# LOG#105. Einstein’s equations.

In 1905,  one of Einstein’s achievements was to establish the theory of Special Relativity from 2 single postulates and correctly deduce their physical consequences (some of them time later).  The essence of Special Relativity, as we have seen, is that  all the inertial observers must agree on the speed of light “in vacuum”, and that the physical laws (those from Mechanics and Electromagnetism) are the same for all of them.  Different observers will measure (and then they see) different wavelengths and frequencies, but the product of wavelength with the frequency is the same.  The wavelength and frequency are thus Lorentz covariant, meaning that they change for different observers according some fixed mathematical prescription depending on its tensorial character (scalar, vector, tensor,…) respect to Lorentz transformations.  The speed of light is Lorentz invariant.

By the other hand, Newton’s law of gravity describes the motion of planets and terrrestrial bodies.  It is all that we need in contemporary rocket ships unless those devices also carry atomic clocks or other tools of exceptional accuracy.  Here is Newton’s law in potential form:

$4\pi G\rho = \nabla ^2 \phi$

In the special relativity framework, this equation has a terrible problem: if there is a change in the mass density $\rho$, then it must propagate everywhere instantaneously.  If you believe in the Special Relativity rules and in the speed of light invariance, it is impossible. Therefore, “Houston, we have a problem”.

Einstein was aware of it and he tried to solve this inconsistency.  The final solution took him ten years .

The apparent silly and easy problem is to develop and describe all physics in the the same way irrespectively one is accelerating or not. However, it is not easy or silly at all. It requires deep physical insight and a high-end mathematical language.  Indeed,  what is the most difficult part are  the details of Riemann geometry and tensor calculus on manifolds.  Einstein got  private aid from a friend called  Marcel Grossmann. In fact, Einstein knew that SR was not compatible with Newton’s law of gravity. He (re)discovered the equivalence principle, stated by Galileo himself much before than him, but he interpreted deeper and seeked the proper language to incorporante that principle in such a way it were compatible (at least locally) with special relativity! His  “journey” from 1907 to 1915 was a hard job and a continuous struggle with tensorial methods…

Today, we are going to derive the Einstein field equations for gravity, a set of equations for the “metric field” $g_{\mu \nu}(x)$. Hilbert in fact arrived at Einstein’s field equations with the use of the variational method we are going to use here, but Einstein’s methods were more physical and based on physical intuitions. They are in fact “complementary” approaches. I urge you to read “The meaning of Relativity” by A.Einstein in order to read a summary of his discoveries.

We now proceed to derive Einstein’s Field Equations (EFE) for General Relativity (more properly, a relativistic theory of gravity):

Step 1. Let us begin with the so-called Einstein-Hilbert action (an ansatz).

$S = \int d^4x \sqrt{-g} \left( \dfrac{c^4}{16 \pi G} R + \mathcal{L}_{\mathcal{M}} \right)$

Be aware of  the square root of the determinant of the metric as part of the volume element.  It is important since the volume element has to be invariant in curved spacetime (i.e.,in the presence of a metric).  It also plays a critical role in the derivation.

Step 2. We perform the variational variation with respect to the metric field $g^{\mu \nu}$:

$\delta S = \int d^4 x \left( \dfrac{c^4}{16 \pi G} \dfrac{\delta \sqrt{-g}}{\delta g^{\mu \nu}} + \dfrac{\delta (\sqrt{-g}\mathcal{L}_{\mathcal{M}})}{\delta g^{\mu \nu}} \right) \delta g^{\mu \nu}$

Step 3. Extract out  the square root of the metric as a common factor and use the product rule on the term with the Ricci scalar R:

$\delta S = \int d^4 x \sqrt{-g} \left( \dfrac{c^4}{16 \pi G} \left ( \dfrac{\delta R}{\delta g^{\mu \nu}} +\dfrac{R}{\sqrt{-g}}\dfrac{\delta \sqrt{-g}}{\delta g^{\mu \nu}} \right) +\dfrac{1}{\sqrt{-g}}\dfrac{\delta ( \sqrt{-g}\mathcal{L}_{\mathcal{M}})}{\delta g^{\mu\nu}}\right) \delta g^{\mu \nu}$

Step 4.  Use the definition of a Ricci scalar as a contraction of the Ricci tensor to calculate the first term:

$\dfrac{\delta R}{\delta g^{\mu \nu}} = \dfrac{\delta (g^{\mu \nu}R_{\mu \nu})}{\delta g^{\mu \nu} }= R_{\mu\nu} + g^{\mu \nu}\dfrac{\delta R_{\mu \nu}}{\delta g^{\mu \nu}} = R_{\mu \nu} + \mbox{total derivative}$

A total derivative does not make a contribution to the variation of the action principle, so can be neglected to find the extremal point.  Indeed, this is the Stokes theorem in action. To show that the variation in the Ricci tensor is a total derivative, in case you don’t believe this fact, we can proceed as follows:

Check 1. Write  the Riemann curvature tensor:

$R^{\rho}_{\, \sigma \mu \nu} = \partial _{\mu} \Gamma ^{\rho}_{\, \sigma \nu} - \partial_{\nu} \Gamma^{\rho}_{\, \sigma \mu}+ \Gamma^{\rho}_{\, \lambda \mu} \Gamma^{\lambda}_{\, \sigma \nu} - \Gamma^{\rho}_{\, \lambda \nu} \Gamma^{\lambda}_{\, \sigma \mu}$

Note the striking resemblance with the non-abelian YM field strength curvature two-form

$F=dA+A \wedge A = \partial _{\mu} A_{\nu} - \partial _{\nu} A_{\mu} + k \left[ A_\mu , A_{\nu} \right]$.

There are many terms with indices in the Riemann tensor calculation, but we can simplify stuff.

Check 2. We have to calculate the variation of the Riemann curvature tensor with respect to the metric tensor:

$\delta R^{\rho}_{\, \sigma \mu \nu} = \partial _{\mu} \delta \Gamma^{\rho}_{\, \sigma \nu} - \partial_\nu \delta \Gamma^{\rho}_{\, \sigma \mu} + \delta \Gamma ^{\rho}_{\, \lambda \mu} \Gamma^{\lambda}_{\, \sigma \nu} - \delta \Gamma^{\rho}_{\lambda \nu}\Gamma^{\lambda}_{\, \sigma \mu} + \Gamma^{\rho}_{\, \lambda \mu}\delta \Gamma^{\lambda}_{\sigma \nu} - \Gamma^{\rho}_{\lambda \nu} \delta \Gamma^{\lambda}_{\, \sigma \mu}$

One cannot calculate the covariant derivative of a connection since it does not transform like a tensor.  However, the difference of two connections does transform like a tensor.

Check 3. Calculate the covariant derivative of the variation of the connection:

$\nabla_{\mu} ( \delta \Gamma^{\rho}_{\sigma \nu}) = \partial _{\mu} (\delta \Gamma^{\rho}_{\, \sigma \nu}) + \Gamma^{\rho}_{\, \lambda \mu} \delta \Gamma^{\lambda}_{\, \sigma \nu} - \delta \Gamma^{\rho}_{\, \lambda \sigma}\Gamma^{\lambda}_{\mu \nu} - \delta \Gamma^{\rho}_{\, \lambda \nu}\Gamma^{\lambda}_{\, \sigma \mu}$

$\nabla_{\nu} ( \delta \Gamma^{\rho}_{\sigma \mu}) = \partial _\nu (\delta \Gamma^{\rho}_{\, \sigma \mu}) + \Gamma^{\rho}_{\, \lambda \nu} \delta \Gamma^{\lambda}_{\, \sigma \mu} - \delta \Gamma^{\rho}_{\, \lambda \sigma}\Gamma^{\lambda}_{\mu \nu} - \delta \Gamma^{\rho}_{\, \lambda \mu}\Gamma^{\lambda}_{\, \sigma \nu}$

Check 4. Rewrite the variation of the Riemann curvature tensor as the difference of two covariant derivatives of the variation of the connection written in Check 3, that is, substract the previous two terms in check 3.

$\delta R^{\rho}_{\, \sigma \mu \nu} = \nabla_{\mu} \left( \delta \Gamma^{\rho}_{\, \sigma \nu}\right) - \nabla _{\nu} \left(\delta \Gamma^{\rho}_{\, \sigma \mu}\right)$

Check 5. Contract the result of Check 4.

$\delta R^{\rho}_{\, \mu \rho \nu} = \delta R_{\mu \nu} = \nabla_{\rho} \left( \delta \Gamma^{\rho}_{\, \mu \nu}\right) - \nabla _{\nu} \left(\delta \Gamma^{\rho}_{\, \rho \mu}\right)$

Check 6. Contract the result of Check 5:

$g^{\mu \nu}\delta R_{\mu \nu} = \nabla_\rho (g^{\mu \nu} \delta \Gamma^{\rho}_{\mu\nu})-\nabla_\nu (g^{\mu \nu}\delta \Gamma^{\rho}_{\rho \mu}) = \nabla _\sigma (g^{\mu \nu}\delta \Gamma^{\sigma}_{\mu \nu}) - \nabla_\sigma (g^{\mu \sigma}\delta \Gamma ^{\rho}_{\rho \mu})$

Therefore, we have

$g^{\mu \nu}\delta R_{\mu \nu} = \nabla_\sigma (g^{\mu \nu}\delta \Gamma^{\sigma}_{\mu\nu}- g^{\mu \sigma}\delta \Gamma^{\rho}_{\rho\mu})=\nabla_\sigma K^\sigma$

Q.E.D.

Step 5. The variation of the second term in the action is the next step.  Transform the coordinate system to one where the metric is diagonal and use the product rule:

$\dfrac{R}{\sqrt{-g}} \dfrac{\delta \sqrt{-g}}{\delta g^{\mu \nu}}=\dfrac{R}{\sqrt{-g}} \dfrac{-1}{2 \sqrt{-g}}(-1) g g_{\mu \nu}\dfrac{\delta g^{\mu \nu}}{\delta g^{\mu \nu}} =- \dfrac{1}{2}g_{\mu \nu} R$

The reason of the last equalities is that $g^{\alpha\mu}g_{\mu \beta}=\delta^{\alpha}_{\; \beta}$, and then its variation is

$\delta (g^{\alpha\mu}g_{\mu \nu}) = (\delta g^{\alpha\mu}) g_{\mu \nu} + g^{\alpha\mu}(\delta g_{\mu \nu}) = 0$

Thus, multiplication by the inverse metric $g^{\beta \nu}$ produces

$\delta g^{\alpha \beta} = - g^{\alpha \mu}g^{\beta \nu}\delta g_{\mu \nu}$

that is,

$\dfrac{\delta g^{\alpha \beta}}{\delta g_{\mu \nu}}= -g^{\alpha \mu} g^{\beta \nu}$

By the other hand, using the theorem for the derivation of a determinant we get that:

$\delta g = \delta g_{\mu \nu} g g^{\mu \nu}$

since

$\dfrac{\delta g}{\delta g^{\alpha \beta}}= g g^{\alpha \beta}$

because of the classical identity

$g^{\alpha \beta}=(g_{\alpha \beta})^{-1}=\left( \det g \right)^{-1} Cof (g)$

Indeed

$Cof (g) = \dfrac{\delta g}{\delta g^{\alpha \beta}}$

and moreover

$\delta \sqrt{-g}=-\dfrac{\delta g}{2 \sqrt{-g}}= -g\dfrac{ \delta g_{\mu \nu} g^{\mu \nu}}{2 \sqrt{-g}}$

so

$\delta \sqrt{-g}=\dfrac{1}{2}\sqrt{-g}g^{\mu \nu}\delta g_{\mu \nu}=\dfrac{1}{2}\sqrt{-g}g_{\mu \nu}\delta g^{\mu \nu}$

Q.E.D.

Step 6. Define the stress energy-momentum tensor as the third term in the action (that coming from the matter lagrangian):

$T_{\mu \nu} = - \dfrac{2}{\sqrt{-g}}\dfrac{(\sqrt{-g} \mathcal{L}_{\mathcal{M}})}{\delta g^{\mu \nu}}$

or equivalently

$-\dfrac{1}{2}T_{\mu \nu} = \dfrac{1}{\sqrt{-g}}\dfrac{(\sqrt{-g} \mathcal{L}_{\mathcal{M}})}{\delta g^{\mu \nu}}$

Step 7. The extremal principle. The variation of the Hilbert action will be  an extremum when the integrand is equal to zero:

$\dfrac{c^4}{16\pi G}\left( R_{\mu \nu} - \dfrac{1}{2} g_{\mu \nu}R\right) - \dfrac{1}{2} T_{\mu \nu} = 0$

i.e.,

$\boxed{R_{\mu \nu} - \dfrac{1}{2}g_{\mu \nu} R = \dfrac{8\pi G}{c^4}T_{\mu\nu}}$

Usually this is recasted and simplified using the Einstein’s tensor

$G_{\mu \nu}= R_{\mu \nu} - \dfrac{1}{2}g_{\mu \nu} R$

as

$\boxed{G_{\mu\nu}=\dfrac{8\pi G}{c^4}T_{\mu\nu}}$

This deduction has been mathematical. But there is a deep physical picture behind it. Moreover,  there are a huge number of physics issues one could go into. For instance, these equations bind to particles with integral spin which is good for bosons, but there are matter fermions that also participate in gravity coupling to it. Gravity is universal.  To include those fermion fields, one can consider the metric and the connection to be independent of each other.  That is the so-called Palatini approach.

Final remark: you can add to the EFE above a “constant” times the metric tensor, since its “covariant derivative” vanishes. This constant is the cosmological constant (a.k.a. dark energy in conteporary physics). The, the most general form of EFE is:

$\boxed{G_{\mu\nu}+\Lambda g_{\mu\nu}=\dfrac{8\pi G}{c^4}T_{\mu\nu}}$

Einstein’s additional term was added in order to make the Universe “static”. After Hubble’s discovery of the expansion of the Universe, Einstein blamed himself about the introduction of such a term, since it avoided to predict the expanding Universe. However, perhaps irocanilly, in 1998 we discovered that the Universe was accelerating instead of being decelerating due to gravity, and the most simple way to understand that phenomenon is with a positive cosmological constant domining the current era in the Universe. Fascinating, and more and more due to the WMAP/Planck data. The cosmological constant/dark energy and the dark matter we seem to “observe” can not be explained with the fields of the Standard Model, and therefore…They hint to new physics. The character of this  new physics is challenging, and much work is being done in order to find some particle of model in which dark matter and dark energy fit. However, it is not easy at all!

May the Einstein’s Field Equations be with you!

# LOG#056. Gravitational alpha(s).

The topic today is to review a beautiful paper and to discuss its relevance for theoretical physics. The paper is: Comment on the cosmological constant and a gravitational alpha by R.J.Adler. You can read it here: http://arxiv.org/abs/1110.3358

One of the most intriguing and mysterious numbers in Physics is the electromagnetic fine structure constant $\alpha_{EM}$. Its value is given by

$\alpha_{EM}=7.30\cdot 10^{-3}$

or equivalenty

$\alpha_{EM}^{-1}=\dfrac{1}{\alpha_{EM}}=137$

Of course, I am assuming that the coupling constant is measured at ordinary energies, since we know that the coupling constants are not really constant but they vary slowly with energy. However, I am not going to talk about the renormalization (semi)group in this post.

Why is the fine structure constant important? Well, we can undertand it if we insert the values of the constants that made the electromagnetic alpha constant:

$\alpha_{EM}=\dfrac{e^2}{\hbar c}$

with $e$ being the electron elemental charge, $\hbar$ the Planck’s constant divided by two pi, c is the speed of light and where we are using units with $K_C=\dfrac{1}{4\pi \varepsilon_0}=1$. Here $K_C$ is the Coulomb constant, generally with a value $9\cdot 10^9Nm^2/C^2$, but we rescale units in order it has a value equal to the unit. We will discuss more about frequently used system of units soon.

As the electromagnetic alpha constant depends on the electric charge, the Coulomb’s electromagnetic constant ( rescaled to one in some “clever” units), the Planck’s constant ( rationalized by $2\pi$ since $\hbar=h/2\pi$) and the speed of light, it codes some deep information of the Universe inside of it. The electromagnetic alpha $\alpha_{EM}$ is quantum and relativistic itself, and it also is related to elemental charges. Why alpha has the value it has is a complete mystery. Many people has tried to elucidate why it has the value it has today, but there is no reason of why it should have the value it has. Of course, it happens as well with some other constants but this one is particularly important since it is involved in some important numbers in atomic physics and the most elemental atom, the hydrogen atom.

In atomic physics, there are two common and “natural” scales of length. The first scale of length is given by the Compton’s wavelength of electrons. Usint the de Broglie equation, we get that the Compton’s wavelength is the wavelength of a photon whose energy is the same as the rest mass of the particle, or mathematically speaking:

$\boxed{\lambda=\dfrac{h}{p}=\dfrac{h}{mc}}$

Usually, physicists employ the “reduced” or “rationalized” Compton’s wavelength. Plugging the electron mass, we get the electron reduced Compton’s wavelength:

$\boxed{\lambda_C=\dfrac{\lambda}{2\pi}=\dfrac{\hbar}{m_ec}=\dfrac{\hbar}{m_ec}=3.86\cdot 10^{-13}m}$

The second natural scale of length in atomic physics is the so-called Böhr radius. It is given by the formula:

$\boxed{a_B=\dfrac{\hbar^2}{m_e e^2}=5.29\cdot 10^{-11}m}$

Therefore, there is a natural mass ratio between those two length scales, and it shows that it is precisely the electromagnetic fine structure constant alpha $\alpha_{EM}$:

$\boxed{R_\alpha=\dfrac{\mbox{Reduced Compton's wavelength}}{\mbox{B\"{o}hr radius}}=\dfrac{\lambda_C}{a_B}=\dfrac{\left(\hbar/m_e c\right)}{\left(\hbar^2/m_ee^2\right)}=\dfrac{e^2}{\hbar c}=\alpha_{EM}=7.30\cdot 10^{-3}}$

Furthermore, we can show that the electromagnetic alpha also is related to the mass ration between the electron energy in the fundamental orbit of the hydrogen atom and the electron rest energy. These two scales of energy are given by:

1) Rydberg’s energy ( electron ground minimal energy in the fundamental orbit/orbital for the hydrogen atom):

$\boxed{E_H=\dfrac{m_ee^4}{2\hbar^2}=13.6eV}$

2) Electron rest energy:

$\boxed{E_0=m_ec^2}$

Then, the ratio of those two “natural” energies in atomic physics reads:

$\boxed{R'_E=\dfrac{\mbox{Rydberg's energy}}{\mbox{Electron rest energy}}=\dfrac{m_ee^4/2\hbar^2}{m_ec^2}=\dfrac{1}{2}\left(\dfrac{e^2}{\hbar c}\right)^2=\dfrac{\alpha_{EM}^2}{2}=2.66\cdot 10^{-5}}$

or equivalently

$\boxed{\dfrac{1}{R'_E}=37600=3.76\cdot 10^4}$

R.J.Adler’s paper remarks that there is a cosmological/microscopic analogue of the above two ratios, and they involve the infamous Einstein’s cosmological constant. In Cosmology, we have two natural (ultimate?) length scales:

1st. The (ultra)microscopic and ultrahigh energy (“ultraviolet” UV regulator) relevant Planck’s length $L_P$, or equivalently the squared value $L_P^2$. Its value is given by:

$\boxed{L_P^2=\dfrac{G\hbar}{c^3}\leftrightarrow L_P=\sqrt{\dfrac{G\hbar}{c^3}}=1.62\cdot 10^{35}m}$

This natural length can NOT be related to any “classical” theory of gravity since it involves and uses the Planck’s constant $\hbar$.

2nd. The (ultra)macroscopic and ultra-low-energy (“infrared” IR regulator) relevant cosmological constant/deSitter radius. They are usualy represented/denoted by $\Lambda$ and $R_{dS}$ respectively, and they are related to each other in a simple way. The dimensions of the cosmological constant are given by

$\boxed{\left[\Lambda \right]=\left[ L^{-2}\right]=(\mbox{Length})^{-2}}$

The de Sitter radius and the cosmological constant are related through a simple equation:

$\boxed{R_{dS}=\sqrt{\dfrac{3}{\Lambda}}\leftrightarrow R^2_{dS}=\dfrac{3}{\Lambda}\leftrightarrow \Lambda =\dfrac{3}{R^2_{dS}}}$

The de Sitter radius is obtained from cosmological measurements thanks to the so called Hubble’s parameter ( or Hubble’s “constant”, although we do know that Hubble’s “constant” is not such a “constant”, but sometimes it is heard as a language abuse) H. From cosmological data we obtain ( we use the paper’s value without loss of generality):

$H=\dfrac{73km/s}{Mpc}$

This measured value allows us to derive the Hubble’s length paremeter

$L_H=\dfrac{c}{H}=1.27\cdot 10^{26}m$

Moreover, the data also imply some density energy associated to the cosmological “constant”, and it is generally called Dark Energy. This density energy from data is written as:

$\Omega_\Lambda =\Omega^{data}_{\Lambda}$

and from this, it can be also proved that

$R_{dS}=\dfrac{L_H}{\sqrt{\Omega_\Lambda}}=1.46\cdot 10^{26}m$

where we have introduced the experimentally deduced value $\Omega_\Lambda\approx 0.76$ from the cosmological parameter global fits. In fact, the cosmological constant helps us to define the beautiful and elegant formula that we can call the gravitational alpha/gravitational cosmological fine structure constant $\alpha_G$:

$\boxed{\alpha_G\equiv \dfrac{\mbox{Planck's length}}{\mbox{normalized de Sitter radius}}=\dfrac{L_P}{\dfrac{R_{dS}}{\sqrt{3}}}=\dfrac{\sqrt{\dfrac{G\hbar}{c^3}}}{\sqrt{\dfrac{1}{\Lambda}}}=\sqrt{\dfrac{G\hbar\Lambda}{c^3}}}$

or equivalently, defining the cosmological length associated to the cosmological constant as

$L^2_\Lambda=\dfrac{1}{\Lambda}=\dfrac{R^2_{dS}}{3}\leftrightarrow L_\Lambda=\sqrt{\dfrac{1}{\Lambda}}=\dfrac{R_{dS}}{\sqrt{3}}$

$\boxed{\alpha_G\equiv \dfrac{\mbox{Planck's length}}{\mbox{Cosmological length}}=\dfrac{L_P}{L_\Lambda}=\dfrac{\sqrt{\dfrac{G\hbar}{c^3}}}{\sqrt{\dfrac{1}{\Lambda}}}=\sqrt{\dfrac{G\hbar\Lambda}{c^3}}=L_P\sqrt{\Lambda}=L_P\dfrac{R_{dS}}{\sqrt{3}}}$

If we introduce the numbers of the constants, we easily obtaint the gravitational cosmological alpha value and its inverse:

$\boxed{\alpha_G=1.91\cdot 10^{-61}\leftrightarrow \alpha_G^{-1}=\dfrac{1}{\alpha_G}=5.24\cdot 10^{60}}$

They are really small and large numbers! Following the the atomic analogy, we can also create a ratio between two cosmologically relevant density energies:

1st. The Planck’s density energy.

Planck’s energy is defined as

$\boxed{E_P=\dfrac{\hbar c}{L_P}=\sqrt{\dfrac{\hbar c^5}{G}}=1.22\cdot 10^{19}GeV=1.22\cdot 10^{16}TeV}$

The Planck energy density $\rho_P$ is defined as the energy density of Planck’s energy inside a Planck’s cube or side $L_P$, i.e., it is the energy density of Planck’s energy concentrated inside a cube with volume $V=L_P^3$. Mathematically speaking, it is

$\boxed{\rho_P=\dfrac{E_P}{L_P^3}=\dfrac{c^7}{\hbar G^2}=2.89\cdot 10^{123}\dfrac{GeV}{m^3}}$

It is an huge density energy!

Remark: Energy density is equivalent to pressure in special relativity hydrodynamics. That is,

$\mathcal{P}_P=\rho_P=\tilde{\rho}_P c^2=4.63\cdot 10^{113}Pa$

wiht Pa denoting pascals ($1Pa=1N/m^2$) and where $\tilde{\rho}_P$ represents here matter (not energy) density ( with units in $kg/m^3$). Of course, turning matter density into energy density requires a multiplication by $c^2$. This equivalence between vacuum pressure and energy density is one of the reasons because some astrophysicists, cosmologists and theoretical physicists call “vacuum pressure” to the “dark energy/cosmological constant” term in the study of the cosmic components derived from the total energy density $\Omega$.

2nd. The cosmological constant density energy.

Using the Einstein’s field equations, it can be shown that the cosmological constant gives a contribution to the stress-energy-momentum tensor. The component $T^{0}_{\;\; 0}$ is related to the dark energy ( a.k.a. the cosmological constant) and allow us to define the energy density

$\boxed{\rho_\Lambda =T^{0}_{\;\; 0}=\dfrac{\Lambda c^4}{8\pi G}}$

Using the previous equations for G as a function of Planck’s length, the Planck’s constant and the speed of light, and the definitions of Planck’s energy and de Sitter radius, we can rewrite the above energy density as follows:

$\boxed{\rho_\Lambda=\dfrac{3}{8\pi}\left(\dfrac{E_P}{L_PR^2_{dS}}\right)=4.21 \dfrac{GeV}{m^3}}$

Thus, we can evaluate the ration between these two energy densities! It provides

$\boxed{R_\rho =\dfrac{\mbox{Planck's energy density}}{\mbox{CC energy density}}=\dfrac{\rho_P}{\rho_\Lambda}=\left( \dfrac{3}{8\pi}\right)\left(\dfrac{L_P}{R_{dS}}\right)^2=\left(\dfrac{1}{8\pi}\right)\alpha_G^2=1.45\cdot 10^{-123}}$

and the inverse ratio will be

$\boxed{\dfrac{1}{R_\rho}=6.90\cdot 10^{122}}$

So, we have obtained two additional really tiny and huge values for $R_\rho$ and its inverse, respectively. Note that the power appearing in the ratios of cosmological lengths and cosmological energy densities match the same scaling property that the atomic case with the electromagnetic alpha! In the electromagnetic case, we obtained $R\sim \alpha_{EM}$ and $R_E\sim \alpha_{EM}^2$. The gravitational/cosmological analogue ratios follow the same rule $R\sim \alpha_G$ and $R_\rho\sim \alpha_G^2$ but the surprise comes from the values of the gravitational alpha values and ratios. Some comments are straightforward:

1) Understanding atomic physics involved the discovery of Planck’s constant and the quantities associated to it at fundamental quantum level ( Böhr radius, the Rydberg’s constant,…). Understanding the Cosmological Constant value and the mismatch or stunning ratios between the equivalent relevant quantities, likely, require that $\Lambda$ can be viewed as a new “fundamental constant” or/and it can play a dynamical role somehow ( e.g., varying in some unknown way with energy or local position).

2) Currently, the cosmological parameters and fits suggest that $\Lambda$ is “constant”, but we can not be totally sure it has not varied slowly with time. And there is a related idea called quintessence, in which the cosmological “constant” is related to some dynamical field and/or to inflation. However, present data say that the cosmological constant IS truly constant. How can it be so? We are not sure, since our physical theories can hardly explain the cosmological constant, its value, and why it is current density energy is radically different from the vacuum energy estimates coming from Quantum Field Theories.

3) The mysterious value

$\boxed{\alpha_G=\sqrt{\dfrac{G\hbar\Lambda}{c^3}}=1.91\cdot 10^{-61}}$

is an equivalent way to express the biggest issue in theoretical physics. A naturalness problem called the cosmological constant problem.

In the literature, there have been alternative definitions of “gravitational fine structure constants”, unrelated with the above gravitational (cosmological) fine structure constant or gravitational alpha. Let me write some of these alternative gravitational alphas:

1) Gravitational alpha prime. It is defined as the ratio between the electron rest mass and the Planck’s mass squared:

$\boxed{\alpha'_G=\dfrac{Gm_e^2}{\hbar c}=\left(\dfrac{m_e}{m_P}\right)^2=1.75\cdot 10^{-45}}$

$\boxed{\alpha_G^{'-1}=\dfrac{1}{\alpha_G^{'}}=5.71\cdot 10^{44}}$

Note that $m_e=0.511MeV$. Since $m_{proton}=1836m_e$, we can also use the proton rest mass instead of the electron mass to get a new gravitational alpha.

2) Gravitational alpha double prime. It is defined as the ratio between the proton rest mass and the Planck’s mass squared:

$\boxed{\alpha''_G=\dfrac{Gm_{prot}^2}{\hbar c}=\left(\dfrac{m_{prot}}{m_P}\right)^2=5.90\cdot 10^{-39}}$

and the inverse value

$\boxed{\alpha_G^{''-1}=\dfrac{1}{\alpha_G^{''}}=1.69\cdot 10^{38}}$

Finally, we could guess an intermediate gravitational alpha, mixing the electron and proton mass.

3) Gravitational alpha triple prime. It is defined as the ration between the product of the electron and proton rest masses with the Planck’s mass squared:

$\boxed{\alpha'''_G=\dfrac{Gm_{prot}m_e}{\hbar c}=\dfrac{m_{prot}m_e}{m_P^2}=3.22\cdot 10^{-42}}$

and the inverse value

$\boxed{\alpha_G^{'''-1}=\dfrac{1}{\alpha^{'''}_G}=3.11\cdot 10^{41}}$

We can compare the 4 gravitational alphas and their inverse values, and additionally compare them with $\alpha_{EM}$. We get

$\alpha_G <\alpha_G^{'} <\alpha_G^{'''} < \alpha_G^{''}<\alpha_{EM}$

$\alpha_{EM}^{-1}<\alpha^{''-1}_G <\alpha^{'''-1}_G <\alpha^{'-1}_G < \alpha^{-1}_G$

These inequations mean that the electromagnetic fine structure constant $\alpha_{EM}$ is (at ordinary energies) 42 orders of magnitude bigger than $\alpha_G^{'}$, 39 orders of magnitude bigger than $\alpha_G^{'''}$, 36 orders of magnitude bigger than $\alpha_G^{''}$ and, of course, 58 orders of magnitude bigger than $\alpha_G$. Indeed, we could extend this analysis to include the “fine structure constant” of Quantum Chromodynamics (QCD) as well. It would be given by:

$\boxed{\alpha_s=\dfrac{g_s^2}{\hbar c}=1}$

since generally we define $g_s=1$. We note that $\alpha_s >\alpha_{EM}$ by 3 orders of magnitude. However, as strong nuclear forces are short range interactions, they only matter in the atomic nuclei, where confinement, and color forces dominate on every other fundamental interaction. Interestingly, at high energies, QCD coupling constant has a property called asymptotic freedom. But it is another story not to be discussed here! If we take the alpha strong coupling into account the full hierarchy of alphas is given by:

$\alpha_G <\alpha_G^{'} <\alpha_G^{'''} < \alpha_G^{''}<\alpha_{EM}<\alpha_s$

$\alpha_s^{-1}<\alpha_{EM}^{-1}<\alpha^{''-1}_G <\alpha^{'''-1}_G <\alpha^{'-1}_G < \alpha^{-1}_G$

Fascinating! Isn’t it? Stay tuned!!!

ADDENDUM: After I finished this post, I discovered a striking (and interesting itself) connection between $\alpha_{EM}$ and $\alpha_{G}$. The relation or coincidence is the following relationship

$\dfrac{1}{\alpha_{EM}}\approx \ln \left( \dfrac {1}{16\alpha_G}\right)$

Is this relationship fundamental or accidental? The answer is unknown. However, since the electric charge (via electromagnetic alpha) is not related a priori with the gravitational constant or Planck mass ( or the cosmological constant via the above gravitational alpha) in any known way I find particularly stunning such a coincidence up to 5 significant digits! Any way, there are many unexplained numerical coincidences that are completely accidental and meaningless, and then, it is not clear why this numeral result should be relevant for the connection between electromagnetism and gravity/cosmology, but it is interesting at least as a curiosity and “joke” of Nature.