__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:

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

*B**eta 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.

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

Thus,
its

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)

Thus a star of +1 magnitude is

*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.

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