Forces as vectors
… Work done = force x distance (in the direction of the force). The equation given in the formulae sheet for this is:
W = Fscosθ … where F is force, s is displacement and cosθ is required if the force acts at an angle to the displacement.
… Forces can be resolved into two perpendicular components: put your pencil on the force and rotate to the component you want. If you move ‘through the angle’, then it is “cos”.
… Once you have resolved the forces acting on a body into their perpendicular components:
add up the perpendicular components so that you end up with just two perpendicular forces.
Use Pythagoras to calculate the magnitude of the resultant force and,
Use trigonometry to calculate the angle at which the resultant force acts.
… It is also possible to find yet resultant force and direction of two forces acting on an object by drawing a careful scale diagram and using a ‘parallelogram of forces’.
… For bodies in equilibrium, you can work out the magnitude and direction of a force graphically using a scale diagram. Make sure you choose a scale that creates a large diagram (to maximise the accuracy).
Place individual force vectors ‘head to tail’ (arrows flowing in the same direction) and then measure the line which joins the start of the first force vector to the end of the final force vector.
… Stoke’s Law for laminar air flow around a spherical body of radius r and velocity v, moving through a fluid which has a coefficient of viscosity of η:
F = 6π r η v
Note that this relationship is only valid for laminar air flow.
Laminar airflow is when air moves at the same speed and in the same direction, with no or minimal cross-over of air streams (or “lamina”).
By contrast, turbulent flow creates swirls and eddies that deposit particles on surfaces randomly and unpredictably.
So why do golf balls have dimples? It’s actually to create a turbulent boundary layer (the layer between the air and the ball’s surface). Although this increases the ‘skin’ the ball, it has a greater effect in reducing the overall drag force. Here’s more about this effect: https://phys.org/news/2007-06-probing-dimples-golf-balls.html
Newton’s third law of force pairs
… Forces always come in pairs: an ‘action’ force and a ‘reaction’ force (of the same type). Each of these forces have the same magnitude, but are in opposite directions.
… It can be useful to construct a sentence to identify the two forces. For example,
“the gravity force of the Earth acting on the ball downwards… is equal to… the gravity force of the ball acting on the Earth upwards”.
… The action and reaction forces are ALWAYS the same type of force (e.g. contact force, gravity force, magnetic force…)
Kinematics
… Power, force and velocity: P = Fv.
This equation is contrary to our intuition! If an engine has a set power output, P, and is travelling at a velocity v, then:
In 1 second it has travelled a distance v
The work done by the driving force, E = Fv and as this is over 1 second, P = Fv
This means that as the object INCREASES in speed, the driving force due to the constant engine power actually DECREASES… and therefore so does the object’s acceleration!
So it seems that a rocket ship with a thruster giving a constant power output will accelerate less and less, the faster it moves? This is one of those ‘just plain weird’ maths meets physics moments… In fact my gut tells me that something is wrong about this so I’ll double check it… TBC!
… A measuring instrument is linear when its input quantity (eg temperature) is directly proportional to its output quantity (eg the scale reading).
Waves – essentials
… Amplitude is the maximum displacement of a point on a wave from its equilibrium position.
… Frequency is the number of oscillations of a point on a wave per unit time, e.g. oscillations per second (Hz)
… Phase difference is how far ‘out of step’ (out of sync) the oscillations at two points on the wave are.
You can also think of phase difference as the fraction (angle difference) of the cycle between the oscillations of two points on the wave.
… The ‘phase’ of a vibrating particle at a certain time is the fraction of a cycle that it has completed since the start of the cycle.
… A progressive wave transfers energy.
… A stationary wave traps energy in pockets (‘antinodes’) due to the reflection from one end and subsequent superposition of the overlapping waves.
… Coherent wave sources have a constant phase difference between the waves. This phase difference may be zero, i.e. they could be exactly in phase.
Coherent waves must have the same frequency.
… Interference is when two (or more) waves superpose (meet) at a point and there is a change in overall displacement. For example, a peak meeting a peak would superpose to produce a ‘superpeak’.
… Constructive and destructive interference (‘superposition’) occurs when two waves meet. To find the resultant wave amplitude, add up the individual wave amplitudes:
Constructive interference: e.g. crest meets a crest, amplitudes will add.
Destructive interference: e.g. crest meets a trough, amplitudes will subtract.
… Destructive interference occurs when two wave trains of equal amplitude and frequency interfere with a phase difference of 180° or π radians.
… The phase difference, Φ, between two points at distances x and y along a wave:
Φ = 360(x – y) / λ in degrees
or
Φ = 2π(x – y) / λ in radians
… If two sources A and B emit waves of the same wavelength (same frequency) and these waves meet at a point P, then the phase difference in radians at P is given by:
Φ = 2π(path difference) / λ
so
Φ = 2π(AP – BP) / λ
Waves and Polarisation
… Only transverse waves can be polarised so that they vibrate in one plane only.
… The vibrations of a light wave are a varying electrical field and a varying magnetic field which are perpendicular to each other with the same frequency.
Each of these fields is perpendicular to the direction of travel of the wave. Therefore electromagnetic waves are transverse.
… When light interacts with matter, the effects of the electric field usually dominate those of the magnetic field. So we usually only consider the electric field oscillation when polarising an electromagnetic wave.
… A non-polarised light source which is shone through a polarising filter will lose half of its intensity.
… A plane-polarised light wave where the electric field is oscillating at 0° to the vertical can be shone onto a piece of polarising material (such as Polaroid):
- if the Polaroid material is aligned in the same polarising plane, then the wave will transmit (go through) successfully with no loss of intensity.
- if the Polaroid material is rotated through 45° to the vertical, then only part of the polarised wave will transmit successfully. The resulting intensity is actually proportional to (cosA)^2, so in this case we would see (cos45)^2 = 0.5 of the intensity before the filter was applied.
… Television aerials need to be correctly aligned so that they ‘interact’ with the polarised incoming TV waves to generate small electrical currents. The aerial should be aligned in the plane of polarisation of the waves.
… Uses of polarising filters include:
- in a camera – to reduce glare or enhance images of reflections (reflections are partially polarised)
- in a microscope polarimeter to analyse chemicals, concentration or type of sugar.
- stress analysis as light is polarised who transparent materials are stressed. This reveals areas of high/low stress.
- LCD displays (potential differences polarise the light passing through the liquid crystal)
- in cars to reduce glare through windscreens.
- identify minerals/rocks
… The 3 polarising filters trick.
If a beam of light passes through a polarising filter with its axis of polarisation at 0° and then into a second polarising filter with its axis of polarisation at 90°, then NO light is transmitted.
What happens when a third polarising filter is placed BETWEEN the two filters and rotated to 45°?
If light is polarised so that all the waves are vibrating in a vertical plane, then a second polarising filter at 45° to the vertical will ‘squash’ these waves to the 45° component. The amplitude of the waves will then be 1/√2 of its previous value, but they will now be vibrating at 45°.
The third filter does exactly the same thing, so light emerges from the thee filters with 1/2 of the amplitude that emerged from the first filter.
This process is explained very well in more detail here:
http://alienryderflex.com/polarizer/
Waves – diffraction and interference
Single slit diffraction
… Single slit diffraction is a broad spreading out of waves to the first minimum. If the wavelength is about the same as the slit width, then the angle to the first minimum is about 90°.
… The angle to the first MINIMUM for single slit diffraction is given by:
sinθ = λ/b
where b is the slit width.
The angular position of the minima is given by
sinθ = nλ/b
Where n = 1, 2, 3, …
… The maxima either side of the central maximum are much less intense than the central maximum. This is because the central maximum is the only point at which there is constructive interference from every wavefront that reaches it.
The other ‘subsidiary’ maxima lie approximately halfway between the minima.
… The width of the central maximum is DOUBLE the width of each subsidiary maximum.
Multiple slit diffraction and interference
… Young’s double slits geometry and equations. Two-slit interference is observed due to varying path differences between the two wave sources at different angles.
… The amplitude of the minima at either side of the central path are often not entirely zero. This is because the wave source that has travelled a further distance will have lost some of its intensity due to its energy spreading out via diffraction. This means that interference is not totally destructive as the two interfering waves will have slightly different amplitudes.
… Using the equation w/D = nλ/s to predict what will happen when variables are changed.
w is the fringe spacing as seen on the projection screen
D is the distance form the slits to the screen
n is an integer
λ is the wavelength
s is the slit spacing.
… Diffraction gratings (multiple slits) produce very well defined (narrower) fringes as there are more waves interfering constructively (or destructively) at each point. Use the equation:
sinθ = nλ/d
to calculate the angles of each ‘order’ maximum. Here θ is the angle to the nth MAXIMUM.
Also that n=0 is the centre maximum and n = 0,1,2… until the maximum order where sinθ > 1 and the equation stops working.
Each order has two ‘beams’ on either side of the central maximum.
(Note that this equation also works for double slit diffraction interference.)
… If the white light is incident on a diffraction grating, then each orders either side of the central maximum will be a spectrum of the different wavelengths which make up the white light, i.e. ROYGBIV. The red light (R) will be at a greater angle (furthest away from the central maximum) due to sinθ = nλ/d.
… The intensity of the fringes created by 2-source or diffraction grating interference will also follow the single slit diffraction ‘envelope’. This means that the fringes that are seen approaching 90° will hardly be seen if the wavelength is about the same size as the width of the each slit.
… Diffraction grating ‘lines per mm’, N. To find the slit separation, s (in m) use:
s = (1×10^-3) / N
… 2-source interference at the quantum level becomes very strange indeed… Check out this video: https://youtu.be/U7Z_TIw9InA.
Quantum theory suggests that every photon travels along every possible path and that the resultant of the phasers squared = relative probability of the photon taking that path.
Stationary (standing) waves
… Progressive waves travelling along a string (e.g. a guitar string) will reflect off the other end and overlap with the oncoming waves.
… Constructive and destructive interference happens between two waves of similar wavelength which move through each other in opposite directions. This superposition of waves can create a ‘stationary wave’ pattern of nodes (zero amplitude) and antinodes (maximum amplitude), but only at certain wavelengths.
… The wavelength required for a standing wave pattern, λ = 2L/n where n is the harmonic number (1 = fundamental harmonic), and L is the string length.
… Nodes (zero amplitude, destructive interference) and antinodes (maximum amplitude, constructive interference) form in a standing wave. Here is an interesting video on standing waves in a metal plate – i.e. in two dimensions: https://youtu.be/wvJAgrUBF4w
… The equation linking length of a string (L), tension (T), mass per unit length (μ) with frequency of the 1st harmonic (fundamental) is:
f = (1/2L)√(T/μ)
which is given in your formulae booklet.
So if the tension increases, the frequency of the progressive wave will also increase.
What happens to the frequency as the length of the string increases?
Note that:
Mass per unit length = Density x Cross sectional area
μ = ρ x A
“muroa!”
Lasers
“Laser” is an acronym for light amplification by stimulated emission of radiation. A laser is created when the electrons in atoms in special glasses, crystals, or gases absorb energy from an electrical current or another laser and become “excited.” The excited electrons move from a lower-energy orbit to a higher-energy orbit around the atom’s nucleus. When they return to their normal or “ground” state, the electrons emit photons (particles of light).
Parallel mirrors at either end reflect the laser light within the tube and this causes more atoms to return to their ground state and emit more photons, so the light intensifies.
The parallel mirrors also allow any non-parallel light rays to escape from the tube. This means that the laser beam is narrow and hardly spreads out.
One of the mirrors is partially silvered and so lets through a small percentage of the photons as a beam of coherent light.
These photons in this beam are all at the same wavelength and are coherent (in phase) and so laser light stays focused for vast distances, even to the moon and back!
Refraction
… The refractive index of a material,
n = c / v
n = speed of light in air (or a vacuum) / speed of light in the material
So a refractive index of 1.42 means that light travels 1.42 times faster in air compared to in the material.
… When light enters glass from air, both its speed and its wavelength decrease. The frequency of the light stays constant, which means that the COLOUR of the light stays the same.
… Compared to red light, blue light has a higher refractive index in a certain material. This means that blue light travels slower in that material.
When blue light leaves that material, it will refract at a greater angle compared to red light.
… A refractive index of 1n2 means the refractive index of material 2 relative to material 1, so
1n2 = n2/n1
For example, “air n water” = n(water)/n(air) ≈ n(water) … because n(air) ≈ 1
… Snell’s law for light refracting across a boundary between materials 1 and 2:
n1sinθ1 = n2sinθ2
Where n1 and n2 are the refractive indexes of the materials, and θ1 and θ2 are the angles of incidence and refraction.
Note that light of a different colour (i.e. a different frequency) will cause a different refractive index for a certain material.
For example, Red light will have a lower refractive index than blue light in glass.
more usefully, Snell’s law can be written as:
sinθ1 / sinθ2 = n2/n1 = v1/v2
… The critical angle, C, of a ray moving from a higher refractive index n1, to a lower refractive index n2, is given by:
sinθ1 / sin(90) = n2/n1 , where θ1 = C
so
sinC = n2/n1
If the lower refractive index material is air, then n2=1, and so:
sinC = 1/n1
… A step index optical fibre has a uniform refractive index within the core and a sudden (‘step’) decrease in refractive index at the core-cladding interface, caused by a cladding of a lower refractive index.
… Optical fibres come in two main types:
Single (mono) mode fibres have a very small diameter core and work with a single ray of light (a coherent source such as a laser) that travels down the core, totally internally reflecting at large angles of incidence (‘glancing off the boundary’).
This means that a single pulse of light will not spread out very much. Mono mode optical fibres are used for long distances, but the connectors and light sources are more expensive than multi-mode fibres.
Multimode fibres have a much larger diameter core and there are multiple paths that light can take within the fibre, so that means TIR at many different angles (modes).
This means that a single pulse will spread out significantly over distance as some rays will take a longer path and so will arrive after other, more direct rays.
These fibres are cheaper than mono mode fibres but only suitable for shorter distances.
… Optical fibres are made from core (glass or plastic) surrounded by a less optically dense cladding. The reasons for using cladding include:
It protects the core (from scratches, stretching or breakage)
It prevents ‘crossover’ of signal / ensure security of data / prevent loss of information/data/signal
It INCREASES the critical angle. This removes any rays which approach the boundary at a small angle of incidence by being refracted out of the core (these rays take a longer route through the cable due to lots of reflections and so are not wanted). This reduces pulse broadening due to ‘modal dispersion’.
It increases the rate of data transfer (rays which totally internally reflect at large angles of incidence travel a shorter distance).
… Material pulse broadening in an optical fibre is caused when a ray of light contains more than one wavelength. A pulse of light will spread out over a distance down the fibre because the different wavelengths have different refractive indices in the core material and so will travel at different speeds.
The higher the refractive index, the slower the light will travel in the core.
… Uncertainties in lens experiments can be caused by:
- not knowing exactly where the image is in focus (usually causes quite a large uncertainty)
Ultrasound
Ultrasound waves, like electromagnetic radiation, can travel through the surface between two materials. When ALL the energy is transmitted from one material to the other, the materials are said to be acoustically matched.
In this question, you really only needed to show that the ratios of the Young modulii of the two materials was equal to the ratio of the speed of sound in each material (this is the condition for being acoustically matched).
Converting units
… Convert 5.4 N/m2 into N/mm2 (pressure).
1 m2 = 1000 x 1000 = 1×10^6 mm2 (draw a square with a side lengths of 1m = 1000mm), so it looks like we need to multiply by 1×10^6.
HOWEVER, it is the DENOMINATOR which we are changing, so rather than multiplying by 1×10^6, we need to DIVIDE by 1×10^6:
So:
5.4 N/m2 = 5.4 x10^-6 N/mm2
… Convert 43 kg/cm^3 into kg/m3 (density)
How many cm3 are there in a m3? By sketching a cube with side length 1m = 100cm we see that the volume of a 1m3 cube is 100 x 100 x 100 = 1×10^6 cm3
So there are 1,000,000 cm3 in 1 m3, so it looks like we need to divide by 1×10^6.
HOWEVER, it is the DENOMINATOR which is changing, so instead we need to do the opposite and MULTIPLY by 1×10^6.
So:
43 kg/cm3 = 43×10^6 kg/m3
… Convert 76 m/s into km/h
Firstly, we need to convert from m to km. 1000m = 1km so we have to divide the quantity by 1000.
Then, we need to convert from s into h. 3600s = 1 hour, so it looks like we need to divide by 3600, BUT because this is a DENOMINATOR, we have to do the opposite and MULTIPLY by 3600.
So:
76 m/s
(Now ÷ by 1000)
= 0.076 km/s
(Now × by 3600)
= 273.6 km/h
Astrophysics
… Useful AQA Teaching Guide: http://filestore.aqa.org.uk/resources/physics/AQA-7407-7408-TG-A.PDF
Telescopes
… Lenses recap of lens basics, including:
converging vs diverging lenses
Focal length
Principal focus
Principal axis
Power of a lens = 1/f (measured in dioptres)
Lens equation: 1/f = 1/u + 1/v
Magnification of a lens = distance to image / distance to object = v/u
… The light rays from a distant object are essentially parallel when they enter the objective lens.
The refracting telescope
… consists of two converging lenses, an o…?… lens which produces a real image (light rays converge to a focal point), and a converging eyepiece lens, which acts as a magnifying glass.
… The final image after the eyepiece lens appears at infinity (rays are parallel). The observers eye therefore doesn’t have to keep refocusing when changing from looking at the image to looking at the distant object. This is called ‘normal adjustment’ and is when:
the objective lens focal length + eye piece focal length = the length of the telescope
or
fo + fe = D
… Magnification of a refracting telescope, M = α/β = fo/fe
α = angle subtended by image at eye
β = angle subtended by object at unaided eye (angular size of the object)
… Refracting telescopes suffer from chromatic aberration when white light disperses into its component colours which are then focussed by the objective lens over a range of focal lengths. This produces coloured edges to the image.
… Refracting telescopes also suffer from spherical aberration, when the refracted rays at the edges of a spherical lens are refracted more than those near the optical axis. This can be minimised and even eliminated with careful lens geometry corrections (by making both surfaces of the objective lens contribute equally to the ray deviations).
… When drawing ray diagrams to illustrate chromatic and spherical aberration, make sure you draw a minimum of FOUR rays parallel to the principle axis.
… Refracting telescopes are rugged than their more delicate reflecting telescope cousins. After the initial alignment, their optical system is more resistant to misalignment than the reflecting telescopes. The glass surface inside the tube is sealed from the atmosphere and so rarely needs cleaning.
… Reflecting telescopes use parabolic mirrors to focus the parallel light from distant object. They do not suffer from chromatic aberration and are more lightweight than refracting telescopes.
… Parabolic mirror reflecting telescope do not suffer from spherical aberration.
… The Raleigh Criterion is the critical condition in which two sources can only just be resolved (the resolving power of a telescope).
The critical angle (angular resolution) is given by:
θ = λ / D.
D = diameter of the mirror or lens
The criterion is met when the first minimum of one diffraction pattern coincides with the centre of the diffraction pattern of the other.
… The “Airy disc” is the bright central maximum of the diffraction pattern of a point source.
… Charge coupled devices (CCDs) are made from thin wafers of silicon divided into individual picture elements called pixels. Photons that strike a pixel cause electrons to be liberated from the silicon surface and trapped by a positive electrode.
The more photos that hit a pixel, the more electrons are trapped (directly proportional).
CCDs can detect wavelengths beyond visible light from low energy X-rays to infrared.
… X-ray telescopes need to be sited in space because X-rays are absorbed by the Earth’s atmosphere. They use ‘grazing mirrors’ to focus the collected X-rays, rather than a parabolic dish and can have a small aperture because resolving power is proportional to λ/diameter.
… Radio-telescopes collect radio waves from 10m to 0.5mm that are emitted, e.g. by the sun, nebulae and supernovae remnants. They operate in a similar way to reflector telescopes using a parabolic dish but have poor resolving capability due to the large wavelengths of radio waves. To maximise the resolving power, they need to have a large diameter.
… Radio-telescopes can be connected together to improve the angular resolution to beyond that of the best optical telescopes. This is called Interferometry – here’s a video to check out: https://youtu.be/OeyjxNVA8uQ
… Radio telescopes are best sited away from populations centres to avoid interference from radio wave transmissions. They can be sited terrestrially as radio waves are not absorbed by the atmosphere and these telescopes are also very large and heavy!
Stars
… The Luminosity, L, of a star is its power output, measured in units of watts.
… Radiation flux , F (also known as brightness or intensity) is the power per m² measured at a certain distance from a star. As you increase the distance away from the star, it’s radiated power will spread over a larger and larger sphere and so the radiation flux will decrease (becomes less intense).
… Radiation Flux = Luminosity / surface area of a sphere at distance d
F = L / 4πd²
… Astronomers define star ‘brightness’ in terms of apparent magnitude, m (how bright the star appears from Earth) and absolute magnitude, M (how bright the star appears at a standard distance of 32.6 light years, or 10 parsecs).
… The eye perceives the brightest star (apparent magnitude, m = +1) to be roughly 100 times brighter than the faintest star (apparent magnitude, m = +6). There are 5 steps between the faintest and brightest stars, so this ‘backwards’ scale means that a star of m = +2 is 10^(1/5) = 2.51 times brighter than a star of m = 3.
And so:
2.51^5 = 100
–
… Absolute magnitude equation derivation:
If m2 – m1 = magnitude difference between two stars
i2/i1 = ratio of intensity of light measured at the Earth.
Then
m2 – m1 i2/i1
1 2.512
2 2.512^2
3 2.512^3
etc… *Note that an INCREASE of apparent magnitude of 1 means that the intensity of light DECREASES by ×1/2.512
2.512^(m2-m1) = i1/i2 (*rather rather than i2/i1 )
And as 2.512 = 100^(1/5)
100^[(m2-m1)/5] = i1/i2
Taking logs of both sides:
[(m2-m1)/5]log100 = log(i1/i2)
[(m2-m1)/5] × 2 = log(i1/i2)
and so…
m2-m1 = 2.5log(i1/i2)
Now, where m2 = M is the apparent magnitude of a similar star at a distance of 10 parsecs (i.e. it’s absolute magnitude) and i2 = I is the intensity of light received from that star at 10 parsecs.
m and i are the original values for the star:
M – m = 2.5log( i / I )
or
m – M = 2.5log( I / i )
And from the inverse square law:
I = L/4πD² for the star at 10 parsecs away
and
i = L/4πd² for star 1
so
I / i = (d/D)²
and D is the standard distance of 10 parsecs, so:
I / i = (d/10)² , where d is the distance to the star from Earth in parsecs.
Finally, putting the two equations together:
m – M = 2.5log[(d/10)²]
and so
m – M = 5log(d/10).
This can also be written as:
M = m – 5log(d) + 5
So to calculate absolute magnitude, we need to know the distance, d, of the star from the Earth (in parsecs) and its apparent magnitude, m, as viewed from Earth.
… Absolute magnitude, M, is proportional to a star’s luminosity, L (power output), so when comparing two stars:
L1/L2 = M1/M2
… It can be useful to rearrange m – M = 5log(d/10) to compare distances:
d = 10 × 10^[(m-M)/5]
d = 10^[(m-M+1)/5]
… Standard candles: objects with known luminosity (power output) such as cepheid variables and supernovae. If these are nearby to the star of interest, then they can be used to estimate the distance to the star.
… Inverse square law with standard candles means we can calculate distance to the standard candle if we measure the radiation flux (power/unit area) received at the Earth using a sensitive light sensor (charged coupled device).
… Methods of calculating distances to stars include:
1) parallax – for nearby stars
2) using standards candles in the vicinity of the object
3) red shift of the star’s emitted spectrum (using Hubble’s law which links recession velocity to distance)
… The Parallax method for measuring the distance to ‘nearby’ stars against the background of very distant stars by measuring the angle of parallax, θ :
Star distance, d is given by:
d (in AU) = 1 / θ (in radians)
d (in parsecs) = 1 / θ (in seconds of arc)
… Note that the angle of parallax is HALF the angle observed to the star from the two extremes of the Earth’s orbit about the sun.
… 1 parsec is defined as the distance from the Earth to a star which has an angle of parallax of 1 second of arc (1/3600 degree).
… 1 parsec = 2.06×10^5 AU = 3.26 light years = 3.08×10^16m
… The furthest distance that can be measured using the parallax method is about 1000 parsecs (the smallest angle we can measure is about 0.001 seconds of arc).
… The power output, P, of a radiating body depends on its temperature. This is given by Stefan’s law:
P = σA(T^4)
where A is the surface area of the body and σ is the Stefan-Boltz mann constant (5.67 ×10^-8)
… The intensity, or radiation flux, from the Sun as measured at the Earth is sometimes referred to as the solar constant. This is measured in units of W/m²
… A ‘black body’ is a perfect absorber (i.e. it doesn’t reflect) and emitter of radiation. Such a body must be in thermal equilibrium with its surroundings so that the rate of emission is equal to the rate of absorption.
… Hot objects such as stars are effectively black bodies. The wavelength of the radiation they emit depends on their surface temperature. The hotter the temperature, the shorter the predominant wavelength emitted.
Black body curves show the pattern of power radiated versus wavelength for a particular temperature.
… Wien’s law tells us the relationship between the peak intensity wavelength, λmax (in m), and temperature (in K):
λmax T = 2.9×10^-3 mK
So the peak wavelength is inversely proportional to the temperature. A star that has a temperature of several million degrees will emit radiation in the x-ray region!
… Stellar spectroscopy gives us three types of spectra:
- a continuous spectrum due to black body radiation emission (black body curves). An example of this is from the Sun’s photosphere which acts as a source of white light.
This type of spectrum is produced by the motion of charged particles within the hot material – for example, vibration of positive metal ions in a hot solid, or collisions between molecules in a hot gas. So it is actually not due to excitation of electrons. - emission spectra (atoms of an element are excited, e.g. by heating, and electron de-excitations release photons at specific wavelengths).
- absorption spectra (specific wavelengths of a continuous spectrum are absorbed by electrons and re-emitted in random directions resulting in black ‘missing’ lines that a characteristic of the element).
The intensity and position of emission and absorption lines correspond to particular electronic transitions in the atoms of the gas.
For example, in the Balmer series of absorption lines, electrons transition from and back down to the n=2 energy level. Strong Balmer absorption lines means that electrons are mostly excited at the n=2 level to start with due to the particular temperature of the gas (about 11,000K)… This means that the temperature of a star can be deduced from the relative strength of absorption lines of different elements.
… The relative strength of particular absorption lines gives the spectral class of a star, designated by one of seven letters: O B A F G K M (remembered by: “Oh Be A Fine Girl Kiss Me!”):
Prominent Absorption lines
O blue, UV (25000-50000K) H (Balmer lines), He, He+
B blue, UV (11000-25000K) H, He
A blue-white (7500-11000K) H, Ca+, Fe+
F white (6000-7500K) Ca+, Fe+
G yellow-white (5000-6000K) Ca+, Fe+, Fe
K orange (3500-5000K) Ca+, neutral metals
M red (<3500K) Neutral compounds e.g. TiO
It is worth memorising this table (make a flashcard!)
… The spectral classes are subdivided into ten divisions from 0 to 9 (the lower the number, the hotter the star). Our Sun is designated a G2 star.
Here’s a useful (although fast talking!) video about Stellar spectroscopy:
https://youtu.be/ld75W1dz-h0
… The Hertzsprung-Russell diagram has temperature (backwards!) on the x axis and power output (Luminosity) on the y-axis. It’s well worth making a flashcard on this noting the important areas of interest: the main sequence, red giants/super giants and white dwarfs.
Another useful flashcard would be the typical path that a star like our sun takes when it runs out of fuel at the end of its life (note it’s approximate starting coordinates as 5800K = G2, 4.8 absolute magnitude).
… The life cycle of low mass stars such as our sun and red dwarfs:
https://youtu.be/jfvMtCHv1q4
… And how larger mass stars meet their end: https://youtu.be/PWx9DurgPn8
… Red Giants have a similar mass as our Sun, but with an expanded outer shell. They are cooler (fusing helium) but highly luminous due to their large surface area.
… Red supergiants are 10-100 times more massive than the Sun and are more luminous than red Giants due to their higher temperature (core temp = 10^8K) and huge surface area. These stars are able to fuse carbon and heavier elements.
… White dwarfs are the remnants of old stars that have run out of fusion fuel. They are planet-sized and will gradually cool over a few billion years to form black dwarfs, which could account for a significant proportion of the mass of the universe.
… A star like the Sun will spend about 10 billion years on the main sequence. However, a star that has a mass of 15 solar masses will only spend 10 million years also on the main sequence!
… When all the hydrogen in the Sun’s core has become helium, the outer layers expand rapidly to form a hot gas cloud and the star becomes a red giant. The outer layers dissipate as a planetary nebula (a misnomer!) leaving behind the stars core (a white dwarf).
… For stars with a core mass > 1.4 solar masses, fusion of heavier elements is possible up to iron each time the core contracts (when it runs out of one type of fuel) and heats up. These red supergiant stars form an onion-like structure with different elements in each layer.
Once iron has accumulated in the core, no more fusion is energetically feasible and so the gravitational forces overcome the thermal forces and the star collapses very rapidly resulting in a huge explosion called a supernova. Supernovae can produce so much luminosity that they can briefly outshine a whole galaxy!
If the resulting core after the supernovae < 3 solar masses, then a stable neutron star is formed made up of neutrons and an iron crust. The gravitation field around a neutron star is very very strong! To escape, you would need 0.8 the speed of light!
If the resulting core > 3 solar masses, gravitational forces compress the core to a point, forming a black hole from which even light cannot escape.
… The light curve of a typical type 1a supernovae (absolute magnitude vs time) peaks quickly and then gradually falls over 275 days (with a faster fall for the first 50 days)
Red shift and cosmology
… There are two types of red shift
red shift caused by the Doppler effect, i.e. sources moving through space relative to an observer
Cosmological Red Shift caused by the EXPANSION of space, e.g. as observed between galaxies.
… Red Shift is given by the formula:
z = Δλ / λ
Δλ is the change in wavelength due to redshift (positive for a recessional velocity)
λ is the actual wavelength in the lab on Earth.
z = – Δf / f
Δf is the change in frequency due to redshift (negative for a recessional velocity)
For slowly moving galaxies, redshift is also the ratio of the velocity of the galaxy to the velocity of light:
z = – v / c
where v is the relative velocity between the Earth and the observed object (note the negative sign! This is because a negative relative velocity is AWAY from the Earth, i.e. a recession velocity).
… Calculations on binary stars viewed in the plane of orbit, using the variation of light intensity vs time to establish the orbital period.
Also using the Red shift observations of the stars, angular velocity, linear velocity and the distance between the stars can then be calculated.
… Most galaxies we observe are moving away from us – they have a ‘recession velocity’.
… Doppler shift is found by seeing how much spectral absorption lines have shifted when compared to lines measured in the laboratory.
… Hubble Law: v = Ho x d. A galaxy that is twice as far away from us has twice the recession velocity (usually).
… The age of the universe can be estimated using t = 1/Ho
This is because we are ‘winding back time’, assuming that that recession velocity has been constant throughout the life of the universe and so the time taken:
t = diameter of the universe / velocity at which it has expanded
t = d/v = 1/Ho
Note that we have to convert the units of kilometres per second per megaparsec into standard SI units before doing this calculation.
… Estimating the age of the universe like this assumes that the rate of expansion is constant. Evidence now suggests that expansion is accelerating. We have a value of 74.03 for the Hubble constant NOW (the gradient of a v-d graph) but this value would have been lower in the past.
In other words, the rate of expansion would have been smaller in the past, taking more time to expand so the age of the universe is likely to be older than our calculation suggests.
… The cosmic microwave background (CMB) radiation detected by the COBE satellite provides further evidence for the Hot Big Bang model.
The HBB model predicts that the gamma radiation emitted when the universe became transparent (at about t = 300,000 years) should now be observed today red-shifted to the microwave region with a spectrum corresponding to a black body of temperature 2.73K.
The COBE satellite confirmed this spectrum with a peak wavelength corresponding to a the pasture of 2.725K.
… The small variations in temperature of the observed CMB in different directions suggest that the matter and energy in the early universe were not uniformly distributed. This meant that gravitational forces were able to act on ‘clumps’ of matter to form the galaxies we observe today.
… More strong evidence for the HBB theory is provided by observations of the relative abundance of hydrogen (73%) and helium (25%). This high percentage of helium was formed from the fusion of hydrogen during the early, hot universe (other models give very small percentages of helium).
… Quasars (originally called Quasi-stellAr Radio Sources) are believed to be some of the most distant objects in the known universe as they show extremely large optical redshifts. They are very bright radio sources.
Although quasars are optically very faint, the inverse square law shows that they are some of the brightest objects in the universe. For example, a typical quasar has a luminosity of 10^40W which is about 20 trillion Suns!
Quasars sometimes show irregular variation in brightness from hours to months. Their long narrow jets of radiation are formed as gas is drawn into a super massive black hole at the centre of a galaxy. There comes a point where no more gas can fit in the black hole, and so it is ejected in vertical jets.
Here’s a useful short video about quasars: https://youtu.be/qil7bKy1NrQ
Black holes
… Black holes form when stars with cores of > 3 solar masses run out of fuel and explode in a supernova. The core of the star collapses into a tiny radius causing a gravitational field that is so strong that even light cannot escape!
The Schwarzschild radius is the boundary around a black hole (sometimes called the event horizon) within which all matter and information (i.e. light) cannot escape.
This radius can be calculated using the idea of escape velocity:
Potential energy required to ‘lift’ an object to infinitely far away (Ep = 0J) = starting kinetic energy of the object:
GMm/r = 0.5mv²
If v = c (for light) and Rs is the Schwarzschild radius, then this leads to
Rs = 2GM/c²
Note that this expression is only an approximation. We would actually need to use Einstein’s Theory of General Relativity rather than Newton’s Law of Gravitation (which doesn’t work in such intense gravitational fields!)
Detecting Exoplanets
… Planets outside our solar system are known as exoplanets.
Difficulties in detection
… Direct observations of exoplanets are difficult because the light from them tends to be obscured by the much brighter star they orbit, and the planet and star tend to be too close together for the optics to resolve them.
Planets far enough away from the star to be resolved are likely to be very dim as they will not reflect much light.
Radial velocity method
… When a planet orbits a star, the gravitational pull causes the star to “wobble”. This can cause a Doppler shift in the light received from the star. The amount and direction of shift is related to the component of the star’s velocity towards and away from the Earth – ie radially from the Earth’s point of view.
The analysis is similar to that of binary systems.
Transit method
… When a planet passes in front of a star there is very small dimming of its apparent magnitude. Again, the analysis to find the period of rotation is similar to that of eclipsing binary stars.
Direct Observation
The first picture was taken in 2004… 2N1207b is 5x the mass of Jupiter and orbits a brown dwarf. It’s a relatively newly formed solar system and so the planet is HOT and more easily seen using an IR telescope.
Several other planets have been seen in this way, e.g. beta Pictoris, which is 7x mass of Jupiter with a 20 year orbit. Astronomers have actually seen this planet move over the course of a few years!
Circular motion
… A centripetal acceleration is needed for an object to move in a circle. This always acts towards the centre of rotation and is provided by forces such as:
the tension in a string,
the gravitational force (weight) of a mass orbiting a planet,
the electromagnetic force of a charge moving in a magnetic field
… The rotational speed of a body that moves in circular motion is given by:
v = rω
Where r is the radius of curvature and ω is the angular velocity in radians/sec
… The centripetal acceleration required for a particle to move in a circle is:
a = v²/r
or
a = rω²
The centripetal force required to create this acceleration (from F = ma), is:
F = mv²/r
or
F = mrω²
… The following equations are also very useful for circular motion questions:
f = 1/T
ω = 2πf
… Remember that the resultant force on a rotating body NEVER acts outwards (this is sometimes people call this the ‘centrifugal’ force, which is actually an illusion due to Newton’s Third Law of action and reaction).
For example, when you drive around a corner, the contact force of the car on you acts towards the centre of rotation (the ‘centripetal’ force). You might feel as though you are being ‘pushed outwards’ but this is because your mass simply wants to continue in a straight line. Your mass is actually exerting a force on the car outwards (Newton’s third law) – but this is from the car’s point of view!
… You can use “Sum of Forces = ma” to analyse forces for a body describing circular motion (as it is not in equilibrium). Just make sure you define which way is the positive direction.
… Note that the velocity of a body in circular motion is always changing, therefore the acceleration is also always changing and directed towards the centre of rotation.
… Swinging a bucket of water on a string: if you judge it correctly, it is possible to swing it in a vertical circle without getting wet! Also, you may be able to find a minimum speed at which the weight of the bucket and water is just enough to provide the centripetal force required for the circular motion. So no tension will be required in your arm at the vertical position.
… You’ll often need to use the equation given by Gravitational Force = Centripetal force for orbiting bodies.
Simple Harmonic Motion (SHM or SHO for oscillations)
… The definition of SHM:
“An object will describe simple harmonic motion (SHM) if its acceleration is directly proportional to the displacement from an equilibrium position, and directed TOWARDS that equilibrium position.”
a = -kx
… SHM is linked to the x displacement of an object moving in a circle by the equation:
x = Acos(wt)
Differentiating this formula gives velocity and acceleration:
v = -Awsin(wt)
a = -Aw²cos(wt)
… The maximum acceleration and maximum velocity of a body in SHM occur at which displacement positions?
… A pendulum describes SHM only for small angles because sinA ~ A (in radians).
T = 2π√(l/g)
… A mass-spring system moves with SHM with a time period:
T = 2π√(m/k)
… The total energy of a system oscillating with SHM will remain constant as long as there are no external forces such as friction acting. Energy is continually transferred back and forth from potential energy (maximum at x = A) to kinetic energy (maximum at x = 0).
… Acceleration at a certain displacement:
a = -w²x
… Maximum acceleration
a(max) = Aw²
(where A is the amplitude of oscillation)
… Maximum velocity
v(max) = wA
… Calculating speed at any displacement, x:
v = +/- w√(A²-x²)
Forced oscillations and Resonance
… Forced oscillations:
- The oscillating system is forced to vibrate at the same frequency as the driver.
- The amplitude of the oscillating system depends on the frequency of the driver.
- at very low frequencies the amplitude of the driver will be approximately equal to the amplitude of the oscillating system.
… The natural frequency of a body is the frequency at which it will freely oscillate if it is displaced from its equilibrium position and then released. For example, a pendulum will oscillate at its natural frequency in the absence of any driving or damping force.
… Resonance occurs when the driving frequency = natural frequency of the oscillating system. This means that energy is directly transferred from the driver to the oscillating system resulting in a large amplitude.
For example, air in an organ pipe will vibrate at its natural frequency and sound a loud note (large amplitude) when the driver (air flow into the pipe) is at the same frequency.
Note that we can still vibrate oscillating systems (e.g. a mass on a spring) by applying a frequency that is not its natural frequency. However, the resulting maximum amplitude will be smaller than at resonance.
… At resonance the driver and oscillating system have a phase difference of 90°
… The effect of damping will be more noticeable in a pendulum if the pendulum bob is of a lower mass. This is because the frictional forces become more significant compared to the restoring SHM force.
… Increasing the damping on a system causes more energy loss and reduces the amplitude. Increased damping also slightly reduces the resonant frequency.
… Damping can be light, critically or heavily damped (overdamped). Lightly damped systems will have several oscillations before the amplitude returns to zero.
… Critically damped systems return to zero amplitude in the fastest time possible. This is especially useful in the suspension of cars or in moving coil meters where you want to avoid resonance.
… Resonance also occurs usefully in electrical circuits with a capacitor and inductance in parallel – for tuning into radio frequencies.
Standing wave resonance in pipes
… Longitudinal sound waves can create standing waves in pipes due to the reflection and therefore superposition of waves from either a closed end (displacement node) or an open end (displacement antinode).
… At a displacement node, the air is either:
- coming towards the node from either side (causing a maximum positive pressure there), or
- moving away on either side (causing a maximum negative pressure there).
There are therefore pressure antinodes at the sites of the displacement nodes.
… At a displacement antinode, the air in a sounding pipe moves to the maximum extent and the pressure variations are zero.
Electrical circuits
… The definition of resistance is best described using the relationship:
R = V/I
so that:
“resistance is the potential difference required across a component for 1 amp to flow through it.”
… Resistivity (rho, ρ) is a material property. It is best defined by explaining the quantities used in the equation for resistance, R = ρL/A
… The units of resistivity are Ωm (note that this is NOT ohms per metre!)
… The effect of a voltmeter with a low resistance will be to draw current from the circuit. This will reduce the potential difference across the component being measured (V = IR).
For this reason, voltmeters should have a HUGE resistance so that they hardly affect the circuit they are measuring.
… The potential difference (EMF or electromotive force) of a power supply is the chemical energy transferred to electrical energy in each coulomb of charge that flows from the supply.
This potential difference MUST be dropped across the resistances somewhere in the circuit, transferring electrical energy into thermal energy and other energy types.
If there is no resistance in the circuit (a ‘short circuit’), then a VERY large current will flow and the potential difference of the supply will be dropped across the internal resistance of the supply and the very small resistances of the wires. This will cause significant heating (and possibly a fire!).
How to analyse a circuit network (to find currents and potential differences across components) using Kirchhoff’s Laws:
Σ currents into a junction = Σ currents out.
Σ supply EMFs = Σ IR across the components in any series loop that you select.
The first law (law of conservation of charge) means that you can consider just ONE series loop and ignore any other parallel branches to create an equation which you can solve.
… Deriving the formula for resistances in series
1) Conservation of charge means that the current through R1 must be equal to the current through R2.
2) The potential differences across each resistor is then IR1 and IR2
3) Conservation of energy (J/C) means that the potential differences across each resistor must add to the total potential difference across the equivalent total resistance Rt. Therefore:
IR1 + IR2 = IRt
R1 + R2 = Rt
–
Internal resistance and EMF
Vt = EMF – I×(R internal)
… When a power supply has an open circuit, then no current will flow. This means that no voltage will be dropped across the power supply’s internal resistance (as V = IR) and so the output terminal voltage of the supply will equal its EMF:
Vt = EMF – I×(R internal)
If I = 0A
Vt = EMF
… For a power supply of EMF E, and internal resistance r, which supplies a current to a load resistor R:
The EMF of the cell is shared across r and R in the ratio of their resistances (as they form a potential divider).
This means that if R is increased, then the potential difference across R will INCREASE. However, the current flowing in the circuit will decrease.
… Potential divider circuits can act as ‘variable voltage supplies’ or for sensing circuits involving components such as thermistors or light dependent resistors.
The voltage output of a potential divider can be calculated by considering that the supply voltage is shared between the two resistances by the ratio of their resistance values. For example:
V = Vs x R1/(R1+R2) would give the potential difference output across resistance R1.
(Note, however, that the resistance of the switching circuit connected across one of the resistors will affect the voltage output value because its resistance will combine in parallel).
… A potentiometer is often a more versatile ‘voltage selector’ than a fixed resistor / variable resistor potential divider circuit because it can select the full range of voltages from 0V up to the supply voltage.
However, if you add a component from the sliding contact on the potentiometer then it’s resistance will combine in parallel and could significantly change the selected voltage.
… If a component is ‘short circuited’ by a wire connected across it, then current will not flow through the component anymore (as there is a much easier route around it).
… The current flowing through a component will determine the power developed in the component. This means that a higher current flowing through a lamp will make it shine brighter. P = I²R (derived from P=IV and V = IR)
… To explain how a circuit is affected by a change, try considering this ‘story’:
How is the Resistance of the circuit changed?
What effect does this have on the current flowing in the circuit?
How does this change in current effect the potential difference across a component?
… The maximum power theory: the maximum power transfer possible in a circuit is when the resistance of the load equals the internal resistance of the power supply.
A useful application of this theory is when ‘matching’ an amplifier to a loudspeaker. The internal resistance of the amplifier should be about the same as the loud speaker to ensure the maximum power is transferred for the greatest range of volume.
Note that the efficiency (useful Pout / total Pin) increases as the value of the load resistor, R, increases. A smaller and smaller current flowing through the internal resistance, r, means less power is wasted across it.
However, a circuit with hardly any current flowing is probably not going to be very useful!
… Alternating current and voltages.
Root-mean-square (rms) voltage and current is the equivalent dc voltage or current that will deliver the same power to a component as the alternating source. Note that components such as bulbs are often rated using rms values rather than peak alternating current or voltage values.
… V rms = Vpeak / √2
For example, in mains electricity V rms = 320V / √2 = 230V
Similarly, the root mean square current:
I rms =. Ipeak / √2
… To calculate power losses and efficiency, for example over the electricity distribution national grid, first calculate the current flowing in the wires. Then use P = I²R to calculate the power losses.
Note that P = V²/R will NOT work in this situation because the transmission voltage is not the voltage being dropped across the transmission wires (it is being dropped across the town receiving the power).
Superconductivity
… If you reduce the temperature of a conductor to near absolute zero (-273C or 0K) then that material’s electrical resistance disappears – it becomes a superconductor. This means that the material’s resisti vity has dropped to zero and so an electric current can flow through it without transferring any energy (i.e. the current will not heat the conductor).
… The temperature at which the conductor becom a superconductor is known as the Critical Temperature, Tc. For metals, the critical temperature is close to absolute zero (1-4K). More advanced ceramic superconductors may have much higher critical temperatures, for example, -70C.
… Superconductors have important uses for carrying electrical power without losses and for generating huge magnetic fields that can be used to levitate trains and operative MRI scanners.
Capacitors
… Capacitors store charge.
… By applying a potential difference across a capacitor, charge will flow (momentarily) onto one plate of the capacitor… And off the other plate, leaving them with equal and opposite amounts of charge.
… A resistor limits the current that can flow on or off a capacitor to safe levels (without the resistor, the high initial charging or discharging current could damage the capacitor or circuit).
… Capacitors are made of a pair of conducting plates separated by an insulator (called a dielectric, for example, air, oil or paper)
… Capacitance, C, is measured in farads, F. A large (paper) dielectric capacitor may only be 10 microFarads! Paper capacitors have a frequency stability of up to 1 MHz.
… Discharging a capacitor:
Q = Qo( e^(-t/RC) )
Qo is the initial charge on the capacitor and can be calculated using Qo = CVo, where Vo is the capacitor’s initial potential difference.
As Q/C = V, dividing both sides by C gives:
V = Vo( e^(-t/RC) )
It’s also worth noting that the current flowing through the charging capacitor is:
I = Io( e^(-t/RC) )
… Charging a capacitor:
Q = Qo(1 – e^(-t/RC) ).
V = Vo(1 – e^(-t/RC) )
and
I = -Io( e^(-t/RC) ) … The same as charging, but in the opposite direction!
… The constant RC is called the time constant of the circuit and is the time taken (in seconds) for the charge on the capacitor to fall to 1/e (36.8%) of its initial value, when discharging.
When a capacitor is charging, RC is the time taken for Q or V to rise to (1-1/e) of its final value, i.e. when Q = 0.63Qo
… 5RC is the time taken for a capacitor to discharge to less than 1% of its initial charge (practically zero), so this can be used to estimate the time for the capacitor to ‘fully’ discharge.
.. The time for charge (or potential difference) on the capacitor to halve:
T½ = 0.69RC
This can be derived by substituting Q = Qo/2 into
Q = Qo e^(-t/RC)
or
V = Vo e^(-t/RC)
… If the initial discharge current, Io, is maintained at a constant rate (this could be done by reducing the resistance in the circuit at a constant rate), then the time taken for the capacitor to fully discharge is given by t = RC.
This can be proven as follows:
Total charge on capacitor
Qo = CVo (from Q = CV)
Total charge that has flowed due to a constant current Io
Qo = Io t (from Q = It)
These charge values are the same, so:
CVo = Io t
t = CVo/Io
and as R = V/I, we have:
t = CR
… The energy stored on a capacitor is given by E = 0.5QV = 0.5CV².
This is HALF of the energy supplied by battery during charging (which is E = QV).
The other half of the energy supplied by the battery is lost as heat in the resistance of the wires (or across the resistor in the circuit).
… The insulating material between the capacitor plates is called the ‘dielectric’. So for example, you could make a capacitor using two sheets of aluminium foil rolled up with a sheet of grease proof paper between them (kitchen physics!).
… However, some dielectrics are better than others. A very thin dielectric creates a small gap between the plates and so increases the electric field (E = V/d). This pulls more charge onto the capacitor, effectively increasing the capacitance, as C = Q/V.
Even better, you can use a thin POLAR dielectric. This has polar molecules which rotate to line up with the electric field, DEcreasing the electric field strength (because the molecules’ electric fields opposes the field across the plates), and so reducing the overall potential difference across the plates.
As C = Q/V, this means that the capacitance must have increased (we can store more charge for a given potential difference).
… The relative permittivity, εr, of the dielectric tells us how effective it is at establishing an electric field within the capacitor.
εr = Q/Qo
where Q = charge on the capacitor when using a particular material as the dielectric,
and Qo = charge on the capacitor with an vacuum (or air) dielectric (i.e. just a gap)
… Capacitance is related to Area, A, and plate separation, d, by the equation:
C = (A εo εr)/d
Where
εo is the permittivity of free space
and
εr is the relative permittivity of the dielectric material between the plates.
… You may see an experiment where a capacitor is repeatedly charged and discharged via a vibrating reed switch. The relative permittivity of a dielectric material can be found from
εr = I/Io
where I = current flowing from the capacitor when using the material as the dielectric,
and Io = charge on the capacitor which uses just a vacuum (or air) as the dielectric.
… The total capacitance of capacitors in SERIES is given by:
Ctot = (C1C2)/(C1 + C2).
(Note this is like resistors in parallel!)
To prove this, add together the potential differences on each capacitor due to the same charge Q flowing onto each capacitor (V = Q/C).
… The total capacitance of capacitors in PARALLEL is given by:
Ctot = C1 + C2
(Like resistors in series!)
Two similar capacitors in parallel effectively double their surface area – this is why we add their capacitances together.
To prove this formula, add together the charge which flows onto each capacitor due to the same potential difference applied across them (Q = CV).
… Capacitor networks can be analysed by considering the charge that flows onto each capacitor, or onto combined capacitances using Q = CV and the summing formulae.
It is often useful to combine all the capacitances together in a circuit so that you can first work out the total charge that has flowed from the power supply.
Then, capacitors in series will gain the total charge charge and capacitor’s in parallel will split the total charge in ratio of their capacitances.
… Electrolytic capacitors use aluminium borate soaked into the paper. This produces a thin film of aluminium oxide which acts as the dielectric of the capacitor. Electrolytic capacitors can have much higher capacitances (because of the very thin dielectric), but have a poor frequency stability (<10kHz) and must be connected the correct way round (the anode is usually marked).
The Photoelectric Effect and the Ultraviolet Catastrophe… The birth of Quantum Physics
… In the 19th century, classical physics was used to predict the frequency of light emitted by ‘black bodies’ (perfect absorber and emitters of radiation). However, this theory predicted that hot objects would emit INFINITELY HIGH frequencies of radiation – obviously not the case in reality! This problem was called ‘The Ultraviolet Catastrophe’.
… To solve the Ultraviolet Catastrophe, Max Planck in 1900 suggested that radiation energy is always emitted in packets (quanta) of energy. This was the birth of Quantum Physics…
Here’s a useful video about the ‘Ultraviolet Catastrophe’: https://youtu.be/FXfrncRey-4
… The wave theory of light was insufficient to explain the photoelectric effect because it suggested that the energy of a light wave depends on the wave amplitude (intensity). Experiments found that this was not true!
… Einstein took Max Planck’s energy quantum idea and applied it to the photoelectric effect. He proposed that only 1 photon could be absorbed by 1 orbiting electron.
… Einstein’s energy of a photon is given by E = hf, where h is Planck’s Constant (6.63×10^-34)
… If photons of light with a high enough FREQUENCY are absorbed by electrons in a metal, those electrons may be ejected from the surface as ‘photoelectrons’ almost instantaneously.
… The photoelectrons are emitted with a range of kinetic energies from zero to some maximum value. This variation is due to some electrons having to escape from below the surface of the metal.
… The maximum value of KE increases with the frequency of radiation but is unaffected by the INTENSITY of the radiation.
… We can use a vacuum photocell to investigate the relationship between the maximum energy of the photoelectrons and the frequency of incident light.
To find the maximum KE of electrons, a potential difference V can be applied between the metal being illuminated (the cathode) and a collector plate (the anode).
The value of V is varied until the most energetic electrons do not reach the collector (i.e. the current in the circuit falls to zero).
The value of V when this happens is called the ‘stopping potential’, Vs.
The maximum KE of the electrons is then given by:
Ek(max) = ½m(vmax)² = e(Vs)
This is because the work done (energy transferred) on a charge e moving through a potential difference V is: E = eV
We can therefore plot Ek(max) vs frequency and find the gradient, h and the y-axis intercept, -Φ:
y = mx + c
Ek(max) = hf – Φ
… So the stopping potential is the minimum potential difference needed to stop photoelectrons from reaching the anode (which is connected to the negative of the supply), i.e. to reduce the current in the circuit to zero.
… For a given frequency, the maximum KE does not depend on the intensity of the light source.
… The threshold frequency, fo, is the frequency at which photoelectrons only just start to be liberated. Technically they have zero kinetic energy. Below the threshold frequency, NO photoemission takes place, no matter how intense the light is.
… Energy of the photon = minimum energy required to escape from the surface (called the work function) + left over kinetic energy
hf = Φ + ½mv²
… The work function, Φ, of a metal can be calculated when the maximum KE of a photoelectrons = 0 (i.e. it only just escapes), so:
hfo = Φ, where fo is the threshold frequency at which the photoelectric effect just begins.
… We define ionization energy when we have a single atom. On the other hand, the work function is the energy required to remove the outermost shell electrons of a metal’s surface atoms when they are involved in metallic bonding. The work function is the energy needed to take away METALLIC BONDED electrons.
… If the frequency of light > fo, then the number of electrons emitted per second is proportional to the intensity of light.
… Fluorescent tubes involve two excitation processes – one for the mercury vapour and the other for the fluorescent coating.
… Here is a video of the photoelectric being demonstrated: https://youtu.be/v-1zjdUTu0o or https://youtu.be/puT36rd9dkQ
Millikan’s Oil Drop Experiment
Millikan observed oil drops falling through air between two metal plates using a microscope and lamp arrangement. He was able to measure a particular droplet’s terminal velocity and then apply a potential difference across the plates to stop the oil drop from falling – i.e. levitate it in the uniform electric field.
1) Electric field strength E = V/d where V = potential difference across plates and d = distance between plates. E is measured in N/C.
2) The upwards force on an oil drop, F = EQ = VQ/d (from step 1).
3) When the drop is suspended in equilibrium, the upwards force = weight force:
VQ/d = mg
So the charge on an oil drop:
Q = mgd/V
4) When a drop is falling at its terminal velocity v, the drag force = weight:
Stoke’s law states that the drag force on a falling sphere through a fluid, F = 6πηrv
so
6πηrv = mg where η is the viscosity of air and r = radius of the oil drop.
5) Now, as m = (4/3)πr³ρ (volume x density)
we can write:
6πηrv = (4/3)πr³ρg
so
r² = 9ηv/2ρg
6) Millikan was able to measure the terminal velocity, v, of the oil drop.
Now ‘rewind’ these calculations…
From the value of v, he could calculate the radius, r, of the drop.
Using r, he could calculate the mass, m of the drop
From the value of m he could calculate the charge, Q on the drop.
Phew!
7) Millikan carried out this experiment many, many times to give multiple values of Q. He then analysed these values and found that they all had a common factor of 1.6 x10^-19. He concluded that this is an ‘elementary charge’ which cannot be subdivided… i.e. the charge on an electron. (He won a Nobel prize for this work).
Wave-particle duality
… The de Broglie wavelength means that every particle can behave like a wave with a wavelength = h/momentum. The faster the particle (or the more massive it is) then the smaller is wavelength will be.
… Electron diffraction can be used to calculate atomic nucleus diameter. Note that electrons behave as waves and will diffract around and atomic nucleus if their wavelength is similar to the nucleus size.
… The particle nature of light is demonstrated by the photoelectric effect, because if a light source is below the threshold frequency for a particular metal, then increasing the intensity of the light does NOT cause photoelectrons to be emitted.
The wave theory of light was therefore insufficient to explain the photoelectric effect because it suggested that the energy of a light wave depends on the wave amplitude (intensity).
… The reason that you don’t diffract (visibly) when you walk through a doorway is that your de Broglie wavelength is incredibly small compared to the gap size!
However, protons and other subatomic particles can be made to diffract.
… What experiments and observations demonstrate that light can be thought of as:
- a wave?
- a particle?
… Electron microscopes rely on the wave nature of electrons to see much finer detail than is possible using a light microscope. It is defraction that limits our ability to see fine details. Electrons diffract very little because they have a very short wavelength
… Cathode rays (electron beams) are formed by a process called thermionic emission. A metal filament is heated with a current. Electrons are excited to higher energy levels, escaping from their orbitals. Using an accelerating anode, this forms the basis of an electron gun which is how oscilloscopes and (old) TVs work.
The cathode ray oscilloscope
… Formation of an electron beam by ‘boiling off’ electrons from a hot filament.
… These electrons are accelerated and focused into a beam by an anode ring.
… The y plates of the oscilloscope are connected to the voltage source that is under investigation, e.g. a microphone output.
… The x-plates of the oscilloscope control the ‘time base’ of the beam – how fast it sweeps left to right to form the picture.
… Check out this link which is an animation of the process: http://youtu.be/6BgcrRjqnKE
… The time base is often measured in units of ms/cm so the scanning ‘spot’ takes this time to move one horizontal centimetre square across the screen.
… The y-sensitivity, or y-gain tells us how much the spot will move up or down in units of V/cm. So if the y-gain is set to 5mV/cm, the spot will move 2cm if we apply a potential difference of 10mV.
… The units of time base and y-sensitivity can be a great help in calculations, for example:
5mV/cm x 3cm = 15mV
… For an alternating signal, we can measure the peak voltage (from the centre line), or the peak-to-peak voltage (bottom to top).
… When you need to convert units such as ms into s, replace the ‘prefix’ with its power of 10 notation. So for 6ms, we would write 6×10^-3 s
… A useful equation for oscilloscopes is f = 1/T where f is the frequency and T is the time period of the wave (time for one complete oscillation).
Kinetic Theory of Gases
… The relative atomic mass (Ar) of an element is the mass of the atom relative to an atom of carbon-12 (which is defined as having a mass of exactly 12)
… The relative molecular mass (Mr) is the sum of all the relative atomic masses of each atom in a molecule.
… Avogadro’s constant, Na, is the number of particles in 1 mole. For example, 1 mole of oxygen has 6.02×10^23 oxygen molecules.
N/Na = n
Number of particles / Avogadro’s constant = number of moles
… The molar mass is the same as the relative atomic or molecular mass. It tells us the mass (in GRAMS) of 1 mole of the substance.
… So if the molar mass of oxygen is 32g (i.e. the mass of 1 mole of O2 molecules), then:
The mass of 1 molecule = 32/Na = 32 / 6.02×10^23 = 5.3×10^-23 g
Please note that we have calculated mass in GRAMS here! You may need to convert this into kilograms for subsequent calculations.
… 1 mole of gas atoms (or molecules) always occupy a volume of 24dm3 at room temperature and pressure (rtp).
… The empirical (experimental) gas laws can be combined into one single equation:
pV/T = constant (for a constant mass of gas). This is found via:
Boyle’s Law (pV = constant at a fixed temperature)
The Pressure Law (p/T = constant at a fixed volume).
Charles’ Law (V/T = constant at a fixed pressure)
The derivation is straight forward using algebra:
https://en.m.wikipedia.org/wiki/Combined_gas_law
… The ideal gas equation pV = nRT agrees closely with the empirical gas laws (pV/T = constant).
… Assumptions made for an ideal gas – it’s worth having 3 or 4 of these memorised. (For example, ideal gas particles collide perfectly elastically with the walls of the container, particles move randomly and rapidly, the volume of the particles is negligible compared to that of the container…)
… Ideal gas laws: pV = nRT and pV = NkT where k is the Boltzmann constant and R is the molar gas constant.
… k = R / Na (Na = Avogadro’s constant)
… p1V1/T1 = p2V2/T2 for questions involving a change from one situation to another situation. This is true as long as the number of gas molecules remains constant – i.e. the system doesn’t gain or lose any.
… Gas law questions are fairly straightforward if you List your starting data (take care that your temperature is converted to Kelvin!)
… Derivation of pV = (1/3)Nm, the kinetic theory model using a step-by-step derivation starting with a molecule bouncing off a wall with a velocity v:
- change of momentum
Δp = 2mv - time per collision
t = 2L/v (for a container of length L) - average force
F = Δp/t = 2mv/(2L/v) = mv²/L - average pressure
p = F/A = mv²/L³ = mv²/V
(where V is the volume of the container) - extending the formula to N particles:
p = Nmv²/V - and then factoring in three dimensions so that v² = /3 and we get pV = (1/3)Nm
( is called the mean square speed and is a squared average of all the particles’ speeds in all three directions)
… If you have lots of gas molecules moving in the x direction with different velocities u1, u2, u3…, then the mean square velocity can be calculated:
= [ (u1)² + (u2)² +(u3)² + … ] / N
… The mean square speed, , is a type of average for all three dimensions:
= + +
This is Pythagoras in 3D, where , and are the mean square velocities of the molecules in the x, y and z directions respectively.
… As the gas molecules are moving rapidly and randomly in all directions, the the mean square components will be equal: = = , and so:
= 3
and
/3 =
… So now we sub v² = = /3 into:
p = Nmv²/V
to get:
pV = (1/3)Nm
… Note that ½m = (3/2)kT gives the average kinetic energy of one particle of the gas. In other words, the temperature in kelvin, T is proportional to the mean square speed of the particles in the gas (if m is constant).
… From the two equations, pV = nRT and pV = NkT, we see that:
nR = Nk
k = nR/N
Now the Avogadro constant N / Na = n, so we then get:
k = (N/Na)(R/N) = R/Na
Therefore from ½m = (3/2)kT, we also get:
½m = 3RT/2Na
… The ROOT mean square (rms) speed √() is a more meaningful average of a molecule’s speed. The mean square speed is either denoted as or c² with a bar over it.
Root mean square (rms) speed c = √()
Note that the square root and the square do not cancel each other out because is a separate function (the sum of the squares of the molecule velocities in x, y and z components).
… Note that when finding the mass of n moles of a gas (using m = molar mass x n) that you will get an answer in GRAMS! Remember to convert to kilograms when using the ideal gas equation :).
… The pressure at a depth h in a fluid (gas or liquid): p = rho x g x h
The Boltzmann Constant
The average energy, E, of a particle at a certain temperature can be approximated by E = kT
The Boltzmann Factor
… Particles in a gas lose and gain energy at random due to collisions with each other. On average, over a large number of particles, the proportion of particles which have AT LEAST a certain amount of energy E is constant. This is known as the Boltzmann factor and it has a value between 0 and 1. The Boltzmann factor is given by the formula:
N/No = e^(-E/kt)
where:
- N is the number of particles with kinetic energy above an energy level E
- No is the total number of particles in the gas
- T is the temperature of the gas (in kelvin)
- k is the Boltzmann constant (1.38 x 10−23 JK−1).
This energy E could be any sort of energy that a particle can have. For example it could be gravitational potential energy, kinetic energy, or electromagnetic energy.
… E is often referred to as the ‘activation energy’ for a process to occur, for example, for a liquid particle to escape as a gas particle.
Materials
… Tensile forces acting on an object tend to stretch the object, putting the material under tensile stress.
… Stress = force / cross-sectional area, or σ = F/A
The units of stress are…?
… Stress measures the load on a material and removes the effect of geometry because it divides by the cross sectional area.
… Tensile Strain = extension / original length. This is a measure of the fractional increase in length due to applying a tensile force.
ε = Δl / l
Δl = extension
l = original length
Strain therefore has no units!
… Strain gauge calculations involve the equations:
R = ρL/A
ε = Δl / l
V = LA (volume of a strain gauge element is assumed to remain constant as it stretches or compresses).
Using ratios in these calculations are often the easiest way to find the effect of a percentage strain. For example if strain = 1%, then
A = V/L
A(new) = V/1.01L = 0.99V/L … so Area decreases by 1%.
R = ρL/A
R(new) = ρ(1.01L)/(0.99A) = 1.02ρL/A … so Resistance increases by 2%.
… The Young Modulus of a material is a measure of its stiffness (how difficult it is to change the size or shape of the material)
Young’s Modulus = tensile stress / tensile strain
E = σ / ε
or
E = (F l) / (A Δl)
… The units of Young’s Modulus (like stress) are pascals or N/m²
… For reference, Young’s modulus for steel is about 200 × 10^9 Pa
Aluminium is about 71 × 10^9 Pa
… The force acting on some materials is proportional to the extension:
F = kx
This is known as Hooke’s Law, where k is the spring constant (a measure of the stiffness of the spring or material).
For example, the extension of a wire or spring is proportional to the applied load, but only up to some maximum load called the “limit of proportionality”.
… Materials which show ELASTIC behaviour return to their original length when the tensile force is removed.
… Elastic Limit: this is the maximum load which a body can experience and still retain it’s original size and shape when the load is removed – i.e. no permanent deformation.
The elastic limit sometimes coincides with the limit of proportionality.
… Yield Point: stress is increased beyond the elastic limit, a point is reached at which “there is a marked increase in extension for a small increase in stress” (key definition). This is the yield point.
The internal structure of the material has changed at this point and crystal planes have slid across each other – the material is said to be showing PLASTIC behaviour. Few materials show a yield point, but mild steel is one that does.
… Breaking Stress: this is also called the “Ultimate Tensile Stress” and is the maximum stress which can be applied to a material without it breaking.
… Useful example of a stress – strain curve:
https://images.app.goo.gl/AMwoU7Y3NsJwENU19
… Ductile materials can be permanently stretched.
… The term ‘Brittle’ describes the process of fracture. Brittle materials fail by crack propagation due to ‘stress concentration at the crack tip’.
A brittle material cannot be permanently stretched – it breaks soon after the elastic limit has been reached. Brittle materials are often very strong in compression (like concrete).
… The energy stored in an elastic material stretched by a force F causing an extension Δl is:
Energy = ½FΔl = ½kΔl²
This is also given by the area under a force – extension graph.
… Tough materials (opposite of brittle materials) can absorb a lot of energy before breaking. These materials will often be able to deform plastically (e.g. wet chewing gum), however, elastic polymers such as rubber can also absorb large amounts of energy without breaking.
The area under a stress-strain curve represents the energy stored per unit volume of the material, which is a measure of the material’s toughness.
… Some materials such as rubber exhibit ‘hysteresis’. This means that their loading and unloading curves of stress vs. strain are different (and therefore not linear).
These materials do not obey Hooke’s law but they can be elastic (return to their original shape).
On each cycle of loading and unloading, energy is transferred internally to thermal energy due to the ‘friction’ acting between the molecules.
A common example of a material which exhibits hysteresis is rubber.
… Crystalline materials have particles which are arranged in a highly ordered microscopic structure, forming a regular crystal lattice that extends in all directions.
… Amorphous materials (or non-crystalline solids) are solids that lack the long-range order that is characteristic of a crystal.
… Polycrystalline materials are solids that are composed of many crystal grains of varying size and orientation. Under a microscope, grain boundaries can be seen.
… Dislocations are areas were the atoms are out of position in a crystal structure. Dislocations are generated and move when a stress is applied and their motion allows plastic deformation to occur.
Unlike metals, ceramics cannot break and re-form bonds and so dislocations cannot move. Therefore ceramic materials are brittle and do not deform plastically.
Electric Fields, Coulomb’s Law and Electric Potential Energy
… In a uniform electric field, the lines of equipotential are evenly spaced (potential varies linearly with distance from a plate)
… The electric field between two parallel plates is uniform and any charges moving through the region will move in a parabolic path.
… Force on a charge within a uniform electric field: F = EQ
… Electric field strength, E (measured in N/C or V/m) = V/d
… In a radial field between a positive charge Q and a positive ‘test’ charge q:
Force, F = kQq/r² (N)
Electric field strength, E = kQ/r² (N/C)
Electrical potential energy, Ep = kQq/r (J)
Electrical potential (‘voltage’), V = kQ/r (J/C)
where k = 1/4πεo = 8.99×10^9, and εo is the permittivity of free space (8.85×10^-12)
Note that:
F = Eq is the link between the force equation and the field strength equation.
E = Vq is the link between potential and potential energy.
… The permittivity of free space is a number which allows us to describe how easily (or how difficult) it is for electric lines of force to pass through air, water or any other medium.
It’s called permittivity because of how much a given substance “permits” electric, (or magnetic in the case of magnetism ) field lines to pass through them.
… The potential difference (voltage) between two points at distances r1 and r2 in a radial field,
ΔV = kQ(1/r2 – 1/r1)
… dE/dr = -F and dV/dr = -E.
The reason for the minus sign is as follows:
Suppose a positive test charge, q, is moved a distance, dr, away from another positive charge, Q, in an electric field strength, E.
Then the work done in moving this charge, dW is given by:
dW = -qE(dr) because F = EQ and W = Fd. The negative sign shows us that electrical potential energy is being lost by the test charge.
so
dW/q = -E(dr)
and so
dV = -E(dr) because potential, V = work done per coulomb.
this means that
dV/dr = -E
… When finding the resultant force on a test charge due to two or more point charges:
do draw a large, neat diagram showing the forces acting on the test charge.
Then resolve each force into two perpendicular components.
Add up the corresponding components.
If required, form a vector triangle (Pythagoras) to work out the resultant force.
… Positive and negative fields can get confusing when they are combined… So try to keep things simple by drawing a large, clear vector diagram of the fields showing the ACTUAL direction of the force experienced due to each charge source.
To do this, think about the direction in which a positive charge will experience a force.
You can then effectively ignore any negative values that come out of the field equations.
Radioactivity
… The intensity of Gamma radiation is inversely proportional to the distance from the source in air or a vacuum (intensity is measured in W/m²):
I = k / d²
This is also true for alpha and beta radiation, but only if in a vacuum (because alpha and beta ionise air molecules strongly)
Ionising effect of radiation:
… Alpha particles are highly ionising, causing about 10^4 to 10^5 ion pairs / cm in air.
… Beta particles are less intensely ionising at about 10^3 ion pairs / cm in air.
… Gamma rays are weakly ionising at about 10 ion pairs / cm in air.
Radiation experiments:
… Always measure the background count before and after the experiment so:
the count rate measurements can be corrected
Any contamination from the experiment can be detected.
… Geiger counters are very good at detecting beta radiation, however they are less efficient at counting alpha or gamma radiation. This means that they will significantly underestimate the activity of alpha or gamma sources.
… Half life calculation options:
1) take a mean from about 3 measurements of half life from a graph of count rate (or activity) vs time
2) plot ln(count rate) vs time. The resulting straight line will have a gradient of -λ which can then be used to find half life via Half life = ln2/λ
… The decay constant is the probability that a nucleus will decay in one second. The units of λ can be /s , /h or /year
… Half life = (ln2) / λ
… The activity of a sample is measured in becquerels, Bq (decays per second).
A = Ao e^(-λt) where Ao is the initial activity at t=0.
… Activity is proportional to the number of undecayed nuclei remaining:
A = λN
… The number of undecayed nuclei remaining after time t:
N = No e^(-λt). Where No is the number of undecayed nuclei at t=0.
… The measured count rate of a sample after a time t (e.g. via a Geiger-Muller tube and counter):
C = Co e^(-λt) Where Co is the initial count rate at t=0.
This is because the count rate is proportional to the activity of the source.
… Radiocarbon dating relies on finding the ratio of carbon-14 (which is radioactive) to carbon-12 in an organic sample. Living things take in both types of carbon, but when they die, the carbon-14 gradually decays.
If we assume that the ratio of carbon-12 : carbon-14 in the atmosphere has been constant for many thousands of years (not quite true) then we can estimate the age of the sample by using N = No x e^(-decay constant x t).
… Nuclear decay ratios question, for example:
Potassium-40 decays to argon-40 and also decays to calcium-40 via a different decay process which is 8 times more likely to happen.
If a rock is found to contain a ratio of 1:5 atoms of argon for every potassium, then how old is the rock?
The final ratio of elements in the rock must be
Ar : K : Ca
1 : 5 : 8
Because for every decay to Ar, we would expect 8 decays to Ca (on average)
The 5 potassium nuclei have not yet decayed.
This means that originally at t = 0 there would have been 1 + 5 + 8 = 14 nuclei ‘parts’ of Potassium-40, which decay after time, t, to 5 nuclei ‘parts’.
If we know the decay constant, then we can find t using the equation:
N = No e^(-λt)
Where N = 5 and No = 14
Einstein’s Theory of Special Relativity
Einstein postulated that:
- There is no absolute space (frame of reference)
- There is no absolute time (i.e. there is no universal clock that keeps track of everyone’s time)
However, recent experiments such as the Michelson Morley experiment had supported the idea that the speed of light is an absolute value for any observer.
Einstein put these postulates together in a ‘thought experiment’ and using the idea of a ‘light clock’, formed the Time Dilation equation in which the faster a train travels, the slower a clock on the train will run compared to a clock who is watching the train pass on a platform.
The length of the train as measured by the stationary observer on the platform will also seem to contract.
Boost your GCSE physics at