# Calculate Number of NS's in Galaxy

## NS-Capture Rate calculation:

The capture rate will be done starting with a **simple model of the Milky Way Galaxy**, a cylindrical disk, with the following parameters:

Radius of Galaxy: R = 12.5 kpc (kilo-parsecs)

Thickness (Height) of Galaxy: H = 0.5 kpc

Volume of Galaxy: V = pi*(R**2)*H

= 3.14*(12**2)*0.5 kpc**3 (cubic kilo-parsecs)

= 3.14*144*0.5 kpc**3

= 226.1 kpc**3

~= 2*10**11 pc**3 (cubic parsecs)

Number of Stars in Galaxy: Ns = 2*10**11

Average Number Stars per pc**3: ns = 2*10**11/2*10**11 = 1.

So, we are assuming an average of 1 star per pc**3 uniformly across the galaxy. We know that in reality the 2*10**11 stars are not distributed uniformly, but highly concentrated toward the center, but we will start with uniform distribution to get a rough idea of the numbers that will be involved for capture.

**For a single capture**, we will look at a single star located randomly within a cubic parsec. Therefore, if we randomly shoot a NS through the cube, then the odds of a binding collision taking place are:

The chance of 1 capture is the ratio of:

** area of circle of impact parameter b**: pi*b**2

divided by:

** area of one side of the cubic parsec, A** = 1 pc**2

i.e. divide the area of the impact parameter circle of the target star by

the area of the side of the cubic parsec containing the target star,

and the result is the odds of getting one hit.

Let’s stop for a moment to define the impact parameter, b. If there was no gravity attracting the NS and the GS, the NS would simply travel in a straight line and if it came within 3*(radius of GS), it would be captured. Therefore, one might assume that the chance for capture would be the ratio of pi*(3*Rgs)**2. We will refer to this capture radius as d = 3Rgs.

However, there is gravity involved and it turns out that gravity will curve the trajectory of the NS closer to the GS when it is passing by. Therefore, what we really want to know is if we shoot the NS from one edge of the 1pc**3 box, then how far beyond 3Rgs can we go and still have the GS come within 3Rgs when it actually reaches the GS.

There’s an equation for that (see Bhattacharya), and it is:

b = sqrt(2*G*(M+m)*d) / Vinf

where, Vinf is the NS speed when it enters the box, M is the mass of the GS, m is the mass of the NS, and d is the 3*Rgs as mentioned above.

We will take as typical values of M and Rgs based on the value for a giant B-star in:

https://en.wikipedia.org/wiki/Stellar_classification#Harvard_spectral_classification

which gives us:

- Rgs = 6 * Rsun = 6 * 7*10**5 km
- M = 8 * Msun = 8 * 2*10**30 kg
- Vinf = 20 km/s (from other refs: tbd)

Let’s now calculate b and the area ratio:

R-sun = 7*10**5 km = 7*10**8 m

b = [sqrt( 2*(6.6*10**-11)*(8*10**30)*(3*6*10**8) )] / (20,000) (in meters)

b = 2.8*10**11 m = 403*R-sun ~= 3*10**8 km

Area of impact parameter circle, a = pi*b**2

a ~= 3*(3*10**8)**2 km**2

a = 27*10**16 pc**2 ~= 3*10**17 km**2

Now we need to **convert the 1pc**2 square to km**2**

1 pc = 3 ly = 3*c(km/sec)f(sec/yr) (where f converts from seconds to years)

1 pc = 3 * (3*10**5 km/sec) * (3*10**7 sec/yr)

1 pc = 3*3*3 * 10**12 km

1 pc ~= 3* 10**13 km

The area of the side of the 1pc box is then:

A = (3*10**13)**2 km**2

A = 9 * 10**26 km**2

A ~ 1 * 10**27 km**2

Now the **ratio of the impact parameter circle a to the side of the pc**3 cube** is:

a/A = (3*10**17)/(1*10**27)

a/A = 3/(1*10**10)

i.e. the chance of the NS hitting a giant B-star in a random location in the 1 pc**3 box is about three in 10 billion.

In addition we are assuming the that the **velocity of the NS** as it travels through the 1 pc**3 cube is initially 20 km/sec

V = 20 km/sec

We know from above that the length of 1 pc in km is

1pc = 3*10**13 km

Therefore @10km/sec it will take the following **time for the NS to cross the cube**:

T = (3*10**13 km) / (20 km/sec)

T = 1.5*10**12 sec

T = (1.5*10**12 sec) / (3*10**7 sec/year)

T = 0.5*10**5 years.

i.e. it **takes the NS 50,000 years @ 20 km/sec to cross the cube**.

Therefore the **capture rate for a single pc**3** is

three chances in 10 billion in 50,000 years.

If we assume each cubic parsec in our galactic model above, has 1NS passing through it each 50,000 years, and that each star in the galaxy is a giant B-star, we come up with a **total capture rate for the galaxy** as:

CaptureRateGalaxy:

CRG = 2*10**11 pc**3 * (3*10**-10 captures/pc**3) * (1 traversal/ 50,000 years)

CRG = 6*10**1 (captures/50,000 years)

CRG = 60 captures per 50,000 years,

**assuming 200 billion NSs in galaxy**

CRG ~= 1 capture per 1,000 years

Note: that this capture rate is assuming all 200 billion stars are giant B-stars, which is not a good assumption, because the giant B-stars, are in reality, only 0.13 % of the stars, or about 1 B-giant star per 769 regular stars. We will address this further below.

The **SNE rate has been stated to be 1SNE/50 years**. Therefore in order for the NS’s to account for all the SNE’s instead of 1 capture per 1000 yrs we need 1 capture per 50 years.

i.e. we need 20 times more NS’s

Therefore,

Nns = 20 * 2*10**11 NS’s

= 4*10**12 NS’s

= **4 trillion NS’s in the galaxy**

This estimate is actually pretty low, because it is assuming all stars are giant B-stars.

However, we have shown where the estimate comes from on the home page of:

https://secretofthepulsars.com/

**Dark Matter ~= 5 trillion neutron stars**

Note: the purpose of this calculation is primarily to show the “magnitude” of the number of neutron stars that are required to exist based on the NS-Capture theory.

We need to make some assumptions about the progenitors of all the SNEs, and each type will have different impact parameters and different life expectancies, and therefore expected SNE production rates.

However, the main point is that there is “one number” of NSs that are required to exist in the galaxy, and that number must be sufficient to produce all the different SNEs of all the different types We are using “5 trillion NS’s” as a starting point. Using the techniques above, we can modify the assumptions and determine what scale factor might need to be applied to the 5 trillion in order to come up w a number that satisfies all requirements.

## Calculation based on star-types

As indicated above, the numbers were derived assuming all stars are giant B-stars, which is not a reasonable assumption.

Instead, in this section, we will assume that the star types are in ratios consistent with the star classification system reference above: the Harvard Spectral Classification.

In this system we have the following ratios of star types, along with mass and radius to calculate b:

Star_Types = [ M R % Pop O [16.0, 6.6, 0.00003], B [ 8.0, 6.0, 0.13], A [ 2.0, 1.5, 0.6], F [ 1.2, 1.2, 3.0], G [ 0.9, 1.1, 7.5], K [ 0.65, 0.85, 12.1], M [ 0.3, 0.6, 76.4]

The above table shows the mass, M, and radius, R, typical for each star type (O,B,A,F,G,K,M), as well as the percentage of the stellar population in the MWG that each star type represents.

Based on these values and percentages, we calculate (using a python program to be included later) that the total number of captures across all star types is about 1 capture per 50,000 years, which is based on an assumption that there is 1 NS for each regular star, i.e. 200 billion NS’s along with 200 billion RS’s.

One recent estimate of the number of SNE’s events in the MWG is 1 SNE / 50 years. In order to achieve this SNE rate, we would need 1000 times as many NS’s as there are for 1 SNE / 50,000 years, which gives us a rate of 200 Trillion NS’s in the MWG.

Another estimate is that there are 100 High Mass X-ray Binaries (HMXB’s) in the MWG and their life expectancy is about 10 million years. These are the B-Star binaries. Therefore, if we need to produce 100 binaries per 10 million years, that is 1 binary per 100,000 years.

The B-star hit rate, based on the 0.13 % population is therefore 0.0013 * capture rate for 100% population (above) = .0013 captures per 1,000 years or 0.13 captures per 100,000 years. The capture rate we require is 1 binary per 100,000 years, so we would need about 8 times as many NS’s as RS’s to achieve this rate, or about 1.6 Trillion NS’s.

Bottom line, based on these initial estimates, we can say the NS-Capture theory predicts:

- 200 Trillion NS’s required for 1 SNE per 50 years
- 1.6 Trillion NS’s required for 1 HMXB per 100,000 years.

Off by a factor of 100.

For the sake of discussion, let’s say that the SNE rate is really 1/500 years. Then only 20 Trillion NS’s would be required.

Also, if we were to say that the life expectancy of an HMXB is 1,000,000 yrs instead of 10 million yrs, then we would need a factor of 10 more NS’s to produce that rate.

These 2 adjustments would lead us to NS-Capture theory predicting:

- 20 Trillion NS’s required for 1 SNE per 500 years
- 16 Trillion NS’s required for 1 HMXB per 10,000 years.

We are also looking to predict the number of NS’s required for a typical Galactic Cluster of 10 million stars in a sphere of radius 60 light years ~= 20 pc.

Further details will be provided as we provide the python program and calculate the results directly for that. It will also include a simple density function, because the galactic star density is not uniform, which has been the assumption in the above calculations.

So what we are likely looking at there is convergence toward a number of NS’s in the MWG, N = 20 Trillion NS’s = 100 * 200 Billion RS’s.

i.e. 100 times as many NS’s as RS’s in the Galaxy.

Therefore in sphere to closest star to the Sun (4ly), we will probably also have about 100 NS’s.

We therefore, now need to determine whether the evidence for the specific existence of these NS’s can be found, most likely within observational data that has already been collected and available for analysis. The Chandra X-ray data is a likely first place to investigate, since neutron stars interaction with matter tend to be most readily available in X-rays.