25. Astronomy and cosmology

Exam analysis

Most tested topics

25.1 Standard candles

Luminosity L is the total power of electromagnetic radiation emitted by a star in all directions. Its SI unit is the watt (W).

Luminosity is an intrinsic property of the source: moving the observer does not change L, provided the star itself is unchanged.

Radiant flux intensity F is power received per unit area at a location, in W m⁻². It decreases with distance even though the source luminosity remains the same.

Do not define luminosity as apparent brightness or power received by a telescope. Include total, radiation and power in the definition.

For an isotropic source, total power L crosses a sphere of radius d and area 4πd². Power per unit area is therefore the inverse-square radiant flux intensity.

$F=L/(4πd²) L=4πd²F; d=√[L/(4πF)]$

L is in W, d in m and F in W m⁻². Doubling d makes F one quarter; multiplying d by k divides F by k².

For the Sun L=3.85×10²⁶ W at d=1.50×10¹¹ m, F=3.85×10²⁶/[4π(1.50×10¹¹)²]=1.36×10³ W m⁻².

Square the entire distance and retain 4π. This law assumes approximately isotropic emission and no additional absorption between source and observer.

A standard candle is an astronomical object whose luminosity L is known from its identified class or an independent calibration.

Because L is known, measuring its radiant flux intensity F supplies the two quantities needed to infer distance from the inverse-square law.

The object must be correctly identified as belonging to the calibrated class; 'standard' means known intrinsic luminosity, not equal apparent brightness wherever observed.

A bright object is not automatically a standard candle. Its luminosity must be known independently of the distance being inferred.

Identify a standard candle in the galaxy. 2. Use its calibrated luminosity L. 3. Measure its radiant flux intensity F at Earth. 4. Calculate the distance.

$d=√[L/(4πF)]$

For L=1.90×10³⁶ W and F=8.42×10⁻¹⁶ W m⁻², d=√[1.90×10³⁶/(4π×8.42×10⁻¹⁶)]=1.34×10²⁵ m.

A star or event inside a distant galaxy is effectively at the galaxy's distance because the galaxy's size is small compared with its distance from Earth.

Uncorrected absorption by dust makes F too small and therefore makes the inferred d too large. The method assumes the calibrated L applies and propagation losses are corrected or negligible.

25.2 Stellar radii

$λ_max T=b≈2.90×10⁻³ m K T=b/λ_max$

T is the thermodynamic surface temperature of the star and λmax is the wavelength at which its emitted spectral intensity/rate is maximum.

For λmax=624 nm, T=(2.90×10⁻³)/(624×10⁻⁹)=4.65×10³ K.

A cooler star peaks at a longer wavelength and has a lower thermal-emission curve; a hotter star peaks at a shorter wavelength.

Use the emitted/rest-frame peak wavelength. If redshift makes the observed λmax too large and is ignored, Wien's law gives a temperature that is too low. Convert nm to m and use kelvin.

$L=4πr²σT⁴ σ=5.67×10⁻⁸ W m⁻² K⁻⁴$

A star's surface emits power per unit area σT⁴; multiplying by spherical surface area 4πr² gives total luminosity L.

$r=√[L/(4πσT⁴)] T=[L/(4πσr²)]^(1/4)$

For the Sun L=3.85×10²⁶ W and T=5780 K, r=√{3.85×10²⁶/[4π(5.67×10⁻⁸)(5780)⁴]}=6.96×10⁸ m.

At fixed radius, L∝T⁴; at fixed temperature, L∝r². Doubling T multiplies L by 16.

Use radius, not diameter, and thermodynamic temperature in kelvin. Keep 4π and apply the square or fourth root only after isolating r² or T⁴.

Read the star's emitted peak wavelength λmax and use Wien's law for T. 2. Obtain luminosity L independently, for example from measured flux and known distance. 3. Use Stefan-Boltzmann to solve r.

$T=b/λ_max L=4πd²F r=√[L/(4πσT⁴)]$

If λmax=415 nm, T=(2.90×10⁻³)/(415×10⁻⁹)=6.99×10³ K.

With L=5.48×10²⁶ W, r=√{5.48×10²⁶/[4π(5.67×10⁻⁸)(6.99×10³)⁴]}=5.67×10⁸ m.

For equal luminosity, a hotter star must have a smaller radius because each square metre emits much more power (∝T⁴).

Peak wavelength determines T, not r by itself. Do not substitute radiant flux F at Earth where luminosity L is required; first convert using distance if needed.

25.3 Hubble’s law and the Big Bang theory

Identify an element by the distinctive spacing pattern of its emission or absorption lines, then compare the distant object's line wavelengths with the same lines measured in a laboratory/rest spectrum.

If the whole matching pattern appears at larger observed wavelengths than its known values, it is redshifted; the corresponding observed frequencies are lower.

A shift to shorter wavelength/higher frequency is blueshift. The direction of shift is judged from corresponding lines, not the overall colour alone.

Redshift also moves a thermal spectrum's observed peak to a larger wavelength than the emitted peak, so it must be corrected before using Wien's law for true surface temperature.

Do not compare unrelated spectral features. The evidence is a common fractional displacement of an identifiable line pattern from known rest values.

$For recession: Δλ=λ_obs−λ_emit>0 Δf=f_emit−f_obs>0$

$z≈Δλ/λ_emit≈Δf/f_emit≈v/c (v≪c)$

A line shifts from 4.62×10⁻⁷ m to 4.91×10⁻⁷ m: z=(4.91−4.62)/4.62=0.0628, so v=zc=1.88×10⁷ m s⁻¹ away from Earth.

For a source approaching the observer, λobs<λemit and the wavelength change is negative; its speed magnitude may be calculated from |Δλ|/λ.

State how Δf is defined: redshift lowers frequency, so f_emit−f_obs is positive. This approximation is for speeds much smaller than c and uses the emitted/rest value in the denominator.

Spectra from many distant galaxies show identifiable lines at larger wavelengths/lower frequencies than their known rest values.

This redshift indicates that those galaxies are receding and that, on large scales, galaxy separations are increasing.

If distant galaxies are generally moving apart, the universe's large-scale geometry is expanding rather than remaining static.

Observers in different galaxies would see the same large-scale pattern: expansion does not require Earth to be a unique central point.

One nearby galaxy can have local motion toward us. The expansion inference comes from the systematic large-scale redshift pattern of distant galaxies.

$v≈H₀d H₀=v/d; d=v/H₀$

The recession speed v of a distant galaxy is approximately proportional to its distance d from the observer. In SI, v is m s⁻¹, d is m and H0 is s⁻¹.

A graph of recession speed v against distance d is approximately a straight line through the origin with gradient H0.

For v=1.9×10⁷ m s⁻¹ and H0=2.3×10⁻¹⁸ s⁻¹, d=v/H0=8.3×10²⁴ m.

More distant galaxies recede faster. Running this expansion backward means separations were smaller in the past, leading to the idea that the universe began in an extremely hot, dense state and expanded—the Big Bang model.

$simple expansion timescale≈1/H₀ for H₀=2.3×10⁻¹⁸ s⁻¹: 1/H₀≈4.3×10¹⁷ s≈1.4×10¹⁰ years$

Use H0 in s⁻¹ with SI data. Backward extrapolation supports a common dense past; it is not evidence for an explosion from one location into pre-existing empty space.

25. Astronomy and cosmology

Exam analysis

Most tested topics

25.1 Standard candles

Luminosity is a star's total emitted radiation power

Radiant flux falls as luminosity spreads over a sphere

A standard candle is an astronomical object with known luminosity

Use a standard candle's known luminosity and measured flux to find galaxy distance

25.2 Stellar radii

Wien's law uses the emission peak to estimate stellar surface temperature

Stefan-Boltzmann links luminosity to stellar area and T⁴

Estimate stellar radius by combining spectrum temperature with luminosity

25.3 Hubble’s law and the Big Bang theory

Distant-object spectral lines are redshifted from their known laboratory wavelengths

Use fractional redshift to estimate radial speed

Systematic galaxy redshift is evidence that the universe is expanding

Hubble's law links recession to distance and points back to a dense early universe