Why are stars treated as black bodies?

Their spectra closely match the continuous black-body curve for their surface temperature, so black-body laws apply well.

What is the difference between luminosity and intensity?

Luminosity is the total power a star emits. Intensity is the power per square metre arriving at a distance, which falls with the inverse square of distance.

Why does parallax fail for distant stars?

The angle becomes too small to measure against the resolution of telescopes and atmospheric blurring.

What makes a good standard candle?

A known luminosity that can be identified from other observations, and brightness high enough to see at great distances — Type Ia supernovae are the best example.

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153–158SpaceStretch

Luminosity and Distance

Everything we know about a star comes from its light. From that light alone, astronomers find its $\textbf{temperature}$ , its total power output or $\textbf{luminosity}$ , its size and its distance. Four equations do almost all of the work.

40 min Topic 10 · Space

What you'll be able to do

Use Wien’s law, $\lambda_{\max}T = 2.898\times10^{-3}\ \mathrm{m\,K}$
Use the Stefan–Boltzmann law, $L = \sigma AT^{4}$
Use the inverse square law, $I = L/4\pi d^{2}$
Describe trigonometric parallax and standard candles for measuring distance

Black-body radiation and Wien’s law

A $\textbf{black body}$ absorbs all radiation falling on it and emits a continuous spectrum that depends only on its temperature. Stars are good approximations.

The spectrum has a peak, and as temperature rises the peak shifts to shorter wavelengths while the total emission increases. That is why hot stars look blue-white and cool stars look red.

$\textbf{Wien’s displacement law}$ states that the peak wavelength is inversely proportional to absolute temperature.

\lambda_{\max}T = 2.898\times10^{-3}\ \mathrm{m\,K}

Wien’s law.

Example 1The spectrum of Betelgeuse peaks at 830 nm. Find its surface temperature.

T = \dfrac{2.898\times10^{-3}}{830\times10^{-9}}

Answer

3.5\times10^{3}\ \mathrm{K}

Tip — Wavelengths in nanometres must be converted to metres before using Wien’s law.

Luminosity and the Stefan–Boltzmann law

$\textbf{Luminosity}$ is the total power radiated by a star, in watts.

The $\textbf{Stefan–Boltzmann law}$ says luminosity is proportional to surface area and to the fourth power of absolute temperature, with $\sigma = 5.67\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$ .

Treating a star as a sphere, $A = 4\pi r^{2}$ . The fourth power means temperature dominates: doubling the temperature multiplies luminosity by 16.

L = \sigma AT^{4} = 4\pi r^{2}\sigma T^{4}

Stefan–Boltzmann law.

Example 1The Sun has radius

6.96\times10^{8}

m and surface temperature 5780 K. Find its luminosity.

L = 4\pi (6.96\times10^{8})^{2} \times 5.67\times10^{-8} \times 5780^{4}

= 6.09\times10^{18} \times 5.67\times10^{-8} \times 1.116\times10^{15}

Answer

3.85\times10^{26}\ \mathrm{W}

Intensity and the inverse square law

The radiation from a star spreads over a sphere of area $4\pi d^{2}$ at distance $d$ . The $\textbf{intensity}$ received — power per unit area, in $\mathrm{W\,m^{-2}}$ — therefore falls as the inverse square of distance.

Astronomers measure intensity at Earth. If either luminosity or distance is known, the other follows.

I = \dfrac{L}{4\pi d^{2}}

Inverse square law.

Example 1Find the intensity of sunlight at Earth,

1.50\times10^{11}

m from the Sun.

I = \dfrac{3.85\times10^{26}}{4\pi (1.50\times10^{11})^{2}}

Answer

1.36\times10^{3}\ \mathrm{W\,m^{-2}}

Tip — That result is the solar constant, about 1.4 kW m⁻². It makes a good sanity check for any calculation involving the Sun.

Measuring distance

$\textbf{Trigonometric parallax}$ : as Earth orbits the Sun, a nearby star appears to shift against the distant background. Measuring its position six months apart gives a parallax angle $p$ , half the total shift. With Earth’s orbital radius $r$ as a baseline, $d = r/\tan p \approx r/p$ for small angles in radians.

A $\textbf{parsec}$ is the distance at which the parallax angle is one arcsecond, about $3.09\times10^{16}$ m. So $d$ in parsecs $= 1/p$ in arcseconds. Parallax only works to a few thousand parsecs, since more distant stars shift by immeasurably small angles.

$\textbf{Standard candles}$ are objects whose luminosity is known — certain variable stars, and Type Ia supernovae, which all explode with nearly the same peak luminosity. Measure the intensity received, and $d = \sqrt{L/4\pi I}$ gives the distance, even across billions of light-years.

d = \dfrac{1}{p}

Distance in parsecs,

p

in arcseconds.

d = \sqrt{\dfrac{L}{4\pi I}}

Distance from a standard candle.

Example 1A star has parallax angle 0.25 arcseconds. Find its distance in parsecs and metres.

d = 1/0.25 = 4.0

pc.

4.0 \times 3.09\times10^{16}

Answer

4.0\ \mathrm{pc} = 1.2\times10^{17}\ \mathrm{m}

Equation recap

\lambda_{\max}T = 2.898\times10^{-3}\ \mathrm{m\,K}

Wien’s law.

L = \sigma AT^{4}

Stefan–Boltzmann.

I = L/4\pi d^{2}

Inverse square law.

d = 1/p

Parallax in parsecs.

Common mistakes to avoid

Confusing luminosity with intensity.

Luminosity is total power emitted; intensity is power per area received.

Using

L \propto T^{2}

Stefan’s law is fourth power.

Forgetting

4\pi

in the surface area or the intensity.

A sphere of radius

r

has area

4\pi r^{2}

Using the full six-month shift as the parallax angle.

The parallax angle is half the total angular shift.

Key takeaways

Wien: $\lambda_{\max}T = 2.898\times10^{-3}$ m K gives surface temperature.
Stefan: $L = \sigma AT^{4}$ relates luminosity, size and temperature.
Inverse square: $I = L/4\pi d^{2}$ .
Parallax measures nearby distances: $d/\mathrm{pc} = 1/p$ .
Standard candles of known luminosity give distances to far galaxies.

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