Towards a new generalized space expansion dynamics applied to the rotation of galaxies and Tully Fisher law

Abstract

Up to now, the rotational velocities of galaxies are not clearly understood and the experimental Tully Fisher rule, linking the total galactic mass to the fourth power of the velocity, through an acceleration coefficient of about 10⁻¹⁰ m/s² has not found a deep theoretical explanation. Tentative proposals (MOND theory of a modified Newton’s law and extraneous dark matter) do not bring a definite clarification. We propose here a new approach to this problem, without exotic matter and using the classical Newton force. But we introduce a new additional universal acceleration, which could represent a universal expansion law valid at the scale level of a galaxy. We show that this hypothesis leads to a good description of the observed variations of the galactic transverse velocity. It can be considered as a consequence of the Scale Expansion Cosmos theory (SEC) introduced by J. Masreliez, but we postulate that the space expansion acceleration universally applies at any scale. We obtain a formal derivation of the Tully Fisher law, linking the constant galactic transverse velocity to its total mass, via the universal minimum acceleration. We derive a good estimate of the TF acceleration coefficient and show that expansion should be proportional to the square root of the local volumic mass density. Our conjecture is in fact a new dynamics principle which could be applied to many other physical problems at different scales. Applying it to the range of the solar planet system confirms the well known Kepler laws, at least as a valid approximation for the order of magnitude of the solar system.

Appendix: Verification of the Euler equation

For a velocity v _θ (r,t) (angular symmetry is assumed) we have:

$$\frac{dv_{\theta}}{dt} = \frac{\partial v_{\theta}}{\partial r}\dot{r} + \frac{\partial v_{\theta}}{\partial t} = 0 $$

or (for $\dot{r} \ne 0$):

$$\frac{\partial v_{\theta}}{\partial r} = - \frac{\frac{\partial v_{\theta}}{\partial t}}{\dot{r}} $$

The radial dependance of the velocity is the ratio of its time evolution over the expansion rate. Postulating that the former is much lower that the latter leads to the result that the transverse velocity remains quasi constant with r.

More generally, we can verify the Euler equation:

$$\frac{d\vec{v}}{dt} = \frac{\partial \vec{v}}{\partial t} + \mathit{rot}\vec{v} \wedge \vec{v} + \frac{1}{2}\operatorname{grad}\vec{v}^{2} = \biggl( - \varGamma_{t}(r) + \frac{\dot{r}^{2}}{r^{2}} \biggr)\vec{i} + \frac{\dot{r}}{r}v_{0}\vec{j} $$

where $\vec{i}$ and $\vec{j}$ are the radial and transverse unitary vectors. We postulate a constant transverse velocity $r\dot{\theta} = v_{0}$:

$$\vec{v} = \dot{r}(r,t)\vec{i} + v_{0}\vec{j} $$

We have:

$$\frac{\partial \vec{v}}{\partial t} = \frac{\partial \dot{r}}{\partial t}\vec{i} $$

and:

$$\begin{aligned} & \frac{d\vec{v}}{dt} = (\ddot{r} - v_{0}\dot{\theta} )\vec{i} + \dot{r} \dot{\theta} \vec{j}\\ & \mathit{rot}\vec{v} = \frac{1}{r} \biggl( \frac{\partial}{\partial r}rv_{0} - \frac{\partial \dot{r}}{\partial \theta} \biggr)\vec{k} = \frac{v_{0}}{r}\vec{k} \end{aligned}$$

since $\frac{\partial \dot{r}}{\partial \theta} = 0$.

$$[\vec{k} = \vec{i} \wedge \vec{j}] $$

and

$$\begin{aligned} & \mathit{rot}\vec{v} \wedge \vec{v} = - \frac{v_{0}^{2}}{r}\vec{i} + \frac{v_{0}\dot{r}}{r} \vec{j}\\ & \frac{1}{2}\operatorname{grad}\vec{v}^{2} = \dot{r}\operatorname{grad}(\dot{r}) = \dot{r} \frac{\partial \dot{r}}{\partial r}\vec{i} \end{aligned}$$

Then the Euler equation is (for the only two gravitation and expansion forces):

$$\frac{d\vec{v}}{dt} = \frac{\partial \vec{v}}{\partial t} + \mathit{rot}\vec{v} \wedge \vec{v} + \frac{1}{2}\operatorname{grad}\vec{v}^{2} = \biggl( - \varGamma_{t}(r) + \frac{\dot{r}^{2}}{r^{2}} \biggr)\vec{i} + \frac{\dot{r}}{r}v_{0}\vec{j} $$

Assuming $\frac{\partial \dot{r}}{\partial \theta} = 0$ (axisymmetry), we obtain, along vectors $\vec{i}$ and $\vec{j}$ respectively, the following equalities:

$$\begin{aligned} & \ddot{r} - v_{0}\dot{\theta} = \frac{\partial \dot{r}}{\partial t} - \frac{v_{0}^{2}}{r} + \dot{r}\frac{\partial \dot{r}}{\partial r} = - \varGamma_{t}(r) + \frac{\dot{r}^{2}}{r} \end{aligned}$$

(34)

$$\begin{aligned} & \dot{r}\dot{\theta} = \frac{v_{0}\dot{r}}{r} = \frac{\dot{r}}{r}v_{0} \end{aligned}$$

(35)

Knowing that $\ddot{r} = \frac{\partial \dot{r}}{\partial t} + \dot{r}\frac{\partial \dot{r}}{\partial r}$ and $r\dot{\theta} = v_{0}$, these equations are verified and Eq. (34) is equivalent to (2d). We have thus proved that a constant transverse velocity is possible, under the condition that acceleration (4) is added.