# Initial Mass Function physics problem

• cluckaluck
In summary: 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.Hi cluckaluckIn summary, the IMF for a globular cluster of stars is dN/dm=am^(-2.35) and the total mass of the cluster is 10^6 solar masses.

#### cluckaluck

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.

Hi

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.

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.

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.

## 1. What is the initial mass function (IMF) in physics?

The initial mass function is a probability distribution function that describes the distribution of initial masses of stars in a stellar population. It is used to determine the relative number of stars with different masses, and is an important tool for studying the formation and evolution of galaxies.

## 2. How is the IMF measured?

The IMF is typically measured by observing the number of stars in a given mass range in a stellar population. This can be done through various methods such as spectroscopy, photometry, and astrometry. The IMF can also be inferred from the luminosity and mass of a galaxy.

## 3. What factors influence the IMF?

The IMF is influenced by various physical processes such as star formation, stellar evolution, and interactions between stars. Other factors, such as the metallicity and age of a stellar population, can also affect the shape of the IMF.

## 4. What is the significance of the IMF in astrophysics?

The IMF is a fundamental concept in astrophysics as it helps us understand the formation and evolution of galaxies. It also provides valuable insights into the physical processes involved in star formation and the properties of stellar populations.

## 5. Are there different types of IMF?

Yes, there are various types of IMF that have been proposed based on different observations and theoretical models. Some examples include the Salpeter IMF, the Chabrier IMF, and the Kroupa IMF. These different IMFs can have varying shapes and can be used to study different types of stellar populations.