Introduction to Cosmology and the Cosmic Microwave Background: Lecture 2

1 Redshift

In the previous section we described the expansion of the universe using physical distance and comoving distance. Two comoving galaxies keep the same coordinates, yet the physical distance between them can grow with time because of the scale factor $a(t)$. That left an open question: how do we know from observations that the universe is expanding?

We cannot lay a ruler between distant galaxies. Almost everything we know about them comes from the light that reaches us, so we need to understand what the expansion does to that light.

Recall first what a wavelength is. In an electromagnetic wave, the distance between two successive crests is the wavelength, written $\lambda$. For light, wavelength and frequency are related by

$$c = \lambda\nu \tag{1}$$

where $c$ is the speed of light, $\lambda$ the wavelength and $\nu$ the frequency.

The local speed of light $c$ is constant, so if the wavelength increases the frequency has to decrease. Let $\lambda_e$ be the wavelength of a light wave when it leaves the source and $\lambda_o$ the wavelength measured when it reaches the observer. If the observed wavelength is larger than the emitted one,

$$\lambda_o > \lambda_e, \tag{2}$$

the light has shifted toward longer wavelengths. Longer wavelengths in the visible spectrum lie toward the red end, so this effect is called redshift.

The size of the shift is given by the parameter $z$:

$$z = \frac{\lambda_o - \lambda_e}{\lambda_e} \tag{3}$$

Rearranging,

$$z = \frac{\lambda_o}{\lambda_e} - 1 \quad \Rightarrow \quad 1 + z = \frac{\lambda_o}{\lambda_e} \tag{4}$$

The interpretation is direct. If

$$z > 0 \quad \Rightarrow \quad \lambda_o > \lambda_e, \tag{5}$$

then the wavelength has grown on the way to us and the light is redshifted.

Take $z = 1$ as an example. Then

$$1 + z = 2 \tag{6}$$

so

$$\lambda_o = 2\lambda_e \tag{7}$$

The wavelength we observe is twice the wavelength at emission.

So far redshift has been defined only through the change in wavelength. Why should the expansion of the universe change the wavelength of a photon at all?

If $a(t)$ describes the scale of the expanding universe, we need to show how the wavelength is tied to this scale factor. The claim is that, as the universe expands, the physical wavelength of a photon measured by comoving observers grows along with $a(t)$.

Saying "the universe stretches, so the wavelength stretches" is a picture, not a derivation. We want to see how the relation follows from the geometry of an expanding spacetime.

In this section we show from scratch, using general relativity, why the frequency of a photon travelling through an expanding universe changes as

$$\omega \propto \frac{1}{a(t)} \tag{8}$$

and then use the relation between frequency and wavelength to recover $\lambda \propto a(t)$. We start with spacetime coordinates.

1.1 Spacetime coordinates

In special and general relativity an event is labelled by four coordinates:

$$x^\mu = \left(x^0, x^1, x^2, x^3\right). \tag{9}$$

[!note] Notation Greek indices $\mu, \nu, \alpha, \beta$ run over $0, 1, 2, 3$. The zeroth component is timelike and the other three are spatial.

We do not use natural units here and keep the speed of light $c$ explicit. The time coordinate is therefore $x^0 = ct$, and the four-position is

$$x^\mu = (ct, x, y, z) \tag{10}$$

With this choice $ct$, $x$, $y$ and $z$ all have dimensions of length.

1.2 The metric and the spacetime interval

With coordinates in place, we need a way to measure the physical separation between two nearby events. In general relativity the metric tensor does this.

[!info] Definition The spacetime interval between two nearby events is ds^2 = g_{\mu\nu},dx^\mu dx^\nu \tag{11} where $g_{\mu\nu}$ is the metric tensor.

We use the signature $(-,+,+,+)$. In flat Minkowski spacetime the metric is

$$g_{\mu\nu} = \begin{pmatrix} -1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \end{pmatrix} \tag{12}$$

so the interval is

$$ds^2 = -(dx^0)^2 + (dx^1)^2 + (dx^2)^2 + (dx^3)^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 \tag{13}$$

Our universe, however, is expanding, and a static Minkowski spacetime cannot describe it. We need the FLRW metric, in which spatial distances can change with time.

1.3 The FLRW metric

The spacetime geometry of a homogeneous and isotropic universe is given by the FLRW metric:

$$ds^2 = -c^2dt^2 + a^2(t),\gamma_{ij},dx^i dx^j. \tag{14}$$

[!note] Notation Latin indices $i, j = 1, 2, 3$ label spatial components only.

[!info] Definition The function $a(t)$ is the scale factor. It tells us how physical distances in the universe change with cosmic time.

[!info] Definition $\gamma_{ij}$ is the time-independent metric that describes the spatial geometry of a slice of constant cosmic time.

For a spatially flat universe $\gamma_{ij} = \delta_{ij}$. Then $\gamma_{ij}dx^idx^j = dx^2 + dy^2 + dz^2$ and the FLRW metric becomes

$$ds^2 = -c^2dt^2 + a^2(t)\left(dx^2 + dy^2 + dz^2\right) \tag{15}$$

Since $x^0 = ct$, we have $dx^0 = c,dt$. The metric components are

$$g_{\mu\nu} = \begin{pmatrix} -1 & 0 & 0 & 0 \ 0 & a^2 & 0 & 0 \ 0 & 0 & a^2 & 0 \ 0 & 0 & 0 & a^2 \end{pmatrix} \tag{16}$$

and the inverse metric is

$$g^{\mu\nu} = \begin{pmatrix} -1 & 0 & 0 & 0 \ 0 & a^{-2} & 0 & 0 \ 0 & 0 & a^{-2} & 0 \ 0 & 0 & 0 & a^{-2} \end{pmatrix} \tag{17}$$

[!info] Definition The inverse metric is the tensor that satisfies $g^{\mu\alpha}g_{\alpha\nu} = \delta^\mu{}_\nu$. For example, $g^{11}g_{11} = a^{-2}a^2 = 1$.

1.4 The comoving observer and proper time

Many different observers could use this metric, so we have to decide which observer we mean in an expanding universe.

FLRW coordinates can be pictured as a grid that expands with the universe. Observers who sit at fixed points of this grid are called comoving observers.

[!info] Definition A comoving observer has constant spatial coordinates: x^i = \text{const}. \tag{18} Hence $dx^i = 0$ and $dx^i/d\tau = 0$.

Constant coordinates do not mean constant physical distance to other galaxies. The physical distance between two comoving points changes approximately as $d_{\text{phys}}(t) = a(t),d_{\text{com}}$: the coordinate distance stays fixed while the physical distance grows with the scale factor.

An observer carries a clock as well as a ruler. What does a comoving observer's clock read?

[!info] Definition The time measured by an ideal clock carried by an observer is the proper time, written $\tau$.

Along a timelike worldline,

$$ds^2 = -c^2d\tau^2 \tag{19}$$

For a comoving observer $dx^i = 0$, so the FLRW metric reduces to $ds^2 = -c^2dt^2$. Comparing with the definition of proper time,

$$-c^2d\tau^2 = -c^2dt^2 \tag{20}$$

and, choosing the future direction of time,

$$d\tau = dt \tag{21}$$

The clock in a comoving observer's pocket therefore measures the cosmic time $t$ that appears in the FLRW metric.

1.5 The four-velocity of the comoving observer

Next we describe the direction in which the observer moves through spacetime, using the four-velocity.

[!info] Definition The four-velocity of an observer is u^\mu = \frac{dx^\mu}{d\tau} \tag{22}

With $x^\mu = (ct, x, y, z)$,

$$u^\mu = \left(\frac{d(ct)}{d\tau}, \frac{dx}{d\tau}, \frac{dy}{d\tau}, \frac{dz}{d\tau}\right). \tag{23}$$

For a comoving observer $dt = d\tau$ and $dx^i = 0$, which gives $u^0 = c,dt/d\tau = c$ and $u^i = 0$:

$$u^\mu = (c, 0, 0, 0) \tag{24}$$

Now we lower the index.

[!info] Definition An index is lowered with the metric: $u_\mu = g_{\mu\nu}u^\nu$.

The zeroth component is

$$u_0 = g_{00}u^0 + g_{01}u^1 + g_{02}u^2 + g_{03}u^3 = (-1)c = -c \tag{25}$$

The metric is diagonal and the spatial components of the four-velocity vanish, so $u_1 = u_2 = u_3 = 0$ and

$$u_\mu = (-c, 0, 0, 0) \tag{26}$$

1.6 Electromagnetic waves and phase

We can now return to the main problem. The observer's measurement is made on photons, which means on electromagnetic waves, so the observer and the metric are not enough by themselves; we also need a description of the wave.

In the geometric-optics approximation an electromagnetic wave is written as

$$\Psi(x) = A(x),e^{i\Phi(x)} \tag{27}$$

where $A(x)$ is the amplitude and $\Phi(x)$ the phase.

The phase tells us where in its oscillation cycle the wave is. For a plane wave in flat spacetime,

$$\Phi(x) = \mathbf{k}\cdot\mathbf{x} - \omega t \tag{28}$$

or explicitly $\Phi = k_x x + k_y y + k_z z - \omega t$.

Since $x^0 = ct$, we can write $t = x^0/c$ and

$$\Phi = k_x x + k_y y + k_z z - \frac{\omega}{c}x^0 \tag{29}$$

How much does the phase change when we move a small step in spacetime? The total differential of a scalar field is

$$d\Phi = \partial_\mu\Phi,dx^\mu \tag{30}$$

which written out is

$$d\Phi = \frac{\partial\Phi}{\partial x^0}dx^0 + \frac{\partial\Phi}{\partial x^1}dx^1 + \frac{\partial\Phi}{\partial x^2}dx^2 + \frac{\partial\Phi}{\partial x^3}dx^3 \tag{31}$$

1.7 The wave four-covector

To package the change of phase in a single geometric object, we define the wave four-covector.

[!info] Definition The wave four-covector is k_\mu = \partial_\mu\Phi \tag{32} so that $d\Phi = k_\mu dx^\mu$.

In words, $k_\mu$ tells us how much a small displacement in spacetime changes the phase of the wave.

We compute the components one at a time. With

$$\Phi = k_x x + k_y y + k_z z - \frac{\omega}{c}x^0 \tag{33}$$

we get

$$k_0 = \frac{\partial\Phi}{\partial x^0} = -\frac{\omega}{c}, \qquad k_1 = \frac{\partial\Phi}{\partial x} = k_x, \qquad k_2 = k_y, \qquad k_3 = k_z. \tag{34}$$

so

$$k_\mu = \left(-\frac{\omega}{c}, k_x, k_y, k_z\right) \tag{35}$$

The time component carries the frequency, and the spatial components carry the direction of propagation and the wavelength. The magnitude of the spatial wave vector is $|\mathbf{k}| = 2\pi/\lambda$.

Now we raise the index.

[!info] Definition An index is raised with the inverse metric: $k^\mu = g^{\mu\nu}k_\nu$.

The zeroth component is

$$k^0 = g^{00}k_0 = (-1)\left(-\frac{\omega}{c}\right) = \frac{\omega}{c} \tag{36}$$

and the spatial components are

$$k^1 = \frac{k_1}{a^2}, \qquad k^2 = \frac{k_2}{a^2}, \qquad k^3 = \frac{k_3}{a^2} \tag{37}$$

so

$$k^\mu = \left(\frac{\omega}{c}, \frac{k_1}{a^2}, \frac{k_2}{a^2}, \frac{k_3}{a^2}\right) \tag{38}$$

Note that $k_i$ and $k^i$ differ: raising a spatial index brings in a factor $a^{-2}(t)$.

1.8 The four-momentum of the photon

What changes when this wave is treated quantum mechanically, as a photon?

Using $\omega = 2\pi\nu$ and $\hbar = h/(2\pi)$, the photon energy is

$$E = h\nu = h\frac{\omega}{2\pi} = \frac{h}{2\pi}\omega = \hbar\omega \tag{39}$$

The spatial momentum of the photon is related to the wave vector by $\mathbf{p} = \hbar\mathbf{k}$. Energy and momentum combine into the four-dimensional relation

$$p^\mu = \hbar k^\mu \tag{40}$$

and likewise $p_\mu = \hbar k_\mu$.

Since $k^0 = \omega/c$, the zeroth component is

$$p^0 = \hbar k^0 = \frac{\hbar\omega}{c} = \frac{E}{c} \tag{41}$$

1.9 The frequency measured by the observer

With the photon and the observer both defined, we can work out how the observer measures the photon's frequency.

As the observer moves along their worldline, they measure how fast the phase of the wave changes. By the chain rule,

$$\frac{d\Phi}{d\tau} = \frac{\partial\Phi}{\partial x^\mu}\frac{dx^\mu}{d\tau} = k_\mu u^\mu. \tag{42}$$

With our convention $\Phi = \mathbf{k}\cdot\mathbf{x} - \omega t$, the time part of the phase decreases toward the future. To get a positive frequency we define

$$\omega_{\text{obs}} = -k_\mu u^\mu \tag{43}$$

The frequency an observer measures is the rate at which the phase of the wave changes according to that observer's own clock. Everything that follows rests on this formula.

For a comoving observer $u^\mu = (c, 0, 0, 0)$, so

$$\omega_{\text{obs}} = -\left(k_0u^0 + k_1u^1 + k_2u^2 + k_3u^3\right) = -c,k_0. \tag{44}$$

With $k_0 = -\omega/c$ this gives $\omega_{\text{obs}} = \omega$. Since $k^0 = \omega/c$, we can also write

$$\omega_{\text{obs}} = c,k^0 \tag{45}$$

The measured frequency is a single number, a scalar. It depends on both the photon's vector $k^\mu$ and the observer's four-velocity $u^\mu$, which is why different observers can measure different frequencies for the same photon.

1.10 Null trajectories and the affine parameter

The photon is moving, so we also need the equation that governs its path through spacetime.

The photon has zero rest mass. From the general energy-momentum relation $E^2 = p^2c^2 + m^2c^4$, a photon has $E = pc$.

The norm of the four-momentum is $p_\mu p^\mu = -m^2c^2$, so for $m = 0$

$$p_\mu p^\mu = 0 \tag{46}$$

and with $p^\mu = \hbar k^\mu$,

$$k_\mu k^\mu = 0 \tag{47}$$

[!info] Definition A vector with zero norm is called null, or lightlike.

The photon's worldline is therefore a null curve, along which $ds^2 = 0$.

A massive particle's trajectory can be parametrized by its proper time $\tau$. For a photon, $ds^2 = -c^2d\tau^2 = 0$ gives $d\tau = 0$, so proper time cannot serve as a parameter. We need something else.

[!info] Definition The parameter used to order points along a photon's trajectory is the affine parameter, usually written $\lambda_{\text{aff}}$.

To keep it apart from the wavelength, we write the affine parameter as $\lambda_{\text{aff}}$ and the physical wavelength as $\lambda_{\text{phys}}$.

The trajectory is $x^\mu = x^\mu(\lambda_{\text{aff}})$, with tangent vector

$$k^\mu = \frac{dx^\mu}{d\lambda_{\text{aff}}} \tag{48}$$

The zeroth component is

$$k^0 = \frac{dx^0}{d\lambda_{\text{aff}}} = c\frac{dt}{d\lambda_{\text{aff}}} \tag{49}$$

so

$$\frac{dt}{d\lambda_{\text{aff}}} = \frac{k^0}{c} \tag{50}$$

With an affine parameter, the photon's geodesic equation is

$$\frac{dk^\mu}{d\lambda_{\text{aff}}} + \Gamma^\mu{}_{\alpha\beta}k^\alpha k^\beta = 0 \tag{51}$$

1.11 The null condition in the FLRW metric

Before using the geodesic equation, we write the null condition explicitly in the FLRW metric:

$$g_{\mu\nu}k^\mu k^\nu = 0. \tag{52}$$

With $g_{00} = -1$ and $g_{ij} = a^2(t)\gamma_{ij}$,

$$-(k^0)^2 + a^2(t),\gamma_{ij}k^ik^j = 0 \tag{53}$$

which gives

$$\gamma_{ij}k^ik^j = \frac{(k^0)^2}{a^2(t)} \tag{54}$$

This links the time and space components of the photon's wave vector. It does not say how $k^0$ evolves in time; for that we need the geodesic equation.

1.12 Christoffel symbols

The geodesic equation contains the objects $\Gamma^\mu{}_{\alpha\beta}$.

[!info] Definition The Christoffel symbols are \Gamma^\mu{}_{\alpha\beta} = \frac{1}{2}g^{\mu\nu}\left(\partial_\alpha g_{\nu\beta} + \partial_\beta g_{\nu\alpha} - \partial_\nu g_{\alpha\beta}\right) \tag{55}

They encode how the metric components vary with the coordinates.

We want $k^0$, the component tied to the frequency, so we take the $\mu = 0$ component of the geodesic equation:

$$\frac{dk^0}{d\lambda_{\text{aff}}} + \Gamma^0{}_{\alpha\beta}k^\alpha k^\beta = 0. \tag{56}$$

The sum can contain terms of type $\Gamma^0{}_{00}$, $\Gamma^0{}_{0i}$ and $\Gamma^0{}_{ij}$. We check first that the ones with time indices vanish.

For $\Gamma^0{}_{00}$,

$$\Gamma^0{}_{00} = \frac{1}{2}g^{00}\left(\partial_0g_{00} + \partial_0g_{00} - \partial_0g_{00}\right) = \frac{1}{2}g^{00}\partial_0g_{00}. \tag{57}$$

In the FLRW metric $g_{00} = -1$ is constant, so $\partial_0g_{00} = 0$ and

$$\Gamma^0{}_{00} = 0 \tag{58}$$

Next, the mixed term $\Gamma^0{}_{0i}$ with one time index and one space index:

$$\Gamma^0{}_{0i} = \frac{1}{2}g^{00}\left(\partial_0g_{0i} + \partial_ig_{00} - \partial_0g_{0i}\right). \tag{59}$$

Here $g_{0i} = 0$ and $g_{00} = -1$ is constant, so every derivative vanishes:

$$\Gamma^0{}_{0i} = 0 \tag{60}$$

The only remaining term is $\Gamma^0{}_{ij}$, which comes from the time dependence of the spatial metric:

$$\Gamma^0{}_{ij} = \frac{1}{2}g^{00}\left(\partial_ig_{0j} + \partial_jg_{0i} - \partial_0g_{ij}\right). \tag{61}$$

With $g_{0i} = 0$ and $g^{00} = -1$,

$$\Gamma^0{}_{ij} = \frac{1}{2}\partial_0g_{ij} \tag{62}$$

The spatial metric is $g_{ij} = a^2(t)\gamma_{ij}$, and $\partial_0$ is the derivative with respect to $x^0 = ct$:

$$\partial_0 = \frac{\partial}{\partial x^0} = \frac{1}{c}\frac{\partial}{\partial t}. \tag{63}$$

Then

$$\partial_0g_{ij} = \frac{1}{c}\frac{\partial}{\partial t}\left(a^2\gamma_{ij}\right) = \frac{2a\dot{a}}{c}\gamma_{ij}. \tag{64}$$

and therefore

$$\Gamma^0{}_{ij} = \frac{a\dot{a}}{c}\gamma_{ij} \tag{65}$$

This is where the physics enters: the time component of the photon's wave vector changes because the spatial metric changes with time through $a(t)$.

We now have everything we need. The time component of the geodesic equation was

$$\frac{dk^0}{d\lambda_{\text{aff}}} + \Gamma^0{}_{\alpha\beta}k^\alpha k^\beta = 0 \tag{66}$$

Since $\Gamma^0{}_{00} = 0$ and $\Gamma^0{}_{0i} = 0$, only the spatial term survives:

$$\frac{dk^0}{d\lambda_{\text{aff}}} + \Gamma^0{}_{ij}k^ik^j = 0. \tag{67}$$

Inserting $\Gamma^0{}_{ij} = (a\dot{a}/c)\gamma_{ij}$,

$$\frac{dk^0}{d\lambda_{\text{aff}}} + \frac{a\dot{a}}{c}\gamma_{ij}k^ik^j = 0 \tag{68}$$

From the null condition, $\gamma_{ij}k^ik^j = (k^0)^2/a^2$. Substituting,

$$\frac{dk^0}{d\lambda_{\text{aff}}} + \frac{\dot{a}}{ac}(k^0)^2 = 0 \tag{69}$$

[!info] Definition The Hubble parameter is $H(t) = \dot{a}/a$.

The equation becomes

$$\frac{dk^0}{d\lambda_{\text{aff}}} + \frac{H}{c}(k^0)^2 = 0 \tag{70}$$

To rewrite this in terms of cosmic time, recall that

$$\frac{dt}{d\lambda_{\text{aff}}} = \frac{k^0}{c} \tag{71}$$

By the chain rule,

$$\frac{dk^0}{d\lambda_{\text{aff}}} = \frac{dk^0}{dt}\frac{dt}{d\lambda_{\text{aff}}} = \frac{k^0}{c}\frac{dk^0}{dt}. \tag{72}$$

Putting this into the geodesic equation,

$$\frac{k^0}{c}\frac{dk^0}{dt} + \frac{H}{c}(k^0)^2 = 0 \tag{73}$$

and dividing by $k^0/c$,

$$\frac{dk^0}{dt} + Hk^0 = 0 \tag{74}$$

With $H = \dot{a}/a$,

$$\frac{1}{k^0}\frac{dk^0}{dt} = -\frac{\dot{a}}{a} \tag{75}$$

which can be written as

$$\frac{d}{dt}\ln k^0 = -\frac{d}{dt}\ln a \tag{76}$$

Integrating both sides in time gives $\ln k^0 = -\ln a + C$, so

$$k^0 = \frac{C'}{a(t)} \tag{77}$$

that is,

$$k^0 \propto \frac{1}{a(t)} \tag{78}$$

For a comoving observer $\omega_{\text{obs}} = c,k^0$, and the constant $c$ does not affect the proportionality:

$$\omega_{\text{obs}} \propto \frac{1}{a(t)} \tag{79}$$

In an expanding universe, then, the photon frequency measured by a comoving observer is inversely proportional to the scale factor.

If the photon is emitted at time $t_e$ and observed at time $t_o$,

$$\frac{\omega_o}{\omega_e} = \frac{a_e}{a_o} \tag{80}$$

In an expanding universe $a_o > a_e$, so $\omega_o < \omega_e$ and the photon arrives at a lower frequency.

1.13 Wavelength and redshift

We know how the frequency changes, and the relation between frequency and wavelength from introductory physics takes us straight back to where we started.

Angular and ordinary frequency are related by $\omega = 2\pi\nu$, and for light $c = \lambda_{\text{phys}}\nu$. Therefore

$$\lambda_{\text{phys}} = \frac{2\pi c}{\omega} \tag{81}$$

and since $\omega \propto a^{-1}$,

$$\lambda_{\text{phys}} \propto a(t) \tag{82}$$

The emitted and observed wavelengths are then related by

$$\frac{\lambda_o}{\lambda_e} = \frac{a_o}{a_e} \tag{83}$$

[!info] Definition The cosmological redshift is z = \frac{\lambda_o - \lambda_e}{\lambda_e} \tag{84}

It follows that

$$1 + z = \frac{\lambda_o}{\lambda_e} = \frac{a_o}{a_e} = \frac{\omega_e}{\omega_o} \tag{85}$$

Normalizing the present-day scale factor to $a_o = 1$,

$$1 + z = \frac{1}{a_e} \tag{86}$$

As the universe expands, the wavelength grows with the scale factor and the frequency falls as its inverse:

$$\lambda_{\text{phys}} \propto a(t), \qquad \omega \propto \frac{1}{a(t)} \tag{87}$$

The local speed of light stays the same throughout. Every local comoving observer still measures $c = \lambda_{\text{phys}}\nu$: the wavelength increases and the frequency decreases by the same factor, so their product remains $c$.

In the next section we combine these results with the relation between energy and temperature.