# 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#106. Basic Cosmology (I).

The next thread is devoted to Cosmology. I will intend to be clear and simple about equations and principles of current Cosmology with a General Relativity background.

First of all…I will review the basic concepts of natural units I am going to use here. We will be using the following natural units:

$\hbar=c=k_B=1$

We will take the Planck mass to be given by

$M_P=\sqrt{8\pi G_N}\approx 1\mbox{.}2\cdot 10^{19}GeV$

The solar mass is $M_\odot=2\cdot 10^{30}kg$ and the parsec is given by the value

$1pc=3\mbox{.}26lyr=3\mbox{.}1\cdot 10^{16}m$

Well, current Cosmology is based on General Relativity. Even if I have not reviewed this theory with detail in this blog, the nice thing is that most of Cosmology can be learned with only a very little knowledge of this fenomenal theory. The most important ideas are: metric field, geodesics, Einstein equations and no much more…

In fact, newtonian gravity is a good approximation in some particular cases! And we do know that even in this pre-relativistic theory

$\mbox{Gravitational force}=\mbox{Matter/Mass density}$

via the Poisson’s equation

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

This idea, due to the equivalence principle, is generalized a little bit in the general relativistic framework

$\mbox{Spacetime geometry}=\mbox{Matter content/Energy-momentum}$

The spacetime geometry is determined by the metric tensor $g_{\mu\nu}(x)$. The matter content is given by the stress-energy-momentum tensor $T_{\mu\nu}$. As we know one of these two elements, we can know, via Eisntein’s field equations the another. That is, given a metric tensor, we can tell how energy-momentum “moves” in space-time. Given the energy-momentum tensor, we can know what is the metric tensor in spacetime and we can guess how the spacetime bends… This is the origin of the famous motto: “Spacetime says matter how to move, energy-momentum says spacetime how to curve”! Remember that we have “deduced” the Einstein’s field equations in the previous post. Without a cosmological constant term, we get

$G_{\mu\nu}=\kappa^2T_{\mu\nu}=8\pi G_NT_{\mu\nu}$

Given a spacetime metric $g_{\mu\nu}$, we can calculate the (affine/Levi-Civita) connection

$\Gamma^\sigma_{\;\;\mu\nu}=\dfrac{1}{2}g^{\sigma\rho}\left(\partial_\mu g_{\nu\rho}+\partial_\nu g_{\rho\mu}-\partial_\rho g_{\mu\nu}\right)$

The Riemann tensor that measures the spacetime curvature is provided by the equation

$R^\rho_{\;\; \sigma\mu\nu}=\partial_\mu \Gamma^\rho_{\;\;\mu\sigma}-\partial_\mu \Gamma^\rho_{\;\; \mu \sigma}+\Gamma^\rho_{\;\;\mu\lambda}\Gamma^\lambda_{\;\;\nu\sigma}-\Gamma^\rho_{\;\;\nu\lambda}\Gamma^\lambda_{\;\;\mu\sigma}$

The Ricci tensor is defined to be the following “trace” of the Riemann tensor

$R_{\mu\nu}=R^\lambda_{\;\;\mu\lambda \nu}$

The Einstein tensor is related to the above tensors in the well-known manner

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

The Einstein’s equations can be derived from the Einstein-Hilbert action we learned in the previous post, using the action principle and the integral

$\boxed{S_{EH}=\int d^4x \sqrt{-g} \left(\kappa^{-2}R+\mathcal{L}_M\right)}$

The geodesic equation is the path of a freely falling particle. It gives a “condensation” of the Einstein’s equivalence principle too and it is also a generalization of Newton’s law of “no force”. That is, the geodesic equation is the feynmanity

$\boxed{\dfrac{d^2 x^\mu}{d\tau^2}+\Gamma^\mu _{\rho\sigma}\dfrac{dx^\rho}{d\tau}\dfrac{dx^\sigma}{d\tau}=0}$

Finally, an important concept in General Relativity is that of isometry. The symmetry of the “spacetime manifold” is provided by a Killing vector that preserves transformations (isometries) of that manifold. Mathematically speaking, the Killing vector fields satisfy certain equation called the Killing equation

$\boxed{\xi_{\mu ; \nu}+\xi_{\nu ; \mu}=0}$

Maximally symmetric spaces have $n(n+1)/2$ Killing vectors in n-dimensional (nD) spacetime. There are 3 main classes or types of 2D maximally symmetric that can be generalized to higher dimensions:

1. The euclidean plane $E^2$.

2. The pseudo-sphere $H^2$. This is a certain “hyperbolic” space.

3. The spehre $S^2$. This is a certain “elliptic” space.

The Friedmann-Robertson-Walker Cosmology

Current cosmological models are based in General Relativity AND  a simplification of the possible metrics due to the so-called Copernican (or cosmological) principle: the Universe is pretty much the same “everywhere” you are in the whole Universe! Remarkbly, the old “perfect” Copernican (cosmological) principle that states that the Universe is the same “everywhere” and “every time” is wrong. Phenomenologically, we have found that the Universe has evolved and it evolves, so the Universe was “different” when it was “young”. Therefore, the perfect cosmological principle is flawed. In fact, this experimental fact allows us to neglect some old theories like the “stationary state” and many other “crazy theories”.

What are the observational facts to keep the Copernican principle? It seems that:

1st. The distribution of matter (mainly galaxies, clusters,…) and radiation (the cosmic microwave background/CMB) in the observable Universe is homogenous and isotropic.

2nd. The Universe is NOT static. From Hubble’s pioneer works/observations, we do know that galaxies are receeding from us!

Therefore, these observations imply that our “local” Hubble volume during the Hubble time is similar to some spacetime with homogenous and isotropic spatial sections, i.e., it is a spacetime manifold $M=\mathbb{R}\times \Sigma$. Here, $\mathbb{R}$ denotes the time “slice” and $\Sigma$ represents a 3D maximally symmetric space.

The geometry of a locally isotropic and homogeneous Universe is represented by the so-called Friedmann-Robertson-Walker metric

$\boxed{ds^2_{FRW}=-dt^2+a^2(t)\left[\dfrac{dr^2}{1-kr^2}+r^2\left(d\theta^2+\sin\theta^2d\phi^2\right)\right]}$

Here, $a(t)$ is the called the scale factor.  The parameter $k$ determines the geometry type (plane, hyperbolic or elliptical/spherical):

1) If $k=0$, then the Universe is “flat”. The manifold is $E^3$.

2) If $k=-1$, then the Universe is “open”/hyperbolic. The manifold would be $H^3$.

3) If $k=+1$, then the Universe is “closed”/spherical or elliptical. The manifold is then $S^3$.

Remark: The ansatz of local homogeneity and istoropy only implies that the spatial metric is locally one of the above three spaces, i.e., $E^3,H^3,S^3$. It could be possible that these 3 spaces had different global (likely topological) properties beyond these two properties.

Kinematical features of a FRW Universe

The first property we are interested in Cosmology/Astrophysics is “distance”. Measuring distance in a expanding Universe like a FRW metric is “tricky”. There are several notions of “useful” distances. They can be measured by different methods/approaches and they provide something called sometimes “the cosmologidal distance ladder”:

1st. Comoving distance. It is a measure in which the distance is “taken” by a fixed coordinate.

2nd. Physical distance. It is essentially the comoving distance times the scale factor.

3rd. Luminosity distance. It uses the light emitted by some object to calculate its distance (provided the speed of light is taken constant, i.e., special relativity holds and we have a constant speed of light)

4th. Angular diameter distance. Another measure of distance using the notion of parallax and the “extension” of the physical object we measure somehow.

There is an important (complementary) idea in FRW Cosmology: the particle horizon. Consider a light-like particle with $ds^2=0$. Then,

$dt=a(t)\dfrac{1}{\sqrt{1-kr^2}}$

or

$\dfrac{dr}{\sqrt{1-kr^2}}=\dfrac{dt}{a(t)}$

The total comoving distance that light have traveled since a time $t=0$ is equal to

$\boxed{\eta=\int_0^{r_H}\dfrac{dr}{\sqrt{1-kr^2}}=\int_0^t\dfrac{dt'}{a(t')}}$

It shows that NO information could have propagated further and thus, there is a “comoving horizon” with every light-like particle! Here, this time is generally used as a “conformal time” as a convenient tiem variable for the particle. The physical distance to the particle horizon can be calculated

$\boxed{d_H(t)=\int_0^{r_H}\sqrt{g_{rr}}dr=a(t)\int_0^t\dfrac{dt'}{a(t')}=a(t)\eta}$

There are some important kinematical equations to be known

A) For the geodesic equation, the free falling particle, we have

$\Gamma^0_{ij}=\dfrac{\dot{a}}{a}\overline{g}_{ij}$

$\Gamma^i_{0j}=\Gamma^i_{j0}=\dfrac{\dot{a}}{a}\delta_{ij}$

$\Gamma^i_{jk}=\overline{\Gamma}^i_{jk}$

for the FRW metric and, moreover, the energy-momentum vector $P^\mu=(E,\mathbf{p})$ is defined by the usual invariant equation

$P^\mu=\dfrac{dx^\mu}{d\lambda}$

This definition defines, in fact, the proper “time” $\lambda$ implicitely, since

$\dfrac{d}{d\lambda}=\dfrac{dx^0}{d\lambda}\dfrac{d}{dx^0}=E\dfrac{d}{dt}$

and the 0th component of the geodesic equation becomes

$E\dfrac{dE}{dt}=-\Gamma^0_{ij}p^ip^j=-\delta_{ij}a\dot{a}p^ip^j$

$g_{\mu\nu}p^\mu p^\nu=-E^2+a^2\delta_{ij}p^ip^j=-m^2$

$EdE=a^2\vert \mathbf{p}\vert d\vert \mathbf{p}\vert$

$a^2 p\dfrac{dp}{dt}=-a\dot{a} p^2$

$\dfrac{1}{\vert \mathbf{p}\vert }\dfrac{d\vert \mathbf{p}\vert}{dt}+\dfrac{\dot{a}}{a}=0$

Therefore we have deduced that $\vert \mathbf{p}\vert \propto a^{-1}$. This is, in fact, the socalled “redshift”.  The cosmological  redshift parameter is more generally defined through the equation

$\boxed{\dfrac{a(t_0)}{a(t)}=1+z=\dfrac{\lambda_0}{\lambda}}$

B) The Hubble’s law.

The luminosity distance measures the flux of light from a distant object of known luminosity (if it is not expanding). The flux and luminosity distance are bound into a single equation

$\boxed{F=\dfrac{L}{4\pi d^2_L}}$

If we use the comoving distance between a distant emitter and us, we get

$\chi (a)=\int_t^{t_0}\dfrac{dt'}{a(t')}=\int_a^1\dfrac{da'}{a'^2 H(a')}$

for a expanding Universe! That is, we have used the fact that luminosity itself goes through a comoving spherical shell of radius $\chi (a)$. Moreover, it shows that

$F=\dfrac{L (\chi)}{4 \pi \chi (a)^2 a_0^2}=\dfrac{L}{4\pi (\chi (a)/a)^2}$

The luminosity distance in the expanding shell is

$d_L=\dfrac{\chi (a)}{a}=\left(\dfrac{L}{4\pi F}\right)^{1/2}$

and this is what we MEASURE in Astrophysics/Cosmology. Knowing $a(t)$, we can express the luminosity distance in terms of the redshift. Taylor expansion provides something like this:

$H_0d_L=z+\dfrac{1}{2}(1-q_0)z^2+\ldots$

where higher order terms are sometimes referred as “statefinder parameters/variables”. In particular, we have

$\boxed{H_0=\dfrac{\dot{a}_0}{a_0}}$

and

$\boxed{q_0=-\dfrac{a_0\ddot{a}_0}{\dot{a}_0^2}}$

C) Angular diameter distance.

If we know that some object has a known length $l$, and it gives some angular “aperture” or separation $\theta$, the angular diameter distance is given by

$\boxed{d_A=\dfrac{l}{\theta}}$

The comoving size is defined as $l/a$, and the coming distance is again $\chi (a)$. For “flat” space, we obtain that

$\theta=\dfrac{l/a}{\chi (a)}$

that is

$d_A=a\chi (a)=\dfrac{\chi}{1+z}$

In the case of “curved” spaces, we get

$d_A=\dfrac{a}{H_0\sqrt{\vert \omega_k\vert}}\cdot\begin{cases}\sinh \left( \sqrt{\Omega_k}H_0\chi\right),\Omega_k>0\\ \sin \left( \sqrt{-\Omega_k}H_0\chi\right),\Omega_k<0\end{cases}$

FRW dynamics

Gravity in General Relativity, a misnomer for the (locally) relativistic theory of gravitation, is described by a metric field, i.e., by a second range tensor (covariant tensor if we are purist with the nature of components). The metric field is related to the matter-energy-momentum content through the Einstein’s equations

$G_{\mu\nu}=-\kappa^2 T_{\mu\ nu}$

The left-handed side can be calculated for a FRW Universe as follows

$R_{00}=-3\dfrac{\ddot{a}}{a}$

$R_{ij}=(a\ddot{a}+2\dot{a}^2+2k)\overline{g}_{ij}$

$R=6\left(\dfrac{\ddot{a}}{a}+\dfrac{\dot{a}^2}{a^2}+\dfrac{k}{a^2}\right)$

The right-handed side is the energy-momentum of the Universe. In order to be fully consistent with the symmetries of the metric, the energy-momentum tensor MUST be diagonal and $T_{11}=T_{22}=T_{33}=T$. In fact, this type of tensor describes a perfect fluid with

$T_{\mu\nu}=(\rho+p)U_\mu U_\nu+pg_{\mu\nu}$

Here, $\rho, p$ are functions of $t$ (cosmological time) only. They are “state variables” somehow. Moreover, we have

$U_\mu =(1,0,0,0)$

for the fluid at rest in the comoving frame. The Friedmann equations are indeed the EFE for a FRW metric Universe

$3\left(\dfrac{\dot{a}^2}{a^2}+\dfrac{k}{a^2}\right)=\kappa^2\rho$ for the 00th compoent as “constraint equation.

$2\dfrac{\ddot{a}}{a}+\dfrac{\dot{a}^2}{a^2}+\dfrac{k}{a^2}=-\kappa^2p$ for the iith components.

Moreover, we also have

$G_{\mu\nu}^{;\nu}=T_{\mu\nu}^{;\nu}=0$

and this conservation law implies that

$\dot{\rho}+3\dfrac{\dot{a}}{a}(\rho+p)=0$

Therefore, we have got two independent equations for three unknowns $(a, \rho, p)$. We need an additional equation. In fact, the equation of state for $p=p(\rho)$ provides such an additional equation. It gives the “dynamics of matter”!

In summary, the basic equations for Cosmology in a FRW metric, via EFE, are the Friedmann’s equations (they are secretly the EFE for the FRW metric) supplemented with the energy-momentum conservations law and the equation of state for the pressure $p=p(\rho)$:

1) $\boxed{\dfrac{\dot{a}^2}{a^2}+\dfrac{k^2}{a^2}=\dfrac{\kappa^2}{3}\rho}$

2) $\boxed{\dot{\rho}+3\dfrac{\dot{a}}{a}(\rho+p)=0}$

3) $\boxed{p=p(\rho)}$

There are many kinds of “matter-energy” content of our interest in Cosmology. Some of them can be described by a simple equation of state:

$\boxed{p=\omega \rho}$

Energy-momentum conservation implies that $\rho\propto a^{-3(\omega +1)}$. 3 special cases are used often:

1st. Radiation (relativistic “matter”). $\omega=1/3$ and thus, $p=1/3\rho$ and $\rho\propto a^{-4}$

2nd. Dust (non-relativistic matter). $\omega=0$. Then, $p=0$ and $\rho\propto a^{-3}$

3rd. Vacuum energy (cosmological constant). $\omega=-1$. Then, $p=-\rho$ and $\rho=\mbox{constant}$

Remark (I): Particle physics enters Cosmology here! Matter dynamics or matter fields ARE the matter content of the Universe.

Remark (II): Existence of a Big Bang (and a spacetime singularity). Using the Friedmann’s equation

$\dfrac{\ddot{a}}{a}=-\dfrac{\kappa^2}{6}(\rho+3p)$

if we have that $(\rho+3p)>0$, the so-called weak energy condition, then $a=0$ should have been reached at some finite time in the past! That is the “Big Bang” and EFE are “singular” there. There is no scape in the framework of GR. Thus, we need a quantum theory of gravity to solve this problem OR give up the FRW metric at the very early Universe by some other type of metric or structure.

Particles and the chemical equilibrium of the early Universe

Today, we have DIRECT evidence for the existence of a “thermal” equilibrium in the early Universe: the cosmic microwave background (CMB). The CMB is an isotropic, accurate and non-homogeneous (over certain scales) blackbody spectrum about $T\approx 3K$!

Then, we know that the early Universe was filled with a hot dieal gas in thermal equilibrium (a temperature $T_e$ can be defined there) such as the energy density and pressure can be written in terms of this temperature. This temperature generates a distribution $f(\mathbf{x},\mathbf{p})$. The number of phase space elements in $d^3xd^3p$ is

$d^3xd^3p=\dfrac{d^3\mathbf{x}d^3\mathbf{p}}{(2\pi\hbar)^3}$

and where the RHS is due to the uncertainty principle. Using homogeneity, we get that, indeed, $f(x,p)=f(p)$, and where we can write the volume $d^3x=dV$. The energy density and the pressure are given by (natural units are used)

$\rho_i=g_i\int \dfrac{d^3p}{(2\pi)^3}f_i(p)E(p)$

$p_i=g_i\int \dfrac{d^3p}{(2\pi)^3}f_i (p)\dfrac{p^2}{3E(p)}$

When we are in the thermal equilibrium at temperature T, we have the Bose-Einstein/Fermi-Dirac distribution

$f(p)=\dfrac{1}{e^{(E-\mu)/T}\pm 1}$

and where the $+$ is for the Fermi-Dirac distribution (particles) and the $-$ is for the Bose-Einstein distribution (particles). The number density, the energy density and the pressure are the following integrals

$\boxed{\mbox{Number density}=n=\dfrac{N}{V}=\dfrac{g}{2\pi^2}\int_m^\infty \dfrac{(E^2-m^2)^{1/2}}{e^{(E-\mu)/T}\pm 1}dE}$

$\boxed{\mbox{Density energy}=\rho=\dfrac{E}{V}=\dfrac{g}{2\pi^2}\int_m^\infty \dfrac{(E^2-m^2)^{1/2}E^2}{e^{(E-\mu)/T}\pm 1}dE}$

$\boxed{\mbox{Pressure}=p=\dfrac{g}{6\pi^2}\int_m^\infty \dfrac{(E^2-m^2)^{3/2}}{e^{(E-\mu)/T}\pm 1}dE}$

And now, we find some special cases of matter-energy for the above variables:

1st. Relativistic, non-degenerate matter (e.g. the known neutrino species). It means that $T>>m$ and $T>>\mu$. Thus,

$n=\left(\dfrac{3}{4}\right)\dfrac{\zeta (3)}{\pi^2}gT^3$

$\rho=\left(\dfrac{7}{8}\right)\dfrac{\pi^2}{30}gT^4$

$p=\dfrac{1}{3}\rho$

2nd. Non-relativistic matter with $m>>T$ only. Then,

$n=g\left(\dfrac{mT}{2\pi}\right)^{3/2}e^{-(m-\mu)/T}$

$\rho= mn+\dfrac{3}{2}p$, and $p=nT<<\rho$

The total energy density is a very important quantity. In the thermal equilibrium, the energy density of non-relativistic species is exponentially smaller (suppressed) than that of the relativistic particles! In fact,

$\rho_R=\dfrac{\pi^2}{30}g_\star T^4$ for radiation with $p_R=\dfrac{1}{3}\rho_R$

and the effective degrees of freedom are

$\displaystyle{\boxed{g_\star=\sum_{bosons}g_b+\dfrac{7}{8}\sum_{fermions}g_f}}$

Remark: The factor $7/8$ in the DOF and the variables above is due to the relation between the Bose-Einstein and the Fermi-Dirac integral in d=3 space dimensions. In general d, the factor would be

$(1-\dfrac{1}{2^d})=\dfrac{2^d-1}{2^d}$

Entropy conservation and the early Universe

The entropy in a comoving volume IS a conserved quantity IN THE THERMAL EQUILIBRIUM. Therefore, we have that

$\dfrac{\partial p_i}{\partial T}=g_i\int \dfrac{d^3p}{(2\pi)^3}\dfrac{df}{dT}\dfrac{p^2}{3E(p)}=g_i\int \dfrac{4\pi pE dE}{(2\pi)^3}\dfrac{df}{dE}\left(-\dfrac{E}{T}\right)\dfrac{p^2}{3E}$

and then

$\dfrac{\partial p_i}{\partial T}=\dfrac{g_i}{2\pi^2}\int \left(-\dfrac{d}{dE}\left(f\dfrac{p^3E}{3T}\right)+f\dfrac{d}{dE}\left(\dfrac{p^3E}{3T}\right)\right)dE$

or

$\dfrac{\partial p_i}{\partial T}=\dfrac{1}{T}(\rho_i+p_i)$

Now, since

$\dfrac{\partial \rho}{\partial t}+3\dfrac{\dot{a}}{a}(\rho+p)=0$

then

$\dfrac{\partial}{\partial t}\left(a^3(\rho+p)\right)-a^3\dfrac{\partial p}{\partial t}=0$

$\dfrac{1}{a^3}\dfrac{\partial (a^3(\rho +p))}{\partial t}-\dfrac{\partial \rho}{\partial t}=0$

if we multiply by $T$ and use the chain rule for $\rho$, we obtain

$\dfrac{1}{a^3}\dfrac{\partial}{\partial t}\left(\dfrac{a^3(\rho+p)}{T}\right)=0$

but it means that $a^3s=\mbox{constant}$, where $s$ is the entropy density defined by

$\boxed{s\equiv \dfrac{\rho+p}{T}}$

Well, the fact is that we know that the entropy or more precisely the entropy density is the early Universe is dominated by relativistic particles ( this is “common knowledge” in the Stantard Cosmological Model, also called $\Lambda CDM$). Thus,

$\boxed{s=\dfrac{2\pi^2}{45}g_\star T^3}$

It implies the evolution of temperature with the redshift in the following way:

$T\propto g_\star^{-1/3}a^{-1}$

Indeed, since we have that $n\propto a^{-3}$, $s\propto a^{-3}$, the yield variable

$Y_i\equiv \dfrac{n_i}{s}$

is a convenient quantity that represents the “abundance” of decoupled particles.

See you in my 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! 🙂