Mr Toogood's Physics

The Hertzsprung-Russell diagram

The HR diagram is a very useful tool for helping understand the differences between the different types of stars and how a star changes throughout its life. It plots spectral class, or temperature on the x-axis and absolute magnitude or luminosity on the y-axis. It is not a graph, in the true sense on the word, and neither axis starts at zero.

Stars tend to fall into one of four distinct areas on the diagram depending on their size and the stage of their lives.

The main stripe down the middle of the diagram is called the main sequence, and this is where stars will stay throughout the majority of their lives. The hottest stars are on the left of the diagram, and will, typically be much brighter. The cooler stars are usually dimmer and appear on the right of the main sequence.

There are two ‘clouds’ in the top right hand corner of the diagram of giant stars and supergiant stars. These are large stars which are approaching the end of their lives. Most typical stars (such as the Sun) will expand towards the end of their lives and move into the giant cloud. These stars and much brighter, so appear higher up on the y-axis, but are also much cooler, class class K or M, due to their surface area increasing, and following from from Stefan’s law. When such a star finishes fusing all of its fuel it will slowly cool to form a stellar remnant called a white dwarf. These are very small dim objects, but are also very hot, class B. The sun was formed from a cloud of dust and gas called a nebula. This was quite cool and very dim, so would not appear on the diagram. So when the Sun was formed it just appears on the HR diagram in its current position. The Sun will follow a path during its life similar to the one on the diagram below, and you are required to know this path.

You may be expected to draw an HR diagram in your exam, and you may be asked to label the x-axis with either spectral class or temperature, so it worth practising it. Some key points to remember when drawing the diagram are:

• The absolute magnitude on the y-axis should start at 15 at the bottom and go to -10. Make sure that you get these numbers the correct way around, from largest to smallest, not the other way.
• If plotting temperature, the x-axis should go from $\quantity{50\,000}{K}$ on the left to $\quantity{2500}{K}$ on the right.
• The main sequence should be drawn as a band, not a line and must have some curvature.
• Neither the giant cloud nor the white dwarf cloud can touch the main sequence.
• The giants should have an absolute magnitude less than 0.
• The dwarfs should have an absolute magnitude greater than 10.
• You do not need to draw the supergiants.

The Hertzsprung-Russell diagram is also useful for comparing two different stars in terms of their size or temperature. For example if two G class stars are observed to have different absolute magnitudes they can be plotted on the HR as below.

If they are both the same spectral class, they must have the same temperature, but Star A has a brighter absolute magnitude, so has a higher power output. We can no compare these two stars using Stefan’s law, $P=σAT^{4}$:

So star A must have a greater surface area, and therefore a greater diameter.

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Worked example

A red giant and a main sequence star have the same absolute magnitude, 0, and surface temperatures of $\quantity{3000}{K}$ $\quantity{15\,000}{K}$ respectively.

1. What spectral classes are the two stars?

You are expected to remember the associated temperatures for the different spectral classes, especially at either end of the scale. An HR diagram can help with this. If a star has a surface temperature of $\quantity{3000}{K}$ it is one of the coolest stars, class M. The main sequence star is much hotter, and its temperature of $\quantity{15\,000}{K}$ puts its in class B.



2. Show that the radius of the red giant is 25 times that of the main sequence star.

As both of the stars have the same absolute magnitude, they must both have the same power output.

• The red giant - $P_{RG}=σA_{RG}{T_{RG}}^{4}$
• The main sequence star - $P_{MS}=σA_{MS}{T_{MS}}^{4}$

Rearranging and cancelling gives:

$$\frac{A_{RG}}{A_{MS}}=\left(\frac{T_{MS}}{T_{RG}}\right)^{4}=\left(\frac{\quantity{15\,000}{K}}{\quantity{3000}{K}}\right)^{4}=5^{4}=625$$

So the red giant has a surface area, 625 times that of the main sequence star. Since $A=4πr^{2}$, the red giant is $625^{\frac{1}{2}}=25$ times larger.

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