CHAPTER study stellar multiplicity as a function

CHAPTER 3Binarity in A Uniform Cluster for Flat / Steep IMFs3.1 IntroductionIn Astronomy, there is two ways of study for the investigation over a group of objects. In the first case, a separate study of individuals in a group is encouraged.

Second case deals with the distribution of the members of the group under study. Since binary system is a special case of multiple systems, in order to study about the formation of binaries it is significant to know about stellar multiplicity 1. Hence in this dissertation we study stellar multiplicity as a function of primary mass in stellar clusters with flatter / steeper IMFs than Salpeter. 3.2 Binary Formation by Capture- The Condition for BindingFormation of a single binary system may happen in accordance with three basic theories.Capture theory: According to the theory, two independent stars may deviate from their own path and will orbit around their common centre of mass under suitable conditions. Fission theory: The theory suggests that, a single star in its initial state or afterwards may breaks into two to form a binary star system.Independent Nuclei Theory: If the constitute of the nebulae condense to form two nuclei separated by appropriate distance to become a binary star system this process takes place as per the independent nuclei theory.

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In this dissertation we are considering binaries formed by capture process. Capture theory was put forward by Dr. G. Johnston Stoney. He says that if two stars come closer to each other by their usual motions then one of the three things will happen 2. Either they will pass each other without interrupting the visibility of the companion, or they caught together and look like the fragments of a single star, or lastly the stars may come face to face without affecting the visibility of each other. In the third case the stars will revolve in elliptical orbits after the event.

Since density is not uniform in the atmosphere of each star the resistance act as forces. This force may act tangentially and reduce the periastron distance or it may act normally and increase the distance 3. The force dominates each other in a special way and one star will fall into the other. This process results in the formation of binaries 2.Total energy of a system is the sum of its kinetic energy and potential energy.

Imagine two stars are seen to be isolated from other stars. Let the total energy of the combined system is Etotal , then If Etotal < 0, stars are bounded or stars forms a binary system. If Etotal > 0, stars are unbounded and excess energy should be dissipated in any way to form a binary system. Another approach to the boundedness which is used as the binding condition in this dissertation is given below, Let mi and mj are the masses of two stars which are expected to form a binary.

For such a system to be bounded its potential energy should be greater than its kinetic energy.Gmimjrij > miVi-Vj2Where mi > mjG is the gravitational constantVi, Vj – velocities of mi mass star and mj mass star respectively.ri , rj – radius of mi mass star and mj mass star respectively.rij = | ri – rj |3.3 Stellar Multiplicity as a Function of Primary MassIn order to study the stellar multiplicity as a function of primary mass, we start with the process of random sampling from Salpeter function. Firstly we designed a program in the R language to draw 1000 random numbers between 0 and 1 from a uniform distribution and the value of unknown mass can be calculated using the equation 2.4. Salpeter’s slope also obtained from the program.

Second step consists of randomly drawing velocity from a Gaussian distribution with mean zero and varying standard deviation from 0.1 to 1.3. Our aim is to study about binary systems in stellar clusters with IMF flatter / steeper than Salpeter. For this study we developed a program for a cluster having Salpeter IMF with a slope of 2.3. Then compared with clusters having steeper (slope = 3.3) and flatter IMF (slope=1.

3, 1.7) than Salpeter IMF. Next we developed a program for a cluster with IMFs 1.3, 2.3, 3.

3 and for a cluster with IMFs 1.7, 2.3, 3.3.All programs are developed in the R language and are given in the appendix.3.4 ResultsCalculation of unknown mass Coefficients:(Intercept) w 2.

945 -1.349 W is the slope of the graph which is the same as Salpeter’s slope.Plot 3.1: Salpeter’s slope Binary frequency as a function of primary mass for a flat IMFCase 1: slope =1.

31. Sd = 0.1″bounded”499500″notbounded”0Plot 3.2: obtained by taking standard deviation = 0.1 2.

Sd = 0.2″bounded” 499198″notbounded” 302Plot 3.3: obtained by taking standard deviation = 0.2Sd = 0.3″bounded”493810″notbounded”5690Plot 3.4: Obtained by taking standard deviation = 0.

3Sd = 0.4″bounded”482970″notbounded”16530Plot 3.5: Obtained by taking standard deviation = 0.4Sd = 0.5″bounded” 467039 “notbounded”32461Plot 3.6: Obtained by taking standard deviation = 0.5Sd = 0.6″bounded”451460″notbounded”48040Plot 3.

7: Obtained by taking standard deviation = 0.6Sd = 0.7″bounded”430286″notbounded”69214Plot 3.8: Obtained by taking standard deviation = 0.7Sd = 0.8″bounded”413555″notbounded”85945Plot 3.9: Obtained by taking standard deviation = 0.

8Sd = 0.9″bounded”395251″notbounded”104249 Plot 3.10: Obtained by taking standard deviation = 0.9Sd = 1″bounded”380665″notbounded”118835 Plot 3.11: Obtained by taking standard deviation = 1Sd = 1.

1″bounded”367988″notbounded”131512 Plot 3.12: Obtained by taking standard deviation = 1.1Sd = 1.2″bounded”350388″notbounded”149112Plot 3.13: Obtained by taking standard deviation = 1.

2Sd = 1.3″bounded”345426″notbounded”154074 Plot 3.14: Obtained by taking standard deviation = 1.3Case 2: Slope = 1.7Sd = 0.1″bounded”499500″notbounded”0Plot 3.15: Obtained by taking standard deviation = 0.12.

Sd = 0.2″bounded” 499198″notbounded” 302Plot 3.16: Obtained by taking standard deviation = 0.23. Sd = 0.3″bounded”493810″notbounded”5690Plot 3.17: Obtained by taking standard deviation = 0.3Sd = 0.

4″bounded”482970″notbounded”16530Plot 3.18: Obtained by taking standard deviation = 0.4Sd = 0.5″bounded” 467039 “notbounded”32461Plot 3.19: Obtained by taking standard deviation = 0.5Sd = 0.6″bounded”451460″notbounded”48040Plot 3.20: Obtained by taking standard deviation = 0.

6Sd = 0.7″bounded”430286″notbounded”69214Plot 3.21: Obtained by taking standard deviation = 0.7Sd = 0.8″bounded”413555″notbounded”85945Plot 3.22: Obtained by taking standard deviation = 0.8Sd = 0.9″bounded”395251″notbounded”104249Plot 3.

23: Obtained by taking standard deviation = 0.9Sd = 1″bounded”380665″notbounded”118835Plot 3.24: Obtained by taking standard deviation = 1Sd = 1.1″bounded”367988″notbounded”131512Plot 3.25: Obtained by taking standard deviation = 1.

1Sd = 1.2″bounded”350388″notbounded”149112Plot 3.26: Obtained by taking standard deviation = 1.2Sd = 1.3″bounded”345426″notbounded”154074Plot 3.

27: Obtained by taking standard deviation = 1.3Binary frequency as a function of primary mass for steep IMFSlope = 3.3Sd = 0.1 “bounded” 499499″notbounded” 1Plot 3.

28: Obtained by taking standard deviation = 0.1Sd = 0.2″bounded”498969″notbounded”531Plot 3.29: Obtained by taking standard deviation = 0.2Sd = 0.3″bounded”489340″notbounded”10160Plot 3.

30: Obtained by taking standard deviation = 0.3Sd = 0.4 “bounded”472884″notbounded”26616Plot 3.

31: Obtained by taking standard deviation = 0.4Sd = 0.5″bounded”444376″notbounded”55124Plot 3.32: Obtained by taking standard deviation = 0.

5Sd = 0.6″bounded”423807″notbounded”75693Plot 3.33: Obtained by taking standard deviation = 0.6Sd = 0.7″bounded”395298″notbounded”104202Plot 3.34: Obtained by taking standard deviation = 0.7Sd = 0.8″bounded”375378″notbounded”124122Plot 3.

35: Obtained by taking standard deviation = 0.8Sd = 0.9″bounded”355984″notbounded”143516Plot 3.36: Obtained by taking standard deviation = 0.9Sd = 1 “bounded”333765 “notbounded” 165735Plot 3.37: Obtained by taking standard deviation = 1Sd = 1.

1″bounded”316659″notbounded”182841Plot 3.38: Obtained by taking standard deviation = 1.1Sd = 1.2″bounded”301037″notbounded”198463Plot 3.39: Obtained by taking standard deviation = 1.2Sd = 1.

3″bounded”287555″notbounded”211945Plot 3.40: Obtained by taking standard deviation = 1.3Binary frequency as a function of stellar mass for a cluster with IMFs 1.3, 2.3, 3.3.

Case1: In these graphs normalization is done by diving each and every y value by sum of all y values.1.Plot 3.41: A cluster with IMFs 1.3, 2.3, 3.

3 having standard deviation = 0.1 2. Plot 3.42: A cluster with IMFs 1.3, 2.3, 3.

3 having standard deviation = 0.23. Plot 3.43: A cluster with IMFs 1.3, 2.

3, 3.3 having standard deviation = 0.3 4.Plot 3.44: A cluster with IMFs 1.3, 2.3, 3.

3 having standard deviation = 0.4 5. Plot 3.

45: A cluster with IMFs 1.3, 2.3, 3.3 having standard deviation = 0.56.Plot 3.

46: A cluster with IMFs 1.3, 2.3, 3.3 having standard deviation = 0.67. Plot 3.47: A cluster with IMFs 1.3, 2.

3, 3.3 having standard deviation = 0.78.Plot 3.48: A cluster with IMFs 1.

3, 2.3, 3.3 having standard deviation = 0.8 9. Plot 3.49: A cluster with IMFs 1.3, 2.

3, 3.3 having standard deviation = 0.910. Plot 3.50: A cluster with IMFs 1.3, 2.

3, 3.3 having standard deviation = 1 11. Plot 3.51: A cluster with IMFs 1.3, 2.3, 3.

3 having standard deviation = 1.1 12 Plot 3.52: A cluster with IMFs 1.3, 2.3, 3.3 having standard deviation = 1.

213. Plot 3.53: A cluster with IMFs 1.3, 2.3, 3.

3 having standard deviation = 1.3 Case 2: In these graphs normalization is done by dividing each and every y value by ym, where ym is the value of y when x = log0 1. Plot 3.54: A cluster with IMFs 1.3, 2.

3, 3.3 having standard deviation = 0.12.

Plot 3.55: A cluster with IMFs 1.3, 2.3, 3.3 having standard deviation = 0.23.Plot 3.56: A cluster with IMFs 1.

3, 2.3, 3.3 having standard deviation = 0.34.

Plot 3.57: A cluster with IMFs 1.3, 2.3, 3.3 having standard deviation = 0.45.

Plot 3.58: A cluster with IMFs 1.3, 2.3, 3.3 having standard deviation = 0.56.Plot 3.59: A cluster with IMFs 1.

3, 2.3, 3.3 having standard deviation = 0.67.Plot 3.60: A cluster with IMFs 1.3, 2.3, 3.

3 having standard deviation = 0.78. Plot 3.61: A cluster with IMFs 1.3, 2.3, 3.3 having standard deviation = 0.

89. Plot 3.62: A cluster with IMFs 1.3, 2.

3, 3.3 having standard deviation = 0.910.

Plot 3.63: A cluster with IMFs 1.3, 2.3, 3.3 having standard deviation = 111.

Plot 3.64: A cluster with IMFs 1.3, 2.3, 3.3 having standard deviation = 1.

112. Plot 3.65: A cluster with IMFs 1.3, 2.3, 3.3 having standard deviation = 1.

213. Plot 3.66: A cluster with IMFs 1.3, 2.3, 3.3 having standard deviation = 1.

3Binary frequency as a function of stellar mass for a cluster with IMFs 1.7, 2.3, 3.

3.Case1: In these graphs normalization is done by diving each and every y value by sum of all y values.1. Plot 3.67: A cluster with IMFs 1.7, 2.3, 3.

3 having standard deviation = 0.12. Plot 3.68: A cluster with IMFs 1.7, 2.3, 3.3 having standard deviation = 0.

23. Plot 3.69: A cluster with IMFs 1.7, 2.

3, 3.3 having standard deviation = 0.34. Plot 3.70: A cluster with IMFs 1.7, 2.

3, 3.3 having standard deviation = 0.45. Plot 3.71: A cluster with IMFs 1.7, 2.3, 3.3 having standard deviation = 0.

56. Plot 3.72: A cluster with IMFs 1.7, 2.3, 3.3 having standard deviation = 0.

67. Plot 3.73: A cluster with IMFs 1.7, 2.3, 3.3 having standard deviation = 0.

78. Plot 3.74: A cluster with IMFs 1.7, 2.3, 3.3 having standard deviation = 0.89. Plot 3.

75: A cluster with IMFs 1.7, 2.3, 3.

3 having standard deviation = 0.910. Plot 3.76: A cluster with IMFs 1.

7, 2.3, 3.3 having standard deviation = 111. Plot 3.77: A cluster with IMFs 1.7, 2.3, 3.3 having standard deviation = 1.

112. Plot 3.78: A cluster with IMFs 1.7, 2.3, 3.3 having standard deviation = 1.

213. Plot 3.79: A cluster with IMFs 1.

7, 2.3, 3.3 having standard deviation = 1.

3Case 2: In these graphs normalization is done by dividing each and every y value by ym, where ym is the value of y when x = log0 1.1.Plot 3.80: A cluster with IMFs 1.7, 2.3, 3.

3 having standard deviation = 0.12.Plot 3.

81: A cluster with IMFs 1.7, 2.3, 3.3 having standard deviation = 0.23. Plot 3.

82: A cluster with IMFs 1.7, 2.3, 3.

3 having standard deviation = 0.34. Plot 3.83: A cluster with IMFs 1.7, 2.3, 3.3 having standard deviation = 0.

45. Plot 3.84: A cluster with IMFs 1.7, 2.3, 3.3 having standard deviation = 0.56. Plot 3.

85: A cluster with IMFs 1.7, 2.3, 3.3 having standard deviation = 0.67.

Plot 3.89: A cluster with IMFs 1.7, 2.3, 3.

3 having standard deviation = 0.78. Plot 3.90: A cluster with IMFs 1.7, 2.3, 3.3 having standard deviation = 0.89.

Plot 3.91: A cluster with IMFs 1.7, 2.3, 3.3 having standard deviation = 0.910. Plot 3.

92: A cluster with IMFs 1.7, 2.3, 3.3 having standard deviation = 111. Plot 3.93: A cluster with IMFs 1.

7, 2.3, 3.3 having standard deviation = 1.112.

Plot 3.94: A cluster with IMFs 1.7, 2.3, 3.3 having standard deviation = 1.

213. Plot 3.95: A cluster with IMFs 1.7, 2.3, 3.3 having standard deviation = 1. 