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When both endpoint bundles have the filled circular-cone geometry of etendue of a circular aperture and cone, conservation of etendue givesFor the same refractive index, quadrupling the area halves the sine of the cone half-angle. In the $10\mathrm{mm}$, $30^{\circ }$ example, an ideal output of diameter $20\mathrm{mm}$ must have $\sin {\theta }_2=1/4$, so ${\theta }_2\approx 14.4775^{\circ }$. The exact result differs from simply halving $30^{\circ }$. Conversely, reducing the diameter to $5\mathrm{mm}$ requires a complete forward hemisphere, ${\theta }_2=90^{\circ }$, as a limiting case.

Equal bounding area and solid angle do not alone prove equal etendue: an actual output can have unoccupied directions or position-direction correlations. The filled-cone equation is an exact model of the specified bundles and a capacity constraint on an optical instrument, not a guarantee of an aberration-free implementation.

Suppose an extended source fills every position of the entrance aperture with a cone of half-angle ${\theta }_{\mathrm{in}}$, and all these light rays must reach a one-sided planar receiver. If the receiver accepts at most a cone of half-angle ${\theta }_{\mathrm{out}}$, its available etendue bounds the admitted etendue:ConsequentlyThis original derivation combines conservation of etendue with the exact cone integral; the second inequality allows the complete output hemisphere. It bounds capacity and does not prove that a chosen shape attains the bound.

For the $30^{\circ }$ input cone with $n_{\mathrm{in}}=n_{\mathrm{out}}=1$, $C\le 4$: a $10\mathrm{mm}$ entrance cannot deliver the entire filled bundle to a circular receiver smaller than $5\mathrm{mm}$ in diameter. If the output half-angle is restricted to $60^{\circ }$, $C\le 3$ and the receiver diameter must be at least $10/\sqrt{3}\approx 5.7735\mathrm{mm}$. For a receiver medium with model refractive index $n_{\mathrm{out}}=1.5$, the full-hemisphere capacity bound instead gives $C\le 9$; that larger value also depends on transmitting the bundle into that medium.

Attempting $C=10$ in the air-to-air example forces rejection of some input. For the uniform radiance already specified, even filling the entire output hemisphere givesThe maximum mean irradiance gain remains $\eta C\le 4$. With nonuniform radiance, rejecting a given fraction of geometric etendue need not reject the same fraction of radiant flux.

These bounds use a filled extended-source bundle in passive reversible geometrical optics, comparing the same wavelength and the same radiation channels. An ideal optical point source does not fill that four-dimensional set. Diffraction, optical coherence, or coupling a Gaussian beam into an optical fiber require a physical optics calculation; the cone bound is not a universal guarantee for a single wave mode. Likewise, optical gain or fluorescence changes the propagation assumptions.

Spatial geometry. An embedding of the three-sphere isTaking differentials and collecting terms gives the induced metric tensorPut $\bar{r}=A\sin \chi$. Then $d\bar{r}=A\cos \chi d\chi$, soThe areal radius $\bar{r}$ is not a global radial coordinate on the entire three-sphere: its maximum occurs at the equator and two hemispheres share each smaller value. Use separate charts there; the coordinate singularity at $\bar{r}=A$ is not a physical singularity.

For Euclidean space and hyperbolic space of curvature $-1/A^2$, respectively,For the latter, $\bar{r}=A\sinh \chi$ gives $d{\ell }_{-}^2=A^2[d{\chi }^2+{\sinh }^2\chi d{\Omega }^2]$. In all three cases the induced two-metric at fixed $\bar{r}=\bar{R}$ is ${\bar{R}}^{2}d{\Omega }^2$. Its surface area is thereforeThis equality compares the same areal radius, not the same proper radial distance. For a proper radial distance $\ell$, the corresponding areas are $4\pi A^2{\sin }^2(\ell /A)$, $4\pi {\ell }^2$, and $4\pi A^2{\sinh }^2(\ell /A)$.

Homogeneity, isotropy and redshift. The FRW metric has constant-time spatial slices of constant curvature $k/a^2(t)$, with $k=+1,0,-1$ selecting the three geometries above. Their isometries act transitively on points and isotropically about every point, implementing the cosmological principle. Multiplication by one common time-dependent scale factor preserves these properties. With the source's $k=\pm 1$ normalization, $r$ is dimensionless and $a$ has units of length; using a dimensionless scale factor instead rescales the spatial curvature parameter. The origin of the spherical coordinates is arbitrary; it does not identify a physical centre. Cosmic time is the proper time of the comoving observers, and the absence of time-space cross terms makes their worldlines orthogonal to the homogeneous slices.

For a radial null geodesic between comoving endpoints,The next wave crest travels between the same endpoints. Subtracting the two path integrals, for wave periods short compared with the expansion time, gives $\delta t_0/a_0=\delta t_E/a_E$. Thus the cosmological redshift isBoth observers' clocks are cosmic proper-time clocks. Peculiar velocities and local gravitational-potential shifts have been left out of this homogeneous calculation.

Angular size and brightness. For a galaxy small compared with its distance, the angular diameter distance is $D_A=a_{E}R$ and its angular diameter isHere the emitted proper diameter is measured transverse to the line of sight; radial proper distance is a different integral when $k\ne 0$. Ignoring absorption, lensing caustics and intrinsic evolution, photon-number conservation spreads an isotropic-equivalent luminosity across the sphere $4\pi a_0^2{R}^2$. Each photon loses energy by $1+z$, and its arrival interval is stretched by the same factor. The received bolometric flux isThe disc's proper projected area is ${\mathcal{A}}_d=\pi d^2/4$, and its observed solid angle is ${\Omega }_d\simeq {\mathcal{A}}_d/D_A^2$. Therefore its surface brightness isThis is cosmological surface-brightness dimming. The normalization uses $L$ as isotropic-equivalent luminosity along the viewing direction; another disc emission law changes that normalization but not the redshift power. This is a bolometric result. In a specified observed frequency band the emitted spectrum is sampled at a shifted frequency and requires a spectral correction.

The distinction in telescope aperture and resolved surface brightness matters here. A larger telescope collects more photons and improves angular resolution, but cannot restore the incoming radiance lost by cosmological dimming. For aperture diameter $D$ and a diffraction resolution element of solid angle ${\Omega }_{\mathrm{res}}\sim (\lambda /D{)}^2$, the collected power from uniform resolved brightness scales asindependent of $D$. This is consistent with conservation of etendue. It explains why increasing aperture while keeping every gain in resolution does not increase counts per native resolution element from a smooth extended source. It does not make larger telescopes useless: at a fixed angular bin the counts grow as $D^2$, so a photon-noise-limited or uniform-sky-noise-limited signal-to-noise ratio grows roughly as $D\sqrt{t_{\exp }}$. They also resolve smaller structures and improve point-source sensitivity. Aperture, exposure and angular binning help measure a dim galaxy, without changing its intrinsic $(1+z{)}^{-4}$ brightness penalty.

For a diffraction-limited telescope, collecting area scales as $D^2$ while a native angular resolution element scales as $(\lambda /D{)}^2$. Counts from a smooth resolved source per native resolution element therefore do not increase with aperture, at fixed radiance, exposure and throughput. Counts in a fixed angular bin instead grow as $D^2$, and the photon-noise-limited signal-to-noise ratio grows as $D\sqrt{t}$. Conservation of etendue and cosmological surface-brightness dimming do not forbid this latter gain.