Stellar radiation

I.

The energy source of stars

A star such as our own sun radiates an enormous amount of energy into space. The source of its energy is nuclear fusion in the interior of the star, in which nuclei of hydrogen fuse to produce helium and release energy in the process. The energy produced is carried away by photons and neutrinos produced in the reactions. As these particles move outwards they collide with the surrounding protons and electrons and give them some of the energy. The motion of the particles inside the star, as a result of the energy they receive, called radiation pressure, can stabilize the star against gravitational collapse. II. Luminosity and brightness Luminosity is the amount of energy radiated by the star per second; that is, it is the power radiated by the star. Luminosity denoted L depends on the surface temperature T and the surface area A of the star (or its radius R as the surface area of a sphere is A=4𝜋 ∙ 𝑅! ) according to Stefan-Boltzmann law: 𝐿 𝑊 = 𝜎 ∙ 𝐴 ∙ 𝑇 ! = 𝜎 ∙ 4𝜋 ∙ 𝑅(𝑚)! ∙ 𝑇(𝐾)! 𝑤𝑖𝑡ℎ 𝜎 = 𝑆𝑡𝑒𝑓𝑎𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 5.67×10!! 𝑊. 𝑚!! . 𝐾 !! The energy radiated by a star is in the form of electromagnetic radiation and is distributed over an infinite range of wavelengths. However, most of the energy is emitted around the peak wavelength. Calling this 𝜆! , we see that the colour of the star is mainly determined by the colour corresponding to 𝜆! . The Wien displacement law relates the wavelength 𝜆! to surface temperature T: 𝜆! 𝑚 ∙ 𝑇(𝐾) = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 2.9×10!! 𝐾. 𝑚 which implies that the higher the temperature, the lower the wavelength at which most of the energy is radiated. Consider a star of luminosity L. Imagine a sphere of radius d centred at the location of the star. If the star is assumed to radiate uniformly in all directions, then the energy radiated in 1s is distributed over the surface of this imaginary sphere. A detector of area denoted a, placed somewhere on this sphere, will receive a small fraction of this total energy. The fraction is equal to the ratio of the detector area a to the total surface area of the sphere 4𝜋 ∙ 𝑑 ! ; that is, the received energy per second will be 𝑎 ∙ 𝑏 where b is called the apparent 𝐛𝐫𝐢𝐠𝐡𝐭𝐧𝐞𝐬𝐬 given by: b(W. m!! ) = III.

𝐿(𝑊) 4𝜋 ∙ 𝑑(𝑚)!

Stellar spectra

A great wealth of information can be gathered about a star from studies of its spectrum. • Temperature: The surface temperature of a star is determined by measuring the wavelength at which most of the radiation is emitted. • Chemical composition: an absorption spectrum is made of dark lines superimposed on a background of continuous colour. Each dark line represents the absorption of a specific frequency by a specific chemical element in the star’s atmosphere. It has been found, however, that most stars have essentially the same chemical composition, yet show different absorption spectra. Consider two stars with the same content of hydrogen. One is hot, about 25,000 K and the other cool, about 10,000 K. The hydrogen in the hot star is ionised, which means the electrons have left the hydrogen atoms. These atoms cannot absorb any light passing through them, since there are no bound electrons that can absorb the photons and make transitions to higher energy states. Thus, the hot star will not show any absorption lines at hydrogen wavelengths. The cooler star, however, has many of its hydrogen atoms in the energy state n=2. Electrons in this state can absorb photons to make transitions to states such as n=3 and n=4, giving rise to the characteristic hydrogen absorption lines. Similarly, an even cooler star of temperature, say, 3,000 K, will have most of its electrons in the ground state of hydrogen atoms and so can only absorb photons corresponding to

•

• •

IV.

ultraviolet wavelengths. These will not result in dark lines in an optical spectrum. In this way, study of absorption spectra gives information about the temperature of the star and its chemical composition. Radial velocity: if a star moves away from or toward us, its spectral lines will show a Doppler shift (similar to what you experience when you hear the siren of an ambulance passing next to you as it is rushing to the hospital). The shift will be toward the red if the star moves away, and toward the blue if it comes toward us. Measurement of the shift allows the determination of the radial velocity (velocity along the line of sight) of a star or a galaxy. If 𝜆! is the wavelength of a spectral line and 𝜆 is the longer wavelength received on Earth, the redshift denoted z of the star or galaxy is defined as: ! !!! 𝑧 = ! ! . If the speed v of the receding galaxy is small compared to the speed of light c then: 𝑧 = . !

!

Rotation: if a star rotates, then part of the star is moving toward the observer and part away from the observer. Thus, light from the different parts of the star will again show Doppler shifts, from which the rotation speed may be determined. Magnetic fields: In a magnetic field a spectral line may split into two or more lines (the Zeeman effect). Measurement of the amount of splitting yields information on the magnetic field of the star. The Hertzsprung-Russel diagram and the types of stars

In the H-R diagram that follows, the right vertical axis represents luminosity in units of the sun’s luminosity. The top horizontal axis shows the surface temperature of the star; the temperature decreases as we move to the right. Also shown at the bottom of the diagram is the spectral class for each star; stars are divided into seven spectral classes according to their colour, which is related to surface temperature.







Three clear features emerge from this diagram: - Most stars (about 90%) fall on a strip extending diagonally across the diagram from top left to bottom right. This is called the main sequence. Our sun is a typical member of the main sequence. - Some large stars (1%), reddish in colour, occupy the top right – these are the read giants (large, cool stars) - The bottom left is a region of small stars (9%) known as white dwarfs (small and hot). Questions: 1. The light from a star a distance of 70 light-years away is received with an apparent brightness of 3.0×10-8 W.m-2. Calculate the luminosity of the star. Speed of light in a vacuum=c=3.00×108 m.s-1. 2. Two stars A and B have the same luminosity. Star B has a surface temperature, which is twice higher than that of star A. a. Which is the larger star and by how much? b. If the apparent brightness of A is double that of B, what is the ratio of the distance of A to that of B? 3. The luminosity of a star is barely the same as the luminosity of the Sun. Moreover its spectrum shows that most of the energy is radiated at a wavelength of about 500 nm. What class of star is it? In which category does it belong? 4. Two stars A and B emit most of their light at wavelengths of 650 nm and 480 nm respectively. It is known that star A has twice the radius of star B; find the ratio of the luminosities of the stars (provide only one calculation). 5. A hydrogen line has a wavelength of 434 nm (measured at the lab). When received from a distant galaxy, this line is measured on Earth at 486 nm. Assume the speed of the galaxy is small compared to the speed of light, is the galaxy receding or approaching? At what speed? Check your hypothesis.

` Correction 1. One light-year is the distance travelled by light during one year. 1 ly=3.00×108×(365.25×24×3,600)=9.47×1015m; the distance d given: d=70 ly. The apparent brightness is known: b=3.0×10-8 W.m-2 𝐿 = b ∙ 4𝜋 ∙ 𝑑 ! = 3.0×10!! ×4×𝜋× 7.0×9.47×10!" ! 𝐿 = 3.0×4×𝜋× 7.0×9.47 ! ×10!" = 𝟏. 𝟕×𝟏𝟎𝟐𝟗 𝑾 (2 significant figures) 2. We know that according to Stefan-Boltzmann law: 𝐿 𝑊 = 𝜎 ∙ 4𝜋 ∙ 𝑅! ∙ 𝑇 ! a. Star B has a twice higher surface temperature than star A but both stars have the same luminosity; the product 𝑅! ∙ 𝑇 ! must take the same value so: 𝑅(𝐴)! ∙ 𝑇 𝐴 ! = 𝑅(𝐵)! ∙ 𝑇 𝐵 ! ! !

Therefore !

!

=

! ! ! !

!

= 4. Star A is 4 times larger than star B. !(!)

b. We know that both stars have the same luminosity. Moreover: b (W. m!! ) = !!∙!(!)! . If the apparent brightness of A is double that of B, it implies that 𝑑 ! 𝐴 = 3. We apply Wien’s displacement law:

!! ! !

𝑜𝑟

𝒅(𝑨) 𝒅(𝑩)

=

𝟏 𝟐

= 𝟎. 𝟕𝟏

2.9×10!! 𝜆(𝑚) 2.9×10!! 𝑇 𝐾 = = 𝟓, 𝟖𝟎𝟎 𝑲 5×10!! 𝑇(𝐾) =

It is a G star. As its luminosity is barely the same as that of the sun, it belongs to the main sequence. 4. Let’s use Stefan-Boltzmann law that relates luminosity, radius and temperature according to: 𝐿 𝑊 = 𝜎 ∙ 4𝜋 ∙ 𝑅! ∙ 𝑇 ! ! 𝐿(𝐴) 𝑅 𝐴 ! 𝑇 𝐴 =( ) ∙ 𝐿(𝐵) 𝑅 𝐵 𝑇 𝐵 ! Moreover according to Wien’s law: 𝜆! 𝑚 ∙ 𝑇 (𝐾) = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 so: ! Therefore

! !

! !

= !!

!

!



!

𝐿(𝐴) 𝑅 𝐴 ! 𝜆! 𝐵 =( ) ∙ 𝐿(𝐵) 𝑅 𝐵 𝜆! 𝐴 𝐿(𝐴) 480 ! ! = (2) ∙ = 𝟏. 𝟐 𝐿(𝐵) 650



5. As the wavelength of the light received is higher than the wavelength of the light emitted, we can deduce that the galaxy is receding. We know that provided its speed is low relative to the speed of light: 𝑣=𝑐∙

Δ𝜆 486 − 434 = 3×10! × = 3.6×10! 𝑚. 𝑠 !! = 𝟑𝟔. 𝟎𝟎𝟎 𝐤𝐦. 𝒔!𝟏 𝜆 434

! !" The result is consistent with the hypothesis that v is much lower than c: ! = !×!"! ≪ 1