I know that the Cosmic Microwave Background Radiation (CMBR) is the leftover radiation from the "surface of last scattering".
However, at every instant the surface is changing (at the rate of flow of time). So how constant is the CMBR?
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3 Answers
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This is a great question, and it also came up at a recent Planck conference: according to this blog post (see the paragraphs above the CMB gif), simulations suggest that minute changes in the CMB might just be detectable in as little as 100 years from now.
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Just on dimensional grounds I would expect that $$\text{angular size of fluctuation in radians} \times \text{time since recombination}$$ is the time scale for significant changes in the CMB (where you choose what makes a fluctuation "significant" and select the angular size on that basis).
Why? Because the size of the regions defined by the fluctuations is the distance to the observed shell, and that distance is given by the time since recombination and the speed of light; and information can travel across those regions no faster than light.
The whole spectrum will cool a bit faster than that, but the cooling can be expected to be uniform. The rate of cooling is given by the Hubble (non-)Constant.
Note that the time since recombination is roughly 13.5 billion years, so even though the domains are pretty small it is still going to be a long wait.
answered Oct 14, 2011 at 20:46
dmckee --- ex-moderator kittendmckee --- ex-moderator kitten
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We can derive an order of magnitude estimate of the rate of decrease of the CMB temperature in the following way. The equation of state for a photon gas is $$ N = \frac{16 \pi k^3 \zeta(3)}{(h c)^3} \cdot V T^3 $$ If the photon gas is confined within a cavity, it interacts with the electrons in the walls of the cavity so that N fluctuates. But here our cavity is the observable universe. In the interest of getting an estimate rapidly, we will make the assumption that N is fixed. How good or bad this assumption is, is not clear to us, but perhaps it is good enough for an initial estimate. The equation of state for the CMB can then be written as $$ V T^3 = constant $$ Taking the time derivative and noting that the present temperature is approximately 3K gives $$ \dot{T} = -\frac{\dot{V}}{V} $$ Finally, using for the volume of the observable universe $$V = \frac{4}{3} \pi (c t)^3$$ where $t = 13.4 \cdot 10^9$ y, we get $$ \dot{T} \approx -0.2 \cdot 10^{-9} K/y $$
or about 1 nanoKelvin per decade. This number is far too small to be resolved by current measurements, which have uncertainties of about 0.00057 K. The open question is, how good or bad is the assumption that $N = constant$?
