Latest revision 05-09-2022

## Size, Mass and Age

$\def \SP {\quad ; \quad} \def \OF {\quad \mbox{or} \quad} \def \hieruit {\quad \Longrightarrow \quad}$ We have seen with Length Contraction that the size of an atom is inversely proportional with the rest mass of the electron, and upon refinement also inversely proportional with some of the rest mass of the proton. In general, it is inversely proportional with elementary particle rest mass. That mass has been smaller in the past. It is concluded herefrom that atoms have been bigger in the past. For things composed of solid materials and fluids, this means that lengths $L$ and volumes $V$ have been bigger in the past than they are now: $$\frac{L}{L_0} = \frac{m_0}{m} \SP \frac{V}{V_0} = \left(\frac{m_0}{m}\right)^3$$ It makes no difference whether lengths are still measured the old fashioned way - with a rod made of some solid material - though according to modern atomic standards, lengths are measured nowadays with the wavelength of light. Length $L$ is anyway inversely proportional with elementary particle rest mass. Density $\,\rho\,$ is defined as mass divided by volume. Therefore: $$\frac{M/V}{M_0/V_0} = \boxed{\large \frac{\rho}{\rho_0} = \left(\frac{m}{m_0}\right)^4}$$

### Cepheids

Pulsating Stars: Stars that Breathe already have been subject of our attention. At page 20 of the PDF article and next a mathematical model is presented that will be of interest to us. An equation, also called the period-mean density relation, shows that the pulsation period of a star is inversely proportional to the square root of its mean density. Mimicking the derivation in more detail: $$\Pi = 2\int_0^R\frac{dr}{\sqrt{\frac{2}{3}\gamma\pi G \rho(R^2-r^2)}} = \frac{2}{\sqrt{\frac{2}{3}\gamma\pi G \rho}} \int_0^1 \frac{d(r/R)}{\sqrt{1-(r/R)^2}} \\ = \frac{2}{\sqrt{\frac{2}{3}\gamma\pi G \rho}} \left[ \arcsin(x) \right]_{x=0}^{x=1} = \frac{2}{\sqrt{\frac{2}{3}\gamma\pi G \rho}} \frac{\pi}{2} = \sqrt{\frac{3\pi}{2\gamma G \rho}}$$ Here $\Pi=$ pulsation period, $\gamma=$ ratio of specific heats for the stellar material ($=5/3$ for a monatomic gas), $G=$ gravitational constant, $\rho=$ density. Hence, in concordance with the above we have: $$\frac{\rho}{\rho_0} = \left(\frac{m}{m_0}\right)^4 \hieruit \frac{\Pi}{\Pi_0} = \left(\frac{m_0}{m}\right)^2$$ More details are found in Radial Stellar Pulsations: A Rough Explanation of the Period-Luminosity Relation. There we see that the luminosity $L$ is roughly proportional to the area of the star surface ($=\pi R^2$ with $R=$ radius) as seen by an astronomer: $$L \sim R^2 \hieruit \frac{L}{L_0}=\left(\frac{R}{R_0}\right)^2 = \left(\frac{m_0}{m}\right)^2$$ Consequently: $$\frac{\Pi}{L} = \frac{\Pi_0}{L_0}\frac{(m_0/m)^2}{(m_0/m)^2} = \frac{\Pi_0}{L_0}$$ This is a very important consequence: the Age of a Cepheid does not alter its Period-Luminosity Relation.
Cepheids are standard candles in Unified Alternative Cosmology as they already are in common cosmology.

### Comets

References and quote mining to begin with:
Space probes have shown that comets, when visited, have a rocky appearance; they are no "dirty snowballs", at all! But from Wikipedia we also learn that Known comets have been estimated to have an average density of $\,0.6\; g/cm^3$. Which seems to be in contradiction with their rocky appearance, because common Rock and Mineral Densities on earth have a mean value of approximately $2.6\;g/cm^3$. We shall demonstrate now that the discrepancy is readily resolved by a little bit of UAC.
Let's assume that the elementary particles in a comet are young with respect to the the elementary particles on earth. Then, according to our basic hypothesis, they carry less mass. Let suffixes $E=$ earth and $C=$ comet, then according to the above we have: $$\frac{\rho_E}{\rho_C} = \frac{2.6}{0.6} = \left(\frac{m_E}{m_C}\right)^4 \hieruit \frac{m_E}{m_C} = \left(\frac{\rho_E}{\rho_C}\right)^{1/4} = \sqrt{\sqrt{\frac{2.6}{0.6}}} \approx 1.44$$ Therefore a decrease of elementary rest mass with a factor $\approx 1.44$ is already sufficient explanation for the fact that comets have a rocky appearance and a low density as well. If space probes eventually will come back on earth with samples of comet material, then we will see.
It will be shown now that it's even possible to obtain an estimate of the age of the Comet's (elementary particles) matter as compared with ours on Earth. According to the second formulation of Narlikar's Law we have: $$\frac{\mbox{VPM}_E}{\mbox{VPM}_C} = \left(\frac{\mbox{Age}_E}{\mbox{Age}_C}\right)^2 \OF \frac{m_E}{m_C} = \left(\frac{T_0-A_E}{T_0-A_C}\right)^2 \hieruit \sqrt{\left(\frac{\rho_E}{\rho_C}\right)^{1/4}} = \frac{T_0-A_E}{T_0-A_C} \\ \hieruit (T_0-A_C) = \left(\frac{\rho_C}{\rho_E}\right)^{1/8}(T_0-A_E)$$ Numerically (with MAPLE):
rho_E := 2.6; rho_C := 0.6;
Age_E := 4.543*10^9;
Age_C := (rho_C/rho_E)^(1/8)*Age_E;
10
Age_C := 0.3782160674 10

If you ask Google: How old is the Kuiper Belt? - because that's the place where the comets seem to come from - then the answer is: about 4.6 billion years ago.
The order of magnitude obtained with UAC is: about 3.8 billion years ago.

### Dinosaurs

Question is: how to explain why there could exist such large reptiles (dinosaurs), plants, insects and mammals on planet earth in the old days. A testimony of this gigantism is still found in the fossil record. Somewhat to my surprise, some of the issues involved have been analysed reasonably well by Barry Setterfield, as exemplified in the section Gravity and Dinosaurs on his website. Though I wouldn't be surprised if that wisdom has been triggered by G. P. Jellison and W. T. Bridgman: see page 41 of their article. Our own approach to Gigantism is different and it proceeds as presented below, for the mathematically inclined only :-(

As we have seen before, the age of the Earth is $(T_0-A_E) \approx 4.543\times 10^9$ years. According to Wikipedia, Dinosaurs first appeared during the Triassic period, between 243 and 233.23 million years ago. We define $(T_0-T_D) \approx 0.243\times 10^9$ years. The first version of Narlikar's Law shall be employed with these data. It is assumed that dinosaurs are composed of Earth material, that's why. $$\frac{m_D}{m_0} = \left(\frac{T_D-A_E}{T_0-A_E}\right)^2 = \left(1-\frac{T_0-T_D}{T_0-A_E}\right)^2$$ At two places it is defended that Force obeys a quadratic proportionality law with the VPM (Varying elementary Particle rest Mass): $$\frac{F}{F_0} = \left(\frac{m}{m_0}\right)^2 \quad \mbox{where} \quad F=F_D \quad \mbox{and} \quad m=m_D$$ It follows that any Force on an animal - such as weight - diminishes rather quickly with the age of the species. $$\frac{F_D}{F_0} = \left(1-\frac{T_0-T_D}{T_0-A_E}\right)^4$$ Numerically (with MAPLE):

Age_E := 4.543*10^9;
Age_D := 0.243*10^9;
force := (1-Age_D/Age_E)^4;

force := 0.8026068742

For the density of the dinosaur's tissue and bones we have, according to the above: $$\frac{\rho_D}{\rho_0} = \left(\frac{m_D}{m_0}\right)^4 = \left(1-\frac{T_0-T_D}{T_0-A_E}\right)^8$$
Age_E := 4.543*10^9;
Age_D := 0.243*10^9;
dense := (1-Age_D/Age_E)^8;

dense := 0.6441777944
force*dense;
0.5170215260

The physical meaning of the latter outcome is that a certain volume of flesh and bones in an extinct creature such as a dinosaur would be subject to only half ($\approx 0.5$) the (gravitational) force as compared with nowadays. This certainly helps to clarify why there could exist such large reptiles on planet earth in the past. It does, of course, not explain everything, if anything. Further research is needed, as they usually say :-)