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From what I understand about the Cosmic Microwave Background Radiation is that it was from the big bang, and since space has streched and become bigger since then the wavelength of the CMBR has increased over time, which means that when the universe stopped being opaque it wasn't microwaves but gamma rays, that must mean that it went from gamma rays to x-rays to ultra violet to visibile light (blue then all the way to red) then infra red and now microwave.

So the question is simply, at what point in time could someone have looked up in the sky and seen light from all directions? And how intense was this light?

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Actually the "last scattering surface" of the CMB corresponds to the transition of the interstellar/intergalactic medium from an ionized plasma to cooler neutral atoms, about 300 000 years after the big bang. Most atoms have excitation and ionization energies in the visible, so the CMB was probably visible when it formed.


We can be a little more precise about this. The brightest photons in a blackbody spectrum have energy $E=xkT$, where $x\approx3$ if you mean "brightest per unit energy," and $x\approx5$ if you mean "brightest per unit wavelength." For example, the sun's surface temperature is 0.5 eV, and so it's brightest around 2 eV.

The primordial blackbody gas of photons would have been tightly coupled to the hydrogen in the universe until photons above 13.6 eV became rare; this would have happened as the energy of the most common photons fell somewhere around 4–8 eV, which is pretty far into the ultraviolet. However, the temperature of a blackbody with this spectrum (kT ≈ 1.5 eV) is comparable to the surface of the star Bellatrix. It's worth noting that at this temperature there's still a small but non-negligible fraction of hydrogen ions and hydrogen in excited states.

There's a very good reason why the temperature of the CMB at last scattering should be comparable to the temperature of a stellar surface: it's the same physics! A star's photosphere is the last scattering surface for a gas of photons coupled to a mixture of hydrogen and helium which (mostly) gets hotter and denser as you deeper into the star; beneath the photosphere the temperatures of the photon gas the matter plasma are coupled, while beyond the photosphere the photons are free.


A commenter points out that I never actually answered the question at hand: when would the CMB have been visible? I answered that it started off visible, but how long that would have lasted is also interesting.

From another Physics.SE question we have the (approximate) relation $$ T(z) = T_0 (1 + z) \approx T_0\left(\frac{t}{t_\text{now}}\right)^{-2/3} $$ between the current CMB temperature $T_0$, the CMB temperature $T(z)$ seen by an observer at redshift $z$, and the age of the universe $t$ as computed by that observer, compared to its current age $t_\text{now} = 13.6\rm\,Gyr$. If we take "visibly glowing" to mean "at the Draper temperature," that was at time $$ t = \left(\frac{T_0}{T_\text{glow}}\right)^{3/2} t_\text{now} %= \left(\frac{2.7\rm\,K}{800\rm\,K}\right)^{3/2} t_\text{now} %= 2.0\times10^{-4} t_\text{now} = 2.7\rm\,Myr. $$ This was before stars, and therefore before metals and planets. The intensity would have been the same as any other barely-visible glowing blackbody, such as a warm electric cooktop.

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The CMB was emitted at an energy of $E_{em}=13.6\text{ eV}$, which is the binding energy of hydrogen. This corresponds to a wavelength of $$ \lambda_{em} = \frac{hc}{E_{em}} \approx 9.12\times 10^{-8}\text{ m}$$

Redshift can be calculated by $$ 1+z = \frac{\lambda_{obs}}{\lambda_{em}} $$

If we observe blue light at 400 nm, we get a corresponding redshift of about $z_{blue} = \lambda_{blue}/\lambda_{em} -1 =3.4$. For red light at 700 nm, we get $z_{red} = 6.7$.

The scale factor of the universe (the amount by which it has expanded) is related to redshift by $$ \frac{a_\text{obs}}{a_\text{em}} = 1+z $$

For a flat, matter-dominated universe we have $\rho a^3=\text{constant}$ so that the Friedmann equation becomes

$$\left(\frac{\dot{a}_\text{obs}}{a_\text{obs}}\right)^2=\frac{8\pi G\rho_\text{obs}}{3}=\frac{8\pi G\rho_\text{em}}{3}a_\text{em}^3a_\text{obs}^{-3}=H_\text{em}^2a^3_\text{em}a_\text{obs}^{-3} $$ where we used $H^2=8\pi G \rho/3$. Solving this differential equation for $a$ and imposing that $a_\text{obs}=0$ at $t=0$ (the Big Bang) yields

$$ \left(\frac{a_\text{obs}}{a_\text{em}}\right) = \left ( \frac{3 H_\text{em} t}{2} \right ) ^{\!2/3} $$ We solve for $t$, obtaining $$ t = \frac{2}{3 H_\text{em}}\left(\frac{a_\text{obs}}{a_\text{em}}\right)^{3/2} $$ Unfortunately, this does not seem to allow us to obtain an actual number for our estimate, since $H_\text{em}$ is not known...

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    $\begingroup$ If the CMB was emitted at the ionization energy of hydrogen, why is it a blackbody spectrum instead of a line spectrum? $\endgroup$ – rob Jul 11 '14 at 21:48
  • $\begingroup$ That is a very good question that I haven't thought about before. $\endgroup$ – johnpaton Jul 11 '14 at 21:51
  • $\begingroup$ I'm not 100% positive about this, but I think the photons would have been around as a blackbody spectrum before recombination, and the high-energy tail of the spectrum was preventing recombination. Recombination would only have 'happened' (it wasn't instantaneous) when the average went below 13.6eV. $\endgroup$ – johnpaton Jul 11 '14 at 21:55
  • $\begingroup$ As the radiaton is cooling, you must have $t(z_{blue}) \lt t(z_{red})$ and generally $z$ increases with $t$ $\endgroup$ – Ross Millikan Jul 11 '14 at 22:39
  • $\begingroup$ I noticed that after posting, leading to the piles of edits. Still trying to sort it out, $\endgroup$ – johnpaton Jul 11 '14 at 22:42

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