Example Problem: Apparent and Absolute Magnitude


 

Example:      A star has apparent magnitude 6 and is 250 light-years from Earth.What is the star’s absolute magnitude?
 

Solution:       Problems involving magnitudes can usually be worked either of two ways: the “physical logic” way and the purely mathematical way.Personally, I like to work them both ways as a check on my math (or logic).So here are the two ways to work this problem.

 

                    First, we have to convert the star’s distance in light-years to parsecs.Why parsecs and not meters?Because absolute magnitude is defined as the magnitude a star WOULD have IF it were at a distance of 10 parsecs (pc).So we convert 250 light-years to parsecs by dividing by 3.3.Thus:

 

                        Distance = 250 Lyrs x (1 parsec/3.3 Lyrs) = 75.75 pc

 

                    The star is a whopping (OK, semi-whopping) 75.75 parsecs from Earth.So would it be BRIGHTER or DIMMER if it were at 10 parsecs?

 

                    Of course, it would be brighter.How many times brighter?Well, how many times farther than 10 pc is the star?75.75/10 = 7.575.So it’s about seven and a half times farther than 10 pc.Now here’s the “logic” part.We know that brightness decreases as the SQUARE of the distance (something twice as far away is four times dimmer).So if this star is 7.575 times farther away than it would be if it were at 10 pc, it must be (7.575)2 = 57.3 times DIMMER than it would be if it were at 10 pc.Put another way, it would be 57.3 times BRIGHTER if it were at a distance of 10 pc instead of its actual distance of 75.75 pc.So what would its magnitude be if it were at 10 pc (because its magnitude at 10 pc IS its absolute magnitude)?

 

                    Well, the star’s apparent magnitude is 6, so if it were closer (and therefore brighter), its magnitude would have to be something LESS than 6 (remember, smaller magnitudes mean brighter objects).But how many steps in magnitude would it gain if it were 57.3 times brighter?

 

                    To figure that out, you have to figure out how many magnitude steps equal a brightness change of 57.3.And since each step in magnitude is a factor of 2.5 in brightness, the question is this: how many times would I have to multiply 2.5 by itself to get 57.3?In other words, to what power would I have to raise 2.5 to get 57.3?

                    If you don’t know about logarithms, you can do this just by trying different powers of 2.5 on your calculator.It’s got to be more than one (since 2.51 is just 2.5), so let’s start with 2.OK, 2.52 is 6.25.Too small (we need 57.3).So let’s try 2.53: nope, that gives 15.63.OK, 2.54: hmmm, 39.1. Getting close.One more ought to do it.2.55 = 97.7, nope, that’s too big.So somewhere between 4 and 5 magnitude steps is equivalent to a brightness factor of 57.3.So the star’s absolute magnitude (the magnitude it would have if it were at a distance of 10 pc) is between 4 and 5 magnitudes smaller than 6.That means it’s absolute magnitude is between 1 and 2.

 

                    If you’re so inclined, you can get a more-precise answer by trying half-magnitude steps, or even tenth-magnitude steps (that is, raising 2.5 to the 4.1 power, then 4.2 power, and so on, until you get close to 57.3).You should find that the answer is something like 4.42 magnitude steps, so the star’s absolute magnitude is 6 – 4.42 =1.58.

 

                    Right about now, you’re probably saying to yourself, OK, this takes too much thinking.What’s the mathematical approach?

 

                    So here it is.There’s a parameter called the “distance modulus” or DM of a star that is defined as simply the difference between the star’s apparent magnitude and absolute magnitude.So

                   

                                    DM = apparent magnitude – absolute magnitude 

 

                    What good is that?Well, the distance to any star is just 10 parsecs times 10 raised to the power of DM/5.That’s right:

 

                                    Distance (pc) = 10 pc x 10DM/5

 

                    So if you know the apparent and absolute magnitudes, you just subtract the absolute from the apparent, divide by 5, raise 10 to that power, and multiply by 10.And that’s the distance.

 

                    But what if, as in this problem, you HAVE the distance but you don’t have one of the magnitude values (the absolute magnitude in this case)?

 

                    No problemo.Just use this form of the equation:

 

                        5 log10(Distance(pc)/10pc) = DM

 

                    where log10 means “logarithm base 10.”

 

If you prefer, you can use

                   

                    DM = apparent magnitude – absolute magnitude = 5 log10(Distance(pc)/10pc)

 

                    So for this problem, just move absolute magnitude to one side of the equation, giving

 

                        Absolute magnitude = apparent magnitude – 5 log10(Distance(pc)/10pc)

 

                    Plugging in the numbers,

 

                        Absolute magnitude = 6 – 5 log10­(75.75/10) = 6 – 5 (0.879) = 1.6

 

                    Same as we got the other way.Was this so bad?