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Initial Mass Function physics problem

  1. Apr 12, 2009 #1
    Hello, I am having a lot of trouble understanding exactly what the initial mass function is.
    I have a problem in which I am given the IMF for a globular cluster of stars, dN/dm=am^(-2.35) and the total mass of the cluster, 10^6 solar masses, and asked to find the constant a. Could someone help me with this as well as describe what units are involved with IMF's and integrating them? Thank you so much.
  2. jcsd
  3. Apr 12, 2009 #2

    Well units is easy because you've already used "solar masses" - which is the mass of the Sun. The minimum stellar mass is ~0.08 solar masses and the maximum is ~150 solar masses, at least for current metallicity levels.

    Now total number for a given IMF is a bit trickier. You know the number distribution for a given mass range - that's the IMF - and you integrate that for the total number. In the range I gave it's then Q = (30.256/1.35)*a/(Msol^1.35), which doesn't readily translate into a number, but allows you to work out relative numbers in different mass ranges.

    But notice the ratio a/(Msol^1.35.) If we take Msol to be 1, then for a given number of stars, Q, working out a is obvious, a = Q/22.4.

    Let's find the total mass. The integral becomes a*S [the integral 'long S'] M*M^(-2.35) dM, thus total mass of stars, m, across the previous mass range is (2.24/0.35)*a/(Msol^0.35), or (6.42*a) solar masses. Thus a is 1,000,000/6.42 (Msol^1.35) for the example you give. Note the unit.
  4. Apr 13, 2009 #3
    Thanks for your help. I understand almost all it, but I not how you found the total mass. Why did you insert another M into the integral? It comes out to the right answer, but I don't understand the logic behind it. Thanks again.
  5. Apr 13, 2009 #4
    Hi cluckaluck

    Isn't it obvious? The distribution function tells you how many objects of a given mass, so you multiply by that mass to get how much mass there is. Think about it. Took me a while, but it made sense eventually. I had to read a text about distribution functions to really understand what was going on.
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