Mathematical physics

The Friedmann equations and relativistic cosmology

How a single number, the scale factor $a(t)$ , describes the size of the entire universe, why Einstein's field equations reduce for a homogeneous, isotropic cosmos to two ordinary differential equations governing $a(t)$ , and how these Friedmann equations let us compute the age, geometry, and ultimate fate of the universe from measured densities of matter, radiation, and dark energy.

IntuitionA loaf of raisin bread with no edge and no center

Imagine raisins baked into a loaf of bread, and picture the loaf rising in the oven: every raisin moves away from every other raisin, and the farther apart two raisins start, the faster they separate — yet no raisin is special, none sits at a 'center' of the expansion. This is exactly how galaxies recede from one another as the universe expands: space itself stretches uniformly everywhere, carrying galaxies apart, rather than galaxies flying outward from some central explosion into pre-existing empty space. The single number tracking how much the 'loaf' has risen since some reference time is the scale factor $a(t)$ , and the whole of relativistic cosmology reduces to finding the differential equation that governs it.

A curved surface standing in for the spatial geometry a Friedmann–Lemaître–Robertson–Walker universe can have: dragging the scale slider mimics uniformly stretching the whole surface, the way

a(t)

stretches every comoving distance in the real universe, without any point on the surface being a preferred center.

SchoolThe scale factor and Hubble's law

Definition: Scale factor, Hubble parameter, redshift

Write the physical distance between two galaxies at rest with respect to the expanding background ('comoving' galaxies) as $d(t) = a(t)\,d_{0}$ , where $a(t)$ is the dimensionless scale factor (normalized so $a(t_0)=1$ today) and $d_0$ is their fixed comoving distance. Differentiating, $\dot d = \dot a\, d_0 = \dfrac{\dot a}{a}\,d$ , which is exactly Hubble's law $v=H(t)d$ with the Hubble parameter $H(t) = \dfrac{\dot a(t)}{a(t)}$ . Light emitted at time $t_{\text{em}}$ and observed today at $t_0$ is stretched by the same factor the universe has expanded, giving the redshift $1+z = \dfrac{a(t_{0})}{a(t_{\text{em}})}$ .

H(t) = \dfrac{\dot a(t)}{a(t)}

The best current measurement gives $H_{0}\approx 70\ \text{km/s/Mpc}$ (megaparsecs), meaning galaxies $1$ megaparsec away recede at roughly $70$ km/s — but the value $H_0$ today is only the boundary condition; the whole point of the Friedmann equations below is to say how $H(t)$ , and hence $a(t)$ , evolves.

Spatial curvature and the sign of $k$
$k$	Spatial geometry	Total volume	Angle sum of a triangle
$k=+1$	3-sphere (closed, positively curved)	Finite	$>180^\circ$
$k=0$	Flat Euclidean space	Infinite	$=180^\circ$
$k=-1$	Hyperbolic space (open, negatively curved)	Infinite	$<180^\circ$

UndergraduateDeriving the two Friedmann equations

Theorem: First Friedmann equation

For a homogeneous, isotropic universe with mass-energy density $\rho(t)$ , cosmological constant $\Lambda$ , and spatial curvature index $k\in\{-1,0,+1\}$ , the scale factor obeys $H^{2} = \left(\dfrac{\dot a}{a}\right)^{2} = \dfrac{8\pi G}{3}\rho - \dfrac{kc^{2}}{a^{2}} + \dfrac{\Lambda c^{2}}{3}$ .

Why is it true?

The full derivation plugs the Friedmann–Lemaître–Robertson–Walker metric into Einstein's field equations, but a remarkable Newtonian argument (Milne & McCrea, 1934) reproduces exactly the same equation (setting $\Lambda=0$ ): treat a test galaxy on the surface of an expanding sphere of comoving matter as an ordinary projectile in the gravitational field of everything enclosed inside it, and apply energy conservation. Gravity outside a uniform sphere behaves as if all its mass sat at the center (a Newtonian shell theorem), so only the enclosed mass matters — matching the way general relativity only lets locally enclosed mass-energy curve spacetime at a point in a homogeneous universe.

Proof

Fix a comoving test galaxy at comoving radius $r_0$ from an arbitrary origin, so its physical distance is $R(t)=a(t)r_0$ and its physical velocity is $\dot R = \dot a\, r_0$ . By the Newtonian shell theorem, the gravitational pull it feels comes only from the mass $M$ enclosed within radius $R$ , and since no comoving matter crosses the comoving sphere of radius $r_0$ as the universe expands, $M = \dfrac{4}{3}\pi (a r_{0})^{3}\rho$ is constant in time (for pressureless matter). Treat the test galaxy, of mass $m$ , as a projectile: its total mechanical energy $\dfrac12 \dot a^{2} r_{0}^{2} - \dfrac{GM m}{a r_{0}} = E$ is conserved.

Substitute $M$ : $\dfrac12 \dot a^2 r_0^2 - \dfrac{G m}{a r_0}\cdot\dfrac{4}{3}\pi (ar_0)^3\rho = E$ , which simplifies to $\dfrac12\dot a^2 r_0^2 - \dfrac{4}{3}\pi G m \rho\, a^2 r_0^2 = E$ . Divide through by $\tfrac12 m a^2 r_0^2$ : $\left(\dfrac{\dot a}{a}\right)^2 - \dfrac{8\pi G}{3}\rho = \dfrac{2E}{m a^2 r_0^2}$ .

The right-hand side must be independent of the arbitrary radius $r_0$ chosen (the equation has to hold for every comoving observer, and $a(t)$ itself does not depend on $r_0$ ), so $\dfrac{2E}{mr_0^2}$ is a constant of the test galaxy's orbit that can only depend on the fixed comoving coordinate structure — define this constant by $\dfrac{2E}{m r_{0}^{2}} \equiv -kc^{2}$ , giving $H^{2} = \dfrac{8\pi G}{3}\rho - \dfrac{kc^{2}}{a^{2}}$ . A separate thermodynamic argument (treating the cosmological constant as a fluid with constant energy density $\rho_\Lambda = \Lambda c^2/(8\pi G)$ and substituting it into this same formula) restores the $\Lambda c^2/3$ term, giving exactly $H^{2} = \left(\dfrac{\dot a}{a}\right)^{2} = \dfrac{8\pi G}{3}\rho - \dfrac{kc^{2}}{a^{2}} + \dfrac{\Lambda c^{2}}{3}$ — and remarkably, the full general-relativistic calculation from Einstein's field equations produces this identical equation, with $k$ now properly identified as the actual sign of spatial curvature rather than just an integration constant.

Theorem: Second Friedmann (acceleration) equation

Under the same hypotheses, if in addition the density and pressure obey the fluid (continuity) equation $\dot\rho + 3\dfrac{\dot a}{a}\left(\rho + \dfrac{p}{c^{2}}\right) = 0$ , then $\dfrac{\ddot a}{a} = -\dfrac{4\pi G}{3}\left(\rho + \dfrac{3p}{c^{2}}\right) + \dfrac{\Lambda c^{2}}{3}$ .

Why is it true?

Unlike Newtonian gravity, in general relativity pressure itself gravitates alongside energy density — a subtle effect invisible in the first Friedmann equation, which only involves $\rho$ . The acceleration equation reveals this: ordinary matter and radiation, having positive pressure, always decelerate the expansion via the $+3p/c^2$ term, while something with sufically negative pressure, $p<-\rho c^2/3$ , flips the sign and drives $\ddot a>0$ — cosmic acceleration, exactly what dark energy (modeled here by $\Lambda$ ) does.

Proof

Start from the first Friedmann equation in the form $\dot a^2 = \dfrac{8\pi G}{3}\rho a^2 - kc^2 + \dfrac{\Lambda c^2}{3}a^2$ and differentiate both sides with respect to $t$ : $2\dot a\ddot a = \dfrac{8\pi G}{3}\left(\dot\rho a^2 + 2\rho a\dot a\right) + \dfrac{2\Lambda c^2}{3}a\dot a$ .

Divide every term by $2a\dot a$ (valid whenever $\dot a\neq0$ ): $\dfrac{\ddot a}{a} = \dfrac{4\pi G}{3}\left(\dfrac{\dot\rho\, a}{\dot a} + 2\rho\right) + \dfrac{\Lambda c^2}{3}$ .

Now eliminate $\dot\rho$ using the fluid equation $\dot\rho + 3\dfrac{\dot a}{a}\left(\rho + \dfrac{p}{c^{2}}\right) = 0$ , rearranged as $\dfrac{\dot\rho\,a}{\dot a} = -3\left(\rho+\dfrac{p}{c^2}\right)$ . Substituting: $\dfrac{\ddot a}{a} = \dfrac{4\pi G}{3}\left(-3\rho-\dfrac{3p}{c^2}+2\rho\right)+\dfrac{\Lambda c^2}{3} = \dfrac{4\pi G}{3}\left(-\rho-\dfrac{3p}{c^2}\right)+\dfrac{\Lambda c^2}{3}$ .

This is exactly $\dfrac{\ddot a}{a} = -\dfrac{4\pi G}{3}\left(\rho + \dfrac{3p}{c^{2}}\right) + \dfrac{\Lambda c^{2}}{3}$ , as claimed. Notice the fluid equation itself is nothing more than the first law of thermodynamics $d(\rho c^2 a^3) = -p\,d(a^3)$ applied to an expanding comoving volume: energy inside a comoving patch changes only because the pressure does work as the patch's volume grows.

AdvancedHow different contents dilute, and the critical density

Writing an equation of state $p = w\rho c^{2}$ relating pressure to density lets the fluid equation be solved directly: pressureless matter ( $w=0$ , e.g. galaxies, dark matter) dilutes as $\rho_{m}\propto a^{-3}$ — pure volume dilution — while radiation ( $w=\tfrac13$ , e.g. photons) dilutes faster, $\rho_{r}\propto a^{-4}$ , because each photon is also redshifted, losing energy as space stretches. A cosmological constant ( $w=-1$ ) does not dilute at all, $\rho_{\Lambda}\propto a^{0}$ , so it inevitably dominates at late times no matter how small it starts. The density at which the universe is exactly spatially flat ( $k=0$ ) is the critical density $\rho_{c}(t) = \dfrac{3H(t)^{2}}{8\pi G}$ , and $\Omega \equiv \dfrac{\rho}{\rho_{c}}$ tells you at a glance whether space is closed ( $\Omega>1$ ), flat ( $\Omega=1$ ), or open ( $\Omega<1$ ).

\rho_{c}(t) = \dfrac{3H(t)^{2}}{8\pi G}

UndergraduateReal-World Applications and Worked Examples

The Friedmann equations are not abstract bookkeeping: astronomers use them daily to convert a handful of measured densities and $H_0$ into the age of the universe, to predict how galaxy counts should scale with redshift, to model Big Bang nucleosynthesis in the radiation-dominated era, and to test whether dark energy is truly a constant $\Lambda$ or a dynamical field by comparing predicted versus observed expansion histories from supernovae and the cosmic microwave background.

Example: The age of a flat, matter-only universe

Assume a spatially flat ( $k=0$ ), matter-only universe ( $\Lambda=0$ , $p=0$ ) with $\rho_{m}\propto a^{-3}$ . Show that $a(t)\propto t^{2/3}$ , and use $H_{0}\approx 70\ \text{km/s/Mpc}$ (i.e. $H_0 \approx 2.27\times10^{-18}\ \text{s}^{-1}$ ) to estimate the age this simplified model predicts.

Solution

With $k=0$ and $\Lambda=0$ , the first Friedmann equation reduces to $\left(\dfrac{\dot a}{a}\right)^2 = \dfrac{8\pi G}{3}\rho$ . Since $\rho\propto a^{-3}$ , write $\rho = \rho_0 a^{-3}$ (with $\rho_0$ the density today, $a_0=1$ ), so $\dot a^2 = \dfrac{8\pi G\rho_0}{3}\,a^{-1}$ , i.e. $\dot a = \sqrt{\dfrac{8\pi G\rho_0}{3}}\,a^{-1/2}$ .

This is separable: $a^{1/2}\,da = \sqrt{\dfrac{8\pi G\rho_0}{3}}\,dt$ . Integrating from $a=0$ at $t=0$ (the Big Bang) to $a(t)$ at time $t$ gives $\dfrac23 a^{3/2} = \sqrt{\dfrac{8\pi G\rho_0}{3}}\,t$ , i.e. $a(t) = \left(\text{const}\right)\cdot t^{2/3}$ , confirming $a(t)\propto t^{2/3}$ .

To relate this to $H_0$ : from $a\propto t^{2/3}$ , $\dot a\propto \tfrac23 t^{-1/3}$ , so $H(t)=\dot a/a = \tfrac23 t^{-1}$ . Evaluating today ( $t=t_0$ ) gives $H_0 = \dfrac{2}{3t_0}$ , i.e. $t_{0} = \dfrac{2}{3H_{0}} \approx 9.3\times10^{9}\ \text{yr}$ .

Numerically, $t_0 = \dfrac{2}{3\times 2.27\times10^{-18}\ \text{s}^{-1}} \approx 2.94\times10^{17}\ \text{s} \approx 9.3$ billion years. This flat matter-only estimate is noticeably younger than the real $\Lambda$ CDM age of $t_{0}\approx 13.8\times10^{9}\ \text{yr}$ — the missing piece is dark energy, which accelerated the expansion in the second half of cosmic history and so requires slightly more elapsed time to reach today's $H_0$ than a purely decelerating matter universe would.

Example: Reading off the fate of the universe from $\Omega$

Suppose a hypothetical matter-only, $\Lambda=0$ universe today has total density parameter $\Omega_0=\rho_0/\rho_c=1.5$ (i.e. above critical). Using the first Friedmann equation, determine the sign of $k$ and explain qualitatively what eventually happens to $a(t)$ .

Solution

Rewrite the first Friedmann equation (with $\Lambda=0$ ) at $t_0$ ( $a_0=1$ ) by dividing through by $H_0^2$ : $1 = \dfrac{\rho_0}{\rho_c} - \dfrac{kc^2}{a_0^2 H_0^2} = \Omega_0 - \dfrac{kc^2}{H_0^2}$ , so $\dfrac{kc^2}{H_0^2} = \Omega_0-1$ .

Since $\Omega_0=1.5>1$ , the right side is positive, and since $c^2/H_0^2>0$ , this forces $k>0$ , i.e. $k=+1$ : the universe is a positively-curved, spatially closed 3-sphere.

Qualitatively (from the theorem 2 acceleration equation with $p=0$ , $\Lambda=0$ ): $\ddot a = -\dfrac{4\pi G}{3}\rho\, a<0$ always, so the expansion is perpetually decelerating, exactly as gravity pulling all the matter back together would suggest. For $k=+1$ with no dark energy to counteract it, $H^2 = \tfrac{8\pi G}{3}\rho - c^2/a^2$ must eventually hit $H=0$ at some maximum $a_{\max}$ (since $\rho\propto a^{-3}\to0$ while $c^2/a^2\to0$ more slowly, the curvature term wins at large $a$ ), after which the universe recollapses toward a 'Big Crunch' — the closed, matter-dominated analogue of a ball thrown up that must eventually fall back down.

Common mistake. Two classic mistakes with the Friedmann equations. (1) Confusing recession velocity with motion through space. Hubble's law

v=Hd

can give

v>c

for sufficiently distant galaxies, which looks like it violates relativity — but

v

here is the rate at which space itself stretches between two points, not a velocity of any object moving through local space, so special relativity's speed limit (which bounds only local relative velocities) is never actually violated. (2) **Assuming

\Omega_0>1

(closed universe) automatically means eventual recollapse.** That conclusion in the second worked example above relies on

\Lambda=0

; with a positive cosmological constant, even a spatially closed (

k=+1

) universe can expand forever once dark energy dominates, since the

+\Lambda c^2/3

term in the acceleration equation never dilutes away and eventually overwhelms the decelerating pull of matter — the sign of

k

alone does not determine the ultimate fate once

\Lambda\neq0

Research frontier as of 2026

The Friedmann framework is textbook physics, but two open problems sit squarely inside it. First, the Hubble tension: measuring $H_0$ from the early universe (cosmic microwave background, assuming $\Lambda$ CDM) gives $\approx 67.4$ km/s/Mpc, while measuring it from the late-time 'distance ladder' (Cepheids and supernovae, via the SH0ES collaboration) gives $\approx 73$ km/s/Mpc — a persistent $\sim5\sigma$ discrepancy as of 2026 that has survived years of systematic-error hunting on both sides, and which may signal new physics beyond the standard $\Lambda$ CDM cosmological model (extra relativistic species, early dark energy, or modified gravity) rather than a measurement error. Second, the nature of dark energy: the Dark Energy Spectroscopic Instrument (DESI) collaboration's 2024–2025 baryon acoustic oscillation results showed mild but growing evidence, when combined with supernova and CMB data, that the dark energy equation of state $w$ may not be the constant $w=-1$ of a true cosmological constant but instead evolves slightly with time — as of 2026 this remains actively debated and is a central target of ongoing DESI data releases and the Euclid and Roman Space Telescope missions, which aim to map the expansion history precisely enough to distinguish a true $\Lambda$ from a dynamical dark energy field.

In a spatially flat ( $k=0$ ), $\Lambda=0$ universe with only pressureless matter, the first Friedmann equation gives $H^2 \propto a^{-3}$ . What is $a(t)$ proportional to?

Radiation density dilutes as $\rho_r\propto a^{-4}$ rather than matter's $a^{-3}$ . What is the extra factor of $a^{-1}$ physically due to?

Measured densities today give $\Omega_0 = 1.0$ (spatially flat). What does the first Friedmann equation say about $k$ ?

Which content, if present today with sufficient density, would be the only one able to make $\ddot a>0$ (accelerating expansion)?

References

John A. Peacock (1999). Cosmological Physics
Scott Dodelson, Fabian Schmidt (2020). Modern Cosmology
DESI Collaboration (2024). DESI 2024 VI: cosmological constraints from the measurements of baryon acoustic oscillations
NASA/JPL Cosmology Group (2024). Hubble Tension