a-cosmology-group/acg#200

When I see a topic about quarks [ .. ]

Can’t resist: >the Standard Model does not allow electron to be ‘inside’ a neutron.

That would be a simple picture: a neutron is an electron mixed with a proton, or glued on a proton. >Now we can’t see quarks because there is a law against it!

Ah, quarks are like gnomes. Every time you want to see them they are quickly hiding behind a tree. Charming, isn’t it? Strangeness too is a very much conserved quantity in contemporary physics. Professor Poppema started to talk about it and I decided not to become a nuclear physicist. >We still don’t get the masses right.

I have a book in my shelves. It is titled Ruimte, Tijd, Materie, (Space, Time, Matter), written by Roel Andringa-Boxum, 548 pages. I had a discussion with the author (nickname Haushofer) in May 2020 about page 389 in the book: Deze proton- en neutronmassa uitrekenen aan de hand van hun quarks blijkt nog een hels karwei te zijn waarvoor we supercomputers nodig hebben. Google translate: Calculating these proton and neutron masses based on their quarks turns out to be a hellish job for which we need supercomputers. Yes of course: brains off, computers on. Anyway, Roel came up with a paper: Nuclear masses calculated from scratch, accessible via the institution that I am not a member of. It says: An exhaustive calculation of proton and neutron masses vindicates the Standard Model. Honestly, we didn’t expect otherwise.

I don’t think it’s a good look for alternative cosmologies to have to bend the laws of physics in other fields.

That said. It seems AGN and their accretion disks plus jets have the temperature it takes to turn any element into the constituents of hydrogen.

@ExpEarth

1. “redshift itself involves transfer of photon energy to spacetime”
2. “spacetime gives back energy to photons of equal magnitude to the Hubble redshift losses”

Are these mechanisms explained by Arto Anilla? Do you know where they propose these actions to physically occur? How does your own model manifest these actions?

My conceptual aether model ( I also referenced Patrick Grahn, Arto Annila, Erkki Kolehmainen) is based on a sea of mutually destructive interfering spin=1 bosons, which presents itself as a boson spin = 2 (or spin=0?) tensor field, mostly undetectable because of the ‘self-interfering’ and leads nicely to understanding the concept of inertia. With a little added effort, it could answer to point 1 above. One shortcoming of my idea is the (unknown) source of these bosons/photons. (point 2 above)

@Francois-Zinserling

Thanks for the interesting post! The mechanisms 1 and 2 are not described by Arto Annila et al. They are my own. My model is like theirs (I use several parts of theirs), but they miss a key feature. The filaments of gravitons (made of photons) have to be attached to masses for my gravity mechanism to work! They imagine gravity to happen in a different way and they also suppose that the universe is expanding.

The Hubble redshift in my model involves transfer of photon energy to these filaments (which I call spacetime) and the Hubble luminosity reverses this process, generating new photonic energy. The reasons for these transfers have to do with the temperature of the CMB and the equivalent ‘temperature’ of spacetime in my model. It’s a flow of photon energy from hotter to cooler regions.

Your model sounds quite similar to Arto’s model and it seems you have extended it beyond what they have done. Very good! Instead of saying spin 1 bosons and spin 2 bosons, you could just say photons and gravitons. Eventually your question relates to where the CMB came from. The CMB temperature calibrates spacetime ‘temperature’ in my model.

@mikehelland : The ongoing disaster of the standard model of particle physics .

That doesn’t seem relevant. All that we need to know has been shown in labs. Hadrons boil at 1.6 trillion K.

I do not think that quasars are what the standard model describes, and there are no consistent observations of the jets moving away from the bright galaxy.

Hydrogen is produced by fission in the IGM

It could be they are something different, but this really close by quasar has been measured to have temps that exceed the known boiling point of nuclei. So at least based on the known-knowns, here seems to be a source of hydrogen.

What follows is loosely related to the (very informative, thanks) video by @mikehelland .

Planck units

Basic hypothesis: (Micro) black holes do not exist. As a consequence: the “radius” of an elementary particle is always greater than its Schwarzschild radius . Where the “radius” is assumed to be the Compton wavelength $\lambda_e$ of the particle. Mathematically, with $m_e=$ elementary particle mass, $\hbar=$ Reduced Planck constant, $c=$ speed of light, $G=$ gravitational constant (and forget about a factor $2$) :

$$ \lambda_e = \frac{\hbar}{m_e.c} \gt \frac{G.m_e}{c^2} \quad \Longrightarrow \quad m_e \lt \sqrt{\hbar c/G} $$

Let the Planck mass $m_P$ be defined as

$$ m_P = \sqrt{\hbar c/G} \quad \Longrightarrow \quad m_e \lt m_P $$

Ipse est (i.e.): an elementary particle mass is always smaller than the Planck mass. The Planck mass is an absolute upper bound for the mass of an elementary particle.

For the Compton wavelength $\lambda_e$ of an elementary particle we have

$$ \lambda_e = \frac{\hbar}{m_e.c} \gt \frac{\hbar}{m_P.c} \quad \Longrightarrow \quad \lambda_e \gt \sqrt{\hbar G/c^3} = l_P $$

Consequently, the “size” $\lambda_e$ of an elementary particle is always greater than the Planck length $l_P$ . Associated with the Compton wavelength is a Compton time, which is greater than the Planck time $t_P$ .

$$ \frac{\lambda_e}{c} \gt \sqrt{\hbar G/c^5} = t_P $$

Additional information: The Limited Validity of Special Relativity .

Last but not least we have the Planck temperature. Two quotes (found with Google): The Planck temperature is the fundamental upper limit of temperature. At this temperature, the wavelength of light emitted by thermal radiation reaches the Planck length. The latter can more or less be derived with Wien’s displacement law, where $k_B=$ Boltzmann constant, $T=$ temperature.

$$ \frac{h c}{\lambda_{peak}} = x k_B T \quad \mbox{with} \quad x = 4.965114231744276304 $$

Apart from less important math constants we thus derive that for any temperature

$$ \frac{\hbar c}{\sqrt{\hbar G/c^3}} \gt k_B T \quad \Longrightarrow \quad T \lt \sqrt{\hbar c^5/(G k_B^2)} $$

Calculations:

hbar := 1.054571817*10^(-34);
c := 299792458;
G := 6.674*10^(-11);
l_P := sqrt(hbar*G/c^3);
m_P := sqrt(hbar*c/G);
t_P := sqrt(hbar*G/c^5);
k_B := 1.380649*10^(-23);
T_P := sqrt(hbar*c^5/(G*k_B^2));

Summarizing :

$$ \begin{matrix} \mbox{Planck mass} & : & m_P=\sqrt{\hbar c/G} & \approx 0.2176483258 \times 10^{-7}\mbox{ }kg \ \mbox{Planck length} & : & l_P=\sqrt{\hbar G/c^3} & \approx 0.1616218699 \times 10^{-34}\mbox{ }m \ \mbox{Planck time} & : & t_P=\sqrt{\hbar G/c^5} & \approx 0.5391125281 \times 10^{-43}\mbox{ }s \ \mbox{Planck temperature} & : & T_P=\sqrt{\hbar c^5/(G k_B^2)} & \approx 0.1416816004 \times 10^{33}\mbox{ }K \ \end{matrix} $$

The Hagedorn temperature in the video is still orders of magnitude less than he Planck temperature.