8
$\begingroup$

What wavelength was the light that became the CMB after factoring out redshift?

Is it safe to assume that what we're seeing in the CMB was emitted at a distance of approx:

c = 299792458 m/s
Mpc = 3.08567758149137 * (10 ^ 22) m
expansionLow = ((74300 - 2100) m/s) / Mpc
expansionHigh = ((74300 + 2100) m/s) / Mpc
distanceHigh = c / expansionLow = 1.28125 * (10 ^ 26) m
distanceLow = c / expansionHigh = 1.21082 * (10 ^ 26) m

For an emission distance of 1.21082 - 1.28125 * (10 ^ 26) m?

(Expansion rate pulled from the story Speed of Universe's Expansion Measured Better Than Ever.)

$\endgroup$
6
$\begingroup$

The redshift of the CMB is 1089, so the wavelength at emmision was 1/(1089+1) of the wavelength we receive today. The distance at which it was emmited was approximately 0.0416 Glyr, while the distance today has expanded to 45 Glyr (a factor of 1090), which is a little bit less than the particle horizon (assuming that H0=67150m/Mpc/sec, Ωr=5e-5, Ωm=0.317, Ωλ=1-Ωr-Ωm).

$\endgroup$
12
$\begingroup$

The present cosmic microwave background was emitted when the initial plasma formed during the Big Bang had cooled down enough to convert into gas - mostly hydrogen and helium, which then was transparent to radiation.

The reason it was at this time is that hot gas does not glow or absorb much at all: think, for example, of the flame from a candle: there is a simple experiment one can do which involves holding (using a suitable tong!) a piece of metal wire just above the visible part of the flame. It will heat up and glow orange, meaning the air above that flame must be at the same temperature as orange-glowing metal, yet that air itself does not glow at all (the bright yellow visible flame glow is actually due to tiny particles of soot - solids - glowing in the same manner as the metal, not the hot air/gases). The only way that hot neutral gas can emit light is by molecular and atomic transitions, which comprise a line spectrum, not a broad spectrum. On the other hand, plasma can emit and absorb at all wavelengths because its free-roaming charged particles can interact with electromagnetic waves at all frequencies.

The temperature where this happens is roughly around 5000 K, I believe (there is no "sharp" phase boundary with plasma, it's a so-called "second order" transition where the two phases "shade" smoothly from one to the other - as a gas gets hotter and hotter, some atoms have enough energy to bump their electrons off, and that amount goes up with increased temperature). Wien's law says the peak wavelength will be around

$$\lambda_\mathrm{peak} \approx \frac{2.898 \times 10^6\ \mathrm{nm \cdot K}}{T}$$

With $T = 5000\ \mathrm{K}$, that $\lambda_\mathrm{peak}$ is about 580 nm, which is visible light.

So the "CMB" was initially a "Cosmic Visible Light Background", and indeed "space itself" was worse than a blast furnace at this point, because when you're surrounded on all sides by light of exactly a thermal spectrum and intensity, you are effectively within a thermal source at that temperature.

ADD: Thanks to the comment given below, the more accurate temperature is 3000 K. The wavelength is thus around 970 nm, which is the near infrared (NIR) band, but there will still be a noticeable tail in the visible light spectrum: this temperature is quite around that of a tungsten bulb filament.

$\endgroup$
  • $\begingroup$ The multiplier given in the other answer suggests it would be a peak of 174.6507THz (the CMB peak reaching us is at 169.23GHz,) this is quite a ways under the PHz range of visible light. Where did you get the equation and that 2.898*10^6 from? $\endgroup$ – CoryG May 4 at 3:54
  • $\begingroup$ The Wien wavelength constant is tabulated at: physics.nist.gov/cgi-bin/cuu/Value?bwien and just converted to nanometer-kelvins as that's a more appropriately sized unit for these cases. Also, visible light is around 400-750 THz (IIRC). 1 PHz or above is UV light, not visible. $\endgroup$ – The_Sympathizer May 4 at 4:16
  • $\begingroup$ Ah, I screwed up the exponent when looking at it, it's 516.8835THz - still quite a ways from the 174.6507THz from the previous answer based on the observed CMB peak though (looks like that corresponds to 1716.5260nm.) $\endgroup$ – CoryG May 4 at 4:26
  • 1
    $\begingroup$ It was about 3000 K rather than 5000. There are still a good amount of ions in a gas with a temperature of 5000 K. Light would have traveled a nice distance before being Thompson scattered by the time the universe had cooled to 5000 K. A nice distance isn't long enough. The universe had to cool to about 3000 K before light could essentially travel forever before being scattered. $\endgroup$ – David Hammen May 4 at 7:22
  • $\begingroup$ @David Hammen : Thanks for added precision, I just pulled that number off the top of my head as a rough guesstimate. (Nonetheless, factor of 2 ain't so bad for astronomy :g:) At least then that gives a peak wavelength in the near infrared, and of about the same order of magnitude, so you still get a sizeable visible tail. $\endgroup$ – The_Sympathizer May 4 at 7:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.