A\chi E_{\mathrm{real}}

E_{\mathrm{real}}(1+\chi)=\frac{q}{A\varepsilon_0}

E_{\mathrm{real}}=\frac{E}{1+\chi}

\varepsilon_r=1+\chi=\kappa

V=-\int\vec E\cdot d\vec s=-\int\frac{E}{1+\chi}\,\hat z\cdot d\vec s=\frac{qd}{\varepsilon_0\kappa A}

q=CV=\frac{\varepsilon_0\kappa A}{d}V

\therefore\ C=\frac{\kappa\varepsilon_0A}{d}

Back to dipoles

p=qd

Observation point at distance

r

from the dipole center; the two charges are separated by

d

, with angle

\theta

to the dipole axis.

Use potentials.

V_+=\frac{1}{4\pi\varepsilon_0}\frac{q}{\sqrt{r^2+\frac{d^2}{4}-rd\cos\theta}}

V_-=\frac{1}{4\pi\varepsilon_0}\frac{-q}{\sqrt{r^2+\frac{d^2}{4}+rd\cos\theta}}

V=\frac{q}{4\pi\varepsilon_0}\left[\frac{1}{\sqrt{r^2+\frac{d^2}{4}-rd\cos\theta}}-\frac{1}{\sqrt{r^2+\frac{d^2}{4}+rd\cos\theta}}\right]

For $r\gg d$ , the correction is not very big.

V\simeq\frac{q}{4\pi\varepsilon_0r^2}\left(1+\frac{d}{2r}\cos\theta-1+\frac{d}{2r}\cos\theta\right)

Using $(1+x)^{1/2}\simeq1+x/2$ .

V\simeq\frac{qd\cos\theta}{4\pi\varepsilon_0r^2}

Dot product:

V=\frac{\vec p\cdot\hat r}{4\pi\varepsilon_0r^2}

\vec E=-\nabla V

\vec E=-\frac{\partial V}{\partial r}\hat r-\frac{1}{r}\frac{\partial V}{\partial\theta}\hat\theta

\boxed{\vec E=\frac{qd}{4\pi\varepsilon_0r^3}\left(2\cos\theta\,\hat r+\sin\theta\,\hat\theta\right)}

V · 03

Magnetostatics

Magnets

Bar magnets

For a charge $q$ moving with velocity $\mathbf v$ , the electric and magnetic forces are:

\mathbf F_E=q\mathbf E

\mathbf F_B=q\,\mathbf v\times\mathbf B

\boxed{\mathbf F=q\left(\mathbf E+\mathbf v\times\mathbf B\right)}\qquad\text{Lorentz force law}

A magnetic field cannot do work. Why? Because $\mathbf F_B\perp\mathbf v$ .

W=\int \mathbf F\cdot d\mathbf s=0

A positive charge entering a magnetic field directed into the page follows a curved path because the magnetic force remains perpendicular to its velocity.

This still holds under relativistic transformations (Maxwell's equations).

There are no magnetic monopoles. The Earth's geographic north pole is magnetically a south pole, and vice versa.

A stable magnetic field is produced by a constant flow of charge.

Force on a current-carrying wire

Let $n_e$ be the number density of charge carriers, $N_e$ the number of electrons in a wire segment of area $A$ and length $\ell$ , and $\mathbf v_d$ their drift velocity.

N_e=n_eA\ell,\qquad q=-eN_e=-en_eA\ell

i=\frac{dq}{dt}=-en_eA v_d

\sum\mathbf F=-eN_e\,\mathbf v_d\times\mathbf B=-en_eA\ell\left(\mathbf v_d\times\mathbf B\right)

\boxed{\mathbf F=i\,\boldsymbol\ell\times\mathbf B}

The direction follows the right-hand rule.

A current-carrying wire produces a magnetic field

Law of Biot and Savart

\boxed{d\mathbf B=\frac{\mu_0}{4\pi}\frac{i\,d\boldsymbol\ell\times\hat{\mathbf r}}{r^2}}

\mu_0=4\pi\times10^{-7}\ \mathrm{H\,m^{-1}}\qquad\text{(permeability of free space)}

This is a differential law.

Geometry used to integrate the Biot–Savart law for an infinite straight wire.

r=\frac d{\sin\theta},\qquad \ell=r\cos\theta=d\cot\theta,\qquad d\ell=-\frac d{\sin^2\theta}\,d\theta

B=\frac{\mu_0i}{4\pi}\int_{0}^{\pi}\frac{\sin\theta\,|d\ell|}{r^2}=\frac{\mu_0i}{4\pi d}\int_0^\pi\sin\theta\,d\theta

\boxed{|\mathbf B|=\frac{\mu_0i}{2\pi d}}

Force between two parallel conductors

For two parallel wires separated by distance $d$ , the field of wire 1 at wire 2 is $B_1=\mu_0i_1/(2\pi d)$ .

F_2=i_2\ell B_1=i_2\ell\frac{\mu_0i_1}{2\pi d}

\boxed{\frac F\ell=\frac{\mu_0}{2\pi}\frac{i_1i_2}{d}}

Ampère's law (deal with symmetry!)

\boxed{\oint_S\mathbf B\cdot d\boldsymbol\ell=\mu_0 i_{\mathrm{enc}}}

To find $\mathbf B$ , symmetry is needed.

If $\mathbf B\parallel d\boldsymbol\ell$ , the dot-product factor is 1.
If $\mathbf B\perp d\boldsymbol\ell$ , the dot-product factor is 0.
A zero circulation need not mean that the field is zero at every point: this is part of the non-ideal nature of reality.

Straight wire

\oint\mathbf B\cdot d\boldsymbol\ell=B\oint d\ell=B(2\pi r)=\mu_0i

\boxed{B=\frac{\mu_0i}{2\pi r}}\qquad\text{(the same result as before)}

Consider a solenoid

Ampèrian rectangle through an ideal long solenoid; the exterior field is taken as zero and the end contributions vanish.

Use more turns to obtain more enclosed current. For an ideal long solenoid, $B=0$ outside; the end contributions cancel (equivalently, $\mathbf B\cdot d\boldsymbol\ell=0$ ). Edge effects are ignored.

BL=\mu_0 i_{\mathrm{enc}},\qquad i_{\mathrm{enc}}=Ni

B=\frac{\mu_0iN}{L}

Let $n=N/L$ be the number of turns per unit length.

\boxed{B=\mu_0in}

The ideal result does not depend on the solenoid radius.

Magnetic dipoles

A bar magnet with north and south poles has the same external field form as a current loop.

Field on the axis of a circular loop

Circular current loop of radius R and an axial observation point a distance z from its centre.

r=\sqrt{R^2+z^2},\qquad \cos\theta=\frac Rr=\frac R{\sqrt{R^2+z^2}}

B=\frac{\mu_0i}{4\pi}\int\frac{d\ell\cos\theta}{r^2}=\frac{\mu_0i}{4\pi}\frac{2\pi R^2}{\left(R^2+z^2\right)^{3/2}}

\boxed{B=\frac{\mu_0iR^2}{2\left(R^2+z^2\right)^{3/2}}}

Dipole moment

\boldsymbol\mu=Ni\mathbf A

If a dipole is placed in a magnetic field:

\boldsymbol\tau=\boldsymbol\mu\times\mathbf B\qquad\text{(principle of the electric motor)}

U=-\boldsymbol\mu\cdot\mathbf B\qquad\text{(potential energy stored in a magnetic dipole)}

V · 04

Alternating Current

Alternating current

The voltage source varies sinusoidally with time.

V=V_0\cos\omega t

\operatorname{Re}\left(e^{i\theta}\right)=\cos\theta,\qquad V=V_0e^{i\omega t}

The physical voltage is not really complex; its real part is used. The current varies with the same angular frequency:

I=I_0e^{i(\omega t+\phi)}

Inductor

V_L=L\frac{dI}{dt}=i\omega LI

\boxed{Z_L=i\omega L}\qquad\text{(phase shift }\pi/2\text{)}

Capacitor

V_C=\frac qC,\qquad \frac{dV_C}{dt}=\frac1C\frac{dq}{dt}=\frac IC

q=\int I\,dt=\frac{I_0}{i\omega}e^{i(\omega t+\phi)}

V_C=\frac I{i\omega C}=-\frac{iI}{\omega C}

\boxed{Z_C=-\frac{i}{\omega C}}

Resistor

V=IR,\qquad \boxed{Z_R=R}

Redefine Ohm's law as:

\boxed{V=IZ}\qquad Z=\text{impedance (AC resistance)}

Reactance

X_L=\omega L,\qquad X_C=\frac1{\omega C},\qquad X_R=R

Reactance puts the current out of phase with the voltage. Impedances can be added like resistances in series and parallel, and Kirchhoff's laws still apply.

Let's draw pretty pictures!

Impedance phasor diagram: resistance lies on the real axis, inductive reactance upward, capacitive reactance downward, and the vector sum is Z.

\phi=0\quad\Longrightarrow\quad \omega L=\frac1{\omega C}\quad\Longrightarrow\quad \omega^2=\frac1{LC}

V_0e^{i\omega t}=I_0e^{i(\omega t+\phi)}Z,\qquad V_0=I_0|Z|

LR series circuit

V-V_R-V_L=0

V_0e^{i\omega t}=I_0e^{i(\omega t+\phi)}(R+i\omega L)

R+i\omega L=\sqrt{R^2+\omega^2L^2}\,e^{i\tan^{-1}(\omega L/R)}

\boxed{I=\frac{V_0}{\sqrt{R^2+\omega^2L^2}}e^{i\left(\omega t-\tan^{-1}(\omega L/R)\right)}}

With more inductance, the current lags farther behind the voltage.

RC circuit

V-IR-IZ_C=0

V_0e^{i\omega t}=I\left(R-\frac{i}{\omega C}\right)

V_0=I_0|Z|,\qquad I_0=\frac{V_0}{\sqrt{R^2+\frac1{\omega^2C^2}}}

\phi_I=-\tan^{-1}\left(-\frac1{\omega RC}\right)=\tan^{-1}\left(\frac1{\omega RC}\right)

\boxed{I=\frac{V_0}{\sqrt{R^2+\frac1{\omega^2C^2}}}e^{i\left(\omega t+\tan^{-1}(1/\omega RC)\right)}}

Current leads the voltage in a mostly capacitive circuit.

Mnemonic: ELI the ICE man. Here $\mathcal E$ is the EMF (voltage).

Power dissipated in a circuit

P=IV

The average must be taken using the real sinusoidal quantities.

P_{\mathrm{avg}}=\frac{V_0I_0}{T}\int_0^T\cos\omega t\,\cos(\omega t+\phi)\,dt

\cos A\cos B=\frac12\left[\cos(A-B)+\cos(A+B)\right]

P_{\mathrm{avg}}=\frac{V_0I_0}{2T}\int_0^T\left[\cos\phi+\cos(2\omega t+\phi)\right]dt

\boxed{P_{\mathrm{avg}}=\frac{V_0I_0}{2}\cos\phi}

$\cos\phi$ is the power factor. If $\phi=0$ , the power is maximal.

For a pure inductor or a pure capacitor, $\phi=\pi/2$ , so the average power dissipated is zero.

RMS — root mean square

Square, average, and take the square root again.

V_{\mathrm{rms}}=\sqrt{\frac1T\int_0^T V_0^2\cos^2\omega t\,dt}=\sqrt{\frac12V_0^2}=\frac{V_0}{\sqrt2}

This result is for sinusoidal oscillations. Thus $V_0=\sqrt2V_{\mathrm{rms}}$ (“more voltage out!”). The same relation holds for current.

\boxed{P_{\mathrm{avg}}=V_{\mathrm{rms}}I_{\mathrm{rms}}\cos\phi}

RLC circuit

V=IZ,\qquad V_0=I_0|Z|

Z=R+i\omega L-\frac{i}{\omega C}

|Z|=\sqrt{R^2+\left(\omega L-\frac1{\omega C}\right)^2}

I_0=\frac{V_0}{\sqrt{R^2+\left(\omega L-\frac1{\omega C}\right)^2}}

For maximum current:

\left(\omega L-\frac1{\omega C}\right)^2=0\quad\Longrightarrow\quad \omega L=\frac1{\omega C}

\boxed{\omega^2=\frac1{LC},\qquad \omega=\frac1{\sqrt{LC}}}\qquad\text{resonant frequency}

\tan\phi=\frac{\omega L-1/(\omega C)}R

At resonance $\phi=0$ , the power factor is 1.

P=\frac{I_0V_0}{2}=\frac{V_0^2}{2R}=\frac{V_{\mathrm{rms}}^2}{R}

Put a capacitive line filter in place to protect the mains from out-of-phase interference.

As $\omega\to0$ , $Z_L=i\omega L\to0$ , so the inductor may be ignored, while $Z_C=-i/(\omega C)\to\infty$ , so the capacitor is an open circuit.
As $\omega\to\infty$ , $Z_L\to\infty$ and $Z_C\to0$ . At very high frequency a real inductor begins to look somewhat like a capacitor and acts like one.

V · 05

Electrodynamics

Faraday's law

\boxed{\mathcal E=-\frac{\partial\Phi_B}{\partial t}}

\boxed{\Phi_B=\iint\mathbf B\cdot\hat{\mathbf n}\,dA}

If $\mathbf B$ is constant and parallel to the surface normal, $\Phi_B=BA$ .

Lenz's law

The direction of the induced EMF opposes any change in magnetic flux. This is a consequence of conservation of energy.

\mathcal E=-N\frac{d\Phi_B}{dt}

Example: sliding conducting rod

A conducting rod of height h slides right with speed v₀ on rails in a uniform field into the page; the rails are closed through resistance R.

\mathcal E=-\frac{d(BA)}{dt}=-\frac{d(Bhx)}{dt}=-Bh\frac{dx}{dt}

\boxed{|\mathcal E|=Bhv_0},\qquad i=\frac{Bhv_0}{R}

\boxed{|\mathbf F|=ihB=\frac{B^2h^2v_0}{R}}

By Lenz's law, the magnetic force on the rod points to the left, opposing its motion.

Example: the solenoid

\boxed{B=\frac{\mu_0iN}{\ell}}\qquad\text{(from Ampère's law)}

Suppose the field collapses in a time $\Delta t$ . For a solenoid of radius $r$ :

\Phi_B=BA=\frac{\mu_0iN}{\ell}\pi r^2

\Phi_{B,f}=0,\qquad \frac{d\Phi_B}{dt}\approx\frac{\Delta\Phi_B}{\Delta t}=-\frac{\Phi_B}{\Delta t}

\boxed{\mathcal E=-N\frac{d\Phi_B}{dt}=\frac{\mu_0iN^2\pi r^2}{\ell\,\Delta t}}

If $\Delta t$ is small, the induced electric field can be very large.

What if there is no current loop? A changing magnetic flux still produces an induced electric field:

\mathcal E=\oint\mathbf E\cdot d\boldsymbol\ell=-\frac{d\Phi_B}{dt}

Induced electric field from changing flux

Inductance

\boxed{\mathcal E_L=-L\frac{di}{dt}}

Inductance depends on geometry and acts as electrical inertia, analogous to the role of mass in $F=ma$ .

N\Phi_B=LI

N\Phi_B=N\left(\frac{\mu_0iN}{\ell}\pi r^2\right)=\frac{\mu_0iN^2\pi r^2}{\ell}=Li

\boxed{L=\mu_0\pi r^2n^2\ell},\qquad n=\frac N\ell

Example: inductance of a rectangular toroid — full doughnut!

By circular symmetry, for a toroid with $N$ turns:

\oint\mathbf B\cdot d\boldsymbol\ell=\mu_0i_{\mathrm{enc}}=\mu_0Ni

B(2\pi r)=\mu_0Ni,\qquad \mathbf B=\frac{\mu_0Ni}{2\pi r}\,\hat{\boldsymbol\phi}

Rectangular toroid with inner radius a, outer radius b, height h, and N turns.

\Phi_B=\int_0^h\int_a^b\frac{\mu_0Ni}{2\pi r}\,dr\,dh=\frac{\mu_0Nih}{2\pi}\ln\frac ba

N\Phi_B=Li

\boxed{L=\frac{\mu_0N^2h}{2\pi}\ln\frac ba}

Inductance is measured in henries. Its dimensions are $[L]=\mathrm{J\,s^2\,C^{-2}}$ . This is self-inductance.

Mutual inductance

N_2\Phi_{21}=Mi_1,\qquad N_1\Phi_{12}=Mi_2

Mutual inductance is symmetric: it is the same either way. It describes transformer coupling between coils and is often written with a coupling coefficient $k$ .

\mathcal E_2=-M\frac{di_1}{dt},\qquad \mathcal E_1=-M\frac{di_2}{dt}

Mutual inductance of coupled solenoids

For two coaxial solenoids, with the smaller cross-sectional area $A_1=\pi r_1^2$ lying within the second solenoid, the field due to coil 1 is:

B_1=\mu_0n_1i_1=\frac{\mu_0N_1i_1}{\ell}

\Phi_{21}=B_1A_1=\mu_0n_1i_1\pi r_1^2

N_2\Phi_{21}=Mi_1

\boxed{M=\mu_0\pi r_1^2N_2n_1=\frac{\mu_0\pi r_1^2N_1N_2}{\ell}}

\mathcal E_2=-\mu_0\pi r_1^2N_2n_1\frac{di_1}{dt}=-M\frac{di_1}{dt}

Magnetic materials: put iron in solenoids

A dielectric reduces an electric field; ferromagnetic materials increase a magnetic field.

Diamagnetism: if a material has no permanent dipole moment, a slight opposing field is induced; water is an example.
Paramagnetism: dipoles align with the applied magnetic field, strengthening it.
Ferromagnetism: the material's dipoles possess large-scale order and group into domains. This produces a semi-permanent field and greatly increases the magnetic field.

Adding a magnetic material with magnetic susceptibility or relative permeability changes the field in an inductor, in a manner analogous to adding a dielectric.

\mu=\mu_r\mu_0,\qquad L=\mu_rL_0

Energy stored in an inductor: stored in the magnetic field

P=IV

dE=IV\,dt=IL\frac{di}{dt}\,dt=Li\,di

\boxed{E=\int_0^I Li\,di=\frac12LI^2}

For a solenoid:

B=\frac{\mu_0IN}{\ell},\qquad B^2=\frac{\mu_0^2I^2N^2}{\ell^2},\qquad L=\frac{\mu_0AN^2}{\ell}

E=\frac12\frac{\mu_0AN^2}{\ell}I^2=\frac{B^2}{2\mu_0}A\ell

\boxed{u=\frac EV=\frac{B^2}{2\mu_0}}

Here $u$ is the magnetic energy density.

V · 06

Electromagnetism

Current density: J

|I|=A|\mathbf J|\quad\text{if }\mathbf J\text{ is uniform}

\mathbf J=\sigma(\mathbf E+\mathbf v\times\mathbf B)\qquad\text{(general)}

\mathbf J=\rho\mathbf v

\mathbf J=\sigma\mathbf E,\qquad \sigma=\rho_{\rm resistivity}^{-1}

The electric-field form is related to $V=IR$ . $\sigma$ is conductivity.

Energy in an electric field (capacitors)

U_E=\frac12CV^2,\qquad V=Ed

U_E=\frac12CE^2d^2,\qquad C=\frac{\epsilon_0A}{d}

U_E=\frac12Ad\,\epsilon_0E^2

\boxed{\frac{U_E}{\mathcal V}=\frac12\epsilon_0E^2},\qquad \mathcal V=Ad

Parallel-plate capacitor used to obtain the electric-field energy density.

Maxwell equations

Gauss's law:

\oiint \mathbf E\cdot\hat{\mathbf n}\,dA=\frac{q}{\epsilon_0}

Ampère's law:

\oint \mathbf B\cdot\hat{\mathbf t}\,ds=\mu_0 i_{\rm enc}

Faraday's law — a change in magnetic flux generates an electric field:

\oint \mathbf E\cdot\hat{\mathbf t}\,ds=-\frac{d\Phi_B}{dt}

The last law (Gauss's other law):

\oiint \mathbf B\cdot\hat{\mathbf n}\,dA=0

Ampère's law should also be able to create a magnetic field from an electric field. Consider a capacitor.

i_D=\epsilon_0\frac{d\Phi_E}{dt}

Let us invent a “displacement current”. A changing $E$ -field can create a $B$ -field, and vice versa.

Charging capacitor: conduction current in the wires and changing electric flux between the plates give the same enclosed current.

V=\frac qC=Ed

q=CV=\frac{\epsilon_0A}{d}(Ed)=\epsilon_0AE

i=\frac{dq}{dt}=\epsilon_0A\frac{dE}{dt}=i_D

Apply Ampère's law:

\oint\mathbf B\cdot\hat{\mathbf t}\,ds=\mu_0i_D=\mu_0\epsilon_0A\frac{dE}{dt}=\mu_0i

C=\frac{\epsilon_0A}{d}

New Ampère's law

\oint\mathbf B\cdot\hat{\mathbf t}\,ds=\mu_0i_{\rm enc}+\mu_0\epsilon_0\frac{d\Phi_E}{dt}

Maxwell's equations

\oiint\mathbf B\cdot\hat{\mathbf n}\,dA=0

\oiint\mathbf E\cdot\hat{\mathbf n}\,dA=\frac q{\epsilon_0}

\oint\mathbf B\cdot\hat{\mathbf t}\,ds=\mu_0i+\mu_0\epsilon_0\frac{d\Phi_E}{dt}

\oint\mathbf E\cdot\hat{\mathbf t}\,ds=-\frac{d\Phi_B}{dt}

Law of Biot and Savart

d\mathbf B=\frac{\mu_0}{4\pi}\frac{I\,d\boldsymbol\ell\times\hat{\mathbf r}}{r^2}

(For non- $E/B$ materials.)

Is it possible to have a wave solution to Maxwell's equations? Does it satisfy wave equations?

Look at the area vectors: they are perpendicular (it is obvious). In free space there is no current and no electric or magnetic material.

An electromagnetic wave: E and B oscillate in mutually perpendicular planes and are perpendicular to the direction of propagation.

Faraday's law: consider a thin rectangular loop of height $h$ and width $dx$ in the electric field.

\oint\mathbf E\cdot\hat{\mathbf t}\,ds=-\frac{d\Phi_B}{dt}

(E+dE)h-Eh=-\frac{\partial B}{\partial t}h\,dx

h\,dE=-h\frac{\partial B}{\partial t}\,dx

\boxed{\frac{\partial E}{\partial x}=-\frac{\partial B}{\partial t}}

Now consider the magnetic field:

Bh-(B+dB)h=\mu_0\epsilon_0\frac{\partial E}{\partial t}h\,dx

\boxed{-\frac{\partial B}{\partial x}=\mu_0\epsilon_0\frac{\partial E}{\partial t}}

With partial differentiation, you can do different things, like…

\frac{\partial}{\partial x}\!\left(\frac{\partial E}{\partial x}\right)=-\frac{\partial}{\partial x}\!\left(\frac{\partial B}{\partial t}\right)

\frac{\partial^2E}{\partial x^2}=-\frac{\partial^2B}{\partial x\,\partial t}

-\frac{\partial^2B}{\partial t\,\partial x}=\mu_0\epsilon_0\frac{\partial^2E}{\partial t^2}

\frac{\partial^2E}{\partial x^2}=\mu_0\epsilon_0\frac{\partial^2E}{\partial t^2}

\boxed{\frac{\partial^2E}{\partial x^2}-\mu_0\epsilon_0\frac{\partial^2E}{\partial t^2}=0}\qquad\text{(wave equation)}

\frac{\partial^2E}{\partial x^2}-\frac1{v^2}\frac{\partial^2E}{\partial t^2}=0

\frac1{c^2}=\mu_0\epsilon_0,\qquad \boxed{c=\frac1{\sqrt{\mu_0\epsilon_0}}}

The same calculation can be done for the magnetic field. The solutions are sinusoidal oscillations.

E=E_0\sin(kx-\omega t)

B=B_0\sin(kx-\omega t-\phi)

Use $\partial E/\partial x=-\partial B/\partial t$ :

kE_0\cos(kx-\omega t)=\omega B_0\cos(kx-\omega t-\phi)

They have to be equal. $\phi=0$ : the waves are in phase.

\boxed{E_0=B_0\frac\omega k=cB_0}

The energy density is equal in $B$ and $E$ .

The Poynting vector (direction of energy transport)

\boxed{\mathbf S=\frac1{\mu_0}\mathbf E\times\mathbf B}\qquad(\mathrm{W\,m^{-2}})

|S|=\frac1{\mu_0}EB

The Poynting vector points along E × B.

E_0=cB_0,\qquad E=cB,\qquad \mu_0c^2\epsilon_0=1

u_B=\frac{B^2}{2\mu_0}=\frac{|S|}{2c},\qquad u_E=\frac{\epsilon_0E^2}{2}=\frac{|S|}{2c}

|E|=c|B|

\langle S\rangle=\frac1{\mu_0}E_0B_0\langle\sin^2(kx-\omega t)\rangle=\frac{E_0B_0}{2\mu_0}=\frac1{\mu_0}E_{\rm RMS}B_{\rm RMS}

\boxed{I=\frac1{\mu_0}E_{\rm RMS}B_{\rm RMS}}

The Poynting vector can be used to work out the direction of energy flow at the surface when $B$ changes, for example in a solenoid.

B=\mu_0ni\,\hat{\mathbf z}

\oint\mathbf E\cdot\hat{\mathbf t}\,ds=-\frac{d\Phi_B}{dt}

\Phi_B=BAN=BA(n\ell)=\mu_0n^2\ell A\,i

2\pi rE=-\mu_0n^2\ell\pi R^2\frac{di}{dt}

\mathbf E=-\frac{\mu_0n^2\ell R^2}{2r}\frac{di}{dt}\,\hat{\boldsymbol\theta}

\mathbf S=\frac1{\mu_0}\mathbf E\times\mathbf B

At the solenoid surface, $\mathbf S$ points inward, in the $-\hat{\mathbf r}$ direction.

\mathbf S\big|_{r=R}=-\frac{\mu_0n^2R}{2}\,i\frac{di}{dt}\,\hat{\mathbf r}

\frac{dE_{\rm in}}{dt}=|\mathbf S|(2\pi R\ell)=\mu_0\pi R^2\ell n^2\,i\frac{di}{dt}

L=\mu_0\pi R^2\ell n^2

\frac{dE_{\rm in}}{dt}=Li\frac{di}{dt}

\boxed{E=\frac12Li^2}

Poynting flux enters radially through the cylindrical surface of the solenoid.

Modern & Quantum Physics

VI · 01

Modern Physics

Modern Physics — The Bohr Atom

HRK ch. 51. Read!

Classical says continuous energy. Quantum says energy is quantised.

Classical model

Rutherford's model of the atom. Problem: accelerating charges emit radiation; decay of the orbit means loss of energy. Atoms are not stable!

Rutherford model: a negatively charged electron accelerates around a positive nucleus and radiates.

Bohr said it can't work

Postulates:

Stationary states. They do not emit radiation in this state. Energy is quantised, and so is angular momentum.
A transition between stationary states emits or absorbs one photon whose energy is the difference between the two levels.

L=\frac{nh}{2\pi}=n\hbar,\qquad n=1,2,3,\ldots

$h$ is Planck's constant and $\hbar=h/(2\pi)$ .

Quantised energy levels and a transition from

E_2

E_1

that emits a photon.

h\nu=E_2-E_1

A photon is a “packet of light”:

E=E_2-E_1=h\nu,\qquad c=\nu\lambda

$\nu$ is frequency. The lowest possible energy state is called the ground state.

But it's dodgy

Only accounts for one electron.
Does not work for molecules and solids.
Mixes quantum and classical indiscriminately.
Explained spectral lines for hydrogen.
Accounted for lots of problems in the Rutherford model.
Roughly the right distance.

Photoelectric Effect — Einstein's 1921 Nobel Prize

The release of electrons from a metal surface by light.

Light of intensity

I

strikes a metal and ejects an electron with speed

v

K=\frac12mv^2

The work function $\phi$ is the energy required to eject an electron.

Discovered accidentally by Hertz in 1887 while radio tuning.

Photoelectric-effect apparatus. A photosensitive cathode is illuminated; emitted electrons are collected by the anode while a variable opposing voltage is measured.

To test the kinetic energy of electrons, run them against a potential.

K_{\max}\propto V_{\mathrm{stopping}},\qquad \frac12mv^2=qV_0

According to classical physics

Light is a continuous wave.
Intensity is proportional to the amplitude of the wave.

Tests

1) Vary intensity, with $\nu$ constant.

Expected: greater intensity means more electrons, more current, and a higher stopping voltage. What they found: the photocurrent changes, but the stopping voltage is the same. One atom can only absorb one photon at once.

Photocurrent against stopping voltage for three intensities. Greater intensity raises the saturation current, while the stopping voltage is unchanged.

2) Vary frequency, with intensity constant.

Expected: $K_{\max}$ is constant. What was found: $K_{\max}$ increases linearly with frequency; the gradient is $h$ , Planck's constant. Below the threshold frequency $\nu_0$ , $K<\phi$ and there is no photocurrent.

Maximum photoelectron kinetic energy against frequency.

3) Expected: a time delay while an electron absorbs energy. Result: no time delay; it either comes out or not at all. No time delay with photons: make or break, but do not wait.

Einstein's postulates (1905)

Light acts like particles, “photons”, with energy $E=h\nu$ .
Electrons can only take one photon at a time.

h=6.626\times10^{-34}\ \mathrm{J\,s}

K_{\max}=h\nu-\phi\qquad\text{(conservation of energy)}

Modern Physics II

E=h\nu\qquad\text{(photon energy)}

De Broglie says waves can be particles; particles can be waves!

A localized wave packet propagating to the right with velocity

v

. Its indicated spatial scale is of order

\lambda

Photons have momentum…

p=mv

p=\frac{E}{c}=\frac{h\nu}{c}

c=\nu\lambda,\qquad \lambda=\frac{c}{\nu}

\boxed{p=\frac{h}{\lambda}}\qquad\text{momentum of photon}

$\lambda$ is the de Broglie wavelength.

\lambda=\frac{h}{p}\qquad\text{works for everything else}

A point-like wave emits circular fronts; after passing an aperture, the recorded intensity profile is wavy. The note emphasizes that this intensity construction works for waves, not classical particles.

Diffraction/interference demonstration.

Electrons (closet wave).

Example question

For an electron, $v=10^7\,\mathrm{m\,s^{-1}}$ , $m=9.1\times10^{-31}\,\mathrm{kg}$ , and $h=6.6\times10^{-34}\,\mathrm{J\,s}$ .

p=9.1\times10^{-24}\,\mathrm{kg\,m\,s^{-1}}

\lambda=\frac{h}{p}=7.25\times10^{-11}\,\mathrm{m}

For a tennis ball, $v=100\,\mathrm{m\,s^{-1}}$ and $m=0.1\,\mathrm{kg}$ .

\lambda=6.6\times10^{-35}\,\mathrm{m}

(Very small, so tennis balls don’t diffract.)

VI · 02

Quantum Physics

Electrons act as waves

Particle and wave descriptions are contrasted: a localized electron trajectory versus a wave/particle duality picture and a particle-like detection line.

1989, one electron at a time; still get interference. So…

$\psi$ : wave function.

A localized wave function with a bell-shaped amplitude. The squared magnitude gives the probability density for finding the particle.

|\psi|^2=\text{probability density of finding the particle}

The wave functions interfere.

Interpretations

Copenhagen — observation collapses the wave function.
Many Worlds — observation decoheres alternate universes.
Measurement at a slit collapses the wave function prematurely. (It can’t interfere with itself anymore.)

Effects & Conundrums

Two potential-energy sketches: a classically trapped state that cannot escape a well, and quantum tunnelling through a finite barrier, with an exponentially diminished oscillation beyond it.

Chance of detecting it inside or beyond the barrier.

Violate conservation of energy? Negative kinetic energy?

A thought experiment: Schrödinger’s cat

Schrödinger’s-cat apparatus: a radioactive particle, Geiger counter, HCN poison, cat, and video camera; the camera tape is entangled too.

Radioactive particle.

If the atom decays, the Geiger counter triggers the cyanide, killing the cat.

Some time, for finite chance, in a superposition of states.

So is the cat in a superposition state too? Yes, but who cares.

Does the cat count as an observer? Does it matter? It’s like Deckard…

Quantum Suicide

Pull lever, 50% chance you die.

So remaining person knows Many Worlds works.

\Delta p_x\,\Delta x\geq \frac{\hbar}{2}

\Delta E\,\Delta t\geq \frac{h}{4\pi}

Heisenberg uncertainty principle

|\psi|^2=\text{probability}

A broader position distribution has $\Delta x_1>\Delta x_2$ . The centre gives the average (expectation) position $\langle x\rangle$ ; similarly $\langle p\rangle$ is momentum.

\boxed{\Delta x\,\Delta p_x\ge\frac\hbar2},\qquad \hbar=\frac h{2\pi}

$\Delta x$ is the spread of the wavefunction. This is a fundamental limit of wavefunctions.

Broad and narrow probability distributions illustrate the position uncertainty around the expectation value.

Consider a sound wave. Measure its frequency over time (an average). This decreases the certainty of when it had that frequency.

A pure tone extends in time.
A pulsed note is localised in time.
You cannot have both at once.

E=h\nu,\qquad p=\frac Ec

\boxed{\Delta E\,\Delta t\ge\frac\hbar2}

A fundamental limit on accuracy. Measurement can collapse a broad wavefunction to a narrow one.

Even for cloned particles (many), $\Delta x\,\Delta p_x\ge\hbar/2$ (entanglement).

Take lots of measurements and average them.

Heisenberg is not a limit of a single wavefunction only.

Einstein's paradox (to kill the Heisenberg uncertainty principle)

A time component emits particles regularly, and the energy in the box is measured (related to the energy by $E=mc^2$ ).

But when the particle leaves, the box gets lighter and the box goes up. Uncertainty in $x$ is transmitted to the box.

Einstein's clock-in-the-box thought experiment.

Einstein: “I cannot believe God plays dice with the universe.”

Bohr: “Einstein, don't tell God what to do.”

VII

Relativity

VII · 01

Special Relativity

Lecture Notes

Name: Casey Hardman

Subject: Lecture Notes

GNS Group Newsagency Supplies
128 page
A4 maths
5 mm grid book
Binder book

Special relativity

this is the 100th anniversary of relativity.

Cosmic rays produce muons that last for about $2.2\,\mu\mathrm s$ , which is not long enough for them to reach the Earth without time dilation.

Time dilation! Read HRK, chapter 21.

Special relativity is based on light: light, or its speed, is fundamental.

p=\frac{h\nu}{c}=\frac Ec

Photons are massless: they have zero rest mass.

Postulates

The speed of light in a vacuum always has the same value, independent of the speed of the observer.
Space and time do not behave intuitively.

Geometry

Geometry is the theory of invariants. In Euclidean geometry, lengths and angles are invariant under translations and rotations. In relativistic Minkowski geometry, lengths and times are not invariant under boosts.

Lengths do not change with movement? Perhaps not… In Minkowski geometry, the interval is invariant.

S^2=c^2T^2-L^2

$T$ is the time between events and $L$ is their spatial separation. For a timelike interval, $S^2\ge0$ ; the interval is proportional to the proper time:

S=c\tau

Lorentz transformations

\boxed{x'=\gamma(x-vt)}

\boxed{t'=\gamma\left(t-\frac{vx}{c^2}\right)}

\boxed{\gamma=\frac1{\sqrt{1-v^2/c^2}}\ge1}\qquad\text{Lorentz factor}

Check that $S^2=c^2T^2-L^2$ is invariant.

Show that the interval is invariant

S^2=c^2t^2-x^2

x'=\gamma(x-vt),\qquad t'=\gamma\left(t-\frac{xv}{c^2}\right)

S'^2=c^2t'^2-x'^2

=c^2\gamma^2\left(t-\frac{xv}{c^2}\right)^2-\gamma^2(x-vt)^2

=\frac{c^2\left(t-\frac{xv}{c^2}\right)^2-(x-vt)^2}{1-v^2/c^2}

=\frac{c^2t^2-2xvt+\frac{x^2v^2}{c^2}-x^2+2xvt-v^2t^2}{1-v^2/c^2}

=\frac{c^2t^2\left(1-\frac{v^2}{c^2}\right)-x^2\left(1-\frac{v^2}{c^2}\right)}{1-v^2/c^2}

Show $S'^2=c^2t'^2-x'^2$

S^2=c^2t^2-x^2

x'=\gamma(x-vt),\qquad t'=\gamma\left(t-\frac{xv}{c^2}\right),\qquad \gamma=\frac1{\sqrt{1-v^2/c^2}}

\gamma^2=\frac{1}{1-v^2/c^2}=\frac{c^2}{c^2-v^2}

c^2t'^2-x'^2=c^2\gamma^2\left(t-\frac{xv}{c^2}\right)^2-\gamma^2(x-vt)^2

=\gamma^2\left[c^2\left(t-\frac{xv}{c^2}\right)^2-(x-vt)^2\right]

=\gamma^2\left[c^2\left(t^2-\frac{2txv}{c^2}+\frac{x^2v^2}{c^4}\right)-\left(x^2-2xvt+v^2t^2\right)\right]

=\frac{c^2}{c^2-v^2}\left[c^2t^2-2txv+\frac{x^2v^2}{c^2}-x^2+2xvt-v^2t^2\right]

=\frac{c^2}{c^2-v^2}\left[c^2t^2+\frac{x^2v^2}{c^2}-x^2-v^2t^2\right]

=\frac{1}{c^2-v^2}\left[c^4t^2+x^2v^2-c^2x^2-c^2v^2t^2\right]

=\frac{1}{c^2-v^2}\left[c^2t^2(c^2-v^2)+x^2(v^2-c^2)\right]

=\frac{(c^2-v^2)(c^2t^2-x^2)}{c^2-v^2}

=c^2t^2-x^2=S^2

Factorisation check:

(c^2t^2-x^2)(c^2-v^2)=c^4t^2-c^2x^2-c^2v^2t^2+x^2v^2

Thus the Lorentz transformation leaves the spacetime interval unchanged.

Relativity of simultaneity

Consider two simultaneous events $(x_1,t_1)$ and $(x_2,t_1)$ .

t'_1=\gamma\left(t_1-\frac{vx_1}{c^2}\right),\qquad t'_2=\gamma\left(t_2-\frac{vx_2}{c^2}\right)

\boxed{\Delta t'=t'_2-t'_1=\gamma\left(\Delta t-\frac{v\Delta x}{c^2}\right)}

Length contraction

Measure the two ends of an object moving with speed $v$ simultaneously at $(x_1,t_1)$ and $(x_2,t_1)$ .

x'_1=\gamma(x_1-vt_1),\qquad x'_2=\gamma(x_2-vt_1)

\boxed{\ell_p=x'_2-x'_1=\gamma(x_2-x_1)=\gamma\ell}

The proper length is the real length — a statement about the nature of measurement.

Time dilation

For two events occurring at the same position $x_1$ , at times $t_1$ and $t_2$ :

t'_2-t'_1=\gamma(t_2-t_1)

\boxed{\Delta t=\gamma\Delta\tau}

Velocity addition

Let $u=\Delta x/\Delta t$ . Lorentz-transform the displacement and elapsed time:

u'=\frac{\Delta x'}{\Delta t'}=\frac{\gamma(\Delta x-v\Delta t)}{\gamma\left(\Delta t-v\Delta x/c^2\right)}

\boxed{u'=\frac{u-v}{1-uv/c^2}}

This composition law does not permit an object to go faster than light.

Light clocks

For a light clock at rest, with mirror separation $L$ , convert distance to time:

T_p=\frac{2L}{c}

A stationary light clock has a vertical light path; in a frame where the clock moves at speed v, the same pulse follows a longer diagonal path.

T_m=\frac2c\sqrt{L^2+\left(\frac{vT_m}{2}\right)^2}

\boxed{T_m=\gamma T_p}

Ha ha! The second postulate says that the result of any self-contained experiment is independent of any uniform motion of the experiment. This is how time dilation occurs: the clocks must agree.

How do we know the length of the clock does not change? Relativity of simultaneity.

The comparison of perpendicular light clocks is closely related to the Michelson–Morley interferometric experiment, which confirmed that the ticks are independent of the clocks' orientation.

Length contraction from a horizontal light clock

cT_h=L'+vT_h\quad\Longrightarrow\quad T_h=\frac{L'}{c-v}

c(T_m-T_h)=L'-v(T_m-T_h)\quad\Longrightarrow\quad L'=(c+v)(T_m-T_h)

T_m=\gamma\frac{2L}{c}

\boxed{L'=\frac L\gamma}

VII · 02

Spacetime & Relativity

Relativity

Extra dimensions.

s^2=c^2t^2-x^2-y^2-z^2=c^2t^2-L^2

L^2=x^2+y^2=x'^2+y'^2

t'=\frac{t-vx/c^2}{\sqrt{1-v^2/c^2}},\qquad x'=\frac{x-vt}{\sqrt{1-v^2/c^2}}

Keep $s^2$ constant.

c^2t^2-x^2=0\qquad\text{(the interval is zero)}

Light divides real and imaginary intervals.

c^2t'^2-x'^2=0\qquad\text{(the same in all frames of reference)}

Nothing goes faster than light.

This gives a causal structure to spacetime, represented using light cones.

Two-dimensional light cone for event A, separating causally connected future and past from spacelike-separated events.

Events within the cones can have a causal effect; events outside cannot have anything to do with event A. Light cones explain gravity.

Light-cone structure of flat spacetime

Objects not experiencing forces evolve into their local future.

In flat spacetime, local future cones are uniformly oriented and free worldlines are straight.

Gravity = curved spacetime

Curved-spacetime light cones are not uniformly oriented. Evolution into the future is curved: geodesics — falling down.

Mass tips future light cones towards itself. This is how general relativity explains gravity. A star's worldline follows the tipped local future.

Mass tips nearby future light cones, bending the geodesic of a freely moving object.

Tipping light cones mixes time and space.

If a cone tips more than $45^\circ$ , no part of its future contains stationary worldlines of the untipped observer. Moving in time corresponds to moving in space.

Black holes

Inside the event horizon all future-directed paths point toward the singularity. You cannot escape, even at the speed of light.

Tipped future cones at and inside a black-hole event horizon.

Rotation tips future light cones into the rotation direction.

An object can travel around the blue circle, ending up where and when it started: a closed timelike curve.

A ring of tipped light cones can admit a closed timelike curve around a rotating region.

Geometry is the theory of invariants.

In relativistic Minkowskian geometry, the interval is invariant.

s^2=c^2t^2-L^2

The interval can be regarded as the relativistic length of the spacetime four-vector $\underline R$ , itself an invariant.

\underline R=(ct,x),\qquad \underline R\cdot\underline R=c^2t^2-x^2

4-velocities

Make new invariants out of old ones.

\underline V=\frac{d\underline R}{d\tau}=\left(\frac{d(ct)}{d\tau},\frac{dx}{d\tau}\right)=\left(\frac{d(ct)}{d(t/\gamma)},\frac{dx}{d(t/\gamma)}\right)

\boxed{\underline V=\gamma(c,v)}

\underline V\cdot\underline V=\gamma^2(c^2-v^2)=\frac{c^2-v^2}{1-v^2/c^2}=c^2\qquad\text{(invariant)}

4-acceleration

\underline A=\frac{d\underline V}{d\tau}=\gamma\left(c\frac{d\gamma}{dt},\frac{d(\gamma v)}{dt}\right)=\gamma\left(c\frac{d\gamma}{dt},v\frac{d\gamma}{dt}+\gamma a\right)

Its length is negative.

\underline A\cdot\underline A=-a_p^2\qquad\text{(invariant)}

4-momentum

\underline P=m\underline V=(\gamma mc,\gamma mv)

The spatial part is relativistic three-momentum.

\underline P\cdot\underline P=c^2m^2\qquad\text{(rest mass)}

4-force

\underline F=\frac{d\underline P}{d\tau}=m\frac{d\underline V}{d\tau}=m\underline A

4-momentum conservation

\sum_{\rm before}\underline P_i=\sum_{\rm after}\underline P_i

Fundamental physical laws should be expressed as relations between invariants.

This contains relativistic three-momentum conservation and conservation of the time component, $\gamma mc$ .

Kinetic energy

$\gamma mc^2$ is rest energy plus kinetic energy.

K=\gamma mc^2-mc^2=(\gamma-1)mc^2

K=mc^2\left[\left(1-\frac{v^2}{c^2}\right)^{-1/2}-1\right]

K=mc^2\left(1+\frac12\frac{v^2}{c^2}+\cdots-1\right)=\frac12mv^2+\cdots

The leading term is the normal kinetic energy. Momentum and energy are linked like space and time; they are essentially the same thing in four-momentum.

Light

\underline N=f(1,\hat{\mathbf k})

$f$ = frequency; $\hat{\mathbf k}$ = unit wave vector. Lorentz-transforming this gives the Doppler and aberration effects.

Photons?

The Planck/Einstein/de Broglie photon hypothesis relates photon four-momentum to four-frequency:

\underline P=\frac hc\underline N=\frac hc f(1,\hat{\mathbf k})

The rest mass of photons is zero.

\underline P\cdot\underline P=\frac{h^2f^2}{c^2}\left(1-\hat{\mathbf k}\cdot\hat{\mathbf k}\right)=0

VII · 03

Relativistic Dynamics

Dynamics — an important application of special relativity to high energy particle collisions

Consider: is it possible?

e^-+e^+\longrightarrow\gamma\ (?)

Proposed two-to-one collision with incoming four-momenta P₁ and P₂ and outgoing P₃.

4-momentum conservation

\underline P_1+\underline P_2=\underline P_3

Squaring:

m_1^2c^2+m_2^2c^2+2\underline P_1\cdot\underline P_2=m_3^2c^2

2\underline P_1\cdot\underline P_2=(m_3^2-m_1^2-m_2^2)c^2

4-vector algebra

\underline P_1\cdot\underline P_1\ge0,\qquad \underline P^{\,0}>0

\underline P_1\cdot\underline P_2\ge |\underline P_1|\,|\underline P_2|\ge0

m_3^2-m_1^2-m_2^2\ge0\quad\Rightarrow\quad m_3^2\ge m_1^2+m_2^2

Not possible for a single photon: a photon has zero rest mass. This failure is due to conservation of four-momentum.

e^-+e^+\not\longrightarrow\gamma

Similarly, $\gamma\to e^-+e^+$ is not possible as an isolated one-to-two process.

How can photons carry the electromagnetic force? Is $e^-+e^+\to\gamma+\gamma$ possible? Yes: electrons cannot emit or absorb a photon individually while conserving four-momentum; interactions instead exchange photons between charged particles.

Four-momentum is not conserved at an isolated single charged-particle/photon vertex; in a complete interaction it is conserved.

Scattering

Consider a collision of two distinguishable particles with four-momenta $P$ and $K$ .

\underline P_1+\underline K_1=\underline P_2+\underline K_2

Two-body scattering with incoming P₁, K₁ and outgoing P₂, K₂.

m_p^2c^2=(\underline P_1+\underline K_1-\underline K_2)^2

m_p^2c^2=m_p^2c^2+(\underline K_1-\underline K_2)^2+2\underline P_1\cdot(\underline K_1-\underline K_2)

2\underline P_1\cdot(\underline K_1-\underline K_2)=-(\underline K_1-\underline K_2)^2

Consider the case where $K$ is a photon:

\underline K_1\cdot\underline K_1=\underline K_2\cdot\underline K_2=0

2\underline P_1\cdot(\underline K_1-\underline K_2)=2\underline K_1\cdot\underline K_2

Compton scattering

\underline P_1=mc(1,\mathbf0),\qquad \underline K=\frac hc f(1,\hat{\mathbf k})

\underline P_1\cdot(\underline K_1-\underline K_2)=mh(f_1-f_2)

\underline K_1\cdot\underline K_2=\frac{h^2}{c^2}f_1f_2(1-\hat{\mathbf k}_1\cdot\hat{\mathbf k}_2)

mc(f_1-f_2)=\frac hc f_1f_2(1-\cos\theta)

\lambda=\frac cf

\boxed{\lambda_2-\lambda_1=\frac h{mc}(1-\cos\theta)}

The relative wavelength change depends on the scattering angle. $h/(mc)$ is the Compton wavelength.

Inverse Compton scattering

Head-on inverse Compton scattering: an incoming photon meets a relativistic massive particle and leaves with a higher frequency.

\underline P_1=\gamma m(c,\mathbf v),\qquad \hat{\mathbf k}_1=-\hat{\mathbf k}_2

\mathbf v\cdot\hat{\mathbf k}_1=-v

f_2=f_1\frac{\gamma m(c+v)}{2hf_1/c+\gamma m(c-v)}

Low photon-energy limit

\frac{2hf_1}{c}\ll\gamma m(c-v)

f_2\simeq f_1\frac{c+v}{c-v}=f_1\frac{1+v/c}{1-v/c}

Suitably fast charged particles will be slowed by the cosmic microwave background as they travel through space.

i\hbar\frac{\partial\psi}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}

Time is treated differently from space — there are no invariants. This equation must be wrong in comparison with relativity.

VII · 04

Relativistic Optics

Visualising relativity

Special relativistic optics

Observation is not sight.

An observer can make simultaneous measurements.
Sight is not a simultaneous measurement.

Effects: aberration, Doppler shift, intensity shift.

Aberration effects

Visual: angular aberration and distortion.

Angular aberration

\boxed{\tan\frac{\alpha'}2=\sqrt{\frac{c-v}{c+v}}\tan\frac\alpha2},\qquad \alpha'<\alpha

You drive into the rays: the apparent direction is pushed toward the direction of motion.

Angular aberration reduces the observed angle for a forward-moving observer.

Seeing angular aberration

Apparent reduction/magnification.
Objects in front get smaller.
Objects behind get larger.

Seeing backwards

Objects behind you can be brought into the field of view. The same side of the object is still seen.

Aberration brings a rear object into the forward field of view while preserving which face is visible.

Passing objects: they turn inside out.

Distance dependence: objects farther away are farther in the past.

Seeing distortion

Curved objects

Moving past may curve, shear, or twist.

Combined with length contraction: objects change length. The farther-back-in-time part retains a different apparent length, while the simultaneously measured shape is contracted.

A curved object can appear rotated, sheared, or twisted as its differently delayed parts are seen together.

The far edge is also farther back in time (shearing). As an object goes past, it appears to be turning.

Different light-travel times from the near and far edges produce apparent shearing and turning.

Doppler shift (relativistic)

Perceived shift in frequency due to relative motion of wave and source. With light, this gives blue- and red-shifting. Red/blue shifting is observed at relativistic velocities.

\boxed{f'=\gamma\left(1+\frac vc\cos\alpha\right)f}

A relativistically moving observer sees an angular colour pattern: blueshift ahead, little shift transverse to the motion, and redshift behind.

A spectral-line shift at about $0.7c$ is quite interesting. For an object such as Mars going past like a car, its apparent colours progress through blue, green, red, and then darkness as the angle changes.

Intensity effect

Angular concentration.
Light concentrated in the direction of motion (brightening in the direction of motion).

\boxed{I'=\frac{\left(1+\frac vc\cos\alpha\right)^2 I}{\gamma}}

Relativistic angular contraction concentrates a star field into the forward direction: the forward region becomes bright, transverse directions remain normal, and rearward directions become dark.

A direction through a star field gets real bright in the middle.

Backlight & ray tracing

Undergraduate research project — Dr Antony Searle.
Ray tracers determine what colour, and so forth, each ray had.
Program: pretty rainbows… but not really! (Intensity.)
Relativistic Rollercoaster — “Woah!”

Lecture / ray tracer (Backlight): http://www.anu.edu.au/Physics/Searle/Obsolete/Seminar.htm

Backward ray tracing follows each viewing ray from the observer through the relativistic scene to determine its source colour and intensity.

Problem Solving in Physics

SI units

Fundamental Units! (5 of them.)

Units!

Dimensional Analysis: Checking & Fudging

Solving Problems with the 7 Ds & the little s.

Example Problem 1

Definitions

Diagnosis

Derivation

Algebra

Dimensions

Limiting cases

Substitutions

Differentiation

a)

b)

Partial Derivatives

Integration

Example

Integration by Parts

Example: ∫tan⁡−1x dx\int\tan^{-1}x\,dx∫tan−1xdx

Differential Equations

Simple Harmonic Motion

Springs

Example: v=ktxv=ktxv=ktx

1) Coordinate Systems

Polar (plane) — r,θr,\thetar,θ (2D)

Cartesian

Cylindrical polar

Spherical polar

Multivariable Integrations

Volumes

Sphere volume

Lecture 3: Labs

Lab Safety

How You Do Labs Here!

Lab book

Error

Example: Pendulum

Systematic errors (uncertainties) → Accuracy

Random error → Precision

A graph

Discussion

Matrices

Adding

Determinant

Problem 2

Definitions

Diagnosis

Derivation

Dimensions

Limiting cases

Substitution!

Kinematic Equations!

Energy, Energy Conservation & Forces

Force vs Energy

Example: Earth’s surface

Difference in gravitational potential energy

Force, Energy & Work

Mass on a spring

Hookian Stuff

Mechanical Energy & Conservative Forces

Non-Conservative Forces

More Forces — Momentum & Newton’s 2nd & 3rd Laws

Normal force — electrostatic

Example: lift accelerating upwards at aaa

Friction — parallel to surface

Kinetic friction

Air resistance

Tension (not elastic)

Momentum

Example: two carts

Newton’s Third Law!

Elastic collisions

Inelastic Collisions

Rotational Mechanics

Rotational Kinematics

Variables

Angular velocity, ω\omegaω

Example: $\int\tan^{-1}x\,dx$

Example: $v=ktx$

Polar (plane) — $r,\theta$ (2D)

Example: lift accelerating upwards at $a$

Angular velocity, $\omega$

Angular acceleration, $\alpha$

Rotational inertia $I$ (moment of inertia)