Curvature and expansion — the Robertson–Walker metric
How the radius of curvature of a closed universe is tied to its scale factor, why curvature is a relativistic effect, and the Robertson–Walker metric that describes the expanding universe.
The 3-sphere describes the space of a closed universe at one moment. But the universe expands. In the last part of Lecture 11 Guth connects the radius of the 3-sphere to the scale factor a(t) — the one place in the course where we meet, without deriving it, how matter curves space.
From the Newtonian model, physical distances are ℓp(t)=a(t)ℓc, with comoving objects at fixed coordinates. Every distance in the 3-sphere metric is proportional to its radius R, so for the geometry to expand with the universe, R(t) must be proportional to a(t).
The units fix the rest. R is in metres, a in metres per notch and k in 1/notch², so the only possibility is
Radius from units
R2(t)=(dimensionless constant)×ka2(t)
In words
R squared (t) equals ( dimensionless constant ) times a squared (t) over k
The constant says how strongly a given amount of matter curves space, and finding it needs Einstein's field equations. Guth simply states the answer — the constant is 1:
This answers a question left open in Lecture 9: a/k is the radius of curvature of space. Guth describes 1/R as the curvature; in general relativity courses the (Gaussian) curvature of a sphere is usually defined as 1/R2.
Recall k=−2E/c2, with E<0 for a closed universe. Then R2=a2c2/(2∣E∣) — the radius contains c. In terms of quantities astronomers measure, using ρ=Ωρc and ρc=3H2/(8πG):
Radius of curvature from H and Ω
R=Ω−1cH−1
In words
R equals c H to the power minus 1 over the square root of omega minus 1
Two limits are revealing. If c were infinite (at fixed density and expansion rate), R would be infinite: space would be flat. Curvature is a relativistic effect. And as Ω→1, R→∞: a universe at exactly the critical density is flat, which is why the critical-density case is called "flat".
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Radius of curvature of a closed universe, computed from R = cH⁻¹/√(Ω − 1). As Ω approaches 1 the radius grows without limit and space becomes flat.
Worked example
How curved could our universe be?
In Lecture 7 Guth quoted measurements putting Ωtotal within about half a percent of 1. If the universe were closed with Ω−1=0.005 and H=67.3 km/s/Mpc, how large would its radius of curvature be?
Hubble length
c/H=299,792/67.3=4450 Mpc.
Radius
R=4450/0.005=4450×14.1=63,000 Mpc, about 63 Gpc.
Meaning
That is several times larger than the horizon distance, so over the part of the universe we can see, space would be only very slightly curved. (The relation R=cH−1/Ω−1 also holds with dark energy included, as long as Ω is the total density parameter.)
This is the Robertson–Walker metric, the standard description of a homogeneous, isotropic, expanding universe. The scale factor multiplies everything, so every distance grows with a(t), and the constant k from the Friedmann equation now appears directly in the geometry. Guth derives it here for the closed case; open universes follow in Lecture 12, "Non-Euclidean Spaces: Open Universes and the Spacetime Metric".
Every distance in a closed universe scales with a(t), so its radius of curvature does too: R2=(const)a2/k, and general relativity fixes the constant to 1.
a/k is the radius of curvature; in observable terms R=cH−1/Ω−1.
Curvature disappears as c→∞ (it is relativistic) and as Ω→1 (the flat universe).
Substituting into the 3-sphere metric gives the Robertson–Walker metric ds2=a2(t)[dr2/(1−kr2)+r2(dθ2+sin2θdϕ2)].