Topic 2.7 starts at page 33 of the Spec.
The spectrum of a star is made up of it’s emission and absorption.
Firstly, we have a continuous emission spectrum, which is usually the rainbow behind it. On top of the continuous spectrum are Fraunhofer lines, caused by cold gas in a star’s atmosphere absorbing light emitted by hotter gases below. Each element absorbs specific wavelengths, and there can be often multiple lines for one element.
All objects absorb and emit radiation, although they won’t absorb and emit all of it. White things tend to reflect more radiation than black objects.
Black bodies absorb all radiation that is incident on it, hence they’re good emitters as well. Stars are very close to being perfect black bodies, so we can assume that they are.
Let’s look at some typical spectra:
On the y-axis is the spectral intensity (you don’t need to worry about that) and on the x-axis lies the wavelength.
It’s important to know that temperature is what varies these curves. Look at the peak wavelength of the 3’500 K curve and the 5’500 K curve. It’s clear that the hotter temperature line has a lower peak wavelength and a higher spectral intensity.
Now let’s say we were observing a star that had a surface temperature of about 4’500 K. What colour would this star be?
Firstly, look at the 4’500 K curve. How many nanometres is the peak wavelength?
On this graph I’d say roughly 650 nm (in the exam they expect you to use a ruler). 650-700 nm is red light, so this star would look red.
Suppose we didn’t know the temperature but wanted to know it i.e we collected some data from a star and put it into a graph with an approximate curve. What then?
There’s a handy formula that goes hand-in-hand with black body spectra, called Wien’s Law. We can use it to find the surface temperature:
W = λ * T
W = Wien’s Constant (2.9e-3)
λ = peak wavelength (in metres)
T = temperature in Kelvin (temperature in ℃ + 273.15)
Give this question a try:
α boötis, better known as Arcturus, is a very bright star in Boötes. After collecting data from Arcturus, its peak wavelength was measured to be 677 nm. What is Arcturus’ surface temperature?
We know the peak wavelength (677 nm) and we know Wien’s constant, so we’re good to go.
Turn the wavelength into metres by multiplying it by 10e-9.
Now put sub it into W = λ * T, and solve for temperature.
Congrats! You’ve found the surface temperature of Arcturus!
Let’s ponder Wien’s Law
Arcturus’ peak wavelength is 677 nm, which would be red. If this were any shorter it would appear yellow, and even shorter it would look blue, but what would the colour do to the temperature?
If W is a constant, then shortening the wavelength would increase the temperature. So by deduction, “blue” stars are hotter than the sun, and conversely “red” stars are cooler.
Above is a HR diagram, showing luminosity (power output) against temperature (at the top in this graph). Don’t worry about absolute magnitude and spectral class. Temperature goes from hot to cold in a HR diagram. The points/stars have been coloured so that cooler ones are red and hotter ones are blue.
Wien’s law seems to fit the large rainbow stripe; colder stars are red and less luminous, hot stars are blue and very luminous.
Yet, there seems to be some exceptions. In the top right there’s cool but very luminous stars. In the bottom left corner there’s less powerful hot stars. When in a star’s life would it be very hot, but not very luminous? Or when is a star cool but very luminous/big?
The stars in the big rainbow strip are part of the main sequence, and so follow normal fusion of hydrogen into helium. Here, there’s a clear correlation between high temperature and luminosity, but as the star becomes a red giant or a white dwarf, that correlation is no longer there.
In a red giant’s core, fusion of larger elements takes place, which actually increases the luminosity. However, the radius of the star increases a lot to stop it collapsing, that the intensity of heat reaching the surface has decreased.
After that, the star’s surface collapses into the core, and is ejected to make a planetary nebula, leaving a tiny white dwarf. All that’s left is a very hot core, but fusion has stopped, so it’s not generating any power/luminosity.
All this talk on luminosity brings us to our next topic, Stefan’s Law.
Stefan’s Law and Inverse Square Law
Stefan’s Law is very useful, as it basically puts the above explanations into a neat and handy formula.
Now we know how to find surface temperature by wavelengths, let’s try finding the luminosity of a star. For this we can use Stefan’s law:
L = A*σ*T^4
L = Luminosity/power (in Watts)
A = surface area of the star (in m^2)
σ = Stefan’s constant (5.67e-8)
T^4 = temperature (Kelvin) to the power of 4.
You’ll also find P = IA useful, where P = power (luminosity), A = area and I = intensity (examiners love to make you use this equation).
Let’s go over a question:
Betelgeuse is a star in the Orion Constellation. We’ve measured its luminosity as 3.828e+31 Watts, and the surface temperature as 3473.15K. Calculate the radius of Betelgeuse to 3 sf.
First, let’s put all the info we know into Stefan’s Law.
Arrange it to get A on its own.
Now we can figure out the radius by dividing it by 4 π and then square rooting it.
That’s roughly 0.6 billion km, 890 x larger than the sun!
It’s no good observing the universe with only visible light. The complete electromagnetic spectrum needs to be used.
Multiwavelength astronomy is observing with all types of light, from radio to gamma. It reveals a lot more about the universe, because different types of radiation can reveal different processes occurring.
In order of left to right. Radio image of the Crab Nebula (Credit: NRAO/AUI), Infrared image of the Crab Nebula (Credit: NASA/JPL-Caltech/R. Gehrz), Optical image of the Crab Nebula (Credit: NASA/ESA/ASU/J. Hester), Ultraviolet image of the Crab Nebula (Credit: NASA), X-ray image of the Crab Nebula (Credit: NASA/CXC/SAO/F. Seward et al.)
Have a look at the Crab Nebula. From the visible light, we can tell the red parts are remains of the star and the blue comes from electrons interacting with a magnetic field, but with only visible light, you couldn’t easily tell that the field belongs to a pulsar.
Look using radio waves, and the star becomes visible. From this image and the dimensions of the nebula, we can tell that the star is most likely a neutron star. The infrared shows electrons trapped in a magnetic field, meaning there’s something large in the middle that’s generating a strong magnetic field.
In the X-ray the pulsar is clear and its pulses can be seen. The nebula looks smaller than in the other images, suggesting that high energy electrons are mostly near the star, hence the star’s probably the object that’s heating them up.