SL PHYSICS NOTES
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TABLE OF CONTENTS
MEASUREMENTS AND UNCERTAINTIES ..........................................................................................................2
WAVES ...........................................................................................................................................................8
MECHANICS .................................................................................................................................................17
THERMAL PHYSICS .......................................................................................................................................26
CIRCULAR MOTION AND GRAVITATION ........................................................................................................45
ATOMIC, NUCLEAR AND PARTICLE PHYSICS...................................................................................................48
ENERGY PRODUCTION/GEOGRAPHY ............................................................................................................70
RELATIVITY ...................................................................................................................................................82
MEASUREMENTS AND UNCERTAINTIES
SI units
Length (Metre, m): Distance travelled by light in a vacuum in 1/299, 792, 458 seconds
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Mass (Kilogram, kg): Standard is the mass of a certain pla num-iridium alloy kept at the
Bureau Interna onal des Poids et Mesures in France
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Time (Seconds, s): Time taken for light to travel 299, 792, 548m in a vacuum
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Electric Current (Ampere, A): The current which, when owing on two parallel conductors 1m
apart, produces a force of 2 x 10^7N on a length of 1m of the conductors
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Temperature (Kelvin, K): 1/273.16 of the thermodynamic triple point of H2O. °K = °C + 273.15
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Mole (mol., n): Avogadro’s number – 6.02 x 10^23
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Brightness/Luminous intensity (Candela): The intensity of a source of frequency 5.40 x 10^14
Hz emi ng 1/683 W per steradian
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Uncertain es
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Half of the smallest increment (qualita ve measurements)
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±1 of the smallest decimal on the digital screen (digital measurements
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Add absolute uncertain es when adding or subtrac ng values
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Add frac onal/percentage uncertain es when dividing or mul plying values
Signi cant gures
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Important when performing calcula ons based on measurements or given values
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Rules
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The le most non-0 digit is the most signi cant digit
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E.g. 0.0032
If there is no decimal point, the rightmost non-0 digit is the least signi cant digit
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E.g. 5800. 20, 300, 000
If there is a decimal point, the rightmost digit is the least signi cant even if it is a 0
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In mul plica on, division, taking a power or nding a root, the result must only have as many
signi cant gures as the gure in the calcula on with the least sig. gs
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No measurement is ever completely error-free
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Two types
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Systema c errors: usually due to instrument’s calibra on, is always in one direc on
of the accurate/accepted value
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Random errors: unbiased spread of values due to human error or uncontrolled
variables
Limits of reading
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Half the smallest measurement division for analog/qualita ve instruments
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±1 of the smallest division for digital instruments
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Accuracy: Systema c error is small
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All digits between most and least signi cant are considered signi cant gures
Errors
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E.g. 3.2590 means the value is exact. 3.2590 is to the nearest 0.001 when
digitally measured or to the nearest 0.0005 when qualita vely measured
with increments on an instrument
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Averages and uncertain es
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Precise: Random error is small
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Average of all values: Best es mate
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Uncertainty: Half the range of values
Vectors and Scalars
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Scalars: quan
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Vectors: Magnitude and direc on (Force, Velocity)
Drawing vectors: shown by arrow, showing direc on and length of arrow depic ng
magnitude
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Represented in bold text or with squiggly line underneath value or arrow above
value
Adding vectors
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Subtrac ng vectors
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Add a nega ve vector (reversed direc on)
Vectors can be broken down into two perpendicular components
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Join head to tail and write values in
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es with only size/magnitude (speed)
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WAVES
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Oscilla on
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Back and forth (regular) mo on
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Is periodic
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Body moves back and forth around an equilibrium posi on
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Period = 1 full oscilla on
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Amplitude = maximum displacement from equilibrium
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E.g.
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Pendulum
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Mass on the end of a spring
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Ver cal mo on of a body in a liquid with waves
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Guitar strings
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Diving board
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Plane wing
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Tree/skyscraper in wind
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Requires restoring force (force towards equilibrium when mass is displaced)
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Simple harmonic mo on
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Magnitude of accelera on of body that has been displaced is propor onal to the
displacement and the direc on of accelera on towards the equilibrium
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D ∝ -A
A is accelera on (ms^-2)
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Rela onship;
Must be xed equilibrium posi on
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Accelera on is propor onal to the nega ve displacement
Energy in SHM is constant within the system
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Kine c energy + Gravita onal/elas c poten al energy
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D is displacement (metres)
To check for SHM
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(0.5mv^2) + (mg∆h) (pendulum)
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(0.5mv^2) + (0.5kx^2) (spring)
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Energy is conserved in absence of fric on
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Max velocity is when x=0 (when mass goes past equilibrium posi on
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When x = ±A, V=0, as it only has poten al energy (spring, gravity, etc.)
Frequency and period are reciprocals of eachother
Travelling waves
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Mechanical (sound)
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Transverse/electromagne c (light)
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Ma er (electron mo on)
Phase di erence
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When objects are oscilla ng out of step, there is a phase di erence
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Phase described using phase angle – Ø
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Ø = (shi /T) x 360°
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T = period of oscilla on
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3 types
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Wavefronts and rays
Wavefronts are lines that link all parts of a wave that are in phase
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Rays are lines or arrows drawn in the direc on of energy transfer (normal to tangent of
wavefront)
Amplitude and intensity
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Power is the rate at which the wave carries energy
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Unit is wa s (W)
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P=J/T (energy/ me)
Intensity is brightness of light on a surface
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J/s/m^2
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Wm^-2
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Intensity if an inverse square rela onship with distance from source
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When two or more waves of the same type occupy the same posi on at the same me in a
medium, the waves interfere with eachother. The displacement of the combined wave is the
algebraic sum of the displacement of each wave
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(Basically, add the amplitudes of both waves at the same point in me and there ya go)
Pulses
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When pulse/wave hits a rigid surface, re ects at 180° out of phase
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When pulse/wave hits a free surface, re ects at 0/360° out of phase
Polarisa on
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All electromagne c waves are transverse w/ two vectors as electric elds (E) and magne c
elds (B)
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A ray of unpolarised light has electric eld vectors randomly oriented
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A ray of polarised light has electric eld vectors on the same axis
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A polariser only transmits light with a speci c orienta on of the electric eld
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I =I(0) cos^2(θ)
Exit intensity = ini al intensity x cos^2(θ)
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Wave behavior
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Superposi on of waves
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I ∝ 1/d^2
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<i = <r when measured against the normal to the point of incidence and are on the same
plane
Refrac on and Snells Law
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Refrac on is the change in direc on
of a wave due to a chance in speed (1
medium to another)
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Frequency does not change
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Velocity changes, Wavelength
changes
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(Sin(θ)2) / (Sin(θ)1) = V2/V1 = n2/n1
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n = refrac ve index
= λ1/ λ2
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n is always >1
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n = c/v (Speed of light in vacuum/ speed of light in medium)
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air = 1.0003
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water = 1.33
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quartz = 1.46
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sugar = 1.38
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diamond = 2.42
Total internal re ec on
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When light goes from a high refrac ve index to low
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op c ber
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The thinner the core, the less blurring of
pulses due to di erences in ray path
distances
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Angle of incidence must be greater than
cri cal angle
Di rac on
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Refrac on – change in speed and direc on
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Di rac on depends on two things
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Re ec on – change in direc on
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i.e. light refracts away from normal
Size of aperture
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Wavelength
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When aperture is comparable to or smaller than the wavelength, di rac on occurs
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This “spreading” (change in direc on) of a wave as it goes past an obstacle or aperture
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a = aperture
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a > λ, small di rac on
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a < λ, large di rac on
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a ≈ λ, di rac on into the shadow region
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Double slit interference
d = distance between slits
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s = distance between fringes of light on screen
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D = distance between slits and screen
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λ = wavelength of wave
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s = (D λ)/d
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λ = ds/D
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Standing waves
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A special case of interference which occurs when two waves meet which;
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Have the same amplitude
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Have the same frequency
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Travel in opposite direc ons
Property
Standing Wave
Travelling Wave
Amplitude
Each point has its own xed max
min amplitude (cannot go further
than that xed amplitude
All points have same amplitude
and vibrate to it
Frequency
All points (except nodes) have
equal frequency
All points have equal frequency
Wavelength
Twice the distance from adjacent
nodes/an nodes
Shortest distance between two
points in phase
Energy
Not transmi ed
Transmi ed
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Three cases;
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Case 1: Nodes at each end
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All frequencies are mul ples of the fundamental frequency (1st harmonic)
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f(n) = nf(1)
Case 2: an nodes at each end
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Same characteris cs and proper es as case 1
Case 3: one node and one an node at each end
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f(1) = V/4L
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f(2) = 3V/4L, = 3f(1)
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f(3) = 5V/4L, = 5f(1)
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Vibrates at odd mul ples of the rst harmonic
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V = velocity, L = length between node and an node at each end
f(n) = (2n-1) x f(1)
MECHANICS
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Kinema c quali es
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Velocity (v, with vector)= ∆s/t
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Displacement (s) = ∆r
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Distance (d)= length of path followed
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Change (∆) = nal value – ini al value
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Accelera on (a, with vector) =rate of ∆v/t
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Posi on = r
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Speed = v (no vector)
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Time = t
Equa ons
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v = u + at (no displacement)
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s/t = (u+v)/2 (no accelera on)
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s = ut + (1/2)at^2 (no nal velocity)
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v^2 = u^2 – 2as (no me)
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s = vt – (1/2)at^2 (no ini al velocity)
Projec le mo on
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A projec le is an object launched by an external force, con nuing its mo on by its iner a,
and only subject to the force of gravity throughout its mo on
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The mo on of a projec le can be regarded as the separate and independent mo ons
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Ver cally: -9.8m/s
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Horizontally: constant mo on (iner a)
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Fluid – gas or liquid
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Exert a fric onal force on objects moving through them
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Fric on a ects mo on
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There is a speed reached where the uid fric on is equal and opposite to the weight force
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The max (constant speed) is called terminal velocity
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Forces
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Ignore the e ects of air resistance
Fluid resistance
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Ini al velocity = u, nal velocity = v
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Object is shown as a point
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Forces shown as arrows with length of arrows represen ng magnitude and direc on
Weight = mg
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g = gravita onal eld strength
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m = mass
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1m/s/s = 1N/kg
Tension
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Force applied by a rope or string when stretched
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Springs will develop a tension force when compressed or stretched
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f = -kx
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k = spring constant
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x = displacement
Drag
Oppose mo on of object ! opposite direc on
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Depends on shape and speed.
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As speed increases, drag increases
Upthrust
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Force experienced by an object when moving upwards in a uid
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If upthrust
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If upthrust < weight, sinks
Normal reac on contact force
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“Normal”: perpendicular to surface
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Occurs when two bodies are in contact
Fric on
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Opposes the mo on of the body
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Does not depend on speed or shape
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≥ weight, oats
Dynamic and Sta c
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Forces on object shown through free-body diagram
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Dynamic (kine c fric on)
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When one body slides over another
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F = μ(d)R
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μ(d) = coe cient of dynamic fric on
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R = normal reac on force
Sta c
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When surfaces are in contact
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F = μ(s)R
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μ(s) = coe cient of sta c fric on
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R = normal reac on force
▪Fric on
occurs
when
there is
a
tendency
for mo on
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Area of contact does not a ect fric onal force
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Newton’s rst law
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When the net force on a body is zero, the body will move with constant velocity
(which may be 0)
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a=0, F=0
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an = accelera on
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F = ∑ of all forces on object
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F= - f(d), if no accelera on
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F> f(d) if accelera ng
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Iner a – the property of an object that keeps the body in the same state of mo on
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Net force on object = 0, called equilibrium
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To solve equilibrium problems, use vector components
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Newton’s third law of mo on
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Work is the change in energy, W = ∆E (both scalar quan
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Unit is Joules
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Energy is not created or destroyed but is transformed and/or transferred from one
form to another
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A force can do work
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Work done by force = component of force in the direc on of work needed
to be done x distance moved
W = Fcosθ x s
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= Fs cosθ
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F = force
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Principle of conserva on of energy
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Work, Energy and power
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If body A exerts a force on body B, then body B will exert an equal and opposite
force on body A
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∑F(x) = 0, ∑F(y) = 0
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s = distance moved
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θ = angle between force and direc on of work/mo on
Kine c energy
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The energy possessed by an object due to its mo on
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W = ∆E(k) = ∆E(f) - ∆E(i)
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W = (½)mv^2 – (½)mu^2
Poten al energy
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Also known as stored energy
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The energy of a system due to its posi on or shape
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W = ∆E(p) = amount of energy used to bring an object to a posi on or shape
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Any elas c object has poten al energy
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Opposite is “plas c”
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Springs have elas c poten al energy through change in shape
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Objects have gravita onal poten al energy through displacement
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Hooke’s Law
F ∝ -∆x
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Force is propor onal to the nega ve displacement
•F = k∆x
ok is the spring constant
•E(p) (elas c)= (½)k(∆x)2
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Gravita onal poten al energy
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∆E(p) = mg∆h
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m=mass
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g=accelera on due to gravity
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∆h = change in height
Work done by gravity,
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If accelera ng, ∆E(k) > 0
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constant speed or decelera ng, ∆E(k) < 0
E(total) = (½)mv2 + mg∆h + (½)k(∆x)2
The rate at which work is done or energy is used
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Rate (unit per me)
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Unit is Wa s (W)
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Work done/ me
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Energy used/ me
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Change in energy/ me
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P = W/t = ∆E/t
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P = W/t = Fd/t = Fv
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v = instantaneous speed
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d = distance moved
E ciency
How much of the input is changed into useful output
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Percentage
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(Output/input) x 100
Momentum and impulse
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∆E(p) > 0
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Power
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Work done against gravity,
Total mechanical energy
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∆E(p) < 0, ∆E(k) > 0
The linear momentum of an object: p = mv
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P = kgm/s
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F(net) = ∆P/t
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Mass constant, Velocity
changes
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Mass changes, Velocity
constant
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F=∆P/t
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F=∆P/t
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F=∆mv/t
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F=∆mv/t
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F=m(∆v/t)
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F=v(∆m/t)
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F=ma
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F=v(kg/s^2)
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F=vσ
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F(net) = ∆P/t
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∆P = F(net) x me = impulse of force
Conserva on of momentum
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The total momentum of a system is constant
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Final = ini al
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ONLY IF no external forces act on the system to in uence mo on
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Impulse and force/ me graphs
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THERMAL PHYSICS
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Par cle state of ma er – molecular theory of solids, liquids and gases
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Par cles can be individual atoms or molecules
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There are a rac ve forces between the molecules which determine their distribu on and mo on
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In gases, the inter-par cle forces are very weak, almost negligible. Only signi cant during
collisions
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In solids, forces are strong, bonds can be modelled with springs. Par cles con ned by a xed
posi on
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In liquids, forces weaker, par cles cannot move far from each other, con ned to volume of
the liquid
Increasing the separa on of par cles that have an a rac ve force between them requires work to be
done (W = ∆E)//
Characteris c
Solid
Liquid
Gas
Kine c energy
Vibra onal
Vibra onal
Mostly transla onal
Rota onal
Higher rota onal
Transla onal
Higher vibra onal
r0
r0
10r0
Thermal energy of
<
ε >, > ε /10
>
1028
1025
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1028
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Molecules per m3
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par cles (ε)
ε /10
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Mean molecular
separa on
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Highest
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Higher
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Poten al energy
ε
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U = Etotal
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PV = nRT
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P = Pascals (when V is in cubic metres) or Kilopascals (when V is in dm^3
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V = Cubic metres (when P is in Pascals) or decimeters cubed (when P is in kilopascals)
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n = number of moles
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R is gas constant (8.31 J/K/mol.)
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T = Kelvin
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Temperature (how hot or cold a body is) is propor onal to the average random kine c energy of the
par cles/molecules
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Absolute or kelvin temperature scale
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Unit: K
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0K ! EK = 0
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K = C+273.25
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The magnitude of a kelvin is the same as that of a degree Celsius
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Temperature is a measure of the random kine c energy of a substance – not its internal energy (which
also includes EP
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A thermal interac on takes place when two bodies at di erent temperatures are in contact
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Heat is the energy that is transferred from one body to another as a result of a di erence in
temperature. Unit: Joule (J)
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Energy transfer con nues un l thermal equilibrium is reached
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Hea ng
Transfer of energy from a hot to a cold body
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Increases the internal energy of the cold body
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Decreases the internal energy of the hot body
Can be done on the par cles of a substance to increase their poten al energy. Work
increases the internal energy
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Speci c heat capacity
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Work
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U = EK + EP
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Relates change in temperature to a body’s mass (m), and the heat energy supplied/change in
heat (Q) and the energy required to increase the temperature of 1kg of a substance by 1K (c)
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Unit J/kg/K
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Examples
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Lead: 128
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Copper: 385
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Aluminium: 900
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Ice: 2200
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Water: 4180
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Steam (at 110°C):2000
Heat energy required to increase the temperature of a body;
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Q=mc∆T
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c = J kg-1K-1
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Q = ∆H
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m = kg
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∆T = °C/K
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∆T > 0, heat gained
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∆T < 0, lost
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Method of mixtures – measurement of speci c heat capacity of a solid using a calorimeter (heat
measuring device)
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Solid put in container of hot water and allowed to reach thermal equilibrium (constant
temperature). Temp. of solid = Temp. of water
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The solid is transferred into a calorimeter of known mass and speci c heat capacity, with a
known mass of liquid and known ini al temperature
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The solid, liquid and calorimeter are allowed to reach thermal equilibrium and the
temperature is recorded
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The change in temperature of the solid and calorimeter allows the speci c heat capacity of
the solid to be calculated
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msolidcsolid∆Tsolid = mcalorimeterccalorimeter∆Tcalorimeter + mwatercwater∆Twater
Change of phase (state)
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Adding or removing heat ! Change in temperature/change in phase at a constant
temperature
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Solid ! liquid (mel ng)
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Liquid ! Gas (vaporiza on)
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Gas ! liquid (Condensa on)
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Liquid ! solid (freezing)
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Internal energy changes because of a change in the inter-par cle poten al energy of the par cles with
no change in random kine c energy of par cles
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The energy is used to ‘break the bonds’ between par cles, Work is done to increase the separa on
between the par cles
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Speci c latent heat: L
L=Q/m
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Q = energy required to change phase
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m = unit of mass
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Lf = speci c latent heat of fusion (mel ng or freezing)
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Lv = speci c latent heat of vaporiza on (or condensa on)
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Lv > Lf
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2467°C
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10,500kJ/kg
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660°C
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395 kJ/kg
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Aluminium
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100°C
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2260kJ/kg
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0°C
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334 kJ/kg
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Water
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Boiling Temp
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Lv
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Mel ng point
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Lf
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Substance
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Lead
23kJ/kg
327°C
850kJ/kg
1740°C
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Gases – di erent to liquids and solids as they have no xed volume
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n = number of par cles (moles)
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V = volume (dm3 when P is in kPa, m3 when P is in Pa)
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P = pressure (kPa when V is in dm3, Pa when V is in m3)
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T = Temperature (Kelvin)
Ideal gases
Point par cles with negligible volume
▪
Obey the laws of mechanics (not quantum mech.)
▪
No forces between par cles except when they collide
▪
Dura on of collisions is negligible compared to the me between collisions
▪
Collisions of par cles with eachother and container walls are elas c (Ek Conserved)
▪
Par cles move randomly with a range of speeds
▪
Cannot be lique ed or solidi ed, no inter-par cle forces
▪
No poten al energy, only kine c energy for par cles
▪
Real gases can be approximated by an ideal gas when the density is low (Par cles
not close and forces negligible). High temp and low pressure.
Number of par cles
▪
Mole: 6.02 x 1023
▪
“Par cle” is either single atom or molecule
▪
“A” grams of XzA is one mole
Pressure
P = F/A
▪
P = Fcosθ, θ is the angle between the force and the normal to the area
▪
Atmospheric pressure: 1.013 x 105 Pa, or 100 kPa
▪
Due to collisions between par cles/molecules and container walls
▪
Par cles velocity changes – it accelerates, due to collision
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Simpli ed theore cal models of real gases: Helps to understand the real behaviour of real
gases
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•
Modelling a gas
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•
Par cle experiences a net force: Normal reac on force
▪
By Newton’s third law, par cle exerts equal and opposite force on the wall
▪
All par cles colliding with container walls exert force on walls
▪
Therefore, Gas par cles exert pressure on walls
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▪
•
Gases – 1Pa = 1N/m^2
o
o
Boyle’s Law
▪
Volume and pressure are inversely propor onal ,V
▪
PV is constant
▪
P1V1 = P2V2
o
▪
Volume is directly propor onal to temperature at constant pressure and number of
▪
V=kT, where k
▪
V1 = kT, V2 = kT
moles, V
∝T
∈ℝ
Therefore k = V1/T1 = V2/T2, when n and P are constant
Pressure Law (Gay-Lussac’s Law)
▪
Pressure in directly propor onal to temperature at constant pressure and no. of
▪
P = kT, where k
▪
P/T = k = P1/T1 = P2/T2
moles, P
∝T
∈ℝ
Avogadro’s Law
V∝T
▪
•
PV = nRT
o
n = number of par cles (moles)
o
V = volume (dm3 when P is in kPa, m3 when P is in Pa)
o
P = pressure (kPa when V is in dm3, Pa when V is in m3)
o
T = Temperature (Kelvin)
o
R = ideal gas constant (8.31 JK-1mol-1)
Par cles/molecules of a gas have a range of speeds which depend on temperature
o
The higher the temperature, the greater the average speed of the par cles/molecules and
the average Ek
o
Ek = (3/2)(R/NA)(T) = (3/2)(kB)(T) = ½ m(v2)(average)
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Par cle Kine c energy and temperature
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1
P
Charles’ Law
•
o
∝
o
•
KB = R/NA = 1.38 x 10-23 = Boltzmann’s constant
▪
N = number of par cles
▪
n = number of moles
▪
NA = Avogadro’s constant
▪
R = ideal gas constant
Internal energy U of an ideal gas
o
U = Ek + EP
▪
o
KB = R/NA
▪
ti
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o
U = Ek, as EP = 0 in an ideal gas
NKB = NR/NA = nR
For N par cles,
▪
NEk = (3/2) NKBT
▪
= 3nRT/2
▪
= 3PV/2
Electricity and Magne sm
•
Electric charge
o
Posi ve and nega ve (+, -)
o
“like” charges repel
▪
o
Exert a force on eachother that tends to increase space between them
“unlike” charges a ract
▪
•
Exert a force on eachother that tends to decrease space between them
o
Neutral objects have equal amounts of posi ve and nega ve charge
o
Objects acquire electric charge due to the transfer of electrons
▪
Excess of electrons is nega ve charge
▪
De ciency of electrons is a posi ve charge
o
Smallest unit of charge is the charge of an electron
o
Unit: qe, or coulombs (C)
o
Charge of an electron is -1.602 x 10-19 C
o
Charge of a proton is +1.602 x 10-19 C
Coulomb’s Law
o
The electric force between two point charges, q1 and q2
o
F is propor onal to the size of either charge and the distance between is given by
o
F = k (q1q2)/r2
o
q1 q2 = size of charges (coulombs)
o
r = distance between (m)
o
k = constant = 8.99 x 109. ONLY IN A VACUUM
o
k depends on
▪
Permi vity
•
K = 1/4πεo
•
εo is the permi vity of a vacuum
o
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If the charges are not in air or vacuum, the medium’s permi vity must be used, ε
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8.85 x 10-12
•
Electric eld
•
o
An area/volume within which an object with charge experiences a force
o
FE = Charge x Field strength, FE = qE
o
Field strength = Force/ unit Charge. (NC-1)
o
Test for an electric eld uses posi ve charge
o
The eld is a vector – the direc on of force on the test charge is
the same direc on of the eld
Electric current and charge dri speed
Analogy to amazon river
▪
Electric current = water current ( ow of water)
▪
Voltage (poten al di erence) = height (al tude di erence)
▪
Increase the height of the mountain then the current increases
▪
If you add rocks to the river (increasing the resistance) then the current ow
decreases
o
I = nAvq, or I = ∆q/∆t
o
N = charge per unit volume
o
A = cross-sec onal area
o
v = velocity of par cles
o
q = charge of each par cle
fl
Measure of coulombs/second = Amperes
ff
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ti
Size = Rate of ow of charge
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•
Electric poten al di erence: Voltage
•
o
Measured in SI units (Volts)
o
The poten al di erence, V, between two points is the work done per charge to move the
charge from one point to another
o
V=W/q = Emf (electromo ve force)
▪
De ned as the work done per unit charge in moving charge across the terminals of
the ba ery
▪
V = volts
▪
W = work (j)
▪
Q = charge (c)
o
There must be an electric eld for an electric poten al di erence to exist. Also, path length
doesn’t a ect W
o
Joules too large, so eV (electron volt) is used
▪
De ned as the work done when a charge equal to one electron is taken across a
poten al di erence of 1 volt
▪
1eV = 1.6 x 10-19 J
▪
Vq = W = ∆E
Hea ng e ect of an electric current
o
•
In a metal, the charge carriers are electrons which have a nega ve charge. These electrons
collide with the atoms in the conductor in an inelas c collision, impar ng some Ek into the
material, hence their temperature rises.
Electrical resistance
o
Electrical resistance impedes the ow of charge
o
Symbol R
o
Equal to how much poten al di erence is required to move the electrons, giving us the
formula
o
R = V/I
▪
o
•
Used to control the ow of current through a circuit
Ohmic and non-ohmic conductors
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Ohmic conductors obey ohm’s law, and non-ohmic does not
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V = IR
o
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Ohmic
▪
Have a constant resistance at constant temperature
▪
V = IR (I on x-axis)
▪
I = (1/R) x V (V on x-axis)
Non-Ohmic
▪
Resistance is not constant
▪
Gradient is not constant, therefore gradient is not 1/R
•
•
Resistance of a wire
o
R∝l
o
R
∝ 1/A
o
R
∝ p (1/A)
o
•
o
ff
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∝ p (l/A)
▪
R
▪
ρ = roh (resis vity)
▪
ρ = RA/l (ohm meter)
Depends on type of metal
Electric power
o
•
Find point on curve of I (y) and V (x) and R = V/I from points
W = qV
▪
W = work done to move charge
▪
Q = charge
▪
V = volts (poten al di erence)
Power = work done/ me taken
▪
= qV/t
▪
= IV
▪
P = IV
▪
Using R = V/I…
▪
P = IV = RI2 = V2/R
Electromo ve force
•
o
Work done per unit charge
o
Poten al di erence across the ba ery terminals when the ba ery has no internal resistance
o
Emf is measured in volts
o
Power provided by the ba ery per unit current
o
ε = emf = W/q = P/I
Simple circuits
o
A con nuous path that a charge may ow around
o
Resistors in series
▪
No junc ons between resistors – current stays the same, voltage changes a er
each resistor
▪
Voltage (Vn) of an nth resistor ;
•
▪
Sum of voltage (poten al di erence) is
•
V = IR1 + IR2 + IR3 + … + IRn
•
Rtotal = R1 + R2 + R3 + … + Rn
▪
Junc ons between each resistor, voltage stays the same, current splits between all
resistors
▪
Current (I) entering circuit Kirchho ’s current law
∑Iin = ∑Iout
•
Itotal = I1 + I2 + I3 + … + In
•
In = V/Rn
•
Therefore, I = (V/R1) + (V/R2) + (V/R3) + … + (V/Rn)
•
1/Rtotal = (1/R1) + (1/R2) + (1/R3) + … + (1/Rn)
ft
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Resistors in Parallel
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Vn = IRn
•
Electrical meters
o
Ammeter
o
▪
Measures voltage (volts)
▪
Ideal voltmeter has R = ∞ to not a ect circuit
Mul -loop circuits
∑V across all circuits = 0 (kirchho ’s loop law)
The poten al divider
o
By varying where the sliding contact connects along XY, the poten al di erence across the
load can be varied from 0V to the maximum across XY
o
REMEMBER: If XY is a length of wire, R = p (l/a), it is ∆l
o
A xed poten al divider cannot be adjusted
Electric cells
o
Provide electrical energy to charges
o
Electrons (nega ve) go from anode to cathode
o
Conven onal current (posi ve charge) goes from posi ve to nega ve
Internal resistance and terminal poten al di erence
o
An ideal ba ery has no internal resistance
o
V = e – IR, where R is the internal resistance of the ba ery.
ff
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Graph of V vs. I, gradient = -r, y-intercept = e
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•
Ideal ammeter has R = 0 to not a ect circuit
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•
▪
Voltmeter
o
•
Measures current (amps)
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•
▪
•
Primary and secondary cells
o
Primary cell
▪
o
A cell that can only be used once
Secondary cell
▪
•
Is rechargeable – can undergo
electrolysis
Discharging of a cell
•
o
----------------------------------------!
o
The amount of charge it can deliver to an
external circuit in its life me
o
This is the Capacity of a cell
o
The bigger the current, the faster the cell discharges
Magne sm
o
Magne c eld
o
▪
Magne c elds are represented by magne c eld lines
▪
Magne c eld lines start at north poles and end at south poles
▪
The direc on of the magne c eld lines show the direc on of the magne c eld
▪
Magne c elds are vectors
▪
The spacing of the magne c eld lines show the strength of the magne c eld. The
closer the lines, the stronger the strength
▪
Unit of magne c eld strength is the Tesla
Magne c eld around a wire
▪
o
Direc on of current a ects magne c eld around a wire
Solenoid
o
▪
A wire wound in a helix
▪
When the wire carries an electric current, solenoid acts like a magnet – a coil
▪
Uses right hand rule, ngers in direc on of current ow, thumb in direc on of
magne c eld
Force on a charge moving in magne c elds
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es involved
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Three vector quan
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▪
▪
Force (Newtons) (N)
θ is the angle between the vectors B and V
o
when θ = 0 or 180°, force = 0
o
when θ = 90°, F = qvB
•
Velocity of moving charge (ms-1) (V)
•
Force is always perpendicular to Velocity and Field direc ons
Direc on is determined using the right hand grip rule
▪
When moving charged par cles enter an external magne c eld, the magne c eld
created by the moving charged par cles interacts with the external magne c eld
▪
When the moving charged par cles enter the magne c eld at right angles to the
eld, they experience a force at right angles to the velocity and to the direc on of
the external eld.
▪
Right hand push rule
•
Lay right hand out at
•
Fingers in direc on of magne c eld
•
thumb in direc on of current ow
•
palm point in direc on of force
Right hand rule
Curl ngers from V towards B. Thumb in direc on of force.
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L = Length of the conductor in B (magne c eld)
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I = Current (A)
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B = magne c eld strength or magne c ux density
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F = BILsinθ
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Force on a current-carrying wire/conductor
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Magne c eld strength (Magne c ux density) (Tesla) (B)
•
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F = qvBsinθ
Moving charged par cles produce magne c elds
▪
o
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•
•
fi
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•
▪
θ = angle between current and direc on
of B eld
Mo on of charges in magne c elds
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▪
When velocity is at right angles to the
magne c eld, path is a circle
▪
qvB = m (V2/R)
CIRCULAR MOTION AND GRAVITATION
•
Circular mo on and angular speed
o
T = period of mo on in a circle of radius r
o
V = constant velocity
o
Distance covered = 2πr
o
V = (2πr)/T (metres per second)
o
Angular speed
▪
ω = ∆θ/∆t (Radians per second)
▪
When ∆θ is one full revolu on, ∆θ = 2π
▪
And t = T
▪
Therefore, ω = (2π)/T
o
Frequency (rota ons per second) = 1/T (1/ me taken for one rota on)
o
ω = (2π)/T), = 2πf
o
in a me ∆t the object moves a distance of V x ∆t around the circle, and the change in angle
is ∆θ (rad). As the distance travelled is along the circumference of the circle, and distance =
r∆θ,
▪
v∆t = r∆θ
▪
v = r (∆θ)/∆t)
•
•
Centripetal accelera on
o
As there is change in direc on of velocity, there is accelera on in circular mo on
o
a = v2/r, and is directed into the centre of the circle
o
v = 2π/T,
a = (4π2r2)/rT2
▪
•
a = (4π2r)/T2
•
a = (4π2rf2)
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All net force towards centre of circle
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•
Centripetal forces
o
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v = rω
•
•
ti
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Using f = mv2,
▪
F = mv2/r
▪
F = mω2r
The Law of Gravita on
o
F = G(M1M2)/d2
o
F = force of gravity
o
G = 6.667 x 10-11 N m2 kg-2
o
d = centre to centre distance of masses
o
F = G(M1M2)/d = M2g. Therefore F = G(M1)/d = g
Gravita onal eld strength
o
ti
▪
g = F/m
•
Orbital mo on
o
o
o
To maintain orbital mo on
▪
Only force is gravity, no fric on
▪
F = G(Mm)/r2
Distance between smaller and bigger mass do not change. Therefore centripetal force is
equal to gravity
▪
(mv2)/r = G(Mm)/r2
▪
Thus, v =
((GM )/r)
However, using V = 2π/T,
▪
T2/R3 = 4π2/GM
▪
Thus T2/R3 is constant for all bodies orbi ng the same object with mass
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▪
ATOMIC, NUCLEAR AND PARTICLE PHYSICS
•
Discrete energy and radioac vity
o
Energy on a microscopic (atomic/molecular/nuclear) scale is discrete
o
The energy of a system in discrete energy cannot be of any arbitrary value, it is quan zed
Spectrometer (Evidence for discrete energy)
o
1/λ = R ((1/nf)-(1-ni))
•
Neils Bohr – argued that energy in atoms was discrete
•
Energy levels
o
eV (electron volt) = 1.602 x 10-19 J
▪
o
Electrons moving though energy levels have a change in energy
▪
o
Nega ve energy
Determining the wavelengths/frequencies of spectral lines
o
E = energy transi on between energy levels (IN JOULES)
o
h = plank’s constant = 6.63 x 10-34 Js (joule seconds)
o
E = hc/λ
▪
o
In hydrogen (H2 Gas)
o
Transi ons to n = 1 give UV light
o
Transi ons to n = 2 give visible light
o
Transi ons to n =3 or bigger gives IR light
Nuclear Structure
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Energy = planck’s constant x frequency
Nucleus nota on
▪
AZX
▪
A = mass number
ff
•
Energy = (planck’s constant x speed of light)/wavelength
E = hf
▪
•
Ef - Ei
ti
•
Energy required to move one electron through a poten al di erence of 1V
ti
•
Z = atomic number
▪
X = element
o
A nucleus with a certain amount of protons and neutrons is called a nucleide
o
Isotopes – di erent no. of neutrons
o
ff
▪
•
Transmuta on and nuclear reac ons and decay
o
When a radioac ve atom emits an alpha par cle or a beta par cle, an atom of a new
element is formed. This process is called transmuta on
o
In all transmuta ons, the sum of the mass number and the sum of the atomic numbers do
not change
o
Alpha decay;
▪
A nucleus of some element X changes into nucleus Y of another element according
to ; (Alpha par cle is helium nucleus)
▪
o
Beta Decay
▪
Beta minus: A neutron in the nucleus turns into a proton and emits an electron and
an an -neutrino
•
▪
Beta plus: A proton becomes a neutron and emits a positron and a neutrino
•
o
Gamma decay
▪
Usually accompanies alpha and beta decay
▪
Nucleus emits a photon (par cle with no mass) in the gamma ray electromagne c
spectrum
•
•
Nucleus does not change when photon is emi ed, stays the same element
ti
Nucleus changes from a high to low nuclear energy level
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▪
Wavelength = hc/E
▪
o
Gamma rays were iden ed through the di rac on of gamma rays, emi ed from
decaying nuclei, into crystals and measuring the di rac on angle/pa erns
Decay summary table
o
o
Radioac ve Decay Series
▪
Uranium
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▪
▪
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Law of radioac ve decay
▪
Is random and spontaneous
•
Random meaning you cannot know which nucleus in a sample will decay
when there is a decay required
•
Spontaneous means the decay happens unassisted, meaning the rate of
decay cannot be a ected in any way
▪
Cannot predict or in uence when a nucleus will decay
▪
Number of nuclei that decay per second is propor onal to the number of nuclei
that have not decayed yet
∆N
∝N
∆t
•
•
A consequence of this law is that the number of radioac ve nuclei
decreases exponen ally, But never reaches perfect 0
▪
Half-life = the me it takes for a speci ed mass of radioac ve material to decay to
half its original amount (from t=0)
▪
Number of parent nuclei decreases at a decreasing rate, looks like hyperbola
▪
Number of daughter nuclei increase at a decreasing rate, looks like a logarithmic
graph
▪
Measurement of radioac ve decay does not approach zero, it approaches the
background radia on of the environment
ti
Rate of decay is measured in Becquerel (Bq) = 1 Bq is 1 decay/second
ti
▪
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•
•
•
Half-life and Probability
o
Any given nucleus has a 50% chance of decaying in one half-life
o
If undecayed, the same nucleus s ll has 50% chance of decaying in the next half-life.
o
This can be illustrated with a tree diagram, or just 0.5n
o
•
▪
n = nth half-life
▪
this shows the probability that a certain nucleus will decay in a certain half-life
The probability that a nucleus will already have decayed by a certain half-life is done by
adding the chances that it will decay in each half-life
▪
E.g. 0.5 + 0.52 + 0.53 …
▪
Sum of an in nite geometric series
Fundamental forces and their proper es
o
Weak nuclear interac on
o
▪
acts on protons, neutrons, electrons and neutrinos to bring about beta minus and
beta plus decay
▪
radius of 10-18m (1 a ometre)
Strong nuclear interac on
o
▪
Mainly a rac ve, acts on protons and neutrons (nucleons) to keep them inside
nucleus
▪
Radius of 10-15m (1 femtometre)
Electromagne c interac on
o
▪
Acts on any par cle that has charge
▪
Magne sm, at the fundamental level, is based on this
▪
In nite range
Gravita onal interac on
▪
Force of a rac on between masses
▪
Weakest force in nature
▪
In nite range
▪
Sub-atomic par cles have too small a mass to be a ected by this
ff
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Rela ve Strength (on atomic
scales)
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Force
Range (m)
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Strong Nuclear force
1
10-15
Electromagne c
10-2 – 10-3
In nite
Weak nuclear force
10-6
10-18
Gravita on
10-38 – 10-41
In nite
•
Nuclear Reac ons
o
Shows that the sum of the cons tuents of the masses do not equal the absolute mass of the
nucleus
▪
o
o
Therefore, some of the mass is converted to energy
The uni ed atomic mass unit
▪
1/12 of the mass of Carbon-12
▪
1 u (universal atomic mass unit) = 1.6605402 x 10-27kg
The mass defect and binding energy
▪
To split a nucleus, energy must be applied
▪
Therefore, when a nucleus splits, it releases energy
▪
E = mc2
•
Energy (J) = mass x speed of light2
•
Using E = mc2
•
Energy equivalence of 1u = 1.661 x 10-27 x (2.998 x 108)2 = 1.493 x 10-10 J
•
Electron volt equivalence, using E = qV
•
Divide E by q to convert from j to eV
•
E = (1.493 x 10-10)/ (1.602 x 10-19) = 9.315 x 108 eV = 931.5 MeV
▪
Therefore, when a nucleus is brought together, it’s mass is less than the cons tuent
nuclei’s masses
▪
The mass of the protons + the mass of the neutrons is less than the mass of the
nucleus
•
This is the Mass Defect
•
δ = total mass of nucleons – mass of nucleus
•
δ = Zmp + (A-Z) mn - Mnucelus
•
Note: A – Z = the number of neutrons in the nucleus
•
The di erence between the mass of a nucleus and the total mass of its
cons tuent par cles is known as the mass defect ∆m/
One can think of this mass defect as a di erence/change in energy
How much less energy does an alpha par cle have from its four cons tuent
par cles? Given the mass change is 0.030374 u
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δ
•
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This is the binding energy – the energy that holds the nucleus together
•
•
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0.030374 x 931.5 MeV = 28.3 MeV less energy
The binding energy of a nucleus is the work (energy, J) required to
completely separate the nucleons of that nucleus
Binding energy = δc2
Binding energy per nucleon = (Binding energy)/(Number of nucleons)
▪
E.g. Deuterium
▪
Mass of nucleons = 2.01594 u
▪
Mass of nucleus = 2.01355 u
▪
Binding energy = 2.23 MeV
•
The binding energy curve
o
The curve showing the varia on between Binding energy (B.E) per nucleon and mass number
A per nucleon
o
Binding energy for H is 0 as there is only one par cle in the nucleus
o
Curve rises sharply for low values of A
o
Curve at maximum at A = 62, which is Ni
o
Peaks at He-4, C-12 and O-16, these nuclei are more stable than their immediate neighbours
o
Curve drops a er A = 62
o
Most nuclei have a binding energy between 7 and 9 MeV
o
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Maximum value of binding energy per nucleon is 8.7 at Fe
o
Helium has a par cularly high value of binding energy per nucleon, much higher than the
light isotopes of hydrogen
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Energy Released in a decay
o
To calculate whether energy is released in a decay or any other reac on, ∆m must be
calculated
o
∆m = mreactants - mproducts
o
OR
o
IF
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▪
∆m > 0, energy is released and decay can take place
▪
∆m < 0, energy is required for reac on to take place
e.g.
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•
•
Nuclear Fission
•
o
The process in which a heavy nucleus splits into lighter, more stable nuclei
o
E.g. Uranium 235, absorbs a neutron, becomes 236, splits to Barium 144 and Krypton 89 and
3 neutrons
o
Change in mass defect (Final mass defect – ini al mass defect)
o
Mass defect x 2 (Palladium 110) – mass defect Uranium 238
▪
= 2 x 1.01056 – 1.93538 = 0.08574
▪
Energy released = 0.08574 x 931.5 = 79.9MeV
o
The binding energy per nucleon increases, and energy is released
o
Chain reac on will only occur if there is enough Uranium 235 and is the right shape,
otherwise too many neutrons will escape from the fuel through its surface area and the
reac on will stop
Nuclear Fusion
o
•
Fusion of light nuclei
The structure of ma er
o
Elementary par cles – par cles that are not made up of other par cles
o
Standard model of ma er – our present understanding
▪
6 types of elementary par cles.
•
Quarks + An quarks
•
Leptons + An leptons
•
Exchange par cles
o
6 quarks + 6 an quarks
o
6 leptons + 6 an leptons
o
Par cles consisted of any number of quarks is called a hadron
Baryon – 3 quarks (Sigma, Omega, Lambda, in capital)
▪
An baryon – 3 an quarks
▪
Meson – quark and an quark (lower case Greek le ers)
Quarks
▪
3 genera ons
Gen 1: up & down
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Gen 2: charm & strange
•
Gen 3: truth/top & beauty/bo om
6 avours
▪
▪
Baryon number (B)
▪
Quarks B = +1/3
▪
An -quarks B = -1/3
▪
When sum of baryon numbers = 1, par cle is a baryon
▪
When sum of baryon numbers = 0, a meson
▪
All baryons are made up of 3 quarks and their baryon number is 1
▪
All an baryons are made up of 3 an quarks and their baryon number is -1
▪
All mesons are made up of 2 quarks (quark and an quark) and their baryon number
is 0
Baryon number is conserved
o
Electric charge is conserved
o
E.g.
▪
Baryon number: 1 ! 1 + 0, conserved
▪
Charge: 0 ! 1 + (-1), conserved
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However, ∆0 takes 10-25 s, Λ0 takes 10-10 s
▪
Short life me = strong interac on (force) involved (10-25 s)
▪
Long life me = weak interac on (force) involved (10-10 s)
Strangeness
▪
Strange quarks have a property labelled as strangeness
▪
Strangeness is conserved when strong nuclear interac ons and electromagne c
interac ons are involved, but may be either conserved or increase/decrease by 1
unit when weak nuclear interac ons are involved
▪
For every “s” quark, the hadron gets -1 unit of strangeness, and +1 for an an strange quark
▪
E.g. Σ+ (uus) ! n (udd) + π+
•
▪
E.g. Σ- (dds) ! n (udd) + π•
o
Strangeness not conserved, weak interac on involved
Strangeness not conserved, weak interac on involved
Leptons
▪
6 types of leptons
•
Electron and its neutrino
•
Muon and its neutrino
•
Tau and its neutrino
▪
All interact with the weak nuclear interac on
▪
Electrons, muons and Taus interact with electromagne c interac on as well
▪
▪
Exchange par cles
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This exchanged par cle is a photon
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Par cle physics interprets an interac on/force between par cles as an exchange of
par cles between them
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In Lepton decay, a tau can decay into a muon which can decay into an electron, but
not the other way (E=mc2)
•
One emits an electron and one absorbs it
▪
▪
Conserva on rules for interac ons
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Mass energy must be conserved (E=mc2)
o
Momentum conserved
o
Charge conserved
o
Lepton number conserved
o
Baryon number conserved
•
Feynman diagrams
o
Pictorial illustra on of par cle interac ons
o
Illustrate that interac ons involve exchange of par cles
o
Built up from “interac on vertexes”
o
Think of me increasing from le to right
▪
An arrow to the right represents a par cle, e.g. an electron or proton
▪
An arrow to the le represents an an par cle e.g. positron or an neutrino
▪
▪
Displays an an par cle as something that moves backwards in me
▪
A wavy line represent a photon
▪
Dashed lines represent other exchange par cles (e.g. gluons, mesons, W+, W-, Z0)
o
E.g.: Electromagne c interac on
o
Electron absorbs a photon
▪
o
Positron absorbs a photon
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Electron emits a photon
▪
o
Photon materializes into a electron/positron pair
▪
o
Positron/electron annihila on, produces photon
▪
o
Electron/positron pair annihilate and then form another par cle/an par cle pair
▪
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Change of quark avour/emission of Z0 Exchange par cle (weak nuclear interac on)
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▪
▪
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Down quark changes to up quark (emits W- bozon)
▪
Electron changes to electron neutrino (emits W- bozon)
▪
Muon changes to muon neutrino (emits W- bozon)
▪
Electron emits Z0 bozon to remain an electron and changes direc on
Con nement of Quarks
o
Quarks only exist within hadrons,
o
Thus it is impossible to observe isolated quarks
The Higgs par cle
o
Predicted by the standard model
o
Bozon
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•
o
Responsible for giving the other elementary par cles their mass
o
The theory of quarks, leptons and exchange par cles de ne what is called the standard
model of par cles and interac ons
o
For this model to work, the Higgs par cle is required to exist, as the mathema cal model of
the electroweak interac on is one of symmetry
o
This means that the existence of the Higgs par cle allows the W and Z leptons. This is done
through the interac ons of the Higgs par cle
o
The higgs par cle is the quantum (fundamental energy/mass level) of the higgs eld, just as
the photon is the quantum of the electromagne c eld
o
Discovered in July 2012, mass of 125 GeVc-2
Explaining beta minus decay using Feynman diagrams
o
n ! p + e + ῡ
o
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•
Explaining beta plus decay using Feynman diagrams
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p ! n + e+ + υ
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ENERGY PRODUCTION/GEOGRAPHY
•
Primary and secondary energy (Energy sources)
o
Primary energy is energy found in nature not subject to any processing
▪
o
Oil, coal, natural gas, solar, nuclear, wind, geothermal, dal
Secondary energy is produced by the processing of primary energy
▪
Kine c, electric, heat
o
Speci c energy (Es) is the net energy that can be extracted per mass of fuel (JKg-1)
o
Energy density (ED) is the net energy that can be extracted per volume of fuel (Jm-3)
o
ED = pES, where p = density (m/v)
o
o
Generally, the higher the ES and/or ED, the more favourable the fuel
o
Non-renewable sources are energy sources that are nite and can be depleted as they are
being used much faster than they are being replenished, and they will run out (e.g. Fossil
fuels, Nuclear fuels, etc.)
o
Renewable sources are energy sources that can either be replenished as quickly as they are
being used or are in nite (if the sun shines)
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Fossil fuels
o
Fuels produced by the decomposi on of organic material over millions of years
o
▪
o
Process
•
Chemical energy released by burning
•
Heats boiled water (kine c/heat energy)
•
Steam rotates turbine (kine c energy)
•
The turbine turns a coil of wire in a magne c eld (more kine c)
•
Induced electrical current (AC electricity)
Sankey Diagram
▪
o
e (e ciency) = Useful energy (EU)/ Total input energy (ET)
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E.g. e ciency of Coal power plant is 35/100 = 35%
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Fossil fuel power plants are reasonably e cient but contribute largely to Global warming and
the human greenhouse e ect.
o
Natural gas plants are up to 60% e cient and contribute about half of the net greenhouse
gases
o
Advantages
o
•
▪
Rela vely cheap (while they last)
▪
High power output
▪
Variety of engines and devices use them directly and easily
▪
Extensive distribu on network in place
Disadvantages
▪
Non-renewable
▪
Pollute heavily
▪
Contribute to greenhouse e ect and climate change by releasing greenhouse gases
into the atmosphere
Nuclear power
o
Nuclear reactor is where nuclear ssion is used to produce energy
o
o
Fuel is Uranium 235, for induced nuclear ssion where
Neutrons must be slowed down with a moderator which can be water or graphite.
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Uranium must be at cri cal mass (15kg for pure U235, and around 130 kg for 10% U235
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Neutrons collide with more U 235 atoms to produce more ssion reac ons and energy
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▪
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Fuel rods are where the ssion reac ons happen
o
Rate of reac on is a ected by how many neutrons are available for the U235 atoms to be hit
by and cause more reac ons
o
▪
Too few neutrons = decline of reac on to no reac on
▪
Too many neutrons = Uncontrollably large energy release/meltdown
Control rods are used to decrease or increase the rate of reac on
o
▪
Rods made of a neutron absorbing substance such as cadmium or boron that are
moved in or out of the core to control the rate of ssion events in the reactor
▪
When inserted, absorb neutrons and reduce rate of reac on
▪
When removed, allow more neutrons to cause more reac ons, increase rate of
reac on
Coolant
▪
o
o
o
▪
Fast neutrons can be used to turn U238 to P239, which can be used for nuclear
weapons
▪
Nuclear waste disposal
▪
Meltdown and leakage from core
▪
Leakage from waste
▪
Carcinogenic
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Large reserves of nuclear fuels
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Advantages
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Substance in a thermal nuclear reactor that slows the ssion neutrons down so they
can go on and produce further ssion
Risks
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A steel vessel containing pipes through which hot coolant in a sealed circuit is
pumped, causing water passing through the steel vessel in separate pipes to turn
into steam which is used to drive turbines
Moderator
▪
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A uid that is used to prevent a machine or device from becoming dangerously hot
Heat exchanger
▪
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Moderator absorbs energy of neutrons, slowing them down, but it itself increases in
temperature, which is transferred through a heat exchanger (gas or cold water)
▪
o
•
No greenhouse gases
Disadvantages
▪
Radioac ve waste is di cult to dispose of
▪
If meltdown occurs, major health hazard
▪
Issues with uranium mining (environmental and social)
▪
Produces materials for nuclear weapons
Solar power
o
The nuclear fusion in the sun emits energy at a rate of 3.9 x 1026W
o
About 1kW/m2 reaches the earth’s surface
o
1380J/second/m2
o
Energy of the sun produced fossil fuels
o
Hydroelectric dams operate using sun-li ed water
o
Wind turbines use sun driven currents
o
Photovoltaic cells converts sunlight directly into electricity
▪
o
Hea ng panels converts sunlight into heat
o
The slower the water is circulated, the ho er it can get
o
The black surface absorbs heat and the water absorbs it
o
Advantages
▪
Free
▪
Inexhaus ble
▪
Clean
Disadvantages
▪
Works during the day only
▪
A ected by weather
▪
Low power output
▪
Requires large areas
▪
Ini al cost is high
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The cell is made of crystalline silicon (a semiconductor) doped with phosphorous
and boron impuri es
•
Hydroelectricity
o
Water in elevated posi on uses gravita onal energy to transform into kine c energy, which
turns a turbine
o
density of water is 1000kg/m3
o
Poten al energy of water is mg∆h
o
Mass of water is given by pave
▪
o
p is density of water, v is volume
Rate of change of energy = power = change in poten al energy/ me;
▪
o
o
Q = volume ow rate (m3s-1)
o
P ∝ Q (Power is propor onal to volume ow rate)
o
P ∝ h (Power is propor onal to height of water) requires large fall of water
o
Advantages
o
Free
▪
Inexhaus ble
▪
Clean
Disadvantages
▪
Very dependent on loca on
▪
Requires dras c changes to environment
▪
Ini al costs high
Wind power
o
Some people prefer the sight of ruined environments over the sight of wind turbines
o
Blades suscep ble to stress fractures and disrepair
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Reportedly produce low-frequency compression waves that may or may not a ect some
people’s sleeping habits
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Modern wind turbines have vanes of up to 30m, and can produce a few megawa s of power
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Very ancient method of gaining power
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▪
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Good power produc on in winds around 6-14ms-1
o
Design must consider gale-force winds
o
About 30% e cient
o
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Formula
▪
Let cross-sec onal area be A
▪
Let Velocity be v
▪
Time = t
▪
p = density of air
▪
mass in an enclosed tube = pAv∆t
•
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EK = ½mv2 = ½ (pAv∆t) v2 = ½ pA∆tv3
▪
EK/∆t = power
▪
Therefore, P (power) = ½ pAv3
Advantages
▪
Free
▪
Inexhaus ble
▪
Clean
Disadvantages
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Mass that goes through tube per unit me
▪
Dependent on local wind condi ons
▪
Aesthe c problems
▪
Noise problems
•
Thermal energy transfer
o
Conduc on
▪
Thermal energy transfer through the transfer of vibra ons in par cles through
collisions
▪
Heat goes from more heat to less heat (high temp to low temp)
▪
•
∆Q = heat
•
A = surface area
•
L = length between two masses
•
∆T = change in temperature
Requires a medium to transfer through
▪
Solid, liquid and gas
Convec on
▪
Applies mainly to uids (gas and liquid)
▪
Ho er uids rise as they have more kine c energy, colder
method of hea ng is at the bo om, so heat gets spread
▪
Requires a medium to transfer through
Thermal Radia on
▪
No medium required
▪
P = power of radia on
▪
T = temperature (K)
▪
P ∝ T4
▪
P∝A
▪
P ∝ e (emissivity)
▪
P = σeAT4
▪
σ = Stefan Boltzmann constant = 5.67 x 10-8 Wm-2k-4
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Infra-red light, electromagne c spectrum
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∆t = me
▪
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uids sink, but the
•
Black Body radia on
o
A body which has a e value of 1
▪
Perfect absorber, perfect radiator
o
At low temperature, a black body absorbs all incident radia on and re ects none, and
radiates very li le, looks black
o
At high temperature, radiates a lot, absorbs li le, re ects none, looks very bright
o
Consider a body of Temperature T, and surrounding temperature of TS. It radiates heat at a
rate of P = σeAT4, and absorbs heat at a rate of P = σeAT4S.
o
Therefore, the net rate at which energy leaves the body is the rate at which it radiates – the
rate at which it absorbs;
▪
o
When the temperature of the object is the same as its surroundings, (T4 – T4S) = 0
o
Therefore, no emission of radia on
o
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Energy radiated by a body is electromagne c and is over an in nite range of wavelengths, but
the most common one is λmax;
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λmax x T = 2.90 x 103 Kelvin metres = Wien’s displacement law
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Solar constant
o
Sun emits a total power of P = 3.9 x 1026 W (Js-1)
o
Earth-sun average distance = 150,000,000 km = 1.5 x 1011m
o
Use Intensity = power/area
▪
As the energy from the sun radiates in a spherical volume, the intensity spreads out
the further it goes
▪
Therefore, the ‘area’ = 4πd2, where d = earth-sun distance on average
▪
Thus, I (Js-1m-2) = (3.9 x 1026)/4π(1.5 x 1011)
▪
•
= 1.4 x 103 Js-1m-2, (Wm-2) (I)
•
This is the solar constant at the top of the atmosphere
I = P/A, P = IA, ε/t = IA, ε = IAt
Albedo
o
Unit = α
o
De ned as the total sca ered power divided by the total incident power
▪
o
•
Pre
ected/Ptotal
For opaque objects, emissivity + albedo = 1
▪
Albedos: Snow = 0.85, water = 0.1, Earth = 0.3
The power incident upon a disk with SA the size of the earth
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Disk area = πr2, with r = radius of earth
o
I = P/A, thus P = IA, thus P = Sπ r 2
•
Reference frames
•
o
A set of coordinate axes and a set of clocks at every point in space
o
If the reference frame is not accelera ng, it’s called an iner al reference frame (iner a
concept expressed in Newton’s rst law)
o
An iner al reference frame is one in which N (Newton’s rst law) holds true
Nota on
S′
S
Reference frame at rest rela ve to
the observer
Rela on
Reference frame moving at a
uniform velocity, ‘v’, rela ve to
observe in S
Assume that when the origins of
each frames coordinate system
coincide, the clocks in each IRF
(iner al reference frame) are set
to 0
x′
x
Posi on of event/object in S
(distance from the origin)
Rela on
x′ = x − vt
Posi on of event/object in S’
t
t′
me interval measured in S
Rela on
t' me interval measured in S’
u
u′
Speed of object in S
Newton: t
= t′
Einstein: t
≠ t′
Rela on
u′ = u − v | u = u′ + v
t' me interval measured in S’
o
•
Newton’s postulate
o
•
Two observers always agree on the me coordinates
Galilean rela vity
o
x′ = x − vt, t = t′, Galilean transforma ons
o
Laws of physics (mo on) are the same in all iner al reference frames (1632)
o
It is impossible for one observer to claim that they “really” are at rest and the other is
“really” moving
No experiment can dis nguish between iner al reference frames
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RELATIVITY
o
E.g. A ball rolls on the oor of a train at 2ms-1 rela ve to the oor. The train moves rela ve to
the ground A) right at 12ms-1, B)le at 12ms-1 What is the velocity of the ball rela ve to the
ground?
▪
▪
•
U’ = +2ms-1
•
V = +12ms-1
•
U=?
•
Therefore, U = U’ + V
•
12 + 2 = 14ms-1 to the right
•
Same data, except V = -12ms-1
•
U = U’ + V
•
2 – 12 = -10ms-1
B)
Maxwell and the constancy of the speed of light
o
The failure of the Galilean transforma ons
o
1864 – James Clerk Maxwell “discovers” electromagne c waves
o
Maxwell’s equa ons
▪
o
Perpendicular electric and magne c elds caused by an accelera ng electric charge
Maxwell’s equa ons predicted that the speed of light in a vacuum, c, is a universal constant
which does not depend on the speed of its source
▪
o
Light has momentum but no rest mass
Unlike the speed of the ball on the train, c does not depend on the reference frame of the
train (veri ed by Hertz)
1
εμ
c=
▪
ε = electric constant = permi vity of free space = 8.85 x 10-12
▪
μ = magne c constant = permeability of free space = 1.256 x 10-6
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= 3 x 108 ms-1
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A)
E.g. a car is moving at a speed V rela ve to the ground, with an observer watching. Light
leaves the headlights of the car, at a speed c rela ve to the driver, towards the observer. The
light arrives at the observer at speed c.
o
The theory of rela vity a empts successfully to work out what has gone “wrong”
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If the speed of light has the same value for all observers, then the Galilean transforma on
equa ons cannot work for light
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Lorentz transforma ons
o
Space and me are rela ve to reference frames
▪
o
T1 ≠ T2
Lorentz (1904) developed transforma on equa ons compa ble with Maxwell’s theory
▪
Relates measurements in one reference frame to measurements in another
1
Υ=
o
o
1−
o
Lorentz
o
X’ = X – vt
o
X’ = Υ(x-vt)
o
T’ = T
o
X = Υ( X’ + vt’)
o
T’ = Υ(t – vx/c2)
o
T = Υ(t’ + vx’/c2)
The Lorentz transforma ons predict phenomena incompa ble with Newtonian physics;
▪
1. Length/space contrac on
•
Moving clocks run slow
•
Einstein 1905
o
Choice A – Accept Newtonian mechanics and Galilean
transforma ons, and modify Maxwell’s theory
o
Choice B – Accept Maxwell’s theory, and Lorentz ’s
transforma ons, and modify Newtonian mechanics of space and
me
Einstein released that choice B was required to resolve puzzles about
electric and magne c phenomena.
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Many electromagne c phenomena can only be explained using length contrac on and me
dila on
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Why Einstein accept Maxwell and Lorentz and rejected Newton’s
ti
o
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Forces on a charge or current – electric and magne c elds as measured by observers in rela ve
mo on
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The length of a moving objects gets smaller in the direc on of mo on
2. Time dila on
•
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Galileo
•
•
v2
o
▪
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o
Puzzle 1: Current carrying wire with moving charge (electron) next to it
▪
From an observer’s reference frame, the electron is experiencing a magne c force
▪
From the iner al reference frame of the electron,
▪
•
The posi ve charges move to le at speed
•
The posi ve charges produce a magne c eld B
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Fmagne c is now 0, because v of the electron (in its reference frame) is 0
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Therefore where does the force on the charge come from?
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Lengths contract in the direc on of mo on for moving objects
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Therefore the distance between the posi ve charges in the wire is
contracted, and the distance between the nega ve charges is increased, as
the length contrac on from the observers point of view is non-existent.
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Therefore, wire has overall posi ve charge, and the electron experiences an
electric force from the wire
Hence, the force on the charge is magne c in one reference frame but is electric in
the other
Puzzle 2: Two charged par cles moving with parallel veloci es
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The laws of physics are the same in all iner al reference frames
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The speed of light in a vacuum is the same for all observers in iner al reference frames/all
iner al observers
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Einstein derived the same Lorentz transforma ons based on the postulates
Light takes me t=x/c seconds to reach from C0 to CX
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Recipe
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Start C0 and simultaneously emit a beam of light towards CX
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Set CX to show me t=x/c
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Clock at origin, C0, separated by a distance x from another clock, CX.
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Means that all clocks in an iner al reference frame are synchronized (and keep perfect me)
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Clock synchroniza on
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The postulates of rela vity
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The electrons and the charge are at reset
Explana on – Lorentz length contrac on
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When light reaches CX, start clock
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Synchroniza on
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Using the Lorentz transforma ons to nd the coordinates of events in di erent Iner al Reference
Frames
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Separa on of events in space and me
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Geometry of space- me
IRFS
IRFS’
Event 1
(X1, T1)
(X1’, T1’)
Event 2
(X2, T2)
(X2’, T2’)
∆X = X2 – X1 = Υ (∆X’ + vT’)
∆X’ = X2’ – X1’ = Υ (∆X – vT)
∆T = T2 – T1 = Υ (∆T’ + v∆X’/c2)
∆T’ = T2’ – T1’ = Υ (∆T – v∆X/c2)
OR ∆X/c
Space- me interval between two events
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The posi on and me of events varies from one IRF (S) to another (S’)
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However, the space- me interval does not change, we say that it is invariant
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Suppose an observer measures 2 events separated by a distance ∆X and me ∆T. The spaceme interval between these events is: (c∆T)2 – (∆X)2
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(c∆T’)2 – (∆X’)2 = (c∆T)2 – (∆X)2
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An observer in IRF S’ measures this as: (c∆T’)2 – (∆X’)2
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Invariant quan
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“Proper” = when there is no rela ve mo on between object and IRF
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T0
Rest mass: the mass of an object at rest in rela on to an observer in an IRF
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L0
Proper me: the me interval between two events that occur in the same place
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i.e. at rest in an IRF
Proper length: the length of an object measured by an observer in an IRF when the object is
at rest
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es
M0
Rela vis c velocity addi on
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E.g. Two rockets ying towards eachother, one at 0.8c and one at 0.9c
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Rela ve speed to someone on the ground is 1.7c (no problem as neither object exceeds c)
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The speed of A rela ve to B and the speed of B rela ve to A cannot be greater than c
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We need to relate a speed measured in one IRF (S) to the speed measured in another (S’)
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Object U moving rela ve to S’ at speed U’
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S’ moves rela ve to S at V
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Speed of U rela ve to S is
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U′ + V
U=
1+
U′V
c2
If U’ = C (e.g. Photon)
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(C+V)/1+(CV/C2)
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(C+V)/1+(V/C)
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(C+V)/(C+V)/C
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=C
Rocket example – two rockets ying away from eachother at 0.8c to the le and 0.9c to the
right
U = Speed of rocket A seen by rocket B
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U’ = Rocket A moving 0.8c moving rela ve to ground = -0.8c
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S = Rocket B
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S’ = Ground (0.9c to the le rela ve to rocket B) = -0.9c
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C = speed of light = 3x108ms-1
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Another example
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A rocket moves at 1x108ms-1 rela ve to the ground. An electron has a speed of
2x108ms-1 rela ve to the rocket.
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What is the speed of the electron rela ve to the observer on the ground?
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Using formula, 2.45 x 108ms-1
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Time dila on
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Deriving using Lorentz transforma on
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An observer in S’ measures me using a clock at rest in that IRF. As the clock is in the same
place, ∆X’ = 0. A me interval ∆T’ is measured. i.e. the proper me in S’
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Observer in S measures a me interval ∆T
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∆T = Υ (∆T’ + v∆X’/c2), when ∆X’ = 0, ∆T = Υ∆T’, or ∆T = Υ∆T0
Proper me interval
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The me interval between two events that occur at the same point (posi on) e.g. as
measured by a clock at rest]
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NOTE: the proper
observer in S
me can be measured by a moving observer in S’ or by an
Example
▪
Two spaceships A and n pass at a rela ve speed of 0.7c An observer on A measures
the me for a pendulum to complete one oscilla on as 2.0 seconds. What will an
observer on B measure the period of the pendulum on A to be?
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2.8 seconds
1 ly = 1 light year = speed of light x me of 1 year (in seconds)
Simultaneity
E.g. Observer on ground
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Sees three spaceships passing with equal distance between them, A, B and C
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Ship B sends out a ash of light
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From the spaceships’ IRF, A and C get the signal at the same me
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From the observer on the ground, A receives it rst
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Relies on the fact that light is constant in all IRFs
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Simultaneous events at the same point in space are simultaneous for all observers in rela ve
mo on
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Events that are simultaneous for one observer which happen at di erent points in space are
not simultaneous for an observer in rela ve mo on.
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Time interval measured by observer in S = Υ x proper me measured in S’
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Time measured in S is always greater than me measured in S’, as S’ is moving rela ve to S
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T0 is proper me measured in reference frame S’
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If events are simultaneous in S, ∆T = 0
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If events take place in same point in space in S, ∆X = 0
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Therefore, ∆T’ = 0
Events will only be simultaneous in di erent reference frames if they take point in the same
posi on
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Length Contrac on
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An object’s dimensions decrease in the direc on of its mo on
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Consider a rod at rest in S’. Its length is measured as the proper length L0 = ∆X’
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For an observer in S, to measure the length of the rod, ∆X = L, he must do so by recording the
posi ons of the ends of the rod at the same me, ∆T = 0
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∆X’ = Υ (∆X – v∆T)
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∆T = 0
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∆X’ = Υ∆X
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L0 = ΥL
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L = L0/
E.g. Two spaceships A and B pass in space at a rela ve velocity of 0.7c. If an observer in B
measures her spaceship to be 100m long, what is the length of the spaceship according to
the observer on A
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•
Υ
Using L = L0/
Υ, L = 71.4m
The muon decay experiment
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A natural phenomenon that supports me dila on and length contrac on
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Protons enter the upper atmosphere and interact with the nuclei of O2 and N2 molecules to
give pions which then decay into muons which move o at speeds of 0.994c.
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These muons are formed at a height of 10-15 km above the ground
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The half-life of a muon is 2.2 x 10-6 s and so even moving at 0.994c they are only expected to
travel 660m before half of them decayed.
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Muons formed at 12km would take 20 half-lives to reach the ground
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However, more than (½)20 of the muons are reaching the ground, meaning the muons are
“living longer”
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At 0.994c the formula for me dila on gives the half-life for the muons to be 20
microseconds. This means that at 0.994c the propor on of muons reaching the ground
should be about 0.25.
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Hence the number of muons per second detected at the ground is much greater than
expected.
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The half life of muons as measured by an observer on the ground while the muons are
moving is calculated by ∆T = Υ∆T0
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= 2.0 x 10-5s = 20 microseconds
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Space me Diagrams
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A pictorial representa on of rela vis c phenomena that makes concepts of me dila on,
length contrac on and simultaneity par cularly transparent
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In 1908 Minkowski helps visualize this phenomenon
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Space- me diagrams plot ct against x
▪
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Could also be me in years plo ed against distance in light years
Thus, technically both are in metres
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X=5
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ct = 6
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t = 6/3 x 108
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t = 2 x 10-8
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Worldline = a
▪
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A par cle at rest
Red line = worldline of par cle moving to the right
ta n θ =
o
∆x
v
=
∆ ct
c
ta n θ gives speed expressed as a frac on of the speed of light
θ = tan−1
v
(c)
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Thus the worldline of a photon is y = x
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Worldline of any par cle cannot have an angle of greater than 45° between it and the y axis,
as then v would be greater than c
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E.g.
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3 events simultaneous in one IRF, di erent places. Lie on line parallel to space axis
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Not simultaneous in another IRF
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Events at di erent mes, but at the same posi on, lie on a line parallel to the me
axis
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Consider IRF S’ that moves to the right at speed v. Clocks at S and S’ measure 0 when the
origins coincide.
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The wordline of S’ would go from the origin and up and to the right, with an angle of
θ = tan−1
▪
v
(c)
To nd the space me coordinates of an event in S’, you must adjust and lt both
axes by the same angle θ
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= tan−1
v
and draw lines parallel to the new axes
(c)
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This is a graphical representa on of the Lorentz transforms
▪
Scale changes
For an object moving to the le , the angles are the same but they are on the other side of
the axes of the IRF at rest
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The scales on the S and S’ axes are di erent
▪
me dila on, and also remember that
▪
v = speed of S’ rela ve to S
▪
θ = tan−1
▪
Υ = 1.25
▪
Therefore, the S’ axes are o set by 31°
v
. Therefore, θ = 31°
(c)
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E.g. Draw the space me diagram for S and S’ where v is 0.6c
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Remember Length contrac on and
always greater than or equal to 1
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Υ is
▪
Consider x’ = 1. It intercepts x at t = 0
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X’ = Υ (X – VT)
▪
= ΥX (as t = 0)
▪
Therefore, X’ = ΥX (scale)
•
Υ
▪
Similarly, consider ct = 1, and intercepts ct at x = 0
▪
T’ = Υ (T –vx/c2)
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= ΥT, as x = 0
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Therefore, T’ = ΥT
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X = X’/
T = T’/
Υ
Summary of axes
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In comparison to the ct and x axes, the ct’ and x’ axes are rotated and stretched,
▪
Rotated by the angle θ
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Stretched in scale so that the X’=1 line intercepts the X axis at X = 1/
▪
Stretched in scale so that the CT’=1 line intercepts the CT at CT= 1/
= tan−1
v
.
(c)
Υ
Υ
Space me diagrams – explana on of length contrac on and me dila on symmetry
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v = speed of S’ rela ve to S
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θ = tan−1
▪
Υ = 1.25
▪
Therefore, the S’ axes are o set by 31°
v
. Therefore, θ = 31°
(c)
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Say the length of something is 1 as measured in S. Draw a line from X = 1 intersec ng X’ but
parallel to CT. It will be at 0.8 (which is what the S’ IRF measures)
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Then say the length of something is 1 as measured in S’. Draw a line from X’ = 1 intersec ng X
but parallel to CT’. It will be at 0.8 (which is what the S IRF measures)
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Then say the me of something is measured in T’ as 1 second. Draw a line from T’ = 1 to the T
axis but parallel to the X’ axis. It will be at 0.8 (This is the me that S’ sees pass in S)
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Then say the me of something is measured in T as 1 second. Draw a line from T = 1 to the T’
axis but parallel to the X axis. It will be at 0.8 (This is the me that S sees pass in S’
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THEREFORE, OBSERVERS IN BOTH REFERENCE FRAMES SEE THE CLOCKS IN OTHER
REFERENCE FRAMES RUN SLOW
This is explained due to the laws of special rela vity breaking down as there is a change in
direc on and so there is overall accelera on, thus v is not constant
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The other observer, from the S IRF is observed to experience less me
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One observer is in S (e.g. Earth IRF) and experiences true me
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S’ is moving rela ve to S
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Two IRF’s, one being S and the other being S’
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The Twin Paradox
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E.g. Draw the space me diagram for S and S’ where v is 0.6c
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Causality
Events that cause event E and events that are caused by event E are shown by cones above
and below a point in space me, E, on a space me diagram
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The cones are de ned by the melines of photons that leave and arrive at E
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Therefore, nothing outside the ‘past’ light cone could have caused E, and nothing outside the
‘future’ light cone could be caused by E, as nothing can move faster than light (45° on the
diagram)