**Frédéric Dahringer ^{1} and Rainer C. Gaitzsch^{2}**

^{1}Comité de Liaison Enseignants et Astronomes, CLEA (France)

^{2}Teacher's Academy of Bavaria (Germany)

#### Abstract

The main interest of this workshop is to see how we can get values for the Sun's luminosity and surface temperature, by using simple methods.

The participants will share experiments in which the Sun's radiation is measured, leading to the value for the Sun's luminosity. Then, applying the radiation laws of Stefan-Boltzmann and Wien, and looking at the Sun's spectrum, we will achieve the surface temperature of the Sun. By help of a simple experiment we can understand the convection flow beneath the Sun's surface, causing the granules.

A short list of selected exercises closes the workshop.

#### Introduction: Seeing the Sun

If we are to understand the properties of stars, most of which are so far away that they cannot be seen as anything other than points of light, the best place to start is by observing the easiest star to see - our own Sun.

The Sun is sufficiently close to let us observe the existing structures on its surface i.e. sunspots and granules, and allow us to study its output of radiation in detail (see Fig. 1). If you have no special protected telescope for observing the Sun, you can just use binoculars (Fig. 2). The image of the Sun is projected by a flat mirror to a white carton placed laterally. One lens of the binoculars is covered and a pasteboard is used for shading. The image of the Sun moves across the carton because of Earth's daily rotation.

#### How is the Solar energy created?

Solar energy is created deep within the extreme hot core of the Sun where nuclear fusion burns hydrogen to helium. This is mainly performed by the so-called pp-chain, which shortly can be described in the reaction formula:

Four protons are consumed and one helium nucleus containing two protons and two neutrons is produced along with two positrons, two neutrinos and two g-quanta or g photons (Fig. 3).

Nuclear fusion is possible because the Sun's central temperature is 15 × 10^{6} K as a consequence of the extreme high central pressure (10^{11} bar). High temperatures are necessary because the individual hydrogen nuclei (protons) must collide with enough energy to have a reasonable probability of overcoming the repulsive electrical force between these two positively charged particles.

Nuclear reactions must obey the principle of 'conservation of energy'. The total mass of the four protons which enter the fusion process is bigger than the total mass of all the matter-components of the products of the reaction formula. The helium nucleus (also called 'alpha particle') is about 0.7% less massive than the four protons. Therefore, every second our Sun loses more than 4 million tons of mass, and so 700 million tons of hydrogen in the Sun's core are converted into helium ashes.

We must conclude, according to Einstein's law (*E = m c ^{2}*), that a big amount of energy (mainly taken by the γ-quanta) is created in this process to compensate the mass loss. Nonetheless, in total, the Sun has lost only 0.03% of its present mass (1,99 × 10

^{30}kg) since its creation 4,6 × 10

^{9}years ago.

#### Solar constant and Luminosity of the Sun

The expelled energy is carried to the surface of the Sun where it is released as heat and light. The constant flux of radiation then spreads out into space equally with radial symmetry. In the distance of 1 AU (astronomical unit) where the Earth orbits the Sun, only a tiny little fraction of the whole radiation meets us, still big enough to conserve life on Earth.

The so called *Solar constant*, S, is the average flux of solar radiation energy passing at Earth's distance 1 AU during 1 second, through the perpendicular area of 1 m^{2}. By the value of S we can calculate the whole *luminosity*, L, of the Sun. The knowledge of its value is of great importance for understanding the physical conditions of our Sun.

#### Experiments to determine the Solar constant S and Luminosity L

The first experiment to measure the Solar constant has already been presented during the 6^{th} summer school in Finland in 2002. I want to repeat the facts shortly:

A certain amount of water is heated by exposing it to the Sun's radiation during a measured time. The rising of the temperature is analysed and leads to a rough value of the Solar constant (Fig. 4).

The first activities have to be done in the shadow. Measure the effective radius r for sunlight capture of the black layer on the circle-area at the bottom of the flask and calculate its effective area A in m^{2}. Then measure the mass of the flask mG.

Fill the flask up to the top edge with water, close it with a stopper and measure the mass of water mW used to fill it. Insert a thermometer and wait until the temperature of the water is equal to the temperature of the surrounding air. Now go out of the shadow and point the assembly assuring that the sunbeams hit the black bottom area of the flask in perpendicular direction (check it by observing shadows).

Note the starting temperature θ_{1} of the water and simultaneously start the time counting. The whole time for sunlight exposure should be around 10... 15 minutes. Long enough to get sufficient sunlight for heating the water and short enough for only few energy loss to the surrounding air. Measure the temperature of the water every 2 minutes (take care to mingle). Stop the experiment after 12 minutes. Draw the line in the t-θ diagram, it should be about linear. Calculate Δθ/Δt (rising of temperature / duration of time) and write it.

The formula for the amount of energy transferred ΔQ is: ΔQ = c m Δθ, where c is the specific heat-coefficient of the used material, and m is the involved mass. Taking in account that the sunlight energy heats not only the water (W) but also the glass (G), we will use the combined (weighted) expression:

C = c_{W} m_{W} + c_{G} m_{G}

instead of cm.

θ The Solar constant S, as being power devided by area, is:

Our computed value of S will be about 700... 900 W m^{-2}, that is much too low compared with the standard value outside the Earth's atmosphere (measured by satellites).

This value for further consideration is given by:

S = 1,37 × 10^{3} W m^{-2}

The main reason for our low value is that we did not consider the absorption of Earth's atmosphere. Different Solar activity during sunspot cycle also causes S to vary about 1% - 2%. When the Sun's radiation reaches 1 AU, it is spread over a very wide area. To determine L, we can imagine a huge sphere of radius r = 1 AU with the Sun at its center. Each square meter of that sphere gets 1,37 x 10^{3} W of power from the Sun. Therefore the total energy radiated by the Sun per second is equal to the Solar constant S multiplied by the sphere's surface:

Another (more funny) experiment leads directly to a rough estimate of the Sun's luminosity L (Fig. 5):

One cheek of a person is directly orientated to the sunlight. The distance d of an electric bulb (P = 100 W) to the other cheek is shifted until the person (with closed eyes) has exactly the same feeling of warmth on the skin of both cheeks. With the measured distance d and the known distance r to our Sun, we can estimate the Sun's luminosity L, using the radiation-distance-law S ~ 1/r^{2}.

The 'same feeling' means: Solar constant = "Bulb constant"

The value of d will be about 10 cm, and therefore the value of L can be calculated. It is again too low because of the same reasons as in the first experiment described above. The exact value L = 3,85 x 10^{26} W is so extraordinary great that it is hardly imaginable. Yet compared with other stars: our Sun is only average. There are stars that are more luminous by a factor of 10^{3}...x 10^{4}.

Here are some illustrations which might help to get a better idea of the value of L:

- Because we know the size of the Sun (R = 6,96 x 10^{8} m), we can compute the power flux through every square meter on its surface: it is 63 MW (!).

- To produce the same amount of energy which the Sun radiates out into space in every 1 second, all the power stations in whole Germany have to be on duty day and night for about 120 million years.

- Imagine: Somebody would have switched on a 100 W bulb every 1 second since the Big Bang (14 × 10^{9} years). So until today only 5 × 10^{19} W would be installed (10^{-7} of L).

#### Calculating the Surface Temperature of the Sun

The bigger the star (i.e. amount of the emitting surface area) is and the hotter (i.e. surface temperature) it is, the greater is its luminosity. The physics of 'blackbody radiation' displays the exact formula, known as the Stefan-Boltzmann law:

L = A σ T^{4}

where s is the Stefan-Boltzmann-constant (s = 5,67 × 10^{-8} W m^{-2} K^{-4}), A is the emitting surface and T is the surface temperature.

A perfect 'blackbody' does not reflect any light and absorbs all radiation falling on it. So the emitted radiation entirely results of its temperature, not of its chemical composition. The energy of light depends of its wavelength l: High energy means short wavelength and vice versa. In simple words: "blue" means "hot" and "red" means "cold".

The curves for blackbody radiation of various temperatures show the intensity plotted against the emitted wavelength (Fig. 6).

And we can see: our Sun behaves very much like a perfect blackbody (Fig. 7).

Because the Sun (or any star) is a sphere of radius R, we state it has a surface: A = 4 π R^{2}. With the Sun's radius R = 6,96 × 10^{8} m, we can compute its surface temperature T:

There is also another way to get the value of T by using Wien's law:

*λ _{max} T = b*

where *λ _{max}* is the most intensive wavelength, and b the Wien constant (b = 2,90 × 10

^{-3}mK). This method requires some skill in analysing the spectrum to find out

*λ*, which lies in the yellow-green region.

_{max}Try to observe a spectrum of the Sun with the following simple method: Take a thin needle N to get a reflection beam of sunlight. Observe in this direction with your eye B through a straightview glassprism G (Fig. 8).

The needle acts as an optical split. If you arrange everything well, you should be able to see at least the strongest Fraunhofer lines (Fig. 9).

The temperature in a sunspot is low compared with neighbour area, typically 4000 K (70% of normal T).

Therefore sunspots only emit the fraction of 0,70^{4} = 24% of L.

This is why sunspots look black, compared with the surrounding surface (Fig. 10).

#### Energy transport and Granules

From the fusion zone, energy transport starts by g-radiation. These photons diffuse outward, and at the end of the radiation zone (in the distance of 0.8 R from the center) the temperature has cooled down so much, that the gaseous material can follow the stream of the heat (Fig. 11). This convection process reaches the Sun's surface where the convection cells in a synergetic way form the granules (Fig. 10) that are about 1000 km across and last for about 10 minutes. Hot gas rises from beneath, cools off, and sinks down along the dark boundaries of the cells. The motor of convection is the heat.

We will simulate the mechanism of the convection cells (Bénard cells) and the creation of the granules by a simple model experiment: To a thin layer of paraffin-oil in a glass-bowl, some grainy powder of aluminium is added for a better recognition of the convection flux. By careful heating from below, a granule-pattern arises independently and stays as long as a gradient of temperature exists. One can easily observe that the size of the granules depends on the thickness of the oil-layer (Fig. 12).

#### Exercises

1. Use the value of the Sun's luminosity and confirm by calculation that our Sun loses more than 4 million tons of mass in every second.

2. Confirm by calculation the fact that the Sun lost only 0,03% of its present mass (1,99 x 10^{30} kg) since its creation 4,6 x 10^{9} years (1,45 x 10^{17} seconds) ago.

3. If a sunspot has a temperature of 4200 K, and the solar surface has a temperature of 5800 K, how many times brighter is the surface compared with the sunspot?

4. Estimate the Sun's surface temperature from the graphical spectrum of Figure 7.

5. Calculate the peak wavelength of the Planck curve that corresponds to the core temperature of the Sun. In what part of the electromagnetic spectrum does it lie?

#### Supplement: an experiment a bit more sophisticated

**Measure and calculation of the Solar Constant**

We shall use a device (Fig.13) constituted by a small metallic disc insulated with a tube of PCV and a disc of polystyrene, so that the lost of heat will be as small as possible. The metallic disc is painted in black to absorb all the energy that is coming on it; a disc of glass covers the tube to avoid the lost of infrared radiations emitted by the heated disc. A thermometer, placed in a hole made on the disc, gives its temperature.

The tube is oriented to the sun, with the extremity of the tube covered by a sheet of aluminium until the temperature of the disc is stabilized; the protractor gives us the altitude of the sun above the horizon.

The time of exposure to the sun must be sufficient to have a variation of temperature of some degrees, but not too long because when the temperature of the disc is too high, the loss of energy through the tube is important.

The solar constant is calculated in the same way than above. The mass (m) of the disc, its radius and its caloric capacity (c) are written on it.

The metallic disc received the energy:

*Q = m C (θ _{f} - θ_{i})*

(m: mass; C: calorific capacity; θ: temperature)

The solar constant is:

*P = Q / s t*

(s: surface of the disc π r^{2}; t: duration of the experiment)

**Calculation of the energy received out of the atmosphere**

The solar energy is, in part, absorbed by the atmosphere of the Earth. This loss of energy depends on the distance which is covered by the sunbeams and on the limpidity of the atmosphere. We suppose that the atmosphere has a determined thickness h and a uniform limpidity.

To calculate the energy received out of the atmosphere we have to make several measurements with different altitudes (z) of the Sun and some mathematics! (Figures 14 and 15).

The energy received every second of time, at the abscise x is P(x).

We assume that the loss of energy, between x and x + dx is proportional to the energy P(x).

where:

P(0) corresponding to x = 0, out of the atmosphere

x depends on the angle z and the thickness (h) of the atmosphere: x = h/cos(z)

k depends on the limpidity of the atmosphere.

Ln P(x) is a linear function of h/cos(z). It is Bouguer's law (Fig. 16). (h is unknown, but supposed to be the same for every measurement and so, it only introduces a modification on the scale of the abscise).

With an extrapolation to h = 0, we obtain the value of the solar constant out of the atmosphere. We can then calculate the temperature of the Sun as made above, with Stefan-Boltzman law.

#### References

**Dahringer, F.**, Temperature of the Sun, Proceedings of 2nd EAAE International Summer School, Barcelona, Spain, 1999.

**Freedman, R., Kaufmann, W., Universe, Freeman & Co.**, New York 2004.

**Gaitzsch, R.**, Einführung in die Astronomie am Beispiel der Sonne, Physik 10 Addita, Verlag Oldenbourg, 1995.

**Gaitzsch, R.**, Luminosity and Temperature of the Sun, Proceedings of 5th EAAE International Summer School, Barcelona, Spain, 2001.

An introduction to the Sun and Stars, Cambridge University Press, 2003.