﻿ Modelling our galaxy

Modelling our galaxy

Data & facts

To model our galaxy I will use the data from Sofue (2013). The most important original data are copied on a separate page.
 Sofue 2013 Important to my model are:
 Mass (bulge) 1,8 * 1010 M☉ 3,58 * 10 40(kg) Mass (disk) 6,5 * 1010 M☉ 12,92 * 10 40(kg) half-density distance 2,4 kpc
Half-density distance is derived from figure 13: in roughly 8 kpc the density drops with a factor 10. This could be a bit off, but my model is not very sensitive to this value.

Line of reasoning

First I'll explain the way I model our galaxy. All the legwork is included on another page, with a detailed description of every step made. It has the format of 8 Excel sheets. But first of all it is important to understand what I'm doing and how I'm doing it.
1. I distinguish between the disk and the bulge. The gravitational effect of the bulge is at first ignored, to be added later. Sheets 1 and 2 are only concerned with the gravitational effect of the disk
2. Around our galaxy I put I virtual volume in the shape of a very flat cilinder, with a radius of 50 kpc. All mass of the disk is thought to lie in the equatorial plane (eq-lane). Calculations for acceleration perpendicular to the eq-plane are made for values of 0,5 kpc, 0,4 kpc, 0,3 kpc, 0,25 kpc an 0,2 kpc over and below the equatorial plane, for several distances from the centre of the disk..
3. In Δt = 1 (sec) this virtual volume will shrink by 2π MGΔt2 (m3) [from rest].
4. Diminishing the volume of a cilinder can happen in two and only two ways: either the top and/or bottom area‘dent’inward, or the rim moves inward. If not al the volume in (3) is covered by the 'dent', the rest has to be covered by the rim moving inward.
5. In sheet 1 (reference) I divide the disk (with exponentially diminishing density) into rings of 0,05 kpc each. Each ring is treated as if it had a fixed density, resulting in a fixed acceleration above and below each ring, perpendicular to the eq-plane. Density distributrion is chosen conform Sofues data. This disk is not a real model of our galaxy, it's only being used as a point of reference. This is what would happen if all gravity was perpendicualar to the disk, and all change in volume would take place in 'the dents' over and below the disk. To calculate this, I use the formula for the area-mass, as derived before.
6. Given the area of a ring and the accelaration caused by its mass density, the change in volume caused by each individual ring is calculated. All these changes in volume add up to the calculated ΔVol from step 3. (as expected)
7. Since all volume change is accounted for by the calculated 'dent' in 5 and 6, the rim of the cilinder does not move inward – there is no centripetal acceleration in this reference case.
8. In sheet 2 (data) I model our galaxy as it is observed, using Newtonian gravity only. I use the rings from sheet 1 again, but I assess each ring individually. Each ring is divided in 2000 radial sections, each containing a point mass. For several distances to the equatorial plane the contribution in perpendicular acceleration of each point mass of each ring has been calculated. This adds up to a sum total acceleration in perpendicular direction for each point P at a given distance from the center, at a given height above the eq-plane. Since typically 1000 rings are evaluated, this value is te sum total of the acceleration caused by 1000*2000 = 2.000.000 points. The program used to calculate these data can be modified to involve even more rings and segments if needed.
9. In the sheets eq+/- .50 kpc (etc) the data of the previous sheet are interpolated, so that the perpendicular accelaration above each ring is known. This is expressed as a percentage of the acceleration in the reference-case (sheet 1). It turns out that in the centre of the disk the acceleration is quite a bit less than in the reference-case, and further away it is a little bit more than that. This is as expected.
10. From these data I calculate how much less change in volume my numerical approximation causes, compared to the reference-case. This is what I call 'the missing volume'.
11. Since the total change in volume (within a specified time-inteerval, in this case: 1 second) is fixed, this missing volume has to be ‘filled’ by the rim of the cilinder moving inward. I calculate the average distance the outer rim has to move, in order tot get the required total change in volume. This distance is covered (from rest) in Δt = 1 (sec), so the centripetal acceleration due to the disk can be calculated.
12. Only then the centripetal acceleration due to the gravity of the bulge is added to this.
13. On the basis of this total centripetal acceleration, the equilibirum tangential velocity is calculated. The results are visualised in sheet 8(grafics). 14. As you can see, this tangential speed at eq +/- 0,30 kpc stabilases at around 230 km/sec from 5,5 kpc onward, and from there on drops off very slowly, reaching 217 km/sec at 25 kpc.
This fits well with the empirical data. Original research is to be found on this page. Give the crude way our galaxy was modelled here, this is a very good result.
15. As you can see, the prediction of my model comes very close the observed a ~ 1/r, and even comes with a very shallow slope at larger distances. This is also a very natural result: most of the ‘missing’ volume is missing in the centre of the disk.
Distance travelled by the rim is equal to the ‘missing volume’, divided by the area of the rim. Doubling the distance from the centre results in doubling the area of the rim, while the amount of 'missing volume' stays roughly the same.
This results in halving the distance that needs to be covered by the rim to make up for the missing volume. This leads to halving the acceleration when doubling the distance from the centre.
This makes a ~ 1/r the logical and expected result for this distribution of mass.
16. The empirical data indicate a drop in rotational velocity of -1.34 +/- .21 km/sec per kpc. In my calculations this value is -1.596 at altitude .30 kpc over/below the eq-plane, and -1.091 at an altitude of .25 kpc. This slow drop in speed is caused by the fact that at the edges of the disk the calculated perpendicular acceleration is slightly larger then in the reference case. The influence of the bulge (about 1/5 of the total mass) drops with the square of the distance. Up to 3 kpc the acceleration caused by the bulge dominates. Past the 15 kpc it becomes neglectable.

The actual calculations can be seen here. I realize it might be difficult to carefully inspect this all online. If you want to see the original files, please get in touch and I'll send them to you.
 Excel sheet explanation Excel   