Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2012 / S2 / 3

Thermal interstellar dust produces a bright galactic foreground concentrated toward the Milky Way plane, with additional cloud structure. Its emission spectrum differs from the approximately $2.73\mathrm{K}$ blackbody spectrum of the cosmic microwave background, so comparisons across frequencies help separate it. In the sketch below this is the bright equatorial band.

Observer motion produces the cosmic microwave background dipole: to first order in $\beta =v/c$,The direction of motion is hotter and the opposite direction cooler. Its full-sky pattern has one broad positive hemisphere and one broad negative hemisphere; it is not concentrated in the galactic plane.

After subtracting the mean, the motion-induced dipole and galactic foregrounds, intrinsic Cosmic microwave background anisotropy appears as angular hot and cold patches, typically at the $10^{-5}$ fractional scale. These spatial temperature fluctuations encode early density and velocity structure. The displayed patch pattern is illustrative, with no claim to reproduce an observed sky map.

Measure the absolute spectral radiance over a range of frequencies and fit the Planck spectrumFor example, the COBE FIRAS blackbody spectrum experiment used a differential Michelson interferometer to compare the sky with an internal blackbody; an external calibrated blackbody could cover the sky entrance. Cold bolometers measured the difference spectrum, and foreground separation distinguished Milky Way emission from the nearly isotropic background. The fitted background temperature is approximately $\boxed{T_0=2.73\mathrm{K}}$.^[ref] The result relies on an absolute spectral comparison, rather than merely measuring a map's hot-to-cold contrast.

The stated one-sided temperature contrast is $\beta =0.00105$, so the Doppler dipole givesThe motion is toward the hotter region. The contrast already gives one side relative to the mean; the hot-to-cold difference would be twice as large.

Adiabatic expansion preserves the blackbody form while each photon's energy redshifts as $a^{-1}$. Thus the cosmic microwave background temperature obeys $T(z)=T_0(1+z)$, giving

Convert both inputs to energy density. The matter contribution, including rest energy, is $u_{M,0}={\rho }_{M,0}{c}^2=2.25\times 10^{-10}\mathrm{J}{\mathrm{m}}^{-3}$. The given photon value is $u_{\gamma ,0}=5\times 10^5(1.602\times 10^{-19})=8.01\times 10^{-14}\mathrm{J}{\mathrm{m}}^{-3}$. HenceMatter energy density scales as $(1+z{)}^3$, whereas the cosmic microwave background scales as $(1+z{)}^4$. The photon-only matter-radiation equality condition is thereforeThis uses the supplied CMB photon density. Additional relativistic species change equality for total radiation.