# 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!

# 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#057. Naturalness problems.

In this short blog post, I am going to list some of the greatest “naturalness” problems in Physics. It has nothing to do with some delicious natural dishes I like, but there is a natural beauty and sweetness related to naturalness problems in Physics. In fact, they include some hierarchy problems and additional problems related to stunning free values of parameters in our theories.

Naturalness problems arise when the “naturally expected” property of some free parameters or fundamental “constants” to appear as quantities of order one is violated, and thus, those paramenters or constants appear to be very large or very small quantities. That is, naturalness problems are problems of untuning “scales” of length, energy, field strength, … A value of 0.99 or 1.1, or even 0.7 and 2.3 are “more natural” than, e.g., $100000, 10^{-4},10^{122}, 10^{23},\ldots$ Equivalently, imagine that the values of every fundamental and measurable physical quantity $X$ lies in the real interval $\left[ 0,\infty\right)$. Then, 1 (or very close to this value) are “natural” values of the parameters while the two extrema $0$ or $\infty$ are “unnatural”. As we do know, in Physics, zero values are usually explained by some “fundamental symmetry” while extremely large parameters or even $\infty$ can be shown to be “unphysical” or “unnatural”. In fact, renormalization in QFT was invented to avoid quantities that are “infinite” at first sight and regularization provides some prescriptions to assign “natural numbers” to quantities that are formally ill-defined or infinite. However, naturalness goes beyond those last comments, and it arise in very different scenarios and physical theories. It is quite remarkable that naturalness can be explained as numbers/contants/parameters around 3 of the most important “numbers” in Mathematics:

$(0, 1, \infty)$

REMEMBER: Naturalness of X is, thus, being 1 or close to it, while values approaching 0 or $\infty$ are unnatural.  Therefore, if some day you heard a physicist talking/speaking/lecturing about “naturalness” remember the triple $(0,1,\infty)$ and then assign “some magnitude/constant/parameter” some quantity close to one of those numbers. If they approach 1, the parameter itself is natural and unnatural if it approaches any of the other two numbers, zero or infinity!

I have never seen a systematic classification of naturalness problems into types. I am going to do it here today. We could classify naturalness problems into:

1st. Hierarchy problems. They are naturalness problems related to the energy mass or energy spectrum/energy scale of interactions and fundamental particles.

2nd. Nullity/Smallness problems. These are naturalness problems related to free parameters which are, surprisingly, close to zero/null value, even when we have no knowledge of a deep reason to understand why it happens.

3rd. Large number problems (or hypotheses). This class of problems can be equivalently thought as nullity reciprocal problems but they arise naturally theirselves in cosmological contexts or when we consider a large amount of particles, e.g., in “statistical physics”, or when we face two theories in very different “parameter spaces”. Dirac pioneered these class of hypothesis when realized of some large number coincidences relating quantities appearing in particle physics and cosmology. This Dirac large number hypothesis is also an old example of this kind of naturalness problems.

4th. Coincidence problems. This 4th type of problems is related to why some different parameters of the same magnitude are similar in order of magnitude.

The following list of concrete naturalness problems is not going to be complete, but it can serve as a guide of what theoretical physicists are trying to understand better:

1. The little hierarchy problem. From the phenomenon called neutrino oscillations (NO) and neutrino oscillation experiments (NOSEX), we can know the difference between the squared masses of neutrinos. Furthermore, cosmological measurements allow us to put tight bounds to the total mass (energy) of light neutrinos in the Universe. The most conservative estimations give $m_\nu \leq 10 eV$ or even $m_\nu \sim 1eV$ as an upper bound is quite likely to be true. By the other hand, NOSEX seems to say that there are two mass differences, $\Delta m^2_1\sim 10^{-3}$ and $\Delta m^2_2\sim 10^{-5}$. However, we don’t know what kind of spectrum neutrinos have yet ( normal, inverted or quasidegenerated). Taking a neutrino mass about 1 meV as a reference, the little hierarchy problem is the question of why neutrino masses are so light when compared with the remaining leptons, quarks and gauge bosons ( excepting, of course, the gluon and photon, massless due to the gauge invariance).

Why is $m_\nu << m_e,m_\mu, m_\tau, m_Z,M_W, m_{proton}?$

We don’t know! Let me quote a wonderful sentence of a very famous short story by Asimov to describe this result and problem:

“THERE IS AS YET INSUFFICIENT DATA FOR A MEANINGFUL ANSWER.”

2. The gauge hierarchy problem. The electroweak (EW) scale can be generally represented by the Z or W boson mass scale. Interestingly, from this summer results, Higgs boson mass seems to be of the same order of magnitue, more or less, than gauge bosons. Then, the electroweak scale is about $M_Z\sim M_W \sim \mathcal{O} (100GeV)$. Likely, it is also of the Higgs mass  order.  By the other hand, the Planck scale where we expect (naively or not, it is another question!) quantum effects of gravity to naturally arise is provided by the Planck mass scale:

$M_P=\sqrt{\dfrac{\hbar c}{8\pi G}}=2.4\cdot 10^{18}GeV=2.4\cdot 10^{15}TeV$

or more generally, dropping the $8\pi$ factor

$M_P =\sqrt{\dfrac{\hbar c}{G}}=1.22\cdot 10^{19}GeV=1.22\cdot 10^{16}TeV$

Why is the EW mass (energy) scale so small compared to Planck mass, i.e., why are the masses $M_{EW}< so different? The problem is hard, since we do know that EW masses, e.g., for scalar particles like Higgs particles ( not protected by any SM gauge symmetry), should receive quantum contributions of order $\mathcal{O}(M_P^2)$

“THERE IS AS YET INSUFFICIENT DATA FOR A MEANINGFUL ANSWER.”

3. The cosmological constant (hierarchy) problem. The cosmological constant $\Lambda$, from the so-called Einstein’s field equations of classical relativistic gravity

$\mathcal{R}_{\mu\nu}-\dfrac{1}{2}g_{\mu\nu}\mathcal{R}=8\pi G\mathcal{T}_{\mu\nu}+\Lambda g_{\mu\nu}$

is estimated to be about $\mathcal{O} (10^{-47})GeV^4$ from the cosmological fitting procedures. The Standard Cosmological Model, with the CMB and other parallel measurements like large scale structures or supernovae data, agree with such a cosmological constant value. However, in the framework of Quantum Field Theories, it should receive quantum corrections coming from vacuum energies of the fields. Those contributions are unnaturally big, about $\mathcal{O}(M_P^4)$ or in the framework of supersymmetric field theories, $\mathcal{O}(M^4_{SUSY})$ after SUSY symmetry breaking. Then, the problem is:

Why is $\rho_\Lambda^{obs}<<\rho_\Lambda^{th}$? Even with TeV or PeV fundamental SUSY (or higher) we have a serious mismatch here! The mismatch is about 60 orders of magnitude even in the best known theory! And it is about 122-123 orders of magnitude if we compare directly the cosmological constant vacuum energy we observe with the cosmological constant we calculate (naively or not) with out current best theories using QFT or supersymmetric QFT! Then, this problem is a hierarchy problem and a large number problem as well. Again, and sadly, we don’t know why there is such a big gap between mass scales of the same thing! This problem is the biggest problem in theoretical physics and it is one of the worst predictions/failures in the story of Physics. However,

“THERE IS AS YET INSUFFICIENT DATA FOR A MEANINGFUL ANSWER.”

4. The strong CP problem/puzzle. From neutron electric dipople measurements, theoretical physicists can calculate the so-called $\theta$-angle of QCD (Quantum Chromodynamics). The theta angle gives an extra contribution to the QCD lagrangian:

$\mathcal{L}_{\mathcal{QCD}}\supset \dfrac{1}{4g_s^2}G_{\mu\nu}G^{\mu\nu}+\dfrac{\theta}{16\pi^2}G^{\mu\nu}\tilde{G}_{\mu\nu}$

The theta angle is not provided by the SM framework and it is a free parameter. Experimentally,

$\theta <10^{-12}$

while, from the theoretical aside, it could be any number in the interval $\left[-\pi,\pi\right]$. Why is $\theta$ close to the zero/null value? That is the strong CP problem! Once again, we don’t know. Perhaps a new symmetry?

“THERE IS AS YET INSUFFICIENT DATA FOR A MEANINGFUL ANSWER.”

5. The flatness problem/puzzle. In the Stantard Cosmological Model, also known as the $\Lambda CDM$ model, the curvature of the Universe is related to the critical density and the Hubble “constant”:

$\dfrac{1}{R^2}=H^2\left(\dfrac{\rho}{\rho_c}-1\right)$

There, $\rho$ is the total energy density contained in the whole Universe and $\rho_c=\dfrac{3H^2}{8\pi G}$ is the so called critical density. The flatness problem arise when we deduce from cosmological data that:

$\left(\dfrac{1}{R^2}\right)_{data}\sim 0.01$

At the Planck scale era, we can even calculate that

$\left(\dfrac{1}{R^2}\right)_{Planck\;\; era}\sim\mathcal{O}(10^{-61})$

This result means that the Universe is “flat”. However, why did the Universe own such a small curvature? Why is the current curvature “small” yet? We don’t know. However, cosmologists working on this problem say that “inflation” and “inflationary” cosmological models can (at least in principle) solve this problem. There are even more radical ( and stranger) theories such as varying speed of light theories trying to explain this, but they are less popular than inflationary cosmologies/theories. Indeed, inflationary theories are popular because they include scalar fields, similar in Nature to the scalar particles that arise in the Higgs mechanism and other beyond the Standard Model theories (BSM). We don’t know if inflation theory is right yet, so

“THERE IS AS YET INSUFFICIENT DATA FOR A MEANINGFUL ANSWER.”

6. The flavour problem/puzzle. The ratios of successive SM fermion mass eigenvalues ( the electron, muon, and tau), as well as the angles appearing in one gadget called the CKM (Cabibbo-Kobayashi-Maskawa) matrix, are roughly of the same order of magnitude. The issue is harder to know ( but it is likely to be as well) for constituent quark masses. However, why do they follow this particular pattern/spectrum and structure? Even more, there is a mysterious lepton-quark complementarity. The analague matrix in the leptonic sector of such a CKM matrix is called the PMNS matrix (Pontecorvo-Maki-Nakagawa-Sakata matrix) and it describes the neutrino oscillation phenomenology. It shows that the angles of PMNS matrix are roughly complementary to those in the CKM matrix ( remember that two angles are said to be complementary when they add up to 90 sexagesimal degrees). What is the origin of this lepton(neutrino)-quark(constituent) complementarity? In fact, the two questions are related since, being rough, the mixing angles are related to the ratios of masses (quarks and neutrinos). Therefore, this problem, if solved, could shed light to the issue of the particle spectrum or at least it could help to understand the relationship between quark masses and neutrino masses. Of course, we don’t know how to solve this puzzle at current time. And once again:

“THERE IS AS YET INSUFFICIENT DATA FOR A MEANINGFUL ANSWER.”

7. Cosmic matter-dark energy coincidence. At current time, the densities of matter and vacuum energy are roughly of the same order of magnitude, i.e, $\rho_M\sim\rho_\Lambda=\rho_{DE}$. Why now? We do not know!

“THERE IS AS YET INSUFFICIENT DATA FOR A MEANINGFUL ANSWER.”

And my weblog is only just beginning! See you soon in my next post! 🙂