Jan 2016 / Nov 2019

An Introduction to Astronomy

by Frank H. Shu

And, according to that standard book, there is the Missing Mass Problem, as explained here:

In my humble opinion - I'm not an astronomer - it's a bit suspect that anything in the distant
cosmos appears to be oversized. And the more distant, the more oversized. It's not impossible that
flaws in the Cosmic Distance Ladder (Wikipedia) are to be hold responsible for this, as is argued in *Incorrect Assumptions in Astrophysics*. While this might provide an explanation for
the issues (1), (3) and (4) in Box 12.1, it definitely does not clarify issue (2): the flat
Galaxy rotation curves:

At an alternative website, called
*The 100 Year Wrong Turn in Cosmology*, we find a
PDF document
with the intriguing title "*Galactic mass distribution without dark matter or modified Newtonian
mechanics*". It is written by Kenneth F Nicholson, a retired engineer. A more elaborate paper
with analogous content is the PDF titled
"*Mass Distribution in Rotating Thin-Disk Galaxies According to Newtonian Dynamics*",
by James Q. Feng and C. F. Gallo. A very much improved version of the latter paper is:

According to Newton's law of universal gravitation, the attractive force ($F$) between two bodies is directly proportional to the product of their masses ($m_1$ and $m_2$), and inversely proportional to the square of the distance ($r$ : inverse-square law) between them (see Wikipedia): $$ F = G\frac{m_1 m_2}{r^2} $$ The constant of proportionality ($G$) is the gravitational constant: $$ G = 6.674\,08(31) \times 10^{-11}\,m^3\,kg^{-1}\,s^{-2} $$ Newton's law of gravitation shall be applied to a RingWorld of gravitating matter, to begin with.

Let $m$ be a test mass at some distance $d$ from the center $M$ of the RingWorld.
Let $Q$ be the position of an infinitesimal element in the Ring.
The mass $m_1$ in Newton's law shall be identified with the test mass $m$.
Let the mass density (with dimension $kg/m^2$) in the (flat) ring be equal to $\rho$.
Then the infinitesimal element at $Q$ has a mass $\rho\,R\,dR\,d\phi$ , to be identified with $m_2$.
So Newton's law becomes, for the gravitational attraction between the test mass and an infinitesimal
part of the ring:
$$
dF = G\frac{m \times \rho\,R\,dR\,d\phi}{r^2}
$$
Where $r$ is the distance between the test mass $m$ and the infinitesimal element at $Q$.
Due to symmetry, straight lines through the center of the ring must be all equivalent.
For simplicity, select the line through the x-axis for measuring forces. Polar coordinates
are to be employed. Furthermore, let $\overline{QP}$ the perpendicular from $Q$ onto $\overline{mM}$.
Then, according to Pythagoras' theorem and some simple trigonometry:
$$
r^2 = (\overline{Mm}-\overline{MP})^2+(\overline{PQ})^2 =
\left[d-R\cos(\phi)\right]^2+\left[R\sin(\phi)\right]^2 \quad \Longrightarrow \\
r^2 = d^2-2Rd\cos(\phi)+R^2
$$
Due to symmetry, we only need to consider the component of the force that is directed from
$m$ to $M$, being a projection:
$$
\overline{Pm}/\overline{Qm} = \frac{d-R\cos(\phi)}{r}
$$
Herewith Newton's law becomes, with $r^3$ in the denominator:
$$
dF = G\frac{m\times\rho\,R\,dR\,d\phi}{r^2} \times \frac{d-R\cos(\phi)}{r} = m \times
G\frac{\rho\,R\,dR\,\left[d-R\cos(\phi)\right]\,d\phi}{\left[d^2-2Rd\cos(\phi)+R^2\right]^{3/2}}
$$
The total attractive force at the test mass, exerted by the ring, is the integral of this.
Due to symmetry, half of the computational effort can be saved.
$$
F = m \times 2 \int_0^\pi
G\frac{\rho\,R\,dR\,\left[d-R\cos(\phi)\right]\,d\phi}{\left[d^2-2Rd\cos(\phi)+R^2\right]^{3/2}}
$$
Make the result dimensionless as much as possible:
$$
F(d/R) = m \times G \rho\,dR/R \times 2 \int_0^\pi
\frac{\left[(d/R)-\cos(\phi)\right]\,d\phi}{\left[(d/R)^2-2(d/R)\cos(\phi)+1\right]^{3/2}}
$$
This formula is valid even for test masses *inside* the ring: not only for $d > R$
but also for $d < R$. The function $F(d/R)$ is depicted in the figure below. Function value
positive means that the force vector is directed from the left to the right. Function value
negative means that the force vector is directed from the right to the left.

A special case is a test mass *at the ring*. Then the integral becomes, for $d/R=1$:
$$
F = m \times G \rho\,dR/R \times 2 \int_0^\pi
\frac{\left[1-\cos(\phi)\right]\,d\phi}{\left[1-2\cos(\phi)+1\right]^{3/2}} =
m \times G \rho\,dR/R \times \frac{1}{\sqrt{2}} \int_0^\pi \frac{d\phi}{\sqrt{1-\cos(\phi)}}
$$
Quite unfortunately, this special integral is *singular*; it has an infinite outcome,
because the denominator of the integrand is zero for $\phi=0$. How do we deal with that?

The answer is that we don't! Take a look at the above graph. If the test mass $m$ is close
to the ring, then it will experience a strong force towards the ring. This simply means that
a test mass cannot escape from its RingWorld. And that's all. A test mass in the ring does
not contribute to anything else than just staying there. Therefore what we shall do is:
simply *skip / ignore* the infinite contribution.

Employing the end result from the preceding section, we first replace $d$ by $r$ - what's in a name, huh? Then we have, for $\;r/R\ne 1$ and assuming that $\rho(R)$ (the density of the stars) is a function of the radius only: $$ F_\mbox{ring}(r,R) = m \times G\,\rho(R)\,dR/R \times 2 \int_0^\pi \frac{\left[(r/R)-\cos(\phi)\right]\,d\phi}{\left[(r/R)^2-2(r/R)\cos(\phi)+1\right]^{3/2}} $$ This result must be integrated (i.e summed) over all rings with radius $R\in[R_0,R_1]$: $$ F_\mbox{galaxy}(r) = \int_{R_0}^{R_1} F_\mbox{ring}(r,R) = m \times G \int_{R_0}^{R_1} \left[ 2 \int_0^\pi \frac{\left[(r/R)-\cos(\phi)\right]\,d\phi} {\left[(r/R)^2-2(r/R)\cos(\phi)+1\right]^{3/2}}\right] \rho(R)/R\; dR $$ This is the total force exerted by all rings in the galaxy on a test particle in the galaxy. The test particle is part of the ring with radius $r$. That ring is assumed to rotate with uniform velocity $v$, thus creating a centripetal force which is opposite and equal to the gravitational force of the galaxy. The expression for this centripetal force is well known. Thus we have: $$ F = \frac{m\,v^2}{r} = F_\mbox{galaxy}(r) $$ Equating the two shows that the result is independent of the test mass $m$ : $$ v(r)^2 = G \int_{R_0}^{R_1} \left[ 2 \int_0^\pi \frac{\left[(r/R)-\cos(\phi)\right]\,d\phi}{\left[(r/R)^2-2(r/R)\cos(\phi)+1\right]^{3/2}} \right] (r/R)\,\rho(R)\;dR $$There is one important condition, though: $v^2 > 0$ . So the right hand side must be positive. Otherwise it would be impossible for the RingWorld to compensate gravity by means of a centripetal force. Herewith

# Gravity constant (G) G := 6.67408*10^(-11); # Speed of light light := 299792458; # Number of seconds in a year year := 31556926; # Radius of milky way (L) radius := 52850*light*year; # Mass of our sun sun := 1.989*10^(30); # Mass of milky way (W) mass := 1.5*10^(12)*sun; # Speed of sun (V) speed := 250000; # Dimensionless number G*mass/(radius*speed^2); 6.372018778At this stage in the project, it is not quite important though, because we are only interested in velocities relative to each other.

**Arithmetical grid.**

In radial direction: $R = (i+1)/(N+1)$ with $i = 0,1, \cdots ,N$.

In angular direction: $\phi = j\cdot 2\pi/M$ with $j = 0,1 \cdots , M-1$.**Geometrical grid.**

Much more complicated, as is explained in a separate**webpage.**

Crucial for success of the calculations is the (observed) mass density profile of the stars in the galaxy. For a novice in the field, it's not easy to find something sensible in the references available on the internet. After some search I've decided to take this one, on page 19. In the figure below, the graph of this function is sketched in $\color{red}{red}$. There is another scaling factor involved that norms the mass of the galaxy to unity, according to $\int_0^1 \rho(R)\cdot2\pi R\,dR = 1$. Apart from such minor details, this is what we finally have in the program: $$ \rho(R) \sim \left[1+(R/\sigma)^2\right]^{-3/2} \quad \mbox{where} \quad 0 \le R \le 1 \quad \mbox{and} \quad \sigma \approx 0.1 $$

I think that the conclusion is clear. If the adopted mass density profile of the stars is indeed
acceptable, then the galaxy rotation speed profile $v(r)$ is **not** according to the observations.
This means that I've **not** yet ( __2016__ ) been able to reproduce the claims in the papers by
Kenneth F Nicholson and
James Q. Feng and C. F. Gallo.**Note.** These papers
are deviant from the webpage you are reading in that they handle an inverse problem:
how to determine the mass density distribution from a given velocity profile. But still then there is no reason to believe
in a Deficient Reasoning for Dark Matter in Galaxies,
as far as I can see.