Selected Questions -Answers From All Experts Astronomy Forum (Mass-Luminosity Relation)

Question: What is the mass-luminosity relation and how is it used in astronomy?

Answer:  The Mass-Luminosity relation is a quantitative equation, e.g.


L’/L = (M’/M)^3.5

Or Log (L’/L) = 3.5 Log (M’/M)

where L, M refer to solar luminosity and mass  values and L', M' to stellar values.  The equation  enables us to infer the actual brightness of a main sequence star based on its mass, and also to compare stellar luminosities (intrinsic brightnesses) and deduce other stellar properties  such as mass and radius.  I will provide a number of examples to show how this can be done.

Example (1):

The intrinsic brightness (luminosity) of Regulus is greater than the Sun's by a factor 120. Find the approximate mass of Regulus.

Here: L'/L = 120 so Log (120) = 3.5 Log (M'/ M)

and we are seeking to find M' in terms of M.

Log (120) = 2.079 = 3.5 Log (M'/ M)

Or:

0.594 = Log (M'/M)

Taking antilogs of each side:

3.93 = (M'/M) or M' = 3.93 M

So Regulus is approximately 4 times the mass of the Sun.

Aside:  For proper  stellar properties' analysis we need to refine this to deal with "absolute bolometric magnitudes" because the brighter stars (spectral class O and A mainly) require special "color" corrections usually referred to as "bolometric corrections". This refined system of "bolometric magnitudes" is thereby adjusted so the bolometric corrections are small for stars like the Sun (e.g. G class or later) but large for very hot stars where most of the radiated energy is in the unobservable ultraviolet (UV). The Table shown below  gives bolometric corrections for different temperatures and spectral types:                                                                                                   






















Note the difference (B - V) is the "color index" representing the difference in magnitudes m_B and m_V, e.g. (m_B - m_B) where m_B is the apparent magnitude from a bule filter and m_V is the apparent visual magnitude from the  standard yellow or visual filter - most sensitive to wavelengths near 550 nm. (Nanometers)

Now, if the absolute bolometric magnitude (M_bol) of a star is known, then its luminosity can be found as a function of the Sun's luminosity with the relation:

Log (L'/L) = 0.4 (M_bol - M_bol*)

where M_bol is for the Sun and M_bol* is for the star. Note that any given absolute visual magnitude (M_V) can be changed to an absolute bolometric magnitude by applying a bolometric correction such that:

M_bol* = M_V* + B.C.

Example Problem (2):

The star Almach (Gamma Andromeda) has (B - V) = +1.3 and an apparent visual magnitude m_V* = 2.16. What bolometric correction should be applied? Also, find the absolute visual magnitude M_V* and the absolute bolometric magnitude M_bol* of the star. How does it compare in luminosity to the Sun? (The distance of Almach is 80 pc)

Solution:

From the Table:

We find (B - V) = +1.3 corresponds to B.C. = - 0.92.

The absolute visual magnitude can be found from the apparent visual magnitude by using the distance modulus (see question 14 in the Archive).

Thus:

(m - M) = (m_V* - M_V*) = 5 log (D) - 5

and:

M_V *= m_V* - 5 log (D) + 5 = 2.16 - 5 log (8o) + 5

M_V* = 2.16 - 5(1.903) + 5 = -2.36

The absolute bolometric magnitude is:

M_bol* = M_V* + B.C. = (-2.36) + (-0.92) = -3.28

The relative luminosity as a function of absolute bolometric magnitude is:

Log (L'/L) = 0.4 (M_bol - M_bol*) = 0.4 (4.63 - (-3.28))

Log (L'/L) = 0.4(7.91) = 3.16

Finally:

antilog 3.16 =1445 so: L' = 1445 L
 
So Almach has over 1400 times the intrinsic brightness of our Sun.

To get the hang of it here are two problems to work on your own, following the examples I've given:

1)The apparent V-band (filter) magnitude of a star is 8.72, and it requires a bolometric correction of -0.48. Find the apparent bolometric magnitude of the star. (Hint: Apparent bolometric magnitudes are obtained in an analogous way to the absolute forms)

2) A star has a color index (B - V) of +1.0 and its apparent B-band magnitude is 6.4. The corresponding bolometric correction is - 0.5. Find the apparent bolometric magnitude of the star.

Selected Questions- Answers From All Experts Astronomy Forum(Stellar Brightness- Magnitude Scale)

Question: Can you please explain how stellar brightness is discerned and especially the basis fo rthe stellar magnitude scale which still mystifies me?

Answer:

The easiest way to grasp stellar brightness is to use a simple analogy. Say I have  a 100 watt bulb and place it at a distance of 1 m, then what would its relative brightness be if placed twice as far away? By the inverse square law for light (the intensity of a light source decreases as the inverse square of its distance):

(d1/d2)² = B2/ B1

Where in this case, d1 = 1 m, B1 = 100 w and d2 = 2 d1 = 2m. Then, it’s obvious one needs to find B2.

B2 = B1 (d1/d2)²

But (d1/d2)²  = (1/2)² = ¼


Therefore: B2 = 100 w (¼) = 25 watts

In the astronomical context, one ought to also be able to use the same method in applying the inverse square law to stars. 

For example, say you are given this problem:  Beta Crucis has an apparent magnitude of +0.8 and this is from 261 light years. Find its absolute magnitude.

You'd first need to know  the basis of absolute magnitude (M abs) as that it is just  apparent magnitude (m) defined at a distance of 10 parsecs. If there are 3.26 light years per parsec this means, 10 pc (3.26 ly/pc) = 32.6 light years. Then, for Beta Crucis:

(d1/d2) = 261 ly/ 32.6 ly = 8

I.e. Beta Crucis must be eight times nearer for taking M (abs).

Then (by inverse square law for light):

(d2/d1)²   =  (8)² = 64

So:

Thus, B1= 64 B2. In other words, Beta's brightness at 32.6 light years is roughly 64 times more than its brightness at 261 light years . This converts to a magnitude difference of: 4.5. (See the diagram below for a stellar magnitude scale - with the Sun at the extreme left or brightest end, with an apparent magnitude of (-27):

The layout of the scale is arithmetical, like the algebraic number line, but when related to brightness one obtains ratios not differences. And these ratios are based on a logarithmic relation. For example, the apparent magnitude of the Sun makes it apparently the brightest object in the sky. But this is only because it is at 1 A.U. distance. IF “moved” to 32.6 light years (the standard candle distance to obtain absolute magnitude) then one obtains +4.8 magnitude.

So the new magnitude would be: (0.8 – 4.5) = -3.7.

That is, we locate +0.8 on the scale, then march off 4.5 units on it in direction of increased brightness.

Thus, its real magnitude -  in terms of its actual distance of 261 Ly-    makes it somewhat dimmer than it would be at the absolute magnitude distance.

The basic gist of the  magnitude scale is that every 5 increments (UNITS) in magnitude difference translates into 100 times difference in brightness, because each succeeding magnitude is different from the earlier one by 2.512 times, since:(2.515)⁵ = 100.

Thus a star of +1 magnitude is 100 times brighter than a star of +6 magnitude on the scale, since (6 – 1)= 5.

To make sure you understand the magnitude scale I attach the following problem:

Examine carefully the star magnitude positions on the given magnitude scale.

Which is brighter, Sirius or Castor, and by how many times? Which is brighter, Zosma or Sigma Draconis and by how many times? If a new star Alpha Stellaris is discovered and is found to be 15.6 times brighter than Sirius, estimate how many times brighter this new star is than Sigma Draconis.