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Open CosmologyPhysics & cosmology: four years, plus a graduate year

Matter–radiation equality and the pressure of light

Running today's radiation density backwards dates the end of the radiation era to roughly 75,000 years after the Big Bang — and radiation's pressure forces a pressure term into the second Friedmann equation.

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Full reading path (27 pages)
  1. Vectors (8.01)
  2. Position and velocity in one dimension (8.01)
  3. Acceleration in one dimension (8.01)
  4. Motion in two dimensions and projectiles (8.01)
  5. The homogeneous expanding universe (8.286)
  6. The Doppler effect and special relativity (8.286)
  7. Cosmological redshift (8.286)
  8. Newton's laws of motion (8.01)
  9. The universal law of gravitation (8.01)
  10. Why a static universe cannot work (8.286)
  11. The expanding sphere and the equation for a(t) (8.286)
  12. The Friedmann equation (8.286)
  13. Open, closed and flat universes (8.286)
  14. The closed universe as a cycloid (8.286)
  15. The age of the universe and the Big Bang singularity (8.286)
  16. The age of a closed universe from H and Ω (8.286)
  17. The open universe and the age of any matter universe (8.286)
  18. Non-Euclidean geometry (8.286)
  19. The metric of a sphere and of a closed universe (8.286)
  20. Curvature and expansion — the Robertson–Walker metric (8.286)
  21. The geometry of open universes (8.286)
  22. The spacetime interval (8.286)
  23. The spacetime metric of the universe (8.286)
  24. The geodesic equation (8.286)
  25. Simplifying the geodesic equation (8.286)
  26. Falling into a black hole (8.286)
  27. Mass, energy and the dilution of radiation (8.286)
On this page · 6 sections
  1. How much radiation is there today?
  2. Back to equality
  3. Matter: three equations, two independent
  4. Pressure does work
  5. Which Friedmann equation gets the pressure?
  6. Summary

Radiation's density falls as 1/a41/a^4 and matter's as 1/a31/a^3, so going back in time radiation becomes ever more important. The second half of Lecture 15 asks when the two were equal, and what changes in the equations of the expanding universe once radiation — which, unlike dust, has pressure — is in charge.

How much radiation is there today?#

Guth quotes the present radiation energy density (how it follows from the temperature of the cosmic microwave background comes later in the course):

Radiation today
ur=7.01×10−14 Jm3,ρr=urc2=7.80×10−31 kgm3=7.80×10−34 gcm3u_r = 7.01 \times 10^{-14}\ \tfrac{\text{J}}{\text{m}^3}, \qquad \rho_r = \frac{u_r}{c^2} = 7.80 \times 10^{-31}\ \tfrac{\text{kg}}{\text{m}^3} = 7.80 \times 10^{-34}\ \tfrac{\text{g}}{\text{cm}^3}
In words

u sub r equals 7.01 times 10 to the power minus 14 J over m cubed , rho sub r equals u sub r over c squared equals 7.80 times 10 to the power minus 31 kg over m cubed equals 7.80 times 10 to the power minus 34 g over cm cubed

Source: 8.286 Lecture 15 transcript, pp. 15–16 (board)

This includes neutrinos as well as photons: in the early universe neutrinos behaved just like radiation. The photon part is measured; the neutrino part rests on theory, discussed later. From here on, the subscript γ\gamma means photons and rr means all radiation.

Compared with water, at 1 g/cm³, that is, in Guth's words, 10−3410^{-34} "or maybe 10−3310^{-33}" times as dense. It is tiny even compared with the critical density,

Critical density
ρc=1.88 h02×10−29 gcm3,H0=100 h0 km/sMpc\rho_c = 1.88\,h_0^2 \times 10^{-29}\ \tfrac{\text{g}}{\text{cm}^3}, \qquad H_0 = 100\,h_0\ \tfrac{\text{km/s}}{\text{Mpc}}
In words

rho sub c equals 1.88 h sub 0 squared times 10 to the power minus 29 g over cm cubed , H sub 0 equals 100 h sub 0 km/s over Mpc

Source: 8.286 Lecture 15 transcript, p. 16 (board)

Why write H0H_0 through a dimensionless h0h_0? Partly so astronomers needn't keep saying "kilometres per second per megaparsec", and partly so formulas stay valid as the measured value moves — it changed, Guth notes, every time he taught the course. Dividing gives the radiation share of the critical density:

Radiation density parameter
Ωr=ρrρc=4.15×10−5 h0−2≈9.2×10−5(h0=0.67)\Omega_r = \frac{\rho_r}{\rho_c} = 4.15 \times 10^{-5}\,h_0^{-2} \approx 9.2 \times 10^{-5} \quad (h_0 = 0.67)
In words

omega sub r equals rho sub r over rho sub c equals 4.15 times 10 to the power minus 5 h sub 0 to the power minus 2 is approximately 9.2 times 10 to the power minus 5 (h sub 0 equals 0.67)

Source: 8.286 Lecture 15 transcript, p. 17 (board)

using the 2013 Planck value h0=0.67h_0 = 0.67 (the earlier accepted value was about 0.70). About one part in 10410^4 of today's energy density is radiation.

Back to equality#

With Ωm≈0.30\Omega_m \approx 0.30 for (non-relativistic) matter, today's ratio is Ωr/Ωm≈3.1×10−4\Omega_r/\Omega_m \approx 3.1 \times 10^{-4}. The ratio scales as 1/a1/a, so at any time

Radiation-to-matter ratio
ρr(t)ρm(t)=3.1×10−4  a(t0)a(t)\frac{\rho_r(t)}{\rho_m(t)} = 3.1 \times 10^{-4}\;\frac{a(t_0)}{a(t)}
In words

rho sub r (t) over rho sub m (t) equals 3.1 times 10 to the power minus 4 a(t sub 0 ) over a(t)

Source: 8.286 Lecture 15 transcript, p. 18 (board)

Equality is when this ratio is 1. Since 1+z=a(t0)/a(t)1 + z = a(t_0)/a(t),

Redshift of equality
1+zeq=13.1×10−4⟹zeq≈3,2001 + z_\text{eq} = \frac{1}{3.1 \times 10^{-4}} \quad\Longrightarrow\quad z_\text{eq} \approx 3{,}200
In words

1 plus z sub eq equals 1 over 3.1 times 10 to the power minus 4 which gives z sub eq is approximately 3,200

Source: 8.286 Lecture 15 transcript, p. 19 (board)

[The transcript first has Guth saying "about 32,000"; one sentence later he says 3,200, which is right.] To turn this into a time, Guth treats the whole history since then as matter-dominated, a∝t2/3a \propto t^{2/3}, with t0=13.8×109t_0 = 13.8 \times 10^9 years (a number he admits he had forgotten to give):

Time of equality (estimate)
teq=t0 (1+zeq)−3/2≈75,000 yearst_\text{eq} = t_0\,(1 + z_\text{eq})^{-3/2} \approx 75{,}000\ \text{years}
In words

t sub eq equals t sub 0 (1 plus z sub eq ) to the power minus 3/2 is approximately 75,000 years

Source: 8.286 Lecture 15 transcript, pp. 19–20

This is only an estimate, with two known errors: it ignores the recent acceleration due to dark energy, and close to equality the radiation itself already affects the expansion. Ryden's textbook, including everything, gets 47,000 years — mostly different, Guth says, because she used h0=0.72h_0 = 0.72. Either way: the radiation era lasted tens of thousands of years, long compared with a human life but a tiny fraction of cosmic history.

Densities in units of today's critical density, on logarithmic axes, computed as Ω_m/a³ and Ω_r/a⁴. The radiation line is steeper; the lines cross where a = Ω_r/Ω_m. With Ω_m = 0.30 and Ω_r = 9.24 × 10⁻⁵ (h₀ = 0.67) that is a = 3.08 × 10⁻⁴, 1 + z = 3,245. Left of the crossing the universe was radiation-dominated.
Worked example

Equality with later measurements

Later measurements (Planck 2018, after this lecture) give h0=0.674h_0 = 0.674 and Ωm=0.315\Omega_m = 0.315. Redo Guth's estimate of zeqz_\text{eq} and teqt_\text{eq} with the same method and t0=13.8×109t_0 = 13.8 \times 10^9 years.

Radiation

Ωr=4.15×10−5/0.6742=9.14×10−5\Omega_r = 4.15 \times 10^{-5}/0.674^2 = 9.14 \times 10^{-5}.

Ratio and redshift

Ωr/Ωm=9.14×10−5/0.315=2.90×10−4\Omega_r/\Omega_m = 9.14 \times 10^{-5}/0.315 = 2.90 \times 10^{-4}, so 1+zeq=3,4481 + z_\text{eq} = 3{,}448 and zeq≈3,450z_\text{eq} \approx 3{,}450.

Time

teq=13.8×109×3,448−3/2t_\text{eq} = 13.8 \times 10^9 \times 3{,}448^{-3/2} years ≈68,000\approx 68{,}000 years. More matter means equality came earlier — but this is still the same rough matter-only estimate, good to an order of magnitude. (Planck's own full fit gives zeq≈3,390z_\text{eq} \approx 3{,}390.)

Matter: three equations, two independent#

To follow a radiation-dominated universe, the Friedmann equations must be adapted. (The details are an 8.286 homework problem; the lecture gives the logic.) For non-relativistic matter we have

Friedmann equations for matter
(1) (a˙a)2=8π3Gρ−kc2a2,(2) a¨=−4π3Gρ a\text{(1)}\ \left(\frac{\dot a}{a}\right)^2 = \frac{8\pi}{3}G\rho - \frac{kc^2}{a^2}, \qquad \text{(2)}\ \ddot a = -\frac{4\pi}{3}G\rho\,a
In words

(1) ( a dot over a ) squared equals 8 pi over 3 G rho minus kc squared over a squared , (2) a double dot equals minus 4 pi over 3 G rho a

Source: 8.286 Lecture 15 transcript, p. 22 (board)

together with ρ∝1/a3\rho \propto 1/a^3. Writing ρ=b/a3\rho = b/a^3 with a constant bb and differentiating:

How matter density changes
(3) ρ˙=−3ba4 a˙=−3 a˙a ρ\text{(3)}\ \dot\rho = -\frac{3b}{a^4}\,\dot a = -3\,\frac{\dot a}{a}\,\rho
In words

(3) rho dot equals minus 3b over a to the power 4 a dot equals minus 3 a dot over a rho

Source: 8.286 Lecture 15 transcript, p. 22 (board)

These three are not independent. Differentiate (1) in time, use (3) for ρ˙\dot\rho, and out comes (2); any two give the third. For radiation, ρ∝1/a4\rho \propto 1/a^4 turns (3) into

How radiation density changes
(3′) ρ˙=−4 a˙a ρ\text{(3}'\text{)}\ \dot\rho = -4\,\frac{\dot a}{a}\,\rho
In words

(3 prime ) rho dot equals minus 4 a dot over a rho

Source: 8.286 Lecture 15 transcript, p. 23 (board)

Now the set is inconsistent: if (3) changes, (1) or (2) must change too. Which one?

Pressure does work#

Shouldn't ρ˙\dot\rho be fixed by conservation of energy? It is — but energy conservation includes work done by pressure. A gas in a cylinder pushes on a piston with force pApA; if the piston moves out by dxdx, the gas does work pA dx=p dVpA\,dx = p\,dV and loses that energy:

Work done by an expanding gas
dU=−p dVdU = -p\,dV
In words

dU equals minus p dV

Source: 8.286 Lecture 15 transcript, p. 24 (board)

The formula holds for any container, however irregularly it grows. Dust at rest exerts no pressure; photons racing around do. Apply it to a box fixed in comoving coordinates, whose physical volume is V=a3(t) VcoordV = a^3(t)\,V_\text{coord} and whose energy is U=Vρc2U = V\rho c^2:

Energy balance of a comoving box
ddt(a3ρc2)=−p ddt(a3)⟹ρ˙=−3 a˙a(ρ+pc2)\frac{d}{dt}\left(a^3\rho c^2\right) = -p\,\frac{d}{dt}\left(a^3\right) \quad\Longrightarrow\quad \dot\rho = -3\,\frac{\dot a}{a}\left(\rho + \frac{p}{c^2}\right)
In words

d over dt (a cubed rho c squared ) equals minus p d over dt (a cubed ) which gives rho dot equals minus 3 a dot over a ( rho plus p over c squared )

Source: 8.286 Lecture 15 transcript, p. 25 (board)

With p=0p = 0 this is (3). To get the 4 of (3′), the pressure must be one third of the energy density:

Pressure of radiation
p=13 ρc2p = \tfrac13\,\rho c^2
In words

p equals 1 over 3 rho c squared

Source: 8.286 Lecture 15 transcript, pp. 25–26 (board)

which agrees with any other way of calculating the pressure of light.

Which Friedmann equation gets the pressure?#

A rigorous answer needs general relativity, but Guth motivates it. Equation (1), first order in time, is an energy equation, like 12mv2+V(r)=\frac12mv^2 + V(r) = constant; (2), second order, is a force equation.

Now imagine sticks of TNT spread through all of space, set off at the same cosmic time. The pressure jumps instantly; the energy density does not. If pressure appeared in (1), nothing else in it could jump to match, so (1) would be violated. If it appears in (2), the jump just changes a¨\ddot a suddenly — exactly as a sudden force changes a particle's acceleration. So the pressure belongs in (2). Requiring (1), (2′) and the new ρ˙\dot\rho equation to be consistent fixes the form:

Second Friedmann equation with pressure
(2′) a¨=−4π3 G(ρ+3pc2)a\text{(2}'\text{)}\ \ddot a = -\frac{4\pi}{3}\,G\left(\rho + \frac{3p}{c^2}\right)a
In words

(2 prime ) a double dot equals minus 4 pi over 3 G ( rho plus 3p over c squared )a

Source: 8.286 Lecture 15 transcript, p. 27 (board)

These are the Friedmann equations general relativity gives directly. [Guth momentarily says equation 1 must change before correcting himself to equation 2.] Read off the equation: a positive pressure adds to ρ\rho as a source of deceleration — it makes gravity stronger (a point Guth draws out in Lecture 16).

Summary#

  • Today radiation (photons + neutrinos) has ur=7.01×10−14u_r = 7.01 \times 10^{-14} J/m³, ρr=7.80×10−31\rho_r = 7.80 \times 10^{-31} kg/m³, Ωr=4.15×10−5h0−2≈9.2×10−5\Omega_r = 4.15 \times 10^{-5}h_0^{-2} \approx 9.2 \times 10^{-5}.
  • ρr/ρm∝1/a\rho_r/\rho_m \propto 1/a; with Ωm≈0.30\Omega_m \approx 0.30, equality was at zeq≈3,200z_\text{eq} \approx 3{,}200, roughly teq≈75,000t_\text{eq} \approx 75{,}000 years (Ryden: 47,000) — an order-of-magnitude estimate.
  • With pressure, energy conservation reads ρ˙=−3(a˙/a)(ρ+p/c2)\dot\rho = -3(\dot a/a)(\rho + p/c^2); radiation's 1/a41/a^4 needs p=13ρc2p = \frac13\rho c^2.
  • Pressure enters the second-order equation: a¨=−4π3G(ρ+3p/c2)a\ddot a = -\frac{4\pi}{3}G(\rho + 3p/c^2)a, while (a˙/a)2=8π3Gρ−kc2/a2(\dot a/a)^2 = \frac{8\pi}{3}G\rho - kc^2/a^2 is unchanged.

Symbols on this page

Show the 26 symbols used here
γ\gamma gamma
As a subscript: photons (u_γ, T_γ, n_γ). Say it: “gamma”.
π\pi pi
The ratio of a circle's circumference to its diameter, π ≈ 3.14159. Say it: “pi”.
ρ\rho rho
Density: mass (or energy divided by c²) per unit volume. ρ_c is the critical density, ρ̄ the mean density. Say it: “rho”.
Ω\Omega capital omega
The density parameter Ω = ρ/ρ_c. Subscripts pick one ingredient: Ω_m matter, Ω_vac (or Ω_Λ) vacuum energy, Ω_r radiation, Ω_k curvature. Say it: “omega”.
≈\approx approximately equal
Is approximately equal to. Say it: “is approximately”.
∝\propto proportional to
Is proportional to: doubling one doubles the other. Say it: “is proportional to”.
⇒\Rightarrow implies
Implies: what follows is a consequence of what came before. Say it: “which gives”.
×\times times
Times (multiplication). Between two vectors: the cross product. Say it: “times”.
a˙\dot a dot
Rate of change with time: ȧ = da/dt. Say it: “dot”.
a¨\ddot a double dot
Second rate of change with time: ä = d²a/dt². Say it: “double dot”.
aa a
The scale factor a(t): how much the universe has expanded. Physical distance = a × comoving distance (Guth's units: metres per notch). Say it: “a”.
HH H
The Hubble parameter H = ȧ/a: the expansion rate of the universe at a given time. Say it: “H”.
zz z
Redshift: 1 + z = a(observed)/a(emitted), how much the universe has stretched since the light left. Say it: “z”.
GG G
Newton's gravitational constant, G ≈ 6.674 × 10⁻¹¹ m³ kg⁻¹ s⁻². Say it: “G”.
cc c
The speed of light, c ≈ 2.998 × 10⁸ m/s. Say it: “c”.
kk k
The curvature constant: k > 0 closed, k = 0 flat, k < 0 open (Guth's units: 1/notch²). Say it: “k”.
H0H_{0} Hubble constant
The Hubble constant: today's value of H, about 67–73 km/s/Mpc. Say it: “H nought”.
h0h_{0} little h
H₀ in units of 100 km/s/Mpc: h₀ ≈ 0.67–0.7. Say it: “h nought”.
t0t_{0} today
Today: the present time, about 13.8 billion years after the Big Bang in the real universe (simple models on some pages give other ages). Say it: “t nought”.
zeqz_{\text{eq}} redshift of equality
The redshift of matter–radiation equality, about 3400. Say it: “z equality”.
teqt_{\text{eq}} time of equality
The time of matter–radiation equality, some 50 000–75 000 years after the Big Bang. Say it: “t equality”.
ρc\rho_{c} critical density
The critical density, 3H²/(8πG): the density of a flat universe (about 9 × 10⁻²⁷ kg/m³ today). Say it: “rho c”.
ρm\rho_{m} matter density
The mass density of matter (dark matter and baryons). Say it: “rho m”.
ρr\rho_{r} radiation density
The mass density of radiation (its energy density divided by c²). Say it: “rho r”.
Ωm\Omega_{m} matter density parameter
The matter density as a fraction of the critical density, ρ_m/ρ_c; ',0' means today (about 0.3). Say it: “omega m”.
Ωr\Omega_{r} radiation density parameter
The radiation density as a fraction of the critical density (about 9 × 10⁻⁵ today). Say it: “omega r”.

Tap or click a symbol in a formula to see the same explanation. All symbols: symbols and notation.

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Check your understanding

  1. Today ρr/ρm≈3.1×10−4\rho_r/\rho_m \approx 3.1 \times 10^{-4}. Why does this ratio grow as we look back in time?
  2. Using today's ratio 3.1×10−43.1 \times 10^{-4}, what is the redshift of matter–radiation equality?
  3. When a gas expands by dVdV, what happens to its energy UU?
  4. What pressure turns the 3 in ρ˙=−3(a˙/a)(ρ+p/c2)\dot\rho = -3(\dot a/a)(\rho + p/c^2) into the 4 for radiation?
  5. Why does pressure appear in the a¨\ddot a equation and not in the a˙2\dot a^2 equation?
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