Contents: 10 sections
All three subtopics here are printed under "A Level subject content" in the 9702 syllabus, and every objective carries the tier "A Level". None of it is AS. It is examined on Paper 4. Paper 1 is the AS multiple-choice paper, and the whole 9702 bank on this site comes from Paper 1, so no practice is tagged to this topic.
Syllabus points
25.1 Standard candles
- Understand the term luminosity as the total power of radiation emitted by a star.
- Recall and use the inverse square law for radiant flux intensity F in terms of the luminosity L of the source, F = L/(4πd²).
- Understand that an object of known luminosity is called a standard candle.
- Understand the use of standard candles to determine distances to galaxies.
25.2 Stellar radii
- Recall and use Wien's displacement law λ_max ∝ 1/T to estimate the peak surface temperature of a star.
- Use the Stefan-Boltzmann law L = 4πσr²T⁴.
- Use Wien's displacement law and the Stefan-Boltzmann law to estimate the radius of a star.
25.3 Hubble's law and the Big Bang theory
- Understand that the lines in the emission and absorption spectra from distant objects show an increase in wavelength from their known values.
- Use Δλ/λ ≈ Δf/f ≈ v/c for the redshift of electromagnetic radiation from a source moving relative to an observer.
- Explain why redshift leads to the idea that the Universe is expanding.
- Recall and use Hubble's law v ≈ H₀d and explain how this leads to the Big Bang theory (candidates will only be required to use SI units).
Luminosity and radiant flux intensity
Luminosity L is the total power of radiation emitted by a star, in watts. It is a property of the star itself and does not depend on how far away you are.
Radiant flux intensity F is the power received per unit area at a distance d, in W m⁻². It does depend on distance, and it is what a telescope actually measures.
The two are related by the inverse square law:
F = L / (4πd²)
The 4πd² is the surface area of a sphere of radius d. The star's power spreads out over that sphere, so the same power is shared over a larger area the further you go, and the intensity falls as 1/d².
Keep the two quantities apart. A dim star nearby and a brilliant star far away can give the same flux intensity at Earth, which is precisely the problem astronomy has to solve.
Worked example. A star of luminosity 3.8 × 10²⁶ W is 4.0 × 10¹⁶ m away. Find the flux intensity at Earth.
The denominator first:
4 × 3.142 × (4.0 × 10¹⁶)² = 2.011 × 10³⁴
F = 3.8 × 10²⁶ / (2.011 × 10³⁴) = 1.890 × 10⁻⁸
so about 1.9 × 10⁻⁸ W m⁻².
Standard candles
A standard candle is an object of known luminosity.
That is the whole idea, and it solves the problem above. If you know L from the nature of the object, and you measure F with a telescope, then rearranging the inverse square law gives the distance:
d = √(L / (4πF))
So a standard candle turns a brightness measurement into a distance measurement, which is otherwise very hard to make beyond the nearest stars.
The most useful example is a type Ia supernova. These occur when a white dwarf in a binary system accretes matter until it reaches a fixed critical mass and detonates. Because the mass at which it explodes is always the same, the peak luminosity is always the same, so measuring the peak flux intensity gives the distance. They are also extremely bright, so they can be seen in very distant galaxies. Cepheid variable stars serve the same purpose closer to home.
Standard candles are what made Hubble's law measurable, since it needs an independent distance to each galaxy.
Worked example. A type Ia supernova of luminosity 1.0 × 10³⁶ W gives a flux intensity of 2.0 × 10⁻¹⁴ W m⁻² at Earth.
d² = 1.0 × 10³⁶ / (4 × 3.142 × 2.0 × 10⁻¹⁴) = 3.978 × 10⁴⁸
Taking the square root, d = 2.0 × 10²⁴ m.
Wien's displacement law
A star radiates approximately as a black body, with a continuous spectrum whose intensity peaks at a particular wavelength. Wien's displacement law says that peak wavelength is inversely proportional to the surface temperature:
λ_max ∝ 1/T
or, with the constant,
λ_max T = 2.9 × 10⁻³ m K
Hotter stars therefore peak at shorter wavelengths, which is why hot stars look blue and cool stars look red. That is the opposite of the everyday association of red with hot, and it catches people out.
Note that T here is the thermodynamic temperature in kelvin, as always.
Worked example. The Sun's spectrum peaks at 500 nm.
T = 2.9 × 10⁻³ / (500 × 10⁻⁹) = 5800
so the surface temperature is about 5800 K.
The Stefan-Boltzmann law
The luminosity of a star depends on its surface area and the fourth power of its surface temperature:
L = 4πσr²T⁴
where r is the radius of the star and σ is the Stefan-Boltzmann constant, 5.67 × 10⁻⁸ W m⁻² K⁻⁴. The 4πr² is the surface area of the star.
The fourth power is the part to respect. Doubling the surface temperature multiplies the luminosity by 16. That is why a small difference in temperature between two stars of the same size gives a very large difference in their luminosities.
Estimating the radius of a star
The two laws combine into a method, and the method is the point of the subtopic:
- Measure the peak wavelength of the star's spectrum and use Wien's law to find its surface temperature T.
- Measure the flux intensity F at Earth, and use a standard candle or another distance method to find d, then use F = L/(4πd²) to find the luminosity L.
- Substitute L and T into the Stefan-Boltzmann law and rearrange for r.
Worked example. A star has a peak wavelength of 290 nm and a luminosity of 4.0 × 10²⁸ W.
Temperature from Wien's law:
T = 2.9 × 10⁻³ / (290 × 10⁻⁹) = 10000
so T is 1.0 × 10⁴ K.
Now rearrange the Stefan-Boltzmann law for r². The denominator:
4 × 3.142 × 5.67 × 10⁻⁸ × 10000⁴ = 7.126 × 10⁹
r² = 4.0 × 10²⁸ / (7.126 × 10⁹) = 5.613 × 10¹⁸
Taking the square root gives r = 2.4 × 10⁹ m, which is a few times the radius of the Sun.
Notice that nothing in this method requires the star to be resolved as a disc. No telescope can resolve the disc of an ordinary star, and yet its radius can be found from its colour and its brightness alone. That is worth appreciating rather than merely memorising.
Redshift
The absorption and emission lines in the spectrum of a distant galaxy appear at longer wavelengths than their known laboratory values. The pattern of lines is unmistakably that of a known element, so the shift can be measured exactly.
For speeds much less than c:
Δλ / λ ≈ Δf / f ≈ v / c
where Δλ is the increase in wavelength, λ the laboratory wavelength, and v the speed of recession.
An increase in wavelength is a redshift and means the source is moving away. A decrease would be a blueshift and would mean approach. Almost every galaxy shows a redshift.
Worked example. A line with a laboratory wavelength of 656.3 nm is observed at 672.0 nm.
Δλ = 672.0 - 656.3 = 15.7
v = 3.00 × 10⁸ × 15.7 / 656.3 = 7.177 × 10⁶
so about 7.18 × 10⁶ m s⁻¹, roughly 2.4 per cent of the speed of light.
Why redshift implies an expanding Universe
The observation is that almost all galaxies are redshifted, and that the further away a galaxy is, the greater its redshift and therefore the faster it is receding.
That combination is what forces the conclusion. If galaxies were simply flying apart from us through a fixed space, we would appear to be at the centre of the Universe, which no other observation supports. The interpretation that fits is that space itself is expanding, carrying the galaxies with it and stretching the wavelength of light in transit.
The consequence is that an observer in any galaxy would see exactly the same thing: everything receding, with speed proportional to distance. There is no centre and no privileged position. The usual analogy is dots on a balloon being inflated, where every dot sees every other dot moving away and the dots further off move away faster.
Hubble's law and the Big Bang
Hubble's law states that the recession speed of a galaxy is proportional to its distance:
v ≈ H₀d
where H₀ is the Hubble constant. In SI units it is about 2.4 × 10⁻¹⁸ s⁻¹, and the syllabus notes that only SI units are required, which spares you the more common but awkward units of km s⁻¹ Mpc⁻¹.
Worked example. A galaxy recedes at 7.18 × 10⁶ m s⁻¹.
d = 7.18 × 10⁶ / (2.4 × 10⁻¹⁸) = 2.992 × 10²⁴
so about 3.0 × 10²⁴ m.
How this leads to the Big Bang theory
Run the expansion backwards. If everything is moving apart and has been doing so at a rate proportional to separation, then in the past everything was closer together, and at some finite time in the past everything was at essentially a single point of enormous density and temperature. That event is the Big Bang, and the Universe has been expanding and cooling ever since.
The time since then follows from Hubble's law itself. Taking v as constant, time is distance divided by speed:
t = d / v = d / (H₀d) = 1 / H₀
so the age of the Universe is approximately 1/H₀.
Worked example.
1 / (2.4 × 10⁻¹⁸) = 4.167 × 10¹⁷
Convert to years:
4.167 × 10¹⁷ / (365 × 24 × 3600) = 1.321 × 10¹⁰
so about 13 billion years, which agrees well with the ages of the oldest stars and is a large part of why the theory is accepted.
Note the assumption in that calculation, because a question may ask for it: it takes the rate of expansion to have been constant, which is only an approximation. The value of 1/H₀ is therefore an estimate, not an exact age.
Common mistakes
- Confusing luminosity with radiant flux intensity. Luminosity is emitted power and is a property of the star; flux intensity is what arrives at a detector and depends on distance.
- Using d rather than 4πd² in the inverse square law.
- Saying a standard candle is an object of known distance. It is an object of known luminosity, from which the distance is deduced.
- Forgetting to take the square root when finding a distance or a radius from a squared quantity.
- Using Celsius in Wien's law or the Stefan-Boltzmann law.
- Saying that a hotter star peaks at a longer wavelength, or that red stars are the hottest.
- Forgetting the fourth power of T in the Stefan-Boltzmann law, or raising only part of the expression to that power.
- Saying redshift means the galaxy is moving away through space, rather than that space between us is expanding.
- Concluding from the redshift of every galaxy that we are at the centre of the Universe.
- Confusing Δλ with the observed wavelength in Δλ/λ, and dividing by the wrong one. λ is the laboratory value and Δλ is the change.
- Quoting the age of the Universe as H₀ rather than 1/H₀, or leaving the answer in seconds when the question asks for years.
- Stating 1/H₀ as an exact age without noting that it assumes a constant rate of expansion.