Estimating Star Counts from Density Laws

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SUMMARY

This discussion focuses on estimating star counts from density laws, specifically using two models: a constant density law and an exponential density law. The constant density is defined as 0.1 stars per cubic parsec for distances up to 800 parsecs, while the exponential density law uses a scale height of 300 parsecs. Participants clarify the relationship between absolute magnitude and apparent magnitude, emphasizing the importance of knowing the absolute magnitude of the stars being counted, which is given as 3.5, to determine distances and star counts accurately.

PREREQUISITES
  • Understanding of stellar absolute and apparent magnitudes
  • Familiarity with density laws in astrophysics
  • Knowledge of basic astronomical distance measurements in parsecs
  • Proficiency in using logarithmic functions for data analysis
NEXT STEPS
  • Explore the application of the distance modulus formula in stellar astronomy
  • Investigate methods for fitting observed star counts to density laws
  • Learn about the implications of different density laws on star distribution
  • Study techniques for visualizing astronomical data, such as plotting log-star counts
USEFUL FOR

Astronomers, astrophysics students, and researchers involved in stellar population studies and data analysis in astronomy.

June_cosmo
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Homework Statement


(This is a long problem but I think the question is not hard.)

Determining the star density from star counts is tough, but estimating counts from a density law is simpler. In practice, this method of fit-ting observed number counts to an assumed density law is becoming a more common approach given the large datasets available. In this problem we will explore counts along the z-axis. For this problem, consider two density laws (to be clear, number densities being considered in this problem which is the total number of something being counted per unit volume). First, 1(||)=0 out to some distance where 0=0.1 −3 is a non-zero constant for ||≤800 and 0=0 for ||>800 . Second, an exponential law in which (||)=0^(-||/ℎ) the same constant 0 and a scale height ℎ=300 . Assume you have isolated a type of star in your counts that you know has an absolute magnitude =3.5 and that you can count stars to a faint magnitude limit =20.5. Finally, assume that we are conducting the counts in a solid angle of 0.05 steradians centered on a Galactic pole.

With this information you can make a table listing the following information: (i) apparent magnitude from 6.0-20.5 increments of 0.5 mag; (ii) the distance in pc of stars of the type we are counting each magnitude bin; (iii) the volume of space in ^3 contained within each magnitude bin; (iv) the density of stars in ^3 in each magnitude/distance bin according to both density laws and (v) the total number of stars in each magnitude/distance bin according to both density laws. Plot the log of the resulting star counts as a function of magnitude for both density laws.

Homework Equations


m = M+5(logD-1),where M is the absolute magnitude and m the apparent magnitude, D is the distance between the observer and the star

The Attempt at a Solution


for (ii), since we don't know the corresponding M,how do we know the distance? Did I miss something...
 
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June_cosmo said:
for (ii), since we don't know the corresponding M,how do we know the distance? Did I miss something...
How do you interpret this statement:
June_cosmo said:
Assume you have isolated a type of star in your counts that you know has an absolute magnitude =3.5 and that you can count stars to a faint magnitude limit =20.5. F
?
Sounds to me that you have selected stars with absolute magnitude 3.5, i.e. those are the only stars to be counted. You are in a position to count all of those with apparent magnitude 20.5 or less.
 
haruspex said:
How do you interpret this statement:

?
Sounds to me that you have selected stars with absolute magnitude 3.5, i.e. those are the only stars to be counted. You are in a position to count all of those with apparent magnitude 20.5 or less.
Oh I didn't see your reply before. Thanks! You're right. That makes a lot more sense!
 

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