Calculating the Flux Ratio of two stars

[Kellytabares]

Hi there.
 
When calculating the Flux ratio of two stars they give me the following formula:

Everything is good as I jut need to replace the values. However I don't understand why an antilog is needed in both sides plus where the 1/100 come from? I can't seem to get my head around it at all. Find the exercise below.

Can I get an explanation from the antilog onwards?

So, after that, I go to the test practice and I find this:

-Taking into an account the formula from above. Why do I have three values of magnitude for this question?

I'm so sorry if what I'm asking is stupid, but I can't really make it happen inside my head now :/

Thank you so much in advance.

[michael.h.f.wilkinson]

Magnitudes form a logarithmic scale; flux is linear: a certain magnitude difference corresponds to a certain factor by which one flax is larger than the other. Thus, to convert from flux ratio to magnitude difference, you need to take a logarithm. By contrast, going from magnitude difference to flux ratio you need to take the inverse of a logarithm (= the antilog).  In the example of a 5 magnitude difference, you first divide that by -2.5, obtaining -2, and then raise ten to that power (10^-2), which is just 1/100. In the exercise, you are not given three magnitudes, but three magnitude differences, each of which must be converted to flux ratio. Hope this helps.

[Stub Mandrel]

So whatever magnitude the first star has for a given difference in magnitude the relative change in flux will always be the same. When a value always changes by a given mutiplier for a step of a given size this is called a log law because logarithms describe it perfectly (no coincidence, logs are defined so they work this way which is why slide rules work - on a slide rule the gap between 2 and 8 is the same as between 20 and 80 or between 4 and 16, for example).

By definition two stars with a flux difference of 2.5 times is defined as a magnitude difference of 1.

In your example the given difference of 5 magnitudes has to be converted from a log magnitude value to a flux value, so you need the antilog. The 2.5 is just a scaling factor so the magnitude scale coincides more or less with the old magnitude scale used before it was accurately defined.

To go a bit deeper the eye exhibits a (more or less) logarithmic response to light levels so the difference between bright and not so bright stars is  perceived as much less, and that's why the magnitude scale is used for visual observation rather than flux.

F1    F2      Mag difference
1      0.2      -2
1      0.4      -1
1      1.0      0
1      2.5      1
1      6.3      2
1    15.8      3
1    39.8      4
1  100.0      5
1  251.2      6
1  631.0      7
1 1584.9     8
1 3981.1     9
1 10000.0 10

A contrasting example is when you have a square law, such as how light increases by the square of the diameter of a scope or camera's aperture. This is encapsulated by the F-ratio steps on a camera lens which typically go up in steps of the square root of two (1.41) but mean a doubling of the amount of light:

The numbers are usually heavily rounded but might be: 2 2.8, 4, 5.6 8, 11, 16, 22, 32.

Scientifically, it would have probably made more sense to use a square law relationship between star magnitudes but obviously the need to relate it to a pre-existing 'visual' scale won the day.

[acey]

For simplicity write the flux ratio F_m/F_n as R, and write the magnitude difference m-n as D. Then the definition is D = -2.5logR, where log denotes base-10 logarithm. You are asked to calculate R for three values of D.

a) D=1, so 1 = -2.5logR, and you need to solve for R. We have

logR = -1/2.5 = -0.4

so R = 10^-0.4 = 0.398

Do the rest the same way.

[robin_astro]

By definition two stars with a flux difference of 2.5 times is defined as a magnitude difference of 1.

 A bit late, I know but it is a boring Sunday afternoon :-)

Actually although often quoted, this is not quite correct.  The definition of magnitude is based on   2 stars with a flux ratio of 100 corresponding to a magnitude difference of 5.

This means that a magnitude difference of  1 = a flux ratio of 2.512....  ( ie 2.512... ^5 = 100)  

Conversely log (100) = 2.  Multiplying this by the 2.5 factor gives 5 magnitudes

Robin