How bright is a distant supernova?

Start from the Robertson–Walker metric:

Light from the source to us travels radially, so it pays to simplify the radial part. For a flat universe nothing needs doing; for open and closed universes the best variable differs. Guth takes the closed case ($k > 0$) and introduces an angle $\psi$ — which, traced back, is the angle from the $w$-axis used when the closed universe was first built as a 3-sphere in four dimensions:

The radial term of the metric is exactly the square of this, so

The tilde only distinguishes the new scale factor from the old one; starting from this metric, you could drop it. A student asks whether $a$ and $\tilde a$ have different units. They do. In Guth's convention $r$ is a coordinate distance in notches, so $a$ is in metres per notch; $\psi$ is an angle and naturally dimensionless, so $\tilde a$ is simply a length.

Put the source at the centre ($\psi = 0$), and imagine a sphere around it, drawn in comoving coordinates so that our detector, of area $A$, lies on its surface at $\psi = \psi_d$. The fraction of the light that hits the detector is the detector's area divided by the sphere's area — and the sphere's area must be computed from the metric.

On the sphere only $\theta$ and $\phi$ vary. The angular part of the metric is the usual one for a sphere, multiplied by the square of the sphere's radius, which is therefore $\tilde a\sin\psi$. We evaluate it today, $t_0$, because the detector has area $A$ today and the comparison only makes sense if all distances are measured at the same time. So the fraction is

The redshift costs two more factors. Any clock at the source appears slowed by $1 + z_s$, where $z_s$ is the redshift of the source. One clock is the photon's frequency, so each photon arrives with its energy reduced by $1 + z_s$; the other is the rate at which photons arrive, reduced by the same factor:

With $P$ the power emitted by the source, the power received and the measured intensity $J$ (power per unit area of detector) are

That is effectively the answer, but $\tilde a(t_0)$ and $\psi_d$ mean nothing to an astronomer. Both can be traded for measured quantities.

The scale factor. From the definition of the curvature fraction, $\Omega_{k,0} = -kc^2/(a^2(t_0)H_0^2)$:

For a closed universe $k > 0$, so $\Omega_{k,0}$ is negative and the square root is real. [The transcript reads as if the square root multiplies; from the definition of $\Omega_{k,0}$ it divides.]

The angle. Light travels on null geodesics. For a radial ray, $ds^2 = 0$:

Earlier in the course this rule was motivated by intuition; here it follows directly from the metric, even in a curved universe. Integrating from emission to today gives $\psi_d = \int_{t_e}^{t_0} c\,dt/\tilde a(t)$.

To integrate over redshift instead of time — redshift is how astronomers label the past — use $1 + z = 1/x = \tilde a(t_0)/\tilde a(t)$ and differentiate:

The Friedmann equation from the first half of the lecture, $H^2 = H_0^2F(x)/x^4$ with $x = 1/(1+z)$, supplies $H(z)$. Substituting it and the expression for $\tilde a(t_0)$:

Put this $\psi$ and $\tilde a(t_0)$ into the formula for $J$, and the expected intensity depends only on the present Ω's, $H_0$, the redshift of the source and its intrinsic power. That, Guth says, is exactly the formula the supernova teams used in 1998 — and to fit their data they needed a very significant vacuum energy, of order 60 or 70 percent.

Guth works out only the closed case. The other two follow from it (this section is the page's own addition):

• Open ($k < 0$, $\Omega_{k,0} > 0$): the same steps with $\sinh\psi = \sqrt{|k|}\,r$ give the same formula with $\sin\psi_d$ replaced by $\sinh\psi_d$, and $\sqrt{-\Omega_{k,0}}$ by $\sqrt{\Omega_{k,0}}$.
• Flat ($\Omega_{k,0} \to 0$): $\psi_d$ becomes tiny and $\sin\psi_d \approx \psi_d$, so $\tilde a(t_0)\sin\psi_d \to (c/H_0)\int_0^{z_s}dz/E(z)$, where $E(z) = H(z)/H_0$ is the square root in the integral above.

Astronomers package all three cases into a luminosity distance $d_L$, defined so that $J = P/(4\pi d_L^2)$ looks like the inverse-square law of a static, flat space. Comparing with the formula above, $d_L = (1 + z_s)\,\tilde a(t_0)\sin\psi_d$.

• The 1998 teams measured the intensity of type Ia supernovae against redshift; for standard candles, intensity measures distance.
• In a closed universe, with $\sin\psi = \sqrt k\,r$ and $\tilde a = a/\sqrt k$, the light spreads over a sphere of area $4\pi\tilde a^2(t_0)\sin^2\psi_d$, and redshift costs two factors of $1+z_s$: $J = P/[4\pi(1+z_s)^2\tilde a^2(t_0)\sin^2\psi_d]$.
• $\tilde a(t_0) = cH_0^{-1}/\sqrt{-\Omega_{k,0}}$, and the null-geodesic condition $d\psi/dt = c/\tilde a$ gives $\psi_s = \frac{1}{\tilde a(t_0)}\int_0^{z_s}c\,dz/H(z)$, with $H(z)$ from the Ω's.
• This is the formula fitted to the supernova data; it required a vacuum energy of order 60–70% of the critical density.