Complete Relativity

Special Relativity and Flat Spacetime

Core ideas

Special relativity replaces absolute time with Minkowski spacetime. Events are related by Lorentz transformations, proper time is invariant, and energy-momentum forms a four-vector. Causality is controlled by light cones, not by simultaneity in a chosen frame.

For review, be able to use Lorentz transformations, compute proper time, manipulate four-vectors, and classify intervals as timelike, null, or spacelike. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$ds^2=-c^2dt^2+d\bm x^2,\qquad p^\mu=(E/c,\bm p),\qquad E^2=p^2c^2+m^2c^4$$

Worked example

Consider two events in inertial frame $S$: event $A$ at $(t=0,x=0)$ and event $B$ at $(t=10\ \mu\text{s},x=2400\ \text{m})$. With $c=3.00\times10^{8}\ \text{m/s}$,

$$\Delta s^{2}=-c^{2}\Delta t^{2}+\Delta x^{2}=-(3000\ \text{m})^{2}+(2400\ \text{m})^{2}=-3.24\times10^{6}\ \text{m}^{2}.$$

The interval is timelike. The proper time is

$$\Delta\tau=\frac{\sqrt{-\Delta s^{2}}}{c}=\frac{1800\ \text{m}}{3.00\times10^{8}\ \text{m/s}}=6.0\ \mu\text{s}.$$

Now boost to frame $S'$ moving at $v=0.6c$ ($\gamma=1.25$) along $x$. Using the Lorentz transformation,

$$x'=\gamma(x-vt)=1.25\,(2400-0.6\cdot3000)=750\ \text{m},$$ $$t'=\gamma\!\left(t-\frac{vx}{c^{2}}\right)=1.25\!\left(10\ \mu\text{s}-\frac{0.6c\cdot2400}{c^{2}}\right)=12.5\ \mu\text{s}.$$

The invariant interval checks out: $-c^{2}t'^{2}+x'^{2}=-3.24\times10^{6}\ \text{m}^{2}$.

Problems with Solutions

1. Lorentz boost. A spaceship leaves Earth and travels to a star 4.00 ly away in Earth frame. The ship’s clock records a trip time of 5.00 yr. Find the ship’s speed relative to Earth and the time elapsed on Earth.

   Solution. Let the ship speed be $v=\beta c$ and $\gamma=1/\sqrt{1-\beta^{2}}$. In Earth frame the distance is $L_{0}=4.00$ ly and the trip time is $t_{\oplus}=L_{0}/v$. The ship measures the contracted distance $L=L_{0}/\gamma$ and proper time $\tau=L/v=L_{0}/(\gamma v)$. Thus $5.00=\frac{4.00}{\gamma\beta}\quad\Rightarrow\quad\gamma\beta=0.800.$ Since $\gamma\beta=\beta/\sqrt{1-\beta^{2}}$, squaring gives $\beta^{2}=0.64(1-\beta^{2})$, so $\beta\approx0.625$ and $v\approx0.625c$. Then $\gamma\approx1.28$ and $t_{\oplus}=4.00/0.625\approx6.40$ yr.

2. Relativistic electron. An electron has kinetic energy $K=2m_{e}c^{2}$ ($m_{e}c^{2}=511\ \text{keV}$). Find its total energy, momentum, and speed.

   Solution. Total energy $E=K+m_{e}c^{2}=3m_{e}c^{2}=1.533\ \text{MeV}$. From $E^{2}=p^{2}c^{2}+m_{e}^{2}c^{4}$, $p^{2}c^{2}=9m_{e}^{2}c^{4}-m_{e}^{2}c^{4}=8m_{e}^{2}c^{4}\;\Rightarrow\;pc=\sqrt{8}\,m_{e}c^{2}\approx1.445\ \text{MeV}.$ Hence $p\approx1.445\ \text{MeV}/c$. The Lorentz factor is $\gamma=E/(m_{e}c^{2})=3$, so $v=c\sqrt{1-1/\gamma^{2}}=c\sqrt{8/9}\approx0.943c$.

3. Interval classification. Event $A$ is at $(t=0,x=0)$ and event $B$ at $(t=3\ \text{ns},x=1.0\ \text{m})$. Classify the separation and determine whether $A$ could cause $B$.

   Solution. $c\Delta t=(3.00\times10^{8})(3\times10^{-9})=0.90\ \text{m}$. Since $\Delta x=1.0\ \text{m}>c\Delta t$, $\Delta s^{2}=-(0.90)^{2}+(1.0)^{2}=+0.19\ \text{m}^{2}>0.$ The interval is spacelike, so no signal from $A$ can reach $B$; causality is impossible.

Section summary. Flat spacetime unifies space, time, energy, and momentum.

Manifolds

Core ideas

General relativity needs coordinate-independent geometry. A manifold is a space that locally looks like $\mathbb R^n$ but can be curved globally. Vectors live in tangent spaces, covectors in dual spaces, and tensors encode geometric objects independent of coordinates.

For review, be able to distinguish coordinates from geometry, transform tensor components, use tangent vectors and one-forms, and interpret metrics as inner products. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$ds^2=g_{\mu\nu}dx^\mu dx^\nu,\qquad V^{\mu'}=\frac{\partial x^{\mu'}}{\partial x^\nu}V^\nu$$

Worked example

On a 2-sphere of radius $R=6371\ \text{km}$ (Earth), the line element is

$$ds^{2}=R^{2}\bigl(d\theta^{2}+\sin^{2}\theta\,d\phi^{2}\bigr).$$

Consider a curve of constant latitude $\theta=\pi/2$ (the equator) with $\phi(\lambda)=\lambda$ for $\lambda\in[0,\pi/4]$. The tangent vector is $V^{\mu}=(0,1)$, so

$$|V|^{2}=g_{\mu\nu}V^{\mu}V^{\nu}=g_{\phi\phi}(V^{\phi})^{2}=R^{2}\sin^{2}(\pi/2)=R^{2}.$$

The curve length is therefore

$$L=\int_{0}^{\pi/4}\!|V|\,d\lambda=R\frac{\pi}{4}\approx6371\times0.785\approx5000\ \text{km}.$$

This shows how the metric directly encodes physical distance independent of any embedding in $\mathbb{R}^{3}$.

Problems with Solutions

1. Metric length. In 2D polar coordinates the metric is $ds^{2}=dr^{2}+r^{2}d\theta^{2}$. A vector at $(r=2,\theta=\pi/6)$ has components $(V^{r},V^{\theta})=(3,1)$. Compute its squared length.

   Solution. $g_{rr}=1$, $g_{\theta\theta}=r^{2}=4$. Hence $|V|^{2}=g_{rr}(V^{r})^{2}+g_{\theta\theta}(V^{\theta})^{2}=1\cdot9+4\cdot1=13.$

2. Jacobian transformation. Coordinates $(u,v)$ relate to $(x,y)$ by $x=u+v$, $y=u-v$. A vector has components $V^{x}=1$, $V^{y}=2$. Find $V^{u}$ and $V^{v}$.

   Solution. Inverting, $u=\tfrac12(x+y)$, $v=\tfrac12(x-y)$. The Jacobian gives $V^{u}=\frac{\partial u}{\partial x}V^{x}+\frac{\partial u}{\partial y}V^{y}=\tfrac12(1)+\tfrac12(2)=1.5,$ $V^{v}=\frac{\partial v}{\partial x}V^{x}+\frac{\partial v}{\partial y}V^{y}=\tfrac12(1)-\tfrac12(2)=-0.5.$

3. Covector action. In polar coordinates a covector has components $\omega_{r}=r\cos\theta$, $\omega_{\theta}=-r^{2}\sin\theta$. Evaluate $\omega(V)$ for $V=(\sin\theta,\cos\theta/r)$ at $(r=2,\theta=\pi/4)$.

   Solution. $\omega(V)=\omega_{\mu}V^{\mu}=(r\cos\theta)(\sin\theta)+(-r^{2}\sin\theta)(\cos\theta/r)=r\sin\theta\cos\theta-r\sin\theta\cos\theta=0$. At the given point the contraction vanishes.

Section summary. Manifolds provide the coordinate-free stage for spacetime.

Curvature

Core ideas

Curvature measures the failure of vectors to return unchanged after parallel transport. The connection defines covariant derivatives and geodesics; the Riemann tensor measures curvature; Ricci curvature and the scalar curvature are contractions used in Einstein’s equation.

For review, be able to compute Christoffel symbols for simple metrics, write geodesic equations, and interpret Riemann/Ricci curvature geometrically. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$\Gamma^\rho_{\mu\nu}=\frac12g^{\rho\sigma}(\partial_\mu g_{\nu\sigma}+\partial_\nu g_{\mu\sigma}-\partial_\sigma g_{\mu\nu}),\qquad R^\rho{}_{\sigma\mu\nu}V^\sigma=[\nabla_\mu,\nabla_\nu]V^\rho$$

Worked example

For a 2-sphere of radius $R$, $ds^{2}=R^{2}(d\theta^{2}+\sin^{2}\theta\,d\phi^{2})$. The nonzero metric components are $g_{\theta\theta}=R^{2}$, $g_{\phi\phi}=R^{2}\sin^{2}\theta$ with inverses $g^{\theta\theta}=1/R^{2}$, $g^{\phi\phi}=1/(R^{2}\sin^{2}\theta)$. The Christoffel symbols are

$$\Gamma^{\theta}{}_{\phi\phi}=-\tfrac12 g^{\theta\theta}\partial_{\theta}g_{\phi\phi}=-\sin\theta\cos\theta,$$ $$\Gamma^{\phi}{}_{\theta\phi}=\Gamma^{\phi}{}_{\phi\theta}=\tfrac12 g^{\phi\phi}\partial_{\theta}g_{\phi\phi}=\cot\theta.$$

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

$$R^{\theta}{}_{\phi\theta\phi}=\partial_{\theta}\Gamma^{\theta}{}_{\phi\phi}-\Gamma^{\theta}{}_{\phi\phi}\Gamma^{\phi}{}_{\theta\phi}=-\cos^{2}\theta+\sin^{2}\theta-(-\sin\theta\cos\theta)(\cot\theta)=\sin^{2}\theta.$$

Lowering the first index, $R_{\theta\phi\theta\phi}=g_{\theta\theta}R^{\theta}{}_{\phi\theta\phi}=R^{2}\sin^{2}\theta$. The Gaussian curvature is

$$K=\frac{R_{\theta\phi\theta\phi}}{g_{\theta\theta}g_{\phi\phi}}=\frac{R^{2}\sin^{2}\theta}{R^{2}\cdot R^{2}\sin^{2}\theta}=\frac{1}{R^{2}},$$

showing that curvature is uniform and inversely proportional to the square of the radius.

Problems with Solutions

1. Christoffel symbols. For the metric $ds^{2}=dx^{2}+x^{2}dy^{2}$ ($x>0$), compute $\Gamma^{x}{}_{yy}$ and $\Gamma^{y}{}_{xy}$.

   Solution. $g_{xx}=1$, $g_{yy}=x^{2}$, so $g^{xx}=1$, $g^{yy}=1/x^{2}$. Then $\Gamma^{x}{}_{yy}=-\tfrac12 g^{xx}\partial_{x}g_{yy}=-\tfrac12(2x)=-x,$ $\Gamma^{y}{}_{xy}=\tfrac12 g^{yy}\partial_{x}g_{yy}=\tfrac12\frac{1}{x^{2}}(2x)=\frac{1}{x}.$

2. Geodesic equation. Using the results above, write the geodesic equation for $x(\lambda)$.

   Solution. $\ddot{x}+\Gamma^{x}{}_{xx}\dot{x}^{2}+2\Gamma^{x}{}_{xy}\dot{x}\dot{y}+\Gamma^{x}{}_{yy}\dot{y}^{2}=0$. Only $\Gamma^{x}{}_{yy}$ is nonzero, so $\frac{d^{2}x}{d\lambda^{2}}-x\left(\frac{dy}{d\lambda}\right)^{2}=0.$

3. Great-circle geodesic. On a unit sphere a geodesic starts at $(\theta=\pi/2,\phi=0)$ with initial tangent $(\dot\theta,\dot\phi)=(0,1)$. Show it stays on the equator.

   Solution. The $\theta$ geodesic equation is $\ddot\theta-\sin\theta\cos\theta\,\dot\phi^{2}=0$. At the equator $\theta=\pi/2$, $\cos\theta=0$, so $\ddot\theta=0$. Since $\dot\theta=0$ initially, $\theta$ remains $\pi/2$ for all $\lambda$; the curve is a great circle.

Section summary. Curvature is encoded in how covariant derivatives fail to commute.

Gravitation

Core ideas

Einstein gravity identifies gravity with spacetime curvature sourced by stress-energy. Freely falling bodies follow geodesics; the Newtonian potential appears as a weak-field, slow-motion limit. Conservation of stress-energy follows from geometry through the Bianchi identity.

For review, be able to state Einstein’s equation, recover Newtonian gravity in weak fields, identify stress-energy components, and use geodesics for motion. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$G_{\mu\nu}+\Lambda g_{\mu\nu}=\frac{8\pi G}{c^4}T_{\mu\nu},\qquad \frac{d^2x^\mu}{d\tau^2}+\Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau}=0$$

Worked example

The GPS time-dilation correction is a direct application of the weak-field metric $g_{00}\approx-(1+2\Phi/c^{2})$, where $\Phi=-GM/r$ is the Newtonian potential. For Earth, $M_{\oplus}=5.97\times10^{24}\ \text{kg}$ and $r_{s}=2GM_{\oplus}/c^{2}\approx8.87\ \text{mm}$. A GPS satellite orbits at $r_{\text{orb}}=2.656\times10^{7}\ \text{m}$, while a ground station is at $r_{\text{g}}=6.371\times10^{6}\ \text{m}$. The elapsed proper time relative to coordinate time is $d\tau\approx dt\,(1-r_{s}/2r)$. Over one day ($\Delta t=86400\ \text{s}$) the difference is

$$\Delta\tau_{\text{orb}}-\Delta\tau_{\text{g}}\approx\Delta t\,\frac{r_{s}}{2}\!\left(\frac{1}{r_{\text{g}}}-\frac{1}{r_{\text{orb}}}\right)\approx86400\times4.43\times10^{-3}\times(1.57\times10^{-7}-3.77\times10^{-8})\approx45\ \mu\text{s}.$$

Thus GR predicts that the orbital clock gains about $45\ \mu\text{s}$ per day relative to the ground clock. (The full GPS correction includes an opposing special-relativistic term of \sim$$7\ \mu\text{s}, leaving a net \sim$$38\ \mu\text{s} to be corrected.)

Problems with Solutions

1. Schwarzschild radius of the Sun. Compute $r_{s}=2GM_{\odot}/c^{2}$ using $M_{\odot}=1.99\times10^{30}\ \text{kg}$.

   Solution. With $G=6.674\times10^{-11}\ \text{m}^{3}\text{kg}^{-1}\text{s}^{-2}$ and $c=3.00\times10^{8}\ \text{m/s}$, $r_{s}=\frac{2(6.674\times10^{-11})(1.99\times10^{30})}{(3.00\times10^{8})^{2}}\approx\frac{2.656\times10^{20}}{9.00\times10^{16}}\approx2.95\ \text{km}.$

2. Newtonian limit. In the weak-field, slow-motion limit, show that the geodesic equation reduces to $\ddot{x}^{i}=-\partial^{i}\Phi$.

   Solution. For $v\ll c$ the spatial geodesic equation reduces to $d^{2}x^{i}/dt^{2}\approx-c^{2}\Gamma^{i}{}_{00}$. With $g_{00}\approx-(1+2\Phi/c^{2})$, $\Gamma^{i}{}_{00}=-\tfrac12 g^{ij}\partial_{j}g_{00}\approx-\tfrac12\delta^{ij}(-2\partial_{j}\Phi/c^{2})=\partial_{i}\Phi/c^{2}.$ Hence $\ddot{x}^{i}\approx-c^{2}(\partial_{i}\Phi/c^{2})=-\partial_{i}\Phi$, which is Newton’s second law in a gravitational potential.

3. Neutron-star density. Estimate the mean density of a star with $M=1.4\,M_{\odot}$ and $R=10\ \text{km}$ for which $GM/(Rc^{2})\sim0.1$.

   Solution. $M=0.1Rc^{2}/G\approx0.1(10^{4})(9\times10^{16})/(6.67\times10^{-11})\approx1.35\times10^{30}\ \text{kg}$. The mean density is $\rho=\frac{M}{\tfrac43\pi R^{3}}\approx\frac{1.35\times10^{30}}{4.19\times10^{12}}\approx3.2\times10^{17}\ \text{kg/m}^{3},$ comparable to nuclear saturation density.

Section summary. Matter tells spacetime how to curve; curved spacetime tells matter how to move.

The Schwarzschild Solution

Core ideas

The Schwarzschild metric describes the exterior field of a static spherical mass. It predicts gravitational redshift, light bending, perihelion precession, black hole horizons, and the structure of radial and circular geodesics.

For review, be able to use the Schwarzschild radius, identify horizon vs singularity, compute qualitative orbits, and explain classic weak-field tests. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$ds^2=-\left(1-\frac{2GM}{rc^2}\right)c^2dt^2+\left(1-\frac{2GM}{rc^2}\right)^{-1}dr^2+r^2d\Omega^2$$

Worked example

The white dwarf Sirius B provides a classic test of gravitational redshift. Its mass is $M\approx1.02\,M_{\odot}\approx2.03\times10^{30}\ \text{kg}$ and its radius is $R\approx5.57\times10^{6}\ \text{m}$. The Schwarzschild radius is $r_{s}=2GM/c^{2}\approx3.01\ \text{km}$. For light emitted from the surface and observed at infinity,

$$\frac{\lambda_{\infty}}{\lambda_{\text{em}}}=\left(1-\frac{r_{s}}{R}\right)^{-1/2}\approx1+\frac{r_{s}}{2R}.$$

The redshift is therefore

$$z\approx\frac{r_{s}}{2R}\approx\frac{3.01\times10^{3}}{2(5.57\times10^{6})}\approx2.7\times10^{-4}.$$

Light emitted at $\lambda_{\text{em}}=500\ \text{nm}$ would be observed at $\lambda_{\infty}\approx500.14\ \text{nm}$, a shift readily detectable by high-resolution spectrographs.

Problems with Solutions

1. Schwarzschild radius. Find $r_{s}$ for a black hole of mass $M=5\,M_{\odot}$.

   Solution. $r_{s}=2GM/c^{2}=2(6.674\times10^{-11})(5\times1.99\times10^{30})/(3\times10^{8})^{2}\approx6.66\times10^{20}/9\times10^{16}\approx14.8\ \text{km}$.

2. Photon redshift. A photon is emitted radially from $r=3r_{s}$ and observed at infinity. Compute the redshift $z$.

   Solution. For a radial photon, $E=-p_{0}$ is conserved. The observed frequency ratio is $\omega_{\infty}/\omega_{\text{em}}=\sqrt{1-r_{s}/r_{\text{em}}}$. At $r_{\text{em}}=3r_{s}$, $z=\frac{\omega_{\text{em}}}{\omega_{\infty}}-1=\left(1-\frac13\right)^{-1/2}-1=\sqrt{\frac32}-1\approx0.225.$

3. Perihelion precession. Estimate Mercury’s perihelion advance per orbit using $\Delta\phi\approx6\pi GM_{\odot}/[c^{2}a(1-e^{2})]$ with $a=5.79\times10^{10}\ \text{m}$, $e=0.206$. Solution. $\Delta\phi\approx\frac{6\pi(6.674\times10^{-11})(1.99\times10^{30})}{(3\times10^{8})^{2}(5.79\times10^{10})(1-0.206^{2})}\approx5.0\times10^{-7}\ \text{rad}.$ With 415 orbits per century, the accumulated shift is $\approx415\times5.0\times10^{-7}\approx2.1\times10^{-4}\ \text{rad}\approx43''$, matching the historical anomaly.

Section summary. Schwarzschild spacetime is the basic laboratory for relativistic gravity.

More General Black Holes / Perturbations and Gravitational Waves

Core ideas

Rotating and charged black holes add angular momentum and charge; astrophysical black holes are described mainly by Kerr geometry. Small metric perturbations propagate as gravitational waves with two tensor polarizations. Detectors measure strain from compact binary motion.

For review, be able to recognize Kerr parameters, describe horizons and ergospheres qualitatively, derive the linear wave equation, and connect quadrupole radiation to binary inspiral. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu},\qquad \Box \bar h_{\mu\nu}=-\frac{16\pi G}{c^4}T_{\mu\nu},\qquad h\sim \Delta L/L$$

Worked example

LIGO detects gravitational waves by measuring the tiny differential change in arm length of a Michelson interferometer. Consider a passing wave with plus-polarization amplitude $h_{+}=10^{-21}$ and frequency $f=100\ \text{Hz}$. In the transverse-traceless (TT) gauge the spatial metric perturbation is $h_{xx}^{TT}=-h_{yy}^{TT}=h_{+}\cos[2\pi f(t-z/c)]$. For an interferometer with arm length $L=4\ \text{km}$ oriented along the $x$ and $y$ axes, the proper length oscillations are

$$\frac{\Delta L_{x}}{L}\approx\frac12 h_{+},\qquad\frac{\Delta L_{y}}{L}\approx-\frac12 h_{+}.$$

Hence

$$\Delta L_{x}\approx\frac12(10^{-21})(4000\ \text{m})=2\times10^{-18}\ \text{m}.$$

This displacement is roughly $1/400$ the radius of a proton ($r_{p}\approx8.4\times10^{-16}\ \text{m}$), illustrating the extraordinary sensitivity required for ground-based detection.

Problems with Solutions

1. Arm-length change. If a gravitational wave has strain $h=2\times10^{-20}$ and an interferometer arm is $L=3\ \text{km}$, what is the length oscillation? Compare with a proton radius.

   Solution. $\Delta L=\tfrac12 hL\approx\tfrac12(2\times10^{-20})(3000)=3\times10^{-17}\ \text{m}$. The ratio to the proton radius is $3\times10^{-17}/8.4\times10^{-16}\approx0.036$, i.e., about $1/28$ of a proton radius.

2. Chirp mass. For a binary with $m_{1}=m_{2}=1.4\,M_{\odot}$, compute the chirp mass $\mathcal{M}_{c}=\frac{(m_{1}m_{2})^{3/5}}{(m_{1}+m_{2})^{1/5}}$. Solution. $\mathcal{M}_{c}=\frac{(1.4^{2})^{3/5}}{(2.8)^{1/5}}\,M_{\odot}=\frac{(1.96)^{0.6}}{(2.8)^{0.2}}\,M_{\odot}\approx\frac{1.50}{1.23}\,M_{\odot}\approx1.22\,M_{\odot}.$

3. Merger frequency. Estimate the gravitational-wave frequency at ISCO for a neutron-star binary with total mass $M=2.8\,M_{\odot}$ using the test-particle Kepler formula $f_{\text{ISCO}}\approx c^{3}/(6^{3/2}\pi GM)$. Solution. $f_{\text{ISCO}}\approx\frac{(3\times10^{8})^{3}}{6^{3/2}\,\pi\,(6.674\times10^{-11})(2.8\times1.99\times10^{30})}\approx\frac{2.7\times10^{25}}{2.19\times10^{21}}\approx1.2\times10^{3}\ \text{Hz},$ i.e., about $1\ \text{kHz}$, consistent with numerical relativity.

Section summary. Black holes and gravitational waves are dynamical strong-field predictions of GR.

Cosmology

Core ideas

Relativistic cosmology assumes large-scale homogeneity and isotropy, leading to the FLRW metric and Friedmann equations. Matter, radiation, curvature, and dark energy determine expansion. Observations of redshift, CMB, supernovae, and structure constrain the model.

For review, be able to write the FLRW metric, use scale factor and redshift, interpret density parameters, and connect Friedmann equations to cosmic history. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$ds^2=-c^2dt^2+a^2(t)\left[\frac{dr^2}{1-kr^2}+r^2d\Omega^2\right],\qquad H^2=\frac{8\pi G}{3}\rho-\frac{kc^2}{a^2}+\frac{\Lambda c^2}{3}$$

Worked example

For a flat, matter-dominated universe ($k=0$, $\Lambda=0$), the Friedmann equation reduces to $H^{2}=(8\pi G/3)\rho$ with $\rho\propto a^{-3}$. Writing $H=\dot a/a$, we obtain $\dot a=H_{0}a_{0}^{3/2}a^{-1/2}$. Integrating from the Big Bang ($a=0$) to today ($a=a_{0}$) gives the age

$$t_{0}=\frac{2}{3H_{0}}.$$

Using $H_{0}=70\ \text{km}\,\text{s}^{-1}\,\text{Mpc}^{-1}$ and $1\ \text{Mpc}=3.086\times10^{19}\ \text{km}$,

$$\frac{1}{H_{0}}=\frac{3.086\times10^{19}}{70}\ \text{s}\approx4.41\times10^{17}\ \text{s}\approx14.0\ \text{Gyr}.$$

Thus $t_{0}\approx\tfrac23(14.0)\approx9.3\ \text{Gyr}$. (The inclusion of dark energy raises this to the observed \sim$$13.8\ \text{Gyr}.) A galaxy at redshift $z=2$ was emitted when $a=1/(1+z)=1/3$ and $t_{\text{em}}=t_{0}(1/3)^{3/2}\approx t_{0}/5.2\approx1.8\ \text{Gyr}$, giving a lookback time of \sim$$7.5\ \text{Gyr}.

Problems with Solutions

1. Critical density. Compute the critical density $\rho_{c}=3H_{0}^{2}/(8\pi G)$ for $H_{0}=70\ \text{km}\,\text{s}^{-1}\,\text{Mpc}^{-1}$.

   Solution. Convert $H_{0}=70\times10^{3}/(3.086\times10^{22})\ \text{s}^{-1}\approx2.27\times10^{-18}\ \text{s}^{-1}$. Then $\rho_{c}=\frac{3(2.27\times10^{-18})^{2}}{8\pi(6.674\times10^{-11})}\approx\frac{1.55\times10^{-35}}{1.68\times10^{-9}}\approx9.2\times10^{-27}\ \text{kg/m}^{3},$ equivalent to about $5$ or $6$ hydrogen atoms per cubic metre.

2. Scale factor and age fraction. In a flat matter-dominated universe, a quasar is observed at $z=3$. What was the scale factor at emission, and what fraction of the present age had elapsed?

   Solution. $a_{\text{em}}=1/(1+z)=1/4$. Since $t\propto a^{3/2}$, $\frac{t_{\text{em}}}{t_{0}}=\left(\frac14\right)^{3/2}=\frac18.$ The universe was $12.5\%$ of its current age when the light was emitted.

3. Hubble distance. Evaluate the Hubble distance $d_{H}=c/H_{0}$ in Mpc for $H_{0}=70\ \text{km}\,\text{s}^{-1}\,\text{Mpc}^{-1}$. Solution. $d_{H}=\frac{c}{H_{0}}=\frac{3.00\times10^{5}\ \text{km/s}}{70\ \text{km}\,\text{s}^{-1}\,\text{Mpc}^{-1}}\approx4.29\times10^{3}\ \text{Mpc}\approx4.3\ \text{Gpc}.$ This sets the rough scale of the observable universe today.

Section summary. Cosmology applies GR to the expanding universe.