The $\kappa$-model, a minimal model alternative to dark matter: application to the galactic rotation problem

Abstract

The determination of the velocities, accelerations and the gravitational field intensity at a given location in a galaxy could potentially be achieved in an unexpected manner with the environment of the observer, for instance, the local mean mass density in the galaxy. This idea, mathematically supported by the asymmetric distance concept, is illustrated here by a study regarding the rotation of spiral galaxies. This suggestion is new in the astrophysics field (in the following, it is called the $\kappa$-model) and could help to mimic the main effects seen in modified Newtonian dynamics (MOND) theory, modified gravity (MOG) models, or other related models built with the aim of eliminating dark matter that are already well-established theories. Thus, starting from two selected examples of galaxies, in Sect. 5, we show that there is an equivalence between MOND and the $\kappa$-model. In particular, on the opposite side, we have the speculative nature of the dominant paradigm, the elusive dark matter, a matter whose properties always remain undefined despite intense theoretical, experimental and observational efforts for over 50 years.

Appendices

Appendix A

Here, we explain the origin of form (1) for the dynamics equation. We introduce the formal action

$$ S=\int dt[\frac{1}{2}m\left (\frac{d\mathbf{M}}{dt}\right )^{2}-V(M)] $$

(8)

where $m$ is the mass of a test particle and $V(M)$ the potential experienced by this particle located at a given point $M$. An arbitrary variation $\delta {\mathbf{{M}}}$ from $M$ gives

$$\begin{aligned} \delta S=&\delta \int dt\left [\frac{1}{2}m\left ( \frac{d\mathbf{M}}{dt}\right )^{2}-V \left (M\right )\right ] \\ =& \int dt \delta \left [\frac{1}{2}m\left (\frac{d\mathbf{M}}{dt}\right )^{2}-V \left (M\right )\right ] \end{aligned}$$

(9)

$$\begin{aligned} =&\int dt\ \left [m\frac{d\mathbf{M}}{dt}\delta \left (\frac{d\mathbf{M}}{dt} \right )-\delta V\left (M\right )\right ] \end{aligned}$$

(10)

To continue the calculations, we must now exchange $d$ and $\delta$.

The exchange of $d$ and $\delta$

Let three observers $A$, $B$ and $C$ be located in a plane (Fig. 12). We can define this plane by imagining a common direction perpendicular to ${\mathbf{{M}}}_{0}{\mathbf{{M}}}_{1}$ and ${\mathbf{{M}}}_{0}{\mathbf{{M}}}_{2}$. This operation is possible because any orientation is well defined in the $\kappa$-model. With the help of this figure, we write¹⁶

$$ \left . d\mathbf{M} \triangleq \mathbf{M}_{0} \mathbf{M}_{1}\left [d \boldsymbol{\sigma}\right ]\right \vert _{A}\equiv \left . d \mathbf{M}_{\parallel}= \mathbf{M}_{2}\mathbf{M}_{1\parallel}\right \vert _{C} \longrightarrow \kappa d\boldsymbol{\sigma}\quad (a) $$

The first expression signifies that the observer $A$ measures ${\mathbf{{M}}}_{0}{\mathbf{{M}}}_{1}$ and obtains $\kappa d{\boldsymbol{\sigma }}$ and that observer $C$ measures ${\mathbf{{M}}}_{2}{\mathbf{{M}}}_{1\parallel{}}$ and obtains the same value. Other very similar relations follow

$$\begin{aligned} &\delta \mathbf{M} \triangleq \left .\mathbf{M}_{0}\mathbf{M}_{2} \left [\delta \boldsymbol{\sigma}\right ]\right \vert _{A}\\ &\quad \equiv \delta \mathbf{M}_{\parallel}\triangleq \left . \mathbf{M}_{1} \mathbf{M}_{2\parallel}\left [\delta \boldsymbol{\sigma}\right ] \right \vert _{B}\longrightarrow \kappa \delta \boldsymbol{\sigma} \quad (b)\\ &\left . \mathbf{M}_{2} \mathbf{M}_{4}''\ \left [d\boldsymbol{\sigma}'\right ]\right \vert _{C} \longrightarrow (\kappa +\delta \kappa ) d\boldsymbol{\sigma}'\\ &\quad \equiv \left .\mathbf{M}_{2}\mathbf{M}_{4}\left [d \boldsymbol{\sigma}+ \delta d\boldsymbol{\sigma}\right ]\right \vert _{A} \longrightarrow \kappa \left (d\boldsymbol{\sigma}+ \delta d\boldsymbol{\sigma} \right ) \quad (a')\\ &\left . \mathbf{M}_{1} \mathbf{M}_{4}'\ \left [\delta \boldsymbol{\sigma}'\right ]\right \vert _{B} \longrightarrow ( \kappa +d\kappa ) \delta \boldsymbol{\sigma}'\\ &\quad \equiv \left . \mathbf{M}_{1} \mathbf{M}_{4}\left [\delta \boldsymbol{\sigma}+d \delta \boldsymbol{\sigma}\right ]\right \vert _{A} \longrightarrow \kappa \left (\delta \boldsymbol{\sigma}+d\delta \boldsymbol{\sigma} \right ) \quad (b') \end{aligned}$$

We must remark that ${d{\mathbf{{M}}}}_{\parallel{}}$ is $d{\mathbf{{M}}}$ parallelly displaced with respect to itself respecting the conservation of the length during the transport (likewise for ${\delta {\mathbf{{M}}}}_{\parallel{}}$ vs $\delta \mathbf{{M}}$). The observer $C$ located at ${{{M}}}_{2}$ sees ${d{\mathbf{{M}}}}_{\parallel{}}$ exactly as observer $A$ located at ${M}_{0}$ sees $d{\mathbf{{M}}}$; thus, $d\mathbf{{M}}$ and ${d{\mathbf{{M}}}}_{\parallel{}}$ have the same orientation and the same norm but their origins are distinct. However, these two vectors are perceived differently by observer $A$ located at $M_{0}$. Figure 20 is the projection of the full set of vectors on the background of this observer. Let us note that ${\mathbf{{M}}}_{0}{\mathbf{{M}}}_{1}$ is the real path and ${\mathbf{{M}}}_{2}{\mathbf{{M}}}_{4}''$ is the corresponding varied path. Now, we set

$$ d\delta \mathbf{M} \triangleq \mathbf{M}_{2\parallel} \mathbf{M}_{4}' \qquad \delta d \mathbf{M} \triangleq \mathbf{M}_{1\parallel} \mathbf{M}_{4}'' $$

After subtraction $(a')-(a)$ and $(b')-(b)$, we have

$$\begin{aligned} &\left . \mathbf{M}_{1\parallel} \mathbf{M}_{4}''\ \left [\delta d \boldsymbol{\sigma}'\right ]\right \vert _{C} \quad \longrightarrow \kappa \delta d \boldsymbol{\sigma}\\ & \left . \mathbf{M}_{2\parallel} \mathbf{M}_{4}'\left [d\delta \boldsymbol{\sigma}'\right ]\right \vert _{B}\longrightarrow \kappa d \delta \boldsymbol{\sigma} \end{aligned}$$

We naturally have $\delta d{\boldsymbol{\sigma }}=d\delta {\boldsymbol{\sigma }}$, thus (omitting the indexes)

$$ d\delta \mathbf{M}=\delta d \mathbf{M} $$

Let us specify, however, that for observer $A$ located at $M_{0}$, the vectors $d\delta {\mathbf{{M}}}$ and $\delta d{\mathbf{{M}}}$ (projected on the proper background) are parallel, but their (apparent) lengths are different, resp. $\frac{\kappa }{\kappa +d\kappa }d\delta \boldsymbol{\sigma }$ and $\frac{\kappa }{\kappa +\delta \kappa }\delta d \boldsymbol{\sigma }$; for observer $A$, ${\mathbf{{M}}}_{3}{\mathbf{{M}}}_{4}[d\delta \boldsymbol{\sigma }] \longrightarrow{}\kappa d\delta \boldsymbol{\sigma }$, even though these three vectors differ only by a small third order term. Let us remark that $\left (\kappa +\delta \kappa \right )d{\boldsymbol{\sigma }}'- \kappa d\boldsymbol{\sigma }=$ $\kappa \left (d{\boldsymbol{\sigma }}'-d{\boldsymbol{\sigma }} \right )+\delta \kappa d\boldsymbol{\sigma }'=\kappa \delta d{ \boldsymbol{\sigma }}'+\delta \kappa d{\boldsymbol{\sigma }}'$. This quantity is equal to $\kappa \delta d\boldsymbol{\sigma }$ (in both orientation and norm), but $\kappa \delta d{\boldsymbol{\sigma }}'+\delta \kappa d{ \boldsymbol{\sigma }}'$ (origin $M_{1\parallel{}}$) is evaluated by $C$, and $\kappa \delta d\boldsymbol{\sigma }$ (origin $M_{3}$) is evaluated by $A$.

Fig. 20

Diagram of real and virtual paths: the thick full segment represents the real path, the virtual path is indicated by a thick-dotted line

After exchanging $d$ and $\delta$ in equation (10), we obtain

$$\begin{aligned} &\int dt\ \left [m\frac{{d{\mathbf{{M}}}}}{dt}\frac{d}{dt}\delta {\mathbf{{M}}}-{ \boldsymbol{\nabla{}}}_{M} V\left (M\right )\delta {\mathbf{{M}}}\right ] \\ &\quad =\left .m\frac{{d{\mathbf{{M}}}}}{dt}\delta M\right \vert{}_{extremities} \\ &\qquad + \int dt\ \left [-\frac{d}{dt}\left (m\frac{{d{\mathbf{{M}}}}}{dt}\right )-{\nabla{}}_{M} V\left (M\right )\right ]\delta M \end{aligned}$$

(11)

For a stationary value of $S$, we have $\delta S=0$. We also take $\delta M=0$ at both extremities of the portion of the real trajectory. Eventually, we obtain

$$ \frac{d}{dt}\left (m\frac{d\mathbf{M}}{dt}\right )+ \boldsymbol{\nabla}_{M} V\left (M\right )=\mathbf{0} $$

(12)

This is the same expression as the usual dynamics equation in Newtonian mechanics. The physics is left formally unchanged. This is very interesting. For practical (computational) reasons, however, we rewrite this equation

$$ \frac{d}{dt} \left (m\kappa \frac{d\boldsymbol{\sigma}}{dt}\right )+ \boldsymbol{\nabla}_{(\kappa \boldsymbol{\sigma})} V\left (\kappa \boldsymbol{\sigma}\right )=0 $$

(13)

This is equation (1). The potential $V\left (M\right )$ is the gravitational potential. The dressed potential, experienced by an observer located at a point $M$ and produced by a point source (mass $\mathsf{{M}}$) located at an arbitrary origin (labeled by $\boldsymbol{\sigma}=\mathbf{{0}}$), is¹⁷

$$ V\left (\kappa _{M}\boldsymbol{\sigma}\right )=-G\mathsf{M}m \frac{1}{(\kappa _{M}\boldsymbol{\sigma})} $$

(14)

We explicitly indexed the point where the potential is measured by $\kappa$.¹⁸ For a shifted origin (at $O$), we likewise have

$$\begin{aligned} V\left (\kappa _{M}\sigma \right )=-G\mathsf{M}m \frac{1}{(\kappa _{M}\ \left \Vert \boldsymbol{\sigma}-\boldsymbol{\sigma}_{O}\right \Vert )} \end{aligned}$$

(15)

Equations (14) and (15) need to be explained. The coefficient $\kappa _{M}$, which is linked to the observer located at $M$, determines the intensity of the potential measured by this observer. It would seem that it is the measurement process itself that imposes the distance that separates the observer from the attractive mass $\mathsf{{M}}$ on the observer. This circular reasoning may sound irrational. In reality, both the measured distance and the apparent gravitational potential felt by the observer depend on the mean density $\bar{\rho}$ at $M$. It is the environment of the observer (and obviously not the observer himself) that affects the measurements. There is nothing strange about this. Thus, we can equivalently reason that by admitting that the attractive mass $\mathsf{M}$ is perceived by the observer as $\frac{\mathsf{M}}{\kappa _{M}}$, the smaller $\kappa _{M}$ is, the higher the apparent attractive mass and vice versa (however, the true mass is always $\mathsf{M}$).

Appendix B

Magnification formula

We assimilate a galaxy into a steady and axisymmetric thin disk. The stars travel in pure, uniform circular motion, and the coefficient $\kappa$ is independent of time. After removing the index $i$ for a test particle of unit mass and taking ${\kappa }_{E}=1$,¹⁹ we can simplify equation (1) to

$$ \frac{d}{dt}\left (\frac{d{\boldsymbol{\sigma }}}{dt}\right )= \frac{{\mathbf{{F}}}_{New}}{{\kappa }^{3}} $$

(16)

where the Newtonian force ${\mathbf{{F}}}_{New}$ acting on the test particle is

$$ {\mathbf{{F}}}_{New}=-Gm\sum _{j=1}^{N-1} \frac{({\boldsymbol{\sigma }}-{\boldsymbol{\sigma }}_{j})}{{\ \left \Vert{}{\boldsymbol{\sigma }}-{\boldsymbol{\sigma }}_{j}\right \Vert{}}^{3}} $$

(17)

Because the trajectory of the test particle is circular and the force acting on it is purely radial, equation (16) immediately gives

$$ \frac{{(\sigma \dot{\theta{}})}^{2}}{\sigma }= \frac{F_{New}}{{\kappa }^{3}} $$

(18)

where $\theta{}$ designates the polar angle in the galactic plane from a reference direction taken in this plane. This leads to

$$ \sigma \dot{\theta{}}=\frac{1}{{\kappa }^{\frac{3}{2}}}{{(F}_{New} \sigma )}^{\frac{1}{2}}=\frac{1}{{\kappa }^{\frac{3}{2}}}{v}_{New} $$

(19)

where ${v}_{New}$ is the Newtonian velocity. Eventually, the true velocity²⁰$v$ is obtained by multiplying equation (19) by $\kappa$. We obtain magnification formula (6) after reinserting in it the coefficient ${\kappa }_{E}$.