From gravatoms to dark matter

Gravatoms

Imagine a proton an an electron were bound together in a hydrogen atom by gravitational forces and not by electric forces. We have two interesting problems to solve here:

1st. Find the formula for the spectrum (energy levels) of such a gravitational atom (or gravatom), and the radius of the ground state for the lowest level in this gravitational Bohr atom/gravatom.

2nd. Find the numerical value of the Bohr radius for the gravitational atom, the “rydberg”, and the “largest” energy separation between the energy levels found in the previous calculation.

We will take the values of the following fundamental constants:

$\hbar=1\mbox{.}06\cdot 10^{-34}Js$, the reduced Planck constant.

$m_p=1\mbox{.}67\cdot 10^{-27}kg$, the proton mass.

$m_e=9\mbox{.}11\cdot 10^{-31}kg$, the electron mass.

$G_N=6\mbox{.}67\cdot 10^{-11}Nm^2/kg^2$, the gravitational Newton constant.

Let R be the radius of any electron orbit. The gravitational force between the electron and the proton is equal to:

(1) $F_g=G_N\dfrac{m_pm_e}{R^2}$

The centripetal force is necessary to keep the electron in any circular orbit. According to the gravatom hypothesis, it yields the value of the gravitational force (the electric force is neglected):

(2) $F_c=\dfrac{mv^2}{R}$

(3) $F_c=F_g\leftrightarrow \boxed{\dfrac{mv^2}{R}=G_N\dfrac{m_pm_e}{R^2}}$

Using the hypothesis of the Bohr atomic model in this point, i.e., that “the allowed orbits are those for whihc the electron’s orbital angular momentum about the nucleus is an integral multiple of $\hbar$“, we get

(4) $L=m_evR=n\hbar$ $\forall n=1,2,\ldots,\infty$

Then,

(5) $v=\dfrac{n\hbar}{m_eR}$ and $v^2=\dfrac{n^2\hbar^2}{m_e^2R^2}$

From (3), we obtain

(6) $\boxed{v^2=G_N\dfrac{m_p}{R}}$

Comparing (5) with (6), we deduce that

(7) $G_N\dfrac{m_p}{R}=\dfrac{n^2\hbar^2}{m_e^2R^2}$

and thus

(8) $\boxed{R_n=R(n)=n^2\dfrac{\hbar^2}{G_Nm_pm_e^2}}$

This is the gravatom equivalent of Bohr radius in the common Bohr model for the hydrogen atom. To get the spectrum, we recall that total energy is the sum of kinetic and potential energy:

$E=T+U=\dfrac{1}{2}m_ev^2-G_N\dfrac{m_pm_e}{R}$

Using the value we obtained in (5), by direct substitution, we have

(9) $E=\dfrac{1}{2}m_ev^2-G_N\dfrac{m_pm_e}{R}=-G_N\dfrac{m_pm_e}{2R}$

and then

(10) $E=-\dfrac{G_Nm_em_p}{2}\dfrac{G_Nm_pm_e^2}{n^2\hbar^2}$

and so the spectrum of this gravatom is given by

(11) $\boxed{E_n=E(n)=-G_N^2\dfrac{m_p^2m_e^3}{2n^2\hbar^2}}$

For n=1 (the ground state), we have the analogue of the Bohr radius in the gravatom to be:

$R_1=\dfrac{\hbar^2}{G_Nm_pm_e^2}=1\mbox{.}20\cdot 10^{29}m$

For comparison, the radius of the known Universe is about $R_U=4\mbox{.}4\cdot 10^{26}m$. Therefore, $R(gravatom)>R_U$!!!!!! $R_1$ is very huge because gravitational forces are much much weaker than electrostatic forces! Moreover, the energy in the ground state n=1 for this gravatom is:

$E_1=-G_N^2\dfrac{m_p^2m_e^2}{2\hbar^2}=-4\mbox{.}23\cdot 10^{-97}J$

The energy separation between this and the next gravitational level would be about $1-1/4=3/4$ this quantity in absolute value, i.e.,

$\Delta E=\vert E_2-E_1\vert =3\mbox{.}18\cdot 10^{-97}J=1\mbox{.}99\cdot 10^{-78}eV$

This really tiny energy separation is beyond any current possible measurement. Therefore, we can not measure energy splittings in “gravatoms” with known techniques. Of course, gravatoms are a “toy-model” or hypothetical systems (bubble Universes?).

Remark (I): The quantization of angular momentum provided the above gravatom spectrum. It is likely that a full Quantum Gravity theory provides additional corrections to the quantum potential, just in the same way that QED introduces logarithmic (vacuum polarization) corrections and others (due to relativity or additional quantum effects).

Remark (II): Variations in the above quantization rules can modify the spectrum.

Remark (III): In theories with extra dimensions, $G_N$ is changed by a higher value $G_N^{eff}$ as a function of the compactification radius. So, the effect of large enough extra dimensions could be noticed as “dark matter” if it is “big enough”. Can you estimate how large could the compactification radius be in such a way that the separation between n=1 and n=2 for the gravatom could be measured with current technology? Hint: you need to know what is the tiniest energy separation we can measure with current experimental devices.

Remark (IV): In  Verlinde’s entropic approach to gravity, extra corrections arise due to the change of the functional entropy we choose. It can be  due to extra dimensions and the (stringy) Generalized Uncertainty Principle as well.

Gravatoms and Dark Matter: a missing link

I will end this thread of 3 posts devoted to Bohr’s centenary model to recall a connection between atomic physics and the famous Dark Matter problem! The calculations I performed above (and which anyone with a solid, yet elementary, ground knowledge in physics can do) reveals a surprising link between microscopic gravity and the dark matter problem. I mean, the problem of gravatoms can be matched to the problem of dark matter if we substitute the proton mass by the mass of a galaxy! It is not an unlikely option that the whole Dark Matter problem shows to be related to a right infrared/long scale modified gravitational theory induced by quantum gravity. Of course, this claim is quite an statement! I work on this path since months ago…Even when MOND (MOdified Newtonian Dynamics) or MOG (MOdified Gravity) have been seen as controversial since Milgrom’s and Moffat’s pioneer works, I believe it is yet to come its “to be or not to be” biggest test. Yes, even when some measurements like the Bullet Cluster observations and current simulations of galaxy formation requires a component of dark matter, I firmly believe (similarly, I think, to V. Rubin’s opinion) that if the current and the next generation of experiments trying to discover the “dark matter particle/family of particles” fails, we should take this option more seriously than some people are able to accept at current time.

May the Bohr model and gravatoms be with you!

LOG#107. Basic Cosmology (II).

Evolution of the Universe: the scale factor

The Universe expands, and its expansion rate is given by the Hubble parameter (not constant in general!)

$\boxed{H(t)\equiv \dfrac{\dot{a}(t)}{a(t)}}$

Remark  (I): The Hubble “parameter” is “constant” at the present value (or a given time/cosmological age), i.e., $H_0=H(t_0)$.

Remark (II): The Hubble time defines a Hubble length about $L_H=H^{-1}$, and it defines the time scale of the Universe and its expasion “rate”.

The critical density of matter is a vital quantity as well:

$\boxed{\rho_c=\dfrac{3H^2}{\kappa^2}\vert_{t_0}}$

We can also define the density parameters

$\Omega_i=\dfrac{\rho_i}{\rho_c}\vert_{t_0}$

This quantity represents the amount of substance for certain particle species. The total composition of the Universe is the total density, or equivalently, the sum over all particle species of the density parameters, that is:

$\boxed{\displaystyle{\Omega=\sum_i\Omega_i=\dfrac{\displaystyle{\sum_i\rho_i}}{\rho_c}}}$

There is a nice correspondence between the sign of the curvature $k$ and that of $\Omega-1$. Using the Friedmann’s equation

$\displaystyle{\dfrac{\dot{a}^2}{a^2}+\dfrac{k}{a^2}=\dfrac{\kappa^2}{3}\sum_i\rho_i}$

then we have

$\dfrac{k}{H^2a^2}=\dfrac{\displaystyle{\sum_i\rho_i}}{\rho_c}-1=\Omega-1$

Thus, we observe that

1st. $\Omega>1$ if and only if (iff) $k=+1$, i.e., iff the Universe is spatially closed (spherical/elliptical geometry).

2nd. $\Omega=1$ if and only if (iff) $k=0$, i.e., iff the Universe is spatially “flat” (euclidean geometry).

3rd. $\Omega<1$ if and only if (iff) $k=-1$, i.e., iff the Universe is spatially “open” (hyperbolic geometry).

In the early Universe, the curvature term is negligible (as far as we know). The reason is as follows:

$k/a^2\propto a^{-2}<<\dfrac{\kappa\rho}{3}\propto a^{-3}(MD),a^{-4}(RD)$ as $a$ goes to zero. MD means matter dominated Universe, and RD means radiation dominated Universe. Then, the Friedmann’s equation at the early time is given by

$\boxed{H^2=\dfrac{\kappa^2}{3}\rho}$

Furthermore, the evolution of the curvature term

$\Omega_k\equiv \Omega-1$

is given by

$\Omega-1=\dfrac{k}{H^2a^2}\propto \dfrac{1}{\rho a^2}\propto a(MD),a^2(RD)$

and thus

$\vert \Omega-1\vert=\begin{cases}(1+z)^{-1}, \mbox{if MD}\\ 10^4(1+z)^{-2}, \mbox{if RD}\end{cases}$

The spatial curvature will be given by

$\boxed{R_{(3)}=\dfrac{6k}{a^2}=6H^2(\Omega-1)}$

and the curvature radius will be

$\boxed{R=a\vert k\vert ^{-1/2}=H^{-1}\vert \Omega-1\vert ^{-1/2}}$

We have arrived at the interesting result that in the early Universe, it was nearly “critical”. The Universe close to the critical density is very flat!

By the other hand, supposing that $a_0=1$, we can integrate the Friedmann’s equation easily:

$\boxed{\displaystyle{\left(\dfrac{\dot{a}}{a}\right)^2+\dfrac{k}{a^2}=\dfrac{\kappa^2}{3}\sum_i\rho_i=\dfrac{\kappa^2}{3}\sum_i\rho_i(0)a^{-3(1+\omega_i)}}}$

Then, we obtain

$\dot{a}^2=H_0^2\left[-\Omega_k+\sum_i\Omega_ia^{-1-3\omega_i}\right]$

We can make an analogy of this equation to certain simple equation from “newtonian Mechanics”:

$\dfrac{\dot{a}^2}{2}+V(a)=0$

Therefore, if we identify terms, we get that the density parameters work as “potential”, with

$\displaystyle{V(a)=\dfrac{1}{2}H_0^2\left[\Omega_k-\sum_i\Omega_ia^{-1-3\omega_i}\right]}$

and the total energy is equal to zero (a “machian” behaviour indeed!). In addition to this equation, we also get

$\boxed{\displaystyle{H_0t=\int_0^a\left[-\Omega_k+\sum_i\Omega_i\chi^{-1-3\omega_i}\right]^{-1/2}d\chi}}$

The age of the Universe can be easily calculated (symbolically and algebraically):

$\boxed{t_0=H_0^{-1}f(\Omega_i)}$

with

$f(\Omega_i)=\int_0^1\left[-\Omega_k+\sum_i\Omega_i\chi^{-1-3\omega_i}\right]^{-1/2}d\chi$

This equation can be evaluated for some general and special cases. If we write $p=\omega \rho$ for a single component, then

$a\propto t^{2/3(1+\omega)}$ if $\omega\neq -1$

Moreover, 3 common cases arise:

1) Matter dominated Universe (MD): $a\propto t^{2/3}$

2) Radiation dominated Universe (RD): $a\propto t^{1/2}$

3) Vacuum dominated Universe (VD): $e^{H_0t}$ ($w=-1$ for the cosmological constant, vacuum energy or dark energy).

THE MATTER CONTENT OF THE UNIVERSE

We can find out how much energy is contributed by the different compoents of the Universe, i.e., by the different density parameters.

Case 1. Photons.

The CMB temperature gives us “photons” with $T_\gamma=2\mbox{.}725\pm 0\mbox{.}002K$

The associated energy density is given by the Planck law of the blackbody, that is

$\rho_\gamma=\dfrac{\pi^2}{15}T^4$ and $\mu/T<9\cdot 10^{-5}$

or equivalently

$\Omega_\gamma=\Omega_r=\dfrac{2\mbox{.}47\cdot 10^{-5}}{h^2a^4}$

Case 2. Baryons.

There are four established ways of measuring the baryon density:

i) Baryons in galaxies: $\Omega_b\sim 0\mbox{.}02$

ii) Baryons through the spectra fo distant quasars: $\Omega_b h^{1\mbox{.}5}\approx 0\mbox{.}02$

iii) CMB anisotropies: $\Omega_bh^2=0\mbox{.}024\pm ^{0\mbox{.}004}_{0\mbox{.}003}$

iv) Big Bag Nucleosynthesis: $\Omega_bh^2=0\mbox{.}0205\pm 0\mbox{.}0018$

Note that these results are “globally” compatible!

Case 3. (Dark) Matter/Dust.

The mass-to-light ratio from galactic rotation curves are “flat” after some cut-off is passed. It also works for clusters and other bigger structures. This M/L ratio provides a value about $\Omega_m=0\mbox{.}3$. Moreover, the galaxy power spectrum is sensitive to $\Omega_m h$. It also gives $\Omega_m\sim 0\mbox{.}2$. By the other hand, the cosmic velocity field of galaxies allows us to derive $\Omega_m\approx 0\mbox{.}3$ as well. Finally, the CMB anisotropies give us the puzzling values:

$\Omega_m\sim 0\mbox{.}25$

$\Omega_b\sim 0\mbox{.}05$

We are forced to accept that either our cosmological and gravitational theory is a bluff or it is flawed or the main component of “matter” is not of baryonic nature, it does not radiate electromagnetic radiation AND that the Standard Model of Particle Physics has no particle candidate (matter field) to fit into that non-baryonic dark matter. However, it could be partially formed by neutrinos, but we already know that it can NOT be fully formed by neutrinos (hot dark matter). What is dark matter? We don’t know. Some candidates from beyond standard model physics: axion, new (likely massive or sterile) neutrinos, supersymmetric particles (the lightest supersymmetric particle LSP is known to be stable: the gravitino, the zino, the neutralino,…), ELKO particles, continuous spin particles, unparticles, preons, new massive gauge bosons, or something even stranger than all this and we have not thought yet! Of course, you could modify gravity at large scales to erase the need of dark matter, but it seems it is not easy at all to guess a working Modified Gravitational theory or Modified Newtonian(Einsteinian) dynmanics that avoids the need for dark matter. MOND’s, MOG’s or similar ideas are an interesting idea, but it is not thought to be the “optimal” solution at current time. Maybe gravitons and quantum gravity could be in the air of the dark issues? We don’t know…

Case 4. Neutrinos.

They are NOT observed, but we understand them their physics, at least in the Standard Model and the electroweak sector. We also know they suffer “oscillations”/flavor oscillations (as kaons). The (cosmic) neutrino temperature can be determined and related to the CMB temperature. The idea is simple: the neutrino decoupling in the early Universe implied an electron-positron annihilation! And thus, the (density) entropy dump to the photons, but not to neutrinos. It causes a difference between the neutrino and photon temperature “today”. Please, note than we are talking about “relic” neutrinos and photons from the Big Bang! The (density) entropy before annihilation was:

$s(a_1)=\dfrac{2\pi^2}{45}T_1^3\left[2+\dfrac{7}{8}(2\cdot 2+3\cdot 2)\right]=\dfrac{43}{90}\pi^2 T_1^3$

After the annihilation, we get

$s(a_2)=\dfrac{2\pi^2}{45}\left[2T_\gamma^3+\dfrac{7}{8}(3\cdot 2)T_\nu^3\right]$

Therefore, equating

$s(a_1)a_1^3=s(a_2)a_2^3$ and $a_1T_1=a_2T_\nu (a_2)$

$\dfrac{43}{90}\pi^2(a_1T_1)^3=\dfrac{2\pi^2}{45}\left[2\left(\dfrac{T_\gamma}{T_\nu}\right)^3+\dfrac{42}{8}\right](a_2T_\nu (a_2))^3$

$\dfrac{43}{2}\pi^2(a_1T_1)^3=2\pi^2\left[2\left(\dfrac{T_\gamma}{T_\nu}\right)^3+\dfrac{42}{8}\right](a_2T_\nu (a_2))^3$

and then

$\boxed{\left(\dfrac{T_\nu}{T_\gamma}\right)=\left(\dfrac{4}{11}\right)^{1/3}}$

or equivalently

$\boxed{T_\nu=\sqrt[3]{\dfrac{4}{11}}T_\gamma\approx 1\mbox{.}9K}$

In fact, the neutrino energy density can be given in two different ways, depending if it is “massless” or “massive”. For massless neutrinos (or equivalently “relativistic” massless matter particles):

I) Massless neutrinos: $\Omega_\nu=\dfrac{1\mbox{.}68\cdot 10^{-5}}{h^2}$

2) Massive neutrinos: $\Omega_\nu= \dfrac{m_\nu}{94h^2 \; eV}$

Case 5. The dark energy/Cosmological constant/Vacuum energy.

The budget of the Universe provides (from cosmological and astrophysical measurements) the shocking result

$\Omega\approx 1$ with $\Omega_M\approx 0\mbox{.}3$

Then, there is some missin smooth, unclustered energy-matter “form”/”species”. It is the “dark energy”/vacuum energy/cosmological cosntant! It can be understood as a “special” pressure term in the Einstein’s equations, but one with NEGATIVE pressure! Evidence for this observation comes from luminosity-distance-redshift measurements from SNae, clusters, and the CMB spectrum! The cosmological constant/vacuum energy/dark energy dominates the Universe today, since, it seems, we live in a (positively!) accelerated Universe!!!!! What can dark energy be? It can not be a “normal” matter field. Like the Dark Matter field, we believe that (excepting perhaps the scalar Higgs field/s) the SM has no candidate to explain the Dark Energy. What field could dark matter be? Perhaps an scalar field or something totally new and “unknown” yet.

In short, we are INTO a DARKLY, darkly, UNIVERSE! Darkness is NOT coming, darkness has arrived and, if nothing changes, it will turn our local Universe even darker and darker!

See you in the next cosmological post!