### Cosmology III

The horizon problem: the universe has a finite age, so even light can only have travelled a finite distance by any given time. The CMB from all parts of the sky is at the same temperature (to one part in 105), but points more than 2 degrees apart are not causally connected (calculation to come later), so there seems to be no reason for the whole CMB to be at the same temperature. This is the horizon problem.

The structure problem: Anisotropies in the CMB are the progenitors of the large scale structures we see today. In order to observe such structures, perturbations (amplitude of fluctuations) must have existed in the density spectrum on scales that would have been outside the horizon until recently. Causal processes could not have generated such perturbations.

How can we solve these problems? The proposed solution is-

Inflationary expansion: a period in the evolution of the universe when the scale factor was accelerating, i.e. $\ddot{a}>0$.

Back to the Friedmann equation:

$(\frac{\dot{a}}{a})^2=\frac{8\pi G}{3}\rho-\frac{kc^2}{a^2}$

${\dot{a}}^2=\frac{8\pi G}{3}\rho a^2-kc^2$

Differentiate w.r.t. time-

$2\ddot{a}\dot{a}=\frac{8\pi G}{3}\dot{\rho}a^2+=\frac{16\pi G}{3}\rho a \dot{a}$

Fluid equation: $\dot{\rho}=-\frac{3\dot{a}}{a}(\rho+\frac{p}{c^2})$

Substitute-

$2\ddot{a}=-8\pi Ga(\rho+\frac{p}{c^2})+\frac{16\pi G}{3}\rho a$

$\frac{\ddot{a}}{a}=-4\pi G\rho -\frac{4\pi Gp}{c^2}+\frac{8\pi G\rho}{3}$

$\frac{\ddot{a}}{a}=\frac{4\pi G}{3}(\rho+\frac{3p}{c^2})$

This is the acceleration equation.

Given a positive density, to satisfy $\ddot{a}>0$, we need

$\rho+\frac{3p}{c^2}<0$

$p<\frac{-\rho c^2}{3}$

We can achieve this by including a cosmological constant with an equation of state $p=\frac{-\rho}{c^2}$

Put cosmological constant into Friedmann equation.

$H^2=\frac{8\pi G}{3}\rho-\frac{kc^2}{a^2}+\frac{\Lambda}{3}$

Now suppose the last term dominates

$\Rightarrow (\frac{\dot{a}}{a})^2=\frac{\Lambda}{3}$

$a(t)=e^{\sqrt{\frac{\Lambda}{3}}t}$

an exponential increase in the scale factor.

After some time, inflation must come to an end (end of a phase transition)- then the universe continues to expand in the way we thought it was before.

Inflation is believed to have lasted for 10-35 seconds, but during this time, all cosmic dimensions would be amplified by a factor of 1050.

Note: at the time of inflation, the time required for expansion by a factor of 2 was 10-35 seconds, but today the time required would be 1017 seconds.

Solution to the flatness problem: remember, for $|\Omega-1|=\frac{|k|}{a^2H^2}$ we saw that the LHS is an increasing function of time, driving Ω away from 1. Inflation drives it away-

$\ddot{a}>0 \Rightarrow \frac{d}{dt}(\dot{a})>0 \Rightarrow \frac{d}{dt}(aH)>0$

So inflation makes Ω so close to 1 that all subsequent expansion between the end of inflation and the present is insufficient to move it away again.

Solution to the horizon problem: a region which is the same order of size as the horizon before inflation could encompass the entire observable horizon today. The particle horizon expands superluminally during inflation.

How much expansion do we need from inflation? Inflation ends at 10-34 seconds. The expansion is perfectly exponential and Ω is now near to 1. We’ll assume that the universe is radiation dominated at this point, and that now $|\Omega_0-1|\leq 0.01$.

Age of Universe ~1017 sec

Radiation dominated $\Omega-1 \propto t$

$|\Omega(t_0)-1|\leq 0.01 \Rightarrow |\Omega(10^{-34}s)-1\leq 10^{-54}$

For constant H during inflation, $|\Omega(t_0)-1|\propto \frac{1}{a^2}$

This basic calculation gives us that a is increased by a factor of 1027 during inflation.

The principle of equivalence: Einstein tells us that “all local freely falling, non-rotating laboratories are fully equivalent for the performance of all physical experiments”. In other words, it is always possible to choose a local co-ordinate system such that all the laws of physics have the same form as they do in an unaccelerated gravity-free Cartesian system (in easy to understand terms, basically we don’t need to keep learning new laws of physics every time we go somewhere). When it comes to space-time, this simplifies things greatly- acceleration=gravitation=curvature.

Event: a location in space-time at (x,y,z,t).

World line: the trajectory of a body moving through space-time. A body at a fixed location in space follows a world-line parallel to the time axis, in the direction of increasing time. A body moving in space follows a world line making a slope with respect to the time axis. Since a body or signal cannot exceed the speed of light, there is a maximum slope to such world lines.

For a given point in time, we have a light cone (see the diagram on the Cosmology II page). Light cones act as cosmological horizons, separating the knowable from the unknowable.

Proper time: if we choose a frame in which A and B occur at the same spatial location (xA=xB), then the time measured between the two events is the proper time Δτ.

$\Delta\tau=\frac{\Delta S}{c}$

Proper distance is measured between A and B in a frame in which they occur simultaneously.

Geodesics: when space-time is curved by the presence of mass, the straightest possible world-lines are called geodesics. In curved space, a geodesic has a maximum or minimum interval; a massless particle like a photon follows a null-geodesic with $\int\sqrt{(ds)^2}=0$

Effect of mass on space-time: three fundamentals-

1. Mass acts on space-time, telling it how to curve.
2. Space-time acts on mass, telling it how to move.

The metric explains how to calculate distance intervals.

$ds^2=\sum_{a,b}g_{ab}dx^adx^b$

ds=line element of proper time

dxa, dxb=co-ordinate intervals

gab is the metric; it is a symmetric second-rank tensor (NxN matrix).

Euclidean 2-space

$ds^2=1.dx^2+0.dxdy+0.dydx+1.dy^2$

i.e. Pythagoras’ theorem.

$g_{ab}= \left[ \begin{array}{ cc} 1&0 \\ 0&1 \end{array} \right] = diag(1,1)$

Minkowski space-time

$ds^2=-c^2dt^2+dx^2+dy^2+dz^2$

$ds^2=-c^2dt^2+dr^2+r^2(d\theta^2+\sin^2\theta d\phi^2)$

A light pulse will expand spherically at speed c-

$dx^2+dy^2+dz^2=dr^2=c^2dt^2$

But the motion is purely radial, so dθ=dФ=0 and ds2=0.

No proper time elapses- characteristic of a light ray (so everything works as we thought it did).

Homogeneous and isotropic metrics: a special class of metrics are those which describe isotropic space-time. For these, the purely spatial part of the metric dl2 may be written-

$dl^2=f(r,t)(dx^2+dy^2+dz^2)$

$dl^2=f(r,t)(dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2)$

An example of a homogeneous and isotropic metric is-

$g_{00}=g_{00}(t); g_{0i}=0$

$f(r,t)=a^2(t)(1+\frac{kr^2}{4})^{-2}$

Friedmann-Robertson-Walker metric (FRW metric) for homogeneous and isotropic space-time.

$ds^2=-c^2dt^2+(a(t))^2(1+\frac{kr^2}{4})^{-2}[dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2]$

or equivalently

$ds^2=-c^2dt^2+(a(t))^2[d\chi^2+\chi^2(d\theta^2+\sin^2\theta d\phi^2)]$

(this is the called the ‘co-moving form’ of the metric).

χ=sin r for k>0

χ=r for k=0

χ=sinh r for k<0

Another alternative-

$ds^2=-c^2dt^2+(a(t))^2[\frac{dr^2}{1-kr^2}+r^2d\theta^2+r^2\sin^2\theta d\phi^2]$

Here, k can be identified with the Gaussian curvature of space, 1/radius2

Deriving cosmological redshift from the FRW metric (go on, you know you want to): For a light ray, ds=0; also, dθ=dФ=0 for a radial path travelled from the point of the light’s emission at χe to its arrival at the Earth at χ=0.

The FRW metric looks like:

$ds^2=-c^2dt^2+(a(t))^2[\frac{dr^2}{1-kr^2}+r^2d\theta^2+r^2\sin^2\theta d\phi^2]$

For a light ray

$ds^2=-c^2dt^2+(a(t))^2[\frac{dr^2}{1-kr^2}$

Suppose that one crest of the light wave was emitted at time te and received at to and the next crest was emitted at tete and received at toto.

$\frac{dt}{a(t)}=\frac{1}{c}\frac{d\chi}{\sqrt{1-k\chi^2}}$

First crest $\int_{t_e}^{t_o} \frac{dt}{a(t)}= \frac{1}{c} \int_0^{\chi_e} \frac{d\chi}{\sqrt{1-k\chi^2}}$

Next crest $\int_{t_e+\Delta t_e}^{t_o+\Delta t_o} \frac{dt}{a(t)}= \frac{1}{c} \int_0^{\chi_e} \frac{d\chi}{\sqrt{1-k\chi^2}}$

Subtracting these equations ->

$\int_{t_e+\Delta t_e}^{t_o+\Delta t_o} \frac{dt}{a(t)}- \int_{t_e}^{t_o} \frac{dt}{a(t)}=0$

But

$\int_{t_e+\Delta t_e}^{t_o+\Delta t_o} \frac{dt}{a(t)}= \int_{t_e}^{t_o} \frac{dt}{a(t)}+ \int_{t_o}^{t_o+\Delta t_o} \frac{dt}{a(t)}-\int_{t_e}^{t_e+\Delta t_e} \frac{dt}{a(t)}$

So

$\int_{t_o}^{t_o+\Delta t_o} \frac{dt}{a(t)} -\int_{t_e}^{t_e+\Delta t_e}\frac{dt}{a(t)} =0$

Neglect any change in a(t) during Δte and Δto, so regard a(t) as constant in the above integrals.

$\frac{1}{a(t_o)} \int_{t_o}^{t_o+\Delta t_o}dt =\frac{1}{a(t_e)} \int_{t_e}^{t_e+\Delta t_e}dt$

$\frac{\Delta t_o}{a(t_o)}= \frac{\Delta t_e}{a(t_e)}$

$\frac{\Delta t_e}{\Delta t_o}= \frac{a(t_e)}{a(t_o)}$

Set $a(t_o)=1$

$\Delta t_e=\frac{\lambda_e}{c}$ and $\Delta t_o=\frac{\lambda_o}{c}$

Redshift

$1+z=\frac{\lambda_o}{\lambda_e}=\frac{\Delta t_o}{\Delta t_e}=\frac{a(t_o)}{a(t_e)}$ regardless of the form of a.

Active galaxies and quasars are the most spectacular objects in the sky, with high luminosities (~1013 L¤) produced from relatively tiny volumes (<1pc). They form luminous, compact nuclei at the centres of some galaxies; sometimes jets of relativistic material emanate from the active nuclei and propagate over hundreds of thousands of kpc. Active galaxies and quasars were more numerous in the past, and are useful as beacons to the distant universe.

Properties of active galactic nuclei (AGN)

1. Angular size. Most have very small angular size, but this property is also wavelength dependent.
2. Most have >galactic luminosity; but of course, objects with low luminosity are hard to find in the first place (possibly skewing results)- dust, obscuration and beaming can mislead.
3. Most have a broad band continuum; often spectral shape from IR to x-rays is constant.
4. Most have strong emissions, sometimes very broad.
5. Most are variable- short wavelengths are stronger and faster than long.
6. Most are weakly polarised; a small minority are strongly variable and polarised (these are correlated with radio and gamma rays- in some cases emission lines are absent).
7. Many show compact radio emission. A minority show spectacular extended radio emission.

Naïve estimate of size-scale of emitting region from time variability: if an emitting region varies more rapidly than the light-crossing time, then the variability will be washed out. Hence, if a source varies on a timescale of δt, then the size L of the emitting region satisfies

$L

(Actually, relativistic corrections are needed but this will do as the aforementioned naïve estimate).

At radio wavelengths, Very Long Baseline Interferometry is used to measure milli-arcsecond scales.

Hydrogen lines

1. Lyman series- transitions to n=1 (UV)
2. Balmer series- transitions to n=2 (visible); 656.3nm (red), 486.1nm (blue-green), 434.1nm (violet) and 410.2nm (violet)- Hα, Hβ, Hγ and Hδ
3. Paschen series- transitions to n=3 (IR)

Why are some emission lines so broad? Our first thought might be that it is due to thermal broadening. Put $mv^2 \simeq kT$, where v=2000kms-1 – for hydrogen, this gives T~5×108K. But at this temperature, the atoms will most likely be ionised, in which case we won’t be seeing any emission lines anyway.

Also, this would predict that the line widths depend on the mass of the atomic species (for a given T, $v \propto \frac{1}{\sqrt{m}}$), but measurements show that carbon and hydrogen lines are comparable (C should be $\frac{1}{\sqrt{12}}$ H).

So thermal broadening cannot make the dominant contribution- bulk motions of the gas are likely to be responsible instead.

Apparent superluminal motion

To an observer on Earth-

Object travelling at speed v is at A at t=0.

Light from A reaches Earth at $t=\frac{vt_1\cos{\theta}+d}{c}$

Object is at B at t=t1.

Light from B reaches Earth at $t=t_1+\frac{d}{c}$

It looks like the object has travelled distance vt1sinθ in time $t_1+\frac{d}{c}-\frac{vt_1\cos{\theta}}{c}-\frac{d}{c}$

Apparent transverse velocity of object measured on Earth

$v_{app}=\frac{vt_1\sin{\theta}}{t_1-\frac{vt_1\cos{\theta}}{c}}$

$v_{app}=\frac{v\sin{\theta}}{1-\beta \cos{\theta}}$

which can appear to be greater than c.

At B, time=t1

At C, time=β t1cosθ

$\Delta t=t_1(1-\beta \cos\theta)$

$\Delta x=vt_1\sin\theta$

$\Delta s^2=c^2\Delta t^2-\Delta x^2$

$\Delta s^2=c^2t_1^2(1-\beta \cos\theta)^2- v^2t_1^2\sin^2\theta$

If vapp>c, then B and C have spacelike separation.

Active galaxies- models: Any model must explain the high luminosity- a small but efficient energy source is needed. Possibilities are-

1. Star cluster- luminosity due to supernova. But observed variability is too coherent for a supernova, the lifetime is too short and there is no explanation for the radio jets.
2. Supermassive stars- again unlikely, because these are unstable to collapse.
3. Supermassive black hole (fuelled by accretion) provides the best possibility- it is small and efficient with a stable potential well and a stable direction for radio jets. There is evidence for such enormous masses in nearby galaxies.