Latest revision 23-09-2022

## Intrinsic Redshift

$\def \MET {\quad \mbox{with} \quad} \def \EN {\quad \mbox{and} \quad} \def \SP {\quad \mbox{;} \quad} \def \hieruit {\quad \Longrightarrow \quad} \def \slechts {\quad \Longleftrightarrow \quad} \def \half {\frac{1}{2}}$ Suppose for a moment that there indeed does exist an intrinsic red shift. How could such a thing be accomplished? And what would be the consequences? I have decided to use "old" quantum mechanics, especially the Bohr model of the hydrogen atom, knowing that the latter anyway can be considered as a crude model for all atoms. Some Wikipedia references are: The Rydberg formula is given by: $$\frac{1}{\lambda} = R_\infty \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$ Where $\,n_1\,$ and $\,n_2\,$ are any two different positive integers and $\,\lambda\,$ is the wavelength (in vacuum) of the emitted or absorbed light.
The Rydberg constant $\,R_\infty\,$, with $\,m_e\,$ the electron (rest) mass, $\,h\,$ the Planck constant, $\,c\,$ the speed of light and $\,\alpha\,$ the fine-structure constant, is given by: $$R_\infty = \frac{m_e c^2}{h c}\frac{\alpha^2}{2} = 1.097\;373\;156\;8539(55) \times 10^7 \,\text{m}^{-1}$$ The fine-structure constant $\,\alpha\,$, with $\,e\,$ the elementary charge, $\,\hbar = h/2\pi\,$ the reduced Planck constant, $\,\epsilon_0\,$ the electric permittivity of free space, $\,\mu_0\,$ the magnetic permittivity of free space, is given by: $$\alpha = \frac{1}{4\pi\epsilon_0}\frac{e^2}{\hbar c} = \frac{\mu_0}{4 \pi} \frac{e^2 c}{\hbar} \qquad \mbox{where} \qquad \alpha^{-1} \approx 137.035\,999\,139(31)$$ Whatever the picture of the cosmos will be, everybody shall agree upon the fact that what we see in distant space is actually "old" information that originated (light)years ago, due to the speed of light which is known (or assumed) to be finite. Thus what we in fact see is younger matter.
Now take a look again at the Rydberg constant. The only mass in there is the electron rest mass $\,m_e\,$. Suppose that this rest mass has been different / smaller in the past, say with a value $\,m^*_e\,$ , where the asterix $^*$ stands for "young". Then we have: $$m^*_e < m_e \quad \Longrightarrow \quad \frac{m^*_e c^2}{h c}\frac{\alpha^2}{2} < \frac{m_e c^2}{h c}\frac{\alpha^2}{2} \quad \Longrightarrow \quad R^*_\infty < R_\infty \quad \Longrightarrow \\ R^*_\infty \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) < R_\infty \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \quad \Longrightarrow \quad \frac{1}{\lambda^*} < \frac{1}{\lambda} \quad \Longrightarrow \quad \lambda^* > \lambda$$ In common redshift terms, according to Wikipedia: $$z_i = \frac{\lambda_\mbox{obs}-\lambda_\mbox{emit}}{\lambda_\mbox{emit}} \quad \Longrightarrow \quad 1+z_i = \frac{\lambda_\mbox{obs}}{\lambda_\mbox{emit}} = \frac{\lambda^*}{\lambda} > 1$$ In words: the higher the age of an atom, the larger its redshift. So this is an immediate and extremely simple explanation of the Hubble redshift. And it is intrinsic - subscript $i$ - because young atoms are different from present ones: the electron and and other elementary particles (e.g. proton & neutron) have smaller rest mass than they have nowadays. In a more refined approximation, the proton rest mass is involved with the Rydberg constant as well. This has no influence on the end-result.
Thus the Rydberg constant, according to our theory, is not really a constant but varies with time. With a varying Rydberg constant, it's possible to have young and old galaxies close together with different intrinsic red shifts, an idea that has been coined up by Halton Arp.

The Bohr radius is approximately equal to the most probable distance between the nucleus and the electron in a hydrogen atom in its ground state. It is given by: $$a = \frac{4 \pi \varepsilon_0 \hbar^2}{m_e e^2} = \frac{\hbar}{m_e\,c}\frac{1}{\alpha}$$ with the meaning of the constants as explained before. Suppose again that the elementary particle (rest) mass (of the electron) shall be different / smaller in ints own past / greater in its own future, say with a value $m_e^*$. Then we have for the Bohr radius $a^*$ at that time: $$a^* = \frac{\hbar}{m_e^*\,c\,\alpha} = \frac{m_e}{m_e^*}\frac{\hbar}{m_e\,c\,\alpha} \Longrightarrow \quad a^* = \frac{m_e}{m_e^*} a$$ Here $a$ and $m_e$ are the values of the Bohr radius and (electron rest) mass here and now.
Conclusion: the (Bohr) radius of a hydrogen atom is inversely proportional with varying elementary particle rest mass (of the electron).

Results remain valid even if Special Relativity is taken into account. The Bohr–Sommerfeld / Dirac equation solution for the energy levels of the hydrogen atom is: $$E = m_e c^2 \left[\left\{ 1 + \frac{ (Z.\alpha)^2 } { \left[ \sqrt{ (j+ \frac{1}{2} )^2 - (Z.\alpha)^2 } + n \right] ^2 } \right\}^{-1/2}-1\right]$$ Where: $j+\frac{1}{2} \geq 1$ en $Z=1$ , $\alpha \approx 1/137$ . There is an interesting consequence, worth to be noticed: $$\sqrt{ (j+ \frac{1}{2} )^2 - (Z.\alpha)^2 } = \mbox{real} \quad \Longleftrightarrow \quad (1)^2 - (Z.\alpha)^2 \ge 0 \quad \Longleftrightarrow \quad Z \le 1/\alpha$$ It follows that $\,Z \le 137$ : elements with more than $137$ protons are impossible, but there is no nuclear physics involved with this equation.
It is seen again that the energy levels are proportional with (the electron) rest mass. Because of $\,E=hc/\lambda\,$ there is a redshift with decreasing rest mass, just as with the non-relativistic approach.

Now the hydrogen atom stands as a quantum mechanical prototype for all other atoms in the universe. Let's make this argument more precise, starting with the helium atom as an example. The Schrödinger equation for the Helium atom is: $$H\psi = E\psi \MET \psi(\vec{r}_1,\vec{r}_2) \EN \\ H=-{\frac {\hbar ^{2}}{2m_e}}(\nabla _{1}^{2}+\nabla _{2}^{2})+{\frac {e^{2}} {4\pi \epsilon _{0}}}{\Bigg [}{\frac {2}{r_{1}}}+{\frac {2}{r_{2}}}- {\frac {1}{|{\vec {r}}_{1}-{\vec {r}}_{2}|}}{\Bigg ]}$$ Assume for the moment that all size within the helium atom is scaled as $\vec{r'} = \vec{r}/L$, then we have, with a technique known as Nondimensionalization: $$H\psi = \left\{-\frac{\hbar^2}{2m_e}(\nabla_1'^2+\nabla_2'^2)\frac{1}{L^2}+\frac{e^2}{4\pi\epsilon_0} \left[\frac{2}{r'_1}+\frac{2}{r'_2}-\frac{1}{|\vec{r'}_1-\vec{r'}_2|}\right]\frac{1}{L}\right\} \psi = E\psi \hieruit \\ \left\{-\half(\nabla_1'^2+\nabla_2'^2) + \left[\frac{2}{r'_1}+\frac{2}{r'_2}-\frac{1}{|\vec{r'}_1-\vec{r'}_2|}\right] \frac{m_e}{\hbar^2}L\frac{e^2}{4\pi\epsilon_0}\right\} \psi = \frac{m_e}{\hbar^2}L^2 E\psi$$ With: $$\frac{m_e}{\hbar^2}L\frac{e^2}{4\pi\epsilon_0} = \frac{m_e c}{\hbar}L\frac{e^2}{4\pi\epsilon_0.\hbar c} = 1 \hieruit L = \frac{\hbar}{m_e\,c\,\alpha} = a$$ So the characteristic size for the helium atom is the Bohr radius of a hydrogen atom, i.e. the size of the helium atom must be of the same order of magnitude. But, what's far more important in the present context: that characteristic size is inversely proportional with varying elementary particle rest mass (of the electron).
For the dimensionless energy $E'$ we define: $$E' = \frac{2m_e}{\hbar^2}\left(\frac{\hbar}{m_e\,c\,\alpha}\right)^2 E \hieruit E = m_e c^2 \frac{\alpha^2}{2} E' = hc\,R\,E' \hieruit \\ \frac{hc}{\lambda} = hc\,R(E_1'-E_2') \hieruit \frac{1}{\lambda} = R(E_1'-E_2')$$ Where $\,R\,$ is the Rydberg constant again. This means that the energy levels of a helium atom are comparable with those of a hydrogen atom. But, what's far more important in the present context: the energy levels of a non-hydrogen atom are proportional too with varying elementary particle rest mass. And so the wavelength of the radiation emitted (or absorbed) by a non-hydrogen atom is inversely proportional with varying elementary particle rest mass, mainly the electron's mass.

Last but not least, let's consider the Lamb shift. From the Wikipedia reference we have: $$\Delta E_{{\mathrm {Lamb}}}=\alpha ^{5}m_{e}c^{2}{\frac {k(n,0)}{4n^{3}}}\ {\mathrm {for}}\ \ell =0\,\\ \Delta E_{{\mathrm {Lamb}}}=\alpha ^{5}m_{e}c^{2}{\frac {1}{4n^{3}}}\left[k(n,\ell )\pm {\frac {1}{\pi (j+{\frac {1}{2}})(\ell +{\frac {1}{2}})}}\right]\ {\mathrm {for}}\ \ell \neq 0\ {\mathrm {and}}\ j=\ell \pm {\frac {1}{2}},$$ where the proportionality between electron mass and energy is observed once again. Thus resulting in the same intrinsic redshift for all spectral phenomena, provided - of course - that the increasing elementary particle mass hypothesis is real.