The Equation of State of the Universe

A ball of dust

Fill space with a uniform, pressureless “dust” of density \(\rho(t)\), and let it expand uniformly, which means every distance grows by the same factor. Write the physical distance of a particle from any chosen centre as

where \(x\) is a fixed label for that particle, its comoving distance, and \(a(t)\) is the scale factor, normalised to \(a = 1\) today. The expansion rate is the Hubble parameter,

Pick any point as the centre and draw a sphere through a test particle of mass \(m\) at radius \(R\). Newton's shell theorem says a uniform spherical shell exerts no net force on anything inside it, so all the mass outside the sphere drops out. Only the mass inside pulls:

No dust crosses the surface, because everything moves with the flow, so \(M\) stays constant even as \(\rho\) falls and \(R\) grows.

Newton's second law for the test particle is

Multiply both sides by \(\dot R\) and integrate once. The left side becomes the time derivative of \(\tfrac12\dot R^2\) and the right side the time derivative of \(GM/R\), so what comes out is energy conservation per unit mass:

Now substitute \(R = ax\) and \(M = \tfrac43\pi\rho a^3x^3\), and divide through by \(\tfrac12 a^2x^2\):

The left side and the first term on the right are the same for every particle in the universe. For the expansion to stay uniform, \(U/x^2\) has to be the same for every particle too. Call that constant \(-kc^2/2\), and the result is the Friedmann equation:

This is exactly the equation general relativity gives for a homogeneous, isotropic universe. Friedmann found it in 1922 by solving Einstein's equations. Milne and McCrea showed in 1934 that Newtonian gravity gets there too. This is a heuristic rather than a stand-alone derivation, because its justification comes from relativity (see below). But for dust it lands on exactly the right equation. What relativity changes is not the equation but what the symbols mean.

Why the sphere is allowed. For a finite ball, the shell theorem is cleanly true. In infinite space, Newtonian gravity alone cannot settle the question. The force from an infinite uniform medium is not well defined: the answer depends on the order in which you add up the mass (the Seeliger–Neumann paradox). General relativity resolves this. In any spherically symmetric spacetime, a shell's motion is set only by the mass it encloses (the Misner–Sharp mass; for dust, the Lemaître–Tolman–Bondi solution), whatever lies outside. Birkhoff's theorem is the special case for empty space, which is not quite our situation, because the sphere is full of matter. Equivalently, a ball much smaller than the Hubble radius, filled with slow-moving matter, obeys Newtonian gravity to good accuracy, and homogeneity carries the result to every scale. So the Newtonian argument gives the right answer, but it is relativity that justifies it.

What the constant \(k\) is

In Newtonian gravity, \(k\) is just the sign of the energy \(U\), and its meaning is the familiar one for a ball thrown upward. In relativity the same constant is the curvature of space. The fates agree, at least for dust.

Setting \(k = 0\) defines the critical density, \(\rho_\mathrm{c} = 3H^2/8\pi G\). With \(H_0 = 67.4\ \mathrm{km\,s^{-1}\,Mpc^{-1}}\) it is \(8.5 \times 10^{-27}\ \mathrm{kg\,m^{-3}}\). As an energy density that is \(\rho_\mathrm{c}c^2 = 4.8\ \mathrm{GeV\,m^{-3}}\), and since a hydrogen atom weighs \(0.94\ \mathrm{GeV}\), it comes to about five hydrogen atoms per cubic metre. Dividing the Friedmann equation by \(H^2\) and writing each density as a fraction \(\Omega = \rho/\rho_\mathrm{c}\) turns the Friedmann equation into a budget: the \(\Omega\)s plus a curvature term add up to one. The curvature term is measured to be below about 0.002 today, so from here on the universe is flat and \(k = 0\).

Adding thermodynamics

The Friedmann equation gives \(H\) once you know \(\rho\), but says nothing about how \(\rho\) changes. For dust it is easy: the mass in a comoving sphere is fixed, so \(\rho \propto a^{-3}\). For anything with pressure, the first law of thermodynamics is needed.

Take a comoving sphere of volume \(V \propto a^3\) holding energy \(E = \rho c^2 V\). In a uniform universe there is nowhere for heat to flow, so the expansion is adiabatic and

Write this as a rate, using \(\dot V/V = 3\dot a/a\):

This is the fluid equation, also called the continuity equation. If the contents had no pressure, the energy in a comoving volume would stay fixed and the density would simply dilute as \(1/V \propto a^{-3}\). A gas with positive pressure also does work on its surroundings as it expands, so it loses energy while it spreads out, and its density falls faster than \(a^{-3}\). The \(P\) term in the fluid equation is that work. It is also the route by which pressure enters gravity in this derivation: it carries \(P\) into the acceleration equation below.

Mass quietly becomes energy. In this step \(\rho\) changes meaning. It is now the total energy density divided by \(c^2\), including heat and radiation, not just the rest mass of the dust. Newtonian gravity has no reason to do this, but relativity insists on it, because all energy gravitates.

Pressure gravitates

Newton's law on its own gives

so gravity always slows the expansion. Now differentiate the Friedmann equation with respect to time, use the fluid equation to replace \(\dot\rho\), and use the Friedmann equation once more to tidy up. The \(k\) term cancels, and the result is the acceleration equation, often called the second Friedmann equation:

Why pressure doesn't just push. Radiation pushes outward on a wall and dark energy pulls inward on one. But a uniform universe has no walls: every bit of gas is pressed equally from all sides, so pressure exerts no net force on the expansion. Its only effect is through gravity, and there the signs reverse. Radiation's positive pressure adds to the attraction and slows the expansion more, while dark energy's tension makes gravity repel and speeds it up. For radiation the pressure term equals the energy term, so a gas of photons attracts twice as strongly as dust with the same energy density.

Only two of the Friedmann, fluid and acceleration equations are independent: any two give the third. That is why this derivation worked.

The equation of state

One more relation is needed to close the system, between pressure and density. For every ingredient that matters in cosmology it is a straight proportion,

and the dimensionless number \(w\) is the equation of state. Put it into the fluid equation and integrate:

Put that into the flat Friedmann equation and integrate again:

And the acceleration equation becomes \(\ddot a/a = -\tfrac43\pi G\rho\,(1 + 3w)\). The expansion accelerates exactly when \(w < -\tfrac13\).

Radiation's 1/3 comes straight from kinetic theory. A photon carries momentum \(p = E/c\), and an isotropic gas of particles exerts pressure \(\tfrac13\,n\,p\,v\), so \(P = \tfrac13\rho c^2\). It checks out independently: the number of photons dilutes as \(a^{-3}\) and each one's energy is redshifted as \(a^{-1}\), giving \(a^{-4}\), just as \(\rho \propto a^{-3(1+w)}\) says with \(w = \tfrac13\).

A cosmological constant is the opposite extreme. Its density does not dilute at all, so the fluid equation forces \(\rho + P/c^2 = 0\), and \(w = -1\). That is a negative pressure, a tension. Stretching a box of it creates energy, at exactly the rate \(-P\,dV\) supplies.

Constant \(H\) means exponential expansion

If the density does not dilute, the Friedmann equation with \(k = 0\) makes \(H\) a constant. The definition of \(H\) then reads

This is the one case where the scale factor grows exponentially, like money at a fixed interest rate. It is also where a common picture goes wrong. A constant expansion rate does not mean coasting. \(H\) is a fractional rate: every distance grows by the same fraction each second, so a galaxy's speed \(v = HR\) grows as its distance grows. The acceleration equation agrees, since here \(\ddot a/a = H^2 > 0\). Constant \(H\) is accelerating expansion.

Only a cosmological constant can do this. Matter and radiation dilute as space grows, so with them alone \(H\) keeps falling, as \(2/3t\) for matter and \(1/2t\) for radiation. Any dark energy with \(w \neq -1\) changes too.

If dark energy is a true cosmological constant, this is where our universe is heading. As matter thins out, only the cosmological constant is left, and \(H\) settles to \(H_0\sqrt{\Omega_\Lambda}\), so \(1/H = 17.5\ \mathrm{Gyr}\). Distances will then double every \(\ln 2/H \approx 12\ \mathrm{Gyr}\). A galaxy farther away than \(c/H \approx 17.5\) billion light years will recede faster than light, and light it emits will never reach us. The same mathematics, with \(H\) larger by some fifty powers of ten, describes inflation in the first instant after the big bang.

But DESI's data hint that dark energy may not be constant. If \(w\) drifts, \(H\) never settles. Above \(-1\) the expansion rate keeps slowing, above \(-\tfrac13\) the acceleration ends altogether, and below \(-1\) the expansion rate grows without limit. Which future we get is a question the next round of measurements can start to answer.

Every curve passes through \(a = 1\) today with the same slope, because they share today's \(H_0\). They differ in the past. The more negative \(w\) is, the more slowly the universe has been slowing down, and the further back the big bang sits: \(2/3(1+w)\) Hubble times ago. At \(w = -1\) there is no big bang at all, and the curve runs back forever.

The universe's own \(w\)

The real universe is a mixture, and each ingredient dilutes at its own rate. Whichever has the most negative \(w\) wins in the end. The equation of state of the whole universe is the energy-weighted average, and it sets the deceleration parameter \(q \equiv -\ddot a a/\dot a^2\):

Today's budget is \(\Omega_\mathrm{m} = 0.315\) in matter, \(\Omega_\mathrm{r} \approx 9.2 \times 10^{-5}\) in photons and neutrinos, and the rest, about 0.685, in dark energy. The Friedmann equation becomes

with \(f_\mathrm{DE} = 1\) for a cosmological constant. If dark energy is not a constant, the standard way to let it evolve is \(w(a) = w_0 + w_a(1 - a)\). Putting that into the fluid equation and integrating gives

That is the model DESI fits. Combined with the microwave background and supernovae, DESI's second data release prefers \(w_0 > -1\) and \(w_a < 0\) over a cosmological constant, at between about 2.8 and \(4.2\sigma\) depending on which supernova sample is used. The panel below has both.

Read the lower panel from left to right. The universe began radiation-dominated, with \(w_\mathrm{eff} = \tfrac13\). Matter overtook radiation at a redshift of about 3400, some fifty thousand years in, and \(w_\mathrm{eff}\) sank to nearly zero, where it stayed for most of cosmic history while galaxies formed. Only about six billion years ago did it cross \(-\tfrac13\), and the expansion began to speed up. With a cosmological constant it now heads steadily toward −1 and exponential expansion.

Press the DESI button and the history changes in an interesting way. The dark energy is phantom in the past, with \(w < -1\), and has drifted above −1 in the last few billion years. It is less dense early on and more important late, which is what shifts the measured distances. The far right of the panel is the formula extrapolated well past the data, so take the future it shows as illustration rather than prediction. The dark energy in DESI DR2 is analysed in detail in K. Lodha et al. (DESI Collaboration), “Extended dark energy analysis using DESI DR2 BAO measurements,” Phys. Rev. D 112, 083511 (2025), arXiv:2503.14743.

Natural units

Particle physicists and cosmologists usually set \(\hbar = c = 1\), and Boltzmann's constant \(k = 1\) as well. Then mass, energy, temperature, inverse length and inverse time are all measured in one unit, the electron volt. Three numbers take you back to everyday units:

The first turns an energy into an inverse length: \(1\ \mathrm{eV^{-1}} = 1.973 \times 10^{-7}\ \mathrm{m}\). Dividing it by the second gives \(\hbar = 6.58 \times 10^{-16}\ \mathrm{eV\,s}\), which turns an energy into an inverse time. The third turns an energy into a temperature: \(1\ \mathrm{eV}\) is \(11\,600\ \mathrm{K}\).

The Hubble constant is a rate, \(H_0 = 67.4\ \mathrm{km\,s^{-1}\,Mpc^{-1}} = 2.18 \times 10^{-18}\ \mathrm{s^{-1}}\), so multiply it by \(\hbar\). There is no need to remember \(\hbar\) separately: \(\hbar c\) is your friend, and if you want \(\hbar\), just divide by \(c\):

Newton's constant has dimensions of \(\mathrm{energy}^{-2}\), and it defines the Planck mass, \(M_\mathrm{Pl} = \sqrt{\hbar c/G} = 1.22 \times 10^{19}\ \mathrm{GeV}\), so that \(G = 1/M_\mathrm{Pl}^2 = 6.71 \times 10^{-39}\ \mathrm{GeV^{-2}}\). The critical density is then

To get back to an energy per volume, divide by \((\hbar c)^3\): \(3.7 \times 10^{-11}\ \mathrm{eV^4} / (1.973 \times 10^{-7}\ \mathrm{eV\,m})^3 = 4.8\ \mathrm{GeV\,m^{-3}}\), the same value as before.

The same bookkeeping measures how strong a force is. The electric force between two protons is \(F = (e^2/4\pi\varepsilon_0)/r^2\), and the strength in the numerator has the units of \(\hbar c\):

The fine-structure constant \(\alpha\) is the strength of electromagnetism with the units taken out. In natural units, with \(\varepsilon_0 = 1\) as well, it is simply \(\alpha = e^2/4\pi\), so the charge of the proton is the pure number \(e = 0.303\).

Gravity can be written the same way. Newton's force between two masses is \(F = Gm^2/r^2\), and dividing its strength by \(\hbar c\) gives the gravitational counterpart of \(\alpha\):

Both forces fall as \(1/r^2\), so their ratio does not depend on distance: \(\alpha/\alpha_G = 1.2 \times 10^{36}\) for two protons, and \(4.2 \times 10^{42}\) for two electrons.

The Planck mass is to gravity what the charge \(e/\sqrt\alpha \approx 11.7\,e\) is to electromagnetism: the value at which the force's strength, divided by \(\hbar c\), reaches one. The electron's charge falls short of that by a factor of 12, but a proton's mass falls short of the Planck mass by \(10^{19}\). Gravity is weak between particles because their masses are tiny on gravity's own scale.

Gravity is by far the weakest force, yet it is the one that governs the expansion. Electric charges come in both signs and cancel, so on large scales the universe is neutral. Mass comes in only one sign, so gravity adds up.

The dark energy row is the one that puzzles physicists. Compare it with the energy density built from gravity's own scale, the Planck mass, alone:

Both are energy densities, in the same units, and they differ by a factor of \(10^{123}\). If the energy of empty space were set by gravity, the Planck value is what one would expect. The measured value is smaller by 123 powers of ten, and no one knows why. This is the cosmological constant problem.

The same equations from Einstein

The Newtonian derivation needed three patches from relativity: \(\rho\) had to count all energy, pressure had to gravitate, and \(k\) had to be curvature. General relativity delivers all three at once. The full calculation is a page of index algebra, but what goes in and what comes out are short.

What goes in. First, the geometry. Relativity measures it with the spacetime interval \(ds^2\), the squared length of a small step \(dx^\mu = (c\,dt,\, dx,\, dy,\, dz)\) in space and time. In flat, empty spacetime it is

and every observer, however they move, gets the same value. The minus sign is what makes time different from space: a step along a light ray has \(ds^2 = 0\). In general the interval is written \(ds^2 = g_{\mu\nu}\,dx^\mu dx^\nu\), summed over repeated indices, and the coefficients \(g_{\mu\nu}\) are the metric. In flat spacetime \(g_{00} = -1\) and \(g_{11} = g_{22} = g_{33} = +1\). If the universe is the same everywhere and in every direction, the most general interval is

This is the Friedmann–Lemaître–Robertson–Walker metric. Everything about it is fixed by symmetry except one function of time, \(a(t)\), and one constant, \(k\). Here \(k\) is plainly the curvature of space: flat for \(k = 0\), closed like the surface of a sphere for \(k > 0\), and open for \(k < 0\).

Second, the contents. A uniform fluid is described by its energy-momentum tensor. In a small region, where spacetime looks flat, and in a frame moving with the fluid, it is

The first entry is the energy density, all of it: rest mass, heat and radiation alike. The other three are the flow of momentum across a surface in each direction, which is the pressure.

Third, Einstein's equations, which say that the curvature of spacetime is set by its contents:

The left side, the Einstein tensor, is built from the metric in three steps. Each has a Newtonian counterpart, because for weak gravity \(g_{00} \approx -(1 + 2\Phi/c^2)\): the metric plays the part of the potential \(\Phi\).

• First derivatives: the force. The Christoffel symbols say how the coordinate grid stretches and turns from point to point,
  \[ \begin{aligned} \Gamma^\lambda_{\mu\nu} = \tfrac12\, g^{\lambda\sigma}\big(&\partial_\mu g_{\sigma\nu} + \partial_\nu g_{\sigma\mu} \\ & - \partial_\sigma g_{\mu\nu}\big) . \end{aligned} \]
  A freely falling body follows them, and for weak gravity \(\Gamma^i_{00} \approx \partial_i\Phi/c^2\), which is Newton's gravitational acceleration.
• Second derivatives: the tides. Differentiating once more gives the curvature,
  \[ \begin{aligned} R_{\mu\nu} = {} & \partial_\lambda\Gamma^\lambda_{\mu\nu} - \partial_\nu\Gamma^\lambda_{\mu\lambda} \\ & + \Gamma^\lambda_{\lambda\sigma}\Gamma^\sigma_{\mu\nu} - \Gamma^\lambda_{\nu\sigma}\Gamma^\sigma_{\mu\lambda} . \end{aligned} \]
  This Ricci tensor measures how a small ball of free particles is squeezed or stretched, and for weak gravity \(R_{00} \approx \nabla^2\Phi/c^2\), the Newtonian tidal field.
• A correction for conservation. Finally
  \[ \begin{aligned} G_{\mu\nu} &= R_{\mu\nu} - \tfrac12\,R\,g_{\mu\nu}, \\ R &= g^{\mu\nu}R_{\mu\nu} . \end{aligned} \]
  The extra term makes \(\nabla^\mu G_{\mu\nu} = 0\) hold automatically, so Einstein's equations can only be satisfied by contents that conserve energy and momentum.

So Einstein's equations are the relativistic version of \(\nabla^2\Phi = 4\pi G\rho\): second derivatives of the metric on the left, the contents on the right. For the expanding universe the only thing that varies is \(a(t)\), so the first derivatives bring in \(\dot a/a\) and the second bring in \(\ddot a/a\). That is why \(H\) and the acceleration are what come out.

Einstein's cosmological constant adds a term \(\Lambda g_{\mu\nu}\) to the left. Moved to the right, it acts like a fluid with \(\rho_\Lambda = \Lambda c^2/8\pi G\) and \(P_\Lambda = -\rho_\Lambda c^2\), that is, vacuum energy with \(w = -1\). So it can be counted as one more ingredient in \(\rho\).

What comes out. With this metric and this fluid, Einstein's ten equations reduce to just two:

• the time–time component is the Friedmann equation;
• the space–space components, combined with the Friedmann equation, give the acceleration equation.

The fluid equation is energy conservation, \(\nabla^\mu T_{\mu\nu} = 0\), and Einstein's equations guarantee it automatically. This is why it is also called the continuity equation, and it is the relativistic reason that any two of the three give the third.

The name comes from electromagnetism. Charge is conserved locally: the charge in a volume changes only if a current carries it out through the surface,

Field energy obeys the same kind of law, Poynting's theorem. The energy density \(u\) in a volume changes only if the Poynting vector \(\mathbf{S}\) carries energy out through the surface, or if the field does work on the charges inside:

\(\nabla^\mu T_{\mu\nu} = 0\) is four laws of this kind, one for energy and three for momentum, with the components \(T_{0i}\) playing the part of \(\mathbf{S}\).

In a uniform universe nothing flows. A current or a Poynting vector would have to point in some direction, and no direction is special. What changes instead is the volume. Charge simply dilutes, \(\dot\rho_q + 3H\rho_q = 0\), so \(\rho_q \propto a^{-3}\), exactly like dust. The fluid equation has one more term, \(3HP/c^2\), because energy, unlike charge, can be spent as work, much as the field in Poynting's theorem loses energy to the charges through \(-\mathbf{J}\cdot\mathbf{E}\). That is why radiation dilutes faster than charge.

Why pressure counts three times. Einstein's equations can be rewritten with the source on the right in a different form:

where \(T = g^{\mu\nu}T_{\mu\nu}\) is the trace of the energy-momentum tensor. With the signs of this metric, \(T = -\rho c^2 + 3P\): the pressure appears once for each of the three directions of space. For slow motion and weak fields, the time–time component \(R_{00}\) becomes \(\nabla^2\Phi/c^2\), where \(\Phi\) is the Newtonian potential. Putting in \(T_{00} = \rho c^2\) and \(g_{00} = -1\),

This is Newton's law of gravity, \(\nabla^2\Phi = 4\pi G\rho\), with the pressure added in. For dust, \(P\) is negligible and nothing changes. Applied to the sphere of the opening section, it gives the acceleration equation directly.

Show the curvature for this metric

Use \(x^0 = ct\) as the time coordinate, and keep the dot for \(d/dt\). The metric is \(g_{00} = -1\) and \(g_{ij} = a^2\gamma_{ij}\), where \(\gamma_{ij}\) is the metric of the bracket. The Christoffel symbols that involve time are

\[ \Gamma^0_{ij} = \frac{a\dot a}{c}\,\gamma_{ij}, \qquad \Gamma^i_{0j} = \frac{\dot a}{c\,a}\,\delta^i_j , \]

and the rest are those of the curved 3-space \(\gamma_{ij}\) alone. The Ricci tensor follows:

\[ \begin{aligned} R_{00} &= -\frac{3}{c^2}\,\frac{\ddot a}{a}, \\ R_{ij} &= \left(\frac{a\ddot a + 2\dot a^2}{c^2} + 2k\right)\gamma_{ij} , \end{aligned} \]

so the Ricci scalar and the time–time part of the Einstein tensor are

\[ \begin{aligned} R &= 6\left(\frac{\ddot a}{c^2 a} + \frac{\dot a^2}{c^2 a^2} + \frac{k}{a^2}\right), \\ G_{00} &= 3\left(\frac{\dot a^2}{c^2 a^2} + \frac{k}{a^2}\right) . \end{aligned} \]

Setting \(G_{00} = (8\pi G/c^4)\,\rho c^2\) gives the Friedmann equation. The time–time part of the rewritten equation, \(R_{00} = (4\pi G/c^4)(\rho c^2 + 3P)\), gives the acceleration equation.

What is contained in Newtonian gravity, and what is not

The Newtonian derivation shows that the expansion law is ordinary physics: a self-gravitating ball, thrown outward. The equation of state is where Newtonian gravity runs out. Once pressure is a source of gravity, a negative enough pressure makes gravity push, and that is the simplest way to see why an accelerating universe needs something with \(w < -\tfrac13\). Whether that something is a constant, as Einstein's \(\Lambda\) would be, or slowly changing, as the DESI data hint, is now a measurement.

Sources: E. A. Milne, Q. J. Math. 5, 64 (1934), and W. H. McCrea & E. A. Milne, Q. J. Math. 5, 73 (1934), for the Newtonian derivation; Planck 2018 VI for the parameters; DESI DR2 results II, arXiv:2503.14738, for \(w_0\) and \(w_a\). The DESI button uses \(w_0 = -0.75\), \(w_a = -0.86\) but keeps the Planck \(\Omega_\mathrm{m}\), so it shows the shape of the effect rather than the exact best fit. The radiation density assumes three massless neutrino species. Cosmic times come from integrating \(dt = da/aH(a)\) numerically.

Further afield: the expansion for what \(a(t)\) looks like in the data; \(\Lambda\)CDM and the six parameters for the standard model this page assumes; the evolving dark energy calculator to set \(w_0\) and \(w_a\) yourself and see what moves.