Topic 4 Classification of Stars

• In the exam you are expected to know about:

• Classification by luminosity;

• Relation between brightness and apparent magnitude;

• Apparent magnitude, m;

• Relation between intensity and apparent magnitude (Measurement of m from photographic plates and distinction between photographic and visual magnitude not required);

• Absolute magnitude, M;

• Parsec and light year;

• Definition of M, relation to m;

• 

• Classification by temperature, black body radiation;

• Stefan’s law and Wien’s displacement law;

• General shape of black body curves, experimental verification is not required;

• Use of Wien’s displacement law to estimate black-body temperature of sources;

• λmaxT = constant = 0.0029 mK;

• Inverse square law, assumptions in its application;

• Use of Stefan’s law to estimate area needed for sources to have same power output as the sun:

• 

• Assumption that a star is a black body;

• Problem of detector response as a function of wavelength and atmospheric effects.

Astronomical distances

Space is very big.  Kilometres are far too small to be useful.  We use instead:

The diagram shows the idea of the parsec.

1 arcsec is a very small (1/3600) fraction of a degree.  At 10 pc, the angle subtended by the arc second is 10 AU.

Apparent Magnitude

The Greek astronomer Hipparchus  classified stars according to their apparent brightness to the naked eye, about two thousand years ago.  Its scale was 1 to 6.  It is still used today and is called the apparent magnitude scale.  The apparent magnitude is given the code m.  Magnitude 1 stars are about 100 times brighter than magnitude 6 stars.  A change in 1 magnitude is a change of 2.512 (100^1/5 = 2.512).  The scale is logarithmic because each step corresponds to multiplying by a constant factor.

To work out the apparent magnitude we need to:

• calculate the difference in apparent magnitudes= n.

• work out the ratio of the brightnesses = 2.512ⁿ.

Therefore the Sun is about 400 000 time brighter than the Moon.  Notice that bright objects have negative magnitudes.  The scale looks like this:

Question 1 How much brighter is Sirius (m = -1.46) than Betelgeuse (m = +0.50)? ANSWER

Question 2  What do you think is a problem with the apparent magnitude scale? ANSWER

Absolute Magnitude

When we use absolute magnitude, we "place" all stars at an arbitrary distance of 10 parsec.  We give the absolute magnitude the code M.  So we are comparing the brightnesses of all stars as if they were at 10 pc.

Question 3  What is 10 parsec in kilometres?  A supersonic plane is travelling at 3000 km/h.  How long would it take to travel 10 parsec?  ANSWER

The absolute and apparent brightnesses of stars are related by this formula:

[ m - apparent magnitude; M - absolute magnitude; d - distance (pc)]

Worked Example

If the star is less than 10 pc away, then the absolute magnitude tends to move in the positive sense, which means that they become dimmer.  Stands to reason.  If you move a star further away, the dimmer it gets.  Stars over 10 pc will have absolute magnitudes that are brighter than their apparent magnitude.

Question 4.  The apparent magnitude of the Sun is -26.7  What is its absolute magnitude?  How does it compare with the absolute magnitude for the star Alpha Centauri?  ANSWER

The answer to this question should tell you that the Sun viewed from 10 pc would glow with a brightness similar to many other stars.

Question 5 

Bellatrix and Elinath are two stars with the same apparent magnitude. The distance from the

Earth to Bellatrix is 470 light years and its absolute magnitude is -4.2.

(i) Calculate the distance to Bellatrix in parsecs.

(ii) Calculate the apparent magnitude of Bellatrix.

(iii) Elinath has an absolute magnitude of -3.2. State, giving a reason, which of the two stars is closer to the Earth. 

(AQA Past Question)

ANSWER

Classification by Temperature

The American astronomer Anne Cannon was the first to classify stars using the science of spectroscopy.  Astronomers look for Balmer lines which arise from electron transitions in hydrogen atoms (See Module 1).  As the electron drops from high levels to the second  energy level (the one above the ground state), photons are emitted.  Because they are emitted in random directions, against a complete spectrum of colours the emissions of photons would appear black.  This is an absorption spectrum.  Compare the emission spectrum:

with the absorption spectrum:

By studying the spectral absorption lines at wavelengths corresponding to the photons of the Balmer series, astronomers can get an idea of the temperature:

• At low temperatures there are few violent collisions to excite electrons which remain in the ground state.  Energy level changes are rare.

• At high temperatures, there are many violent collisions between atoms.  Electron transitions occur at higher levels so there are comparatively few Balmer transitions.

• At intermediate levels many electrons are performing Balmer transitions, so there are strong absorption lines.

The graph of intensity against temperature looks like this:

Notice that: 

• temperature decreases from left to right;

• for a given intensity, two temperatures are possible.  To overcome this, the spectra of other elements are analysed.  Peak intensities of different elements are found, and this can tie down the temperature.  The idea is shown on the next graph:

Therefore in the Sun, the spectral lines would be seen for iron and calcium, indicating a surface temperature of about 6000 K.  Very hot stars show spectral lines for light elements while cool stars will show up heavy elements, and spectra for molecules as well.

Question 6 What would you not see when looking at the spectrum of the red giant Betelgeuse? What elements would you expect to see?   ANSWER

The table shows the spectral classes for stars:

Question 7 The graph shows part of the visible spectrum for the star Vega: (note that it is an absorption spectrum so the intensity dips to a minimum at the emitted wavelengths.)

The absorption lines are due to excited hydrogen atoms. The wavelength of each absorption is

given in nm.

(a) Explain how Hydrogen atoms produce these absorption lines.

(b) The diagram below shows the first six energy levels of a hydrogen atom. State which is the largest energy transition which produces an absorption line in the visible spectrum of Vega.

(c) State the value of the wavelength corresponding to this transition.

(d) What is the name given to the series which gives rise to the visible region of the hydrogen

spectrum?

(e) For which spectral classes are these lines the dominant feature?

(AQA Past Question)

ANSWER

The classifications of stars according to spectra are also subdivided further with numbers (e.g A5) to allow the surface temperature to be determined within about 5 %.

Black Body Radiation

We look at the temperature of stars by looking at their colours.  A lot of energy is given off as thermal radiation.  Objects that are red hot have a temperature of about 1200 K.  To understand how the colour of an object depends on its temperature, we need to understand the concept of a black body.  A black body is a perfect absorber so that all radiation that falls on it is absorbed.

A perfect absorber is a perfect emitter.  Therefore if we heat it up it will emit radiation including visible light.  This is true (to a first approximation) for stars.  Note the following for black bodies:

• a hot object emits radiation across a wide range of wavelength;

• there is a peak in intensity at a given wavelength;

• the hotter the object the higher the peak;

• the hotter the object the shorter the peak wavelength.

• the area under the graph is the total energy radiated per unit time per unit surface area.

The peak wavelength is lmax which is the wavelength at which maximum energy is radiated.  This is inversely proportional to the Kelvin temperature.  It is called Wien's  Displacement Law (as the peak is displaced towards shorter wavelengths).  We write it as:

lmax T = constant = 0.00289 m K

Worked example

Question 8  Betelgeuse appears to be red.  If red light has a wavelength of about 600 nm, what would the surface temperature be?  ANSWER

You don't get green stars because the light from stars is emitted at a range of wavelengths, so there is mixing of colours.  So those stars with a lmax in the green region will actually appear to be white.

Luminosity of Stars

The area under the graph above is related to the rate at which a black body radiates energy.  The luminosity of a star is the total energy given out per second, so it's the power.  From the graph the luminosity increases rapidly with temperature, which gives rise to Stefan's Law.  Formally this is stated as:

The total energy per unit time radiated by a black body is proportional to the fourth power of its absolute temperature.

In other words double the temperature and the power goes up sixteen times.  In physics code we write:

[P- Power (W); s - Stefan's constant; A - area (m²); T - temperature (K)]

The strange looking symbol s is "sigma", a greek letter lower case 's'.  It Stefan's Constant.

s = 5.67 x 10^-8 W m^-2 K^-4

We can treat a star as a perfect sphere (A = 4pr²) and a perfect black body.  So for any star, radius r, we can write:

(Note: in some text books the power may be represented as luminosity with the physics code L) 

Question 9  If the Sun has a radius of 6.96 x 10⁸ m and a surface temperature of about 6000 K, what is its total power output?  What is the power per unit area?  What is the peak wavelength?  ANSWER

The intensity of the Sun's radiation decreases by an inverse square law.  Therefore Saturn, about 10 times further from the Sun (i.e. 10 AU) receives only 1 % of the intensity of the Sun's radiation.

Problems of Observing Stars

Meaningful astronomy is getting harder.  Research astronomers are generally to be found in the major universities whose observatories were often set up in major cities.  The observatories are often quite old.  Although the instruments are very capable, their situation limits them:

• Light pollution from street lights makes observation of dim objects difficult;

• If the weather is cloudy, observation is impossible;

• Dust from pollutants can interfere with images;

• The atmosphere is turbulent, leading to scintillation.

These problems can be solved:

• by moving observatories to the top of high mountains;

• mounting a telescope in an aeroplane (although good tracking depends on the skill of the pilot and absence of turbulence that can make for a bumpy ride);

• Putting the telescope in orbit.

If other radiations are being investigated, satellites equipped with appropriate receivers are used.

Home            Physics A2            Module 5A