HOW DID MAX PLANCK MEASURE
THE DIAMETER OF A STAR?

A star has various magnitudes: 1. the apparent visual magnitude m_v, as the star appears to the eye; 2. absolute magnitude M_V as the star would appear at a distance of 10 parsecs, or 32,6 light years. At this distance the Sun would be a dim star of apparent visual magnitude 4,85; 3. the bolometric magnitude M_bol which is a measure of the total energy flux from the whole surface of the star at all wavelengths taken together. The total flux energy depends in a large measure on the temperature of the star. Max K. E. L. Planck argued that if the temperature of a star could be determined, the bolometric magnitude will be an indicator of the size of the radiating surface and from this the diameter can be calculated.

Max Planck based his calculation of the temperature of a star on the concept known as a black body. A black body is one which absorbs all the energy which it receives and then re-radiates the same total amount of energy, without any loss. The body is therefore in thermal equilibrium with its environment. The centres of stars comply with the definition of a black body; but their surfaces only approximate to the definition, although nearly so.

By applying the principles of astro-physics, Planck derived three equations, the first of which linked a star's temperature with its colour index C. The colour index of a star is simply equal to B - V, i.e. the star's magnitude B as seen through a blue filter which allows light of wavelength 4200 angstroms to pass through minus the star's magnitude as seen through a yellow filter V, which allows light of wavelength 5400 angstroms to pass through.

1. Planck's first equation gives the star's temperature T in terms of its colour index C

.

Let us apply this to the case of Sirius which is of spectral type A1 of which the colour index is zero.

Therefore = 13717, say 13700° K.

(Compare this to the Sun's temperature 6000°K (degrees Kelvin)).

2. Planck's second equation derives the bolometric correction B which has to be applied to the star's absolute magnitude M to yield its bolometric magnitude M_bol.

Bolometric correction

.

In the case of Sinus whose surface temperature we have found to be 13700°K:

.

= 42,60 - 2,07 - 10 x 4.137

= 40,53 - 41,37 = -0,84.

The absolute magnitude of Sirius is 1,42. Apply to this the correction of B, -0,84.

M_bol = M_V - 0,84 = 1,42 - 0,84 = 0,58

The diameter of Sirius can now be calculated from Planck's third equation:

M_bol = 42,35 - 10 log T - 5 log D,

where D is the diameter of the Star.

We see that 5 log D = 42,35 - M_bol - 10 log T

= 42,35 - 0,58 - 10(4,137)

= 41,77 - 41,37

= 0,40

\ log D = 0,40 ¸ 5 = 0,08

D is the antilog of 0,08 and it equals 1,2.

The diameter of Sirius is therefore 1,2 times that of the Sun., namely 1,2 x 1 392 530 = 1 671 000 km.

As an exercise, see if you can calculate the diameter of Arcturus a K2 I Up star of colour index 1,23; apparent visual magnitude - 0,04 and parallax 0,09".

First you will have to calculate the absolute magnitude of Arcturus from the formula:

MV = mv + 5 + 5 log P

= -0,04 + 5 +5 log 0,09

= -0 04 + 5 + 5(-1,046) = - 0,27

The absolute magnitude of Arcturus is thus -0,27 and its colour index 1,23. You carry on. You ought to get an answer of 24 times the Sun. By interferometry Arcturus subtends an angle of 0,02 seconds of arc That is the angle subtended at Arcturus by the average distance of the Earth from the Sun, i.e. 1 A U or 107 times the diameter of the Sun.

  (because the angle is very small)

(because the angle is very small)

Trigonometrically the diameter of Arcturus works out to 23,78 times the diameter of the Sun, in perfect agreement with the value of 24 times derived by Max Planck.

Jan Eben van Zyl