Before we begin, for those who didn’t see the About page, we’re sorry but we have to limit posts to one per week. This is because we both have piles of schoolwork to do, but we will try to keep up with one post a week.

This post will involve more about light, as we said it’s quite important. In fact, this post will discuss how we can use light to predict distances from Earth. At this point, math should be expected, along with checking units and significant figures. In fact, we will be introducing one of the most important equations for distance calculations in an Astronomer’s arsenal.

Now we go on the next part of our journey, to Greece of course, where we are joined by the man Hipparchus. He developed a system of apparent magnitudes (denoted as m), which determines how bright stars were by looking at them here on Earth. For some reason he decided that it would be more logical to say that as the numbers decreased the stars became brighter, resulting in the scale ranging from m=1, the brightest stars, to m=6, the dimmest. This was mainly because no highly accurate equipment was available, but this is still extremely important as it describes how objects would appear from an observer on Earth. Originally the system was just based on naked-eye observations, but modern astronomers decided to fix it up.

Now the scale is logarithmic and compares ratios of apparent magnitudes for stars. Apparent magnitude is now considered to be brightness or flux measured in Watts per square meter. It was decided on this scale that 100 would correlate to a magnitude difference of m=5. This should be emphasized as a difference since for the brightness ratio of B_{1}/B_{2} should be equal to the magnitude difference of m_{2}-m_{1 }with the formula:

B_{2}/B1=100^{(m1-m2)/5}

Taking the log of both sides we get:

m_{1}-m_{2} = -2.50 log(B1/B2)

With that we can show that when the brightness ratio equals 100 then we take log(100) which equals 2, multiplying by -2.50 to get -5. But this still works since the scale shows that the object is brighter as the magnitude value decreases. In addition, that means that if you were to measure between one magnitude it would be a factor of 10^{0.4 }which is equal to approximately 2.512 since it comes from 100^{1/5}. So, a 1st magnitude star would be 2.512 times as bright as a 2nd magnitude star, and 2.512^{2 }or about 6.310 times as bright as a 3rd magnitude star. Also, Hipparchus’s scale has had an increased range of magnitudes. The Sun for example is now m=-26.83.

Next we need to establish how we can show radiant flux, denoted F, or those brightnesses. Earlier we mentioned it as watts per square meter, which is exactly shown by a familiar manipulation using light and surface area of a sphere. This is the inverse square law (we can call this brightness or flux, we will now be using F for flux):

F=L/(4πr2)

Now that we have explained how we can view objects with the unaided eye and defined brightness we have to show how to find the actual magnitudes of all objects. How would this be done, though? Astronomers decided to create a system of absolute magnitudes, denoted as M or M_{v, }which shows what the magnitude of stars and objects would be at a set distance of 10 parsecs. This works since instead of having all sorts of objects with different actual magnitudes and different distances from the Earth an established sphere of points can be used to better show the magnitudes. With that and the inverse square law in mind we can create a flux ratio to show how much the magnitudes would be from this set distance. Again, 5 magnitudes separate the apparent magnitudes of two stars which would show a flux ratio of 100. This is the very same brightness comparison formula we had earlier. We can actually manipulate that into something called the distance modulus. We can say that:

100^{(m-M)/5}=F_{10}/F=(d/10 pc)^{2}

This shows that for the flux ratio of a star’s apparent magnitude to its absolute magnitude and for F_{10}, which would show how the star would appear from 10 parsecs, would equal the distance to the star if it were at 10 pc away. This can therefore have multiple manipulations to show a star’s distance away:

d=10^{(m-M+5)/5} pc

Or a star’s apparent and absolute magnitudes:

m-M=5log(d/10)=5log(d)-5

If you were wondering how this could be useful if we don’t necessarily know the distance or absolute magnitudes of every star (you could certainly find the apparent magnitude as it is defined by how an observer would see it from Earth), that is a very good question. Later we will discuss that for certain stars the absolute magnitude is extremely consistent and can be used to find distances very well.

There is still more to this story. The apparent and absolute magnitudes mentioned are measured as bolometric magnitudes, which detect flux from a star across all wavelengths of light. It would be nice to do this, but it is generally easier to target specific wavelengths especially since certain objects can be analyzed better in them. For this we have to look at what we use. UBV wavelength filters are used to find a star’s apparent magnitude and color. U is the ultraviolet magnitude with a filter at 365 nm and bandwith of 68 nm, B is blue magnitude with a filter at 440 nm and a bandwith of 98 nm, and V is for the visual magnitude (sometimes considered green) with a filter at 550 nm and a bandwith of 89 nm. This creates multiple color indices which compare the different wavelength filters to show a star’s apparent or absolute magnitude. The actual device, the bolometer, uses an association between temperature and color as mentioned in the last post to show them. The indices are differences between magnitudes of the U, B, and V to equal absolute magnitudes shown as:

U-B = M_{u} – M_{b}

and

B-V = M_{b} – M_{v}

Since we already noted as the magnitudes increase the brightness decreases we can say that a star with a smaller B-V index would be bluer (as the blue was filtered out more) and would show both that the star is brighter and hotter. The same would apply for the U-B in that lower values would be more ultraviolet and therefore be brighter and hotter as well. So, overall the purpose of U-B and B-V is to quantitatively show what the color and temperature of a star is.

With this we can next say that there is the bolometric correlation, or BC, which shows the comparison between bolometric and visual magnitudes (m_{bol}and M_{bol }are really just m and M):

BC = m_{bol}– V = M_{bol} – M_{v}

There is another factor influencing these formulas. It is called interstellar extinction which creates the effect known as interstellar reddening, denoted A as another magnitude. When it isn’t noted in the question you should ignore this, but it is good to know all the factors influencing this important distance equation. Interstellar extinction refers to the presence of interstellar dust that absorbs or scatters light from an object. The effect is stronger at shorter wavelengths, which interact more strongly with dust. Therefore, red light can be seen more, and if something appears more red than it “should”, then dust is present. This was proven after comparison between expected and observed emissions showed that there was an inaccuracy. If a question mentions some amount of reddening the following corrections are made:

The distance modulus becomes d = 10 ^{0.2 (m – M + 5 – AV)}

B-V values since the color is changed it becomes

True color = (Bo-Vo), Observed color is (B-V)

(B-V)=(B+Ab) – (V+Av)

(B-V)=(Bo-Vo) + (Ab – Av) = Intrinsic color + color excess

Since extinction will occur more in the lower wavelengths, this increases the V values relative to the B or U values. A test can also ask about how this can show in ratios, where it would show Ab/Av or Au/Av. In this case you would have to be given the value of the Av, and then multiplying the two ratios by the Av would get Ab and Au, such that you can correct either your B-V or U-B values.

The last thing to note is the color-color diagram. This relates U-B and B-V indices for stars, and it can show temperature and color as well. Stars actually aren’t perfect blackbodies, so even if they get close the diagram won’t form a straight line. Here is an example, but know that a color-color diagram can be applied to objects with many stars, which will look different:

==========

TL;DR: Developing methods of organizing stars and understanding distances is important in Astronomy since this allows Astronomers to better understand our place in the universe and to construct formulas which explain it well quantitatively. Apparent magnitude is how bright something appears to be, while absolute magnitude shows how bright something actually is from a set distance of 10 pc. With this the distance modulus can be derived to find the distance to most objects. Color indices also are studied to show the temperature, color, and characterize stars better. Lastly, a correction must be made for interstellar dust.

==========

Sources and further reading:

- http://curious.astro.cornell.edu/question.php?number=569
- http://www.phys.ksu.edu/personal/wysin/astro/magnitudes.html
- http://www.astronomynotes.com/starprop/s4.htm
- http://www.astro-tom.com/technical_data/magnitude_scale.htm
- http://spiff.rit.edu/classes/phys301/lectures/absolute/absolute.html
- http://csep10.phys.utk.edu/astr162/lect/stars/magnitudes.html
- http://csep10.phys.utk.edu/astr162/lect/stars/cindex.html
- http://astro.wku.edu/labs/m100/mags.html
- http://ugastro.berkeley.edu/infrared/ir_clusters/dustandstarcolors.pdf
- http://burro.astr.cwru.edu/Academics/Astr221/Light/colors.html
- Carroll and Ostlie,
*An Introduction to Modern Astrophysics*(2nd edition), p. 60-63, 75-79