Wednesday, May 8, 2013

energy input into the CMB

the CMB is produced in thermal equilibrium.... would it be possible to notice if there was some extra energy put into it? what consequences could this have, depending on the way it was done?

maths bonus question: what are the asymptotes of $\exp(\tanh(x))$ for $x\rightarrow\pm\infty$ and what is the slope at $x=0$?

physics bonus question: can you derive the energy-time-uncertainty from the momentum-position-uncertainty in quantum mechanics?


  1. physics bonus question: the energy-time uncertainty is merely a consequence of the momentum-position uncertainty and *not* a genuine relation in itself... the reason being that there is no time-operator in classical quantum mechanics as time is a global parameter. if you use the dispersion relation $E=p^2/(2m)$ and the velocity $\upsilon = p/m$ for a particle of mass m, you can write $\Delta E=\partial E/\partial p\times\Delta p = p/m\times\Delta p$ and $\Delta t = \Delta x / \upsilon = m\Delta x / p$, and collecting the results yields:
    \Delta E\Delta t = \frac{p}{m}\Delta p\:\frac{m}{p}\Delta x = \Delta p\Delta x\simeq \hbar

  2. main question: one could notice energy input into the CMB: if that energy is deposited before decoupling, it gives rise to a nonzero chemical potential leading to a different blackbody shape, the so-called $\mu$-distortion. energy input after decoupling does not conserve the Planck spectrum but gives rise to $y$-distortions like in the thermal Sunyaev-Zel'dovich effect, due to incomplete thermalisation.

  3. maths bonus question: the asymptotes are $\mathrm{e}$ for $x\rightarrow+\infty$ and $1/\mathrm{e}$ for $x\rightarrow-\infty$. the slope at the origin is $2/\mathrm{e}$.