Stellar Physics

Our Nearest Star

The nearest star to Earth is, of course, the Sun. Being "only" 150 million km from the Earth, scientists have been able to study the Sun in great detail, making it the most well-understood star. The Sun itself, however, is a fairly unremarkable star; placed within an H-R Diagram, the Sun appears in the middle of the Main Sequence having approximately average Brightness, Temperature and Mass.

These observations have allowed scientists to understand the Physics of stars, to understand their composition, their Chemical and Nuclear process, as well as their full life cycle. By observing the Sun, an understanding of stars in general can be derived.


Stellar Composition

The Sun is a near-perfect sphere of hot Plasma ( Surface T ~ 5,800 K ) comprising mostly of Hydrogen. The structure of the Sun is split into six separate layers, as shown in the diagram below :-

Stellar Fusion Processes

The Sun as a Main Sequence star is currently undergoing Hydrogen fusion at its core. In simple terms, this process fuses Hydrogen to form Helium, releasing Energy. However, there are several stages required for this fusion process, involving several isotopes of Hydrogen. The three isotopes of Hydrogen are shown in the diagram below :-

In order to fuse Hydrogen into stable Helium nuclei a series of fusion processes must occur a sequence called the Proton-Proton chain :-

There are three main stages within the fusion process :-

1H + 1H Fusion to 2H - Two Hydrogen nuclei fuse to form a Deuterium nucleus, A Positron and an Electron Neutrino. Energy is released ( 1.44 MeV ) .

2H + 1H Fusion to 3He - A Deuterium Nucleus and a Hydrogen nucleus fuse to form an unstable Helium nucleus and a Gamma Ray. Energy is released ( 5.49 MeV ) .

3He + 3He Fusion to 4He - Two Helium nuclei fuse to form a stable Helium nucleus and two Hydrogen nuclei. Energy is released ( 12.86 MeV ) .

For a star of equivalent mass of the Sun, this is the main fusion chain. However, more massive stars will follow different fusion chains.

Stellar Equilibrium

As discussed previously, stars spend between millions and billions of years as Main Sequence stars, staying approximately the same size. This is because the inwards force due to Gravity is balanced by the outward force due to thermal Pressure from fusion. This equilibrium is due to a negative feedback cycle as the two Forces work in opposition :-

Stellar Brightness and Luminosity

The above fusion reaction gives relatively small amounts of energy, but when added up of the whole core of the Sun, this process released a huge amount of Energy (the Power output of the Sun is ~ 4x1026 W).

The Power output of a star is more commonly known as the star's Luminosity. However, it is not possible to travel to the Sun to directly measure the Power output of the Sun, another method needs to be used. By observing the Apparent Brightness ( b ) of the Sun from the Earth, and by knowing the distance to the Sun, its Power output ( Luminosity ) can be calculated.

Note - The Apparent Brightness and the Light Intensity of a star describe the same property, Power per unit area. Apparent Brightness is a special wording of Light Intensity, generally only used for stars.

The Apparent Brightness of an light source depends on two factors, The Luminosity of the object and its distance from an observer. As the observer moves further away from a light source, the lower the Apparent Brightness of the object. This can be seen in the diagram below :-

The Apparent Brightness of a star is defined in Physics as "The Energy received per unit area from a star", measured in Wm-2 . This value will be largest at the surface of a star, and follows and inverse square law relationship with distance.

In order to calculate the Luminosity of the Sun, the Light Intensity at the Earth's surface is calculated, and assuming that this intensity is constant across a spherical shell with radius equal to 1 A.U., the Total Power ( i.e. the Luminosity ) can be calculated :-

The Light Intensity at a distance of 1 A.U. = 1367 Wm-2

Surface Area of a Sphere = 4 π r2

Light Intensity = Power / Unit Area

P = 1367 x 4 π ( 1.496x1011) 2

P = 3.84x1026 W

Luminosity of the Sun = 3.84x1026 W

Note - The above calculation uses the Light Intensity in Earth orbit for the calculation. This is due to the value at the Earth's surface being lower than 1367 Wm-2 due to atmospheric reflection and absorption ( ~ 30% of incoming Radiation is lost to these factors ) .


Apparent Brightness

Apparent Brightness of a star is therefore calculated using the following formula :-

Where :-

b = Apparent Brightness ( Wm-2 )

L = Stellar Luminosity ( W )

r = distance to Star ( m )

Note - Apparent Brightness can be a useful method of description of stars, however, it can be misleading. This is because a nearby dim star could have the same apparent brightness as a bright, distant star.

Stefan-Boltzmann Law

By knowing the size of the Sun (accomplished using Geometry) , the above formula can be used to find the Apparent Brightness at its surface. In 1879 the Austrian scientist Josef Stefan, building on previous work by Ludwig Boltzmann, derived a relationship between the Power per unit Area at a star's surface and its surface Temperature in Kelvin. This became known as the Stefan-Boltzmann Law:-

Power per Unit Area at Surface ( bsurface ) = σT4

Where :-

σ = The Stefan-Boltzmann Constant ( 5.67x10-8 W m-2 K-4 )

T = Surface Temperature of the Star ( K )

By combining the Apparent Brightness and Stefan-Boltzmann formulae , a formula for Stellar Luminosity can be derived :-

Where :-

L = Stellar Luminosity ( W )

r = Radius of the Star ( m )

σ = The Stefan-Boltzmann Constant ( 5.67x10-8 W m-2 K-4 )

T = Surface Temperature of the Star ( K )

Note - be careful when dealing with the several equations above. in the Apparent Brightness calculation, r gives the distance to the star, whereas in the Luminosity calculation r gives the radius of the star.

Example 1 -

Image result for capella

The star Capella is found within the constellation Auriga and is the third brightest star in the northern night sky. If Capella has a radius of 11.98 Solar radii and a surface Temperature of 4970 K, calculate :-

1.The Luminosity of Capella

2.The Apparent Brightness of Capella at 1 A.U.



r = 11.98 x 6.957x108 = 8.3x109 m

T = 4970 K

σ = 5.67x10-8 W m-2 K-4

1 A.U. = 1.496x1011 m


Luminosity :-

L = 4 π r2 σT4

L = 4 x π x ( 8.3x109 )2 x 5.67x10-8 x ( 4970 )4

L = 2.995x1028 W


Apparent Brightness :-

b = L / 4 π r2

b = 2.995x1028 / ( 4 x π x ( 1.496x1011 )2 )

b = 106.494 kW m-2 ( ~ 78 x bSun )

Note - A planet orbiting Capella at 1 A.U. would experience approximately 78 times the Power per unit area than the Earth does due to the Sun. Obviously due to Capella's much greater distance from Earth ( ~ 13.2 parsecs ) the Apparent Brightness of Capella is much lower ( ~ 14 nWm-2 ) from the Earth. This makes the Sun appear ~ 9x1010 times brighter than Capella from Earth, even though Capella is much more Luminous. This again demonstrates the limitations of Apparent Brightness as a quantity.