Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Why is the sprectum of the cosmic microwave background radiation (or seems to be) that of a black body?

share|cite|improve this question
up vote 3 down vote accepted

because this is the most probable configuration. Black-body light is the thermal equilibrium for light, so that anything that produces light which has a finite energy produces light that knocks around eventually to become blackbody light. The light we see was scrambled during the first 300,000 years by innumerable collisions with electrons and nuclei, and thermalized itself into equilibrium. There is no surprise when you see something in thermal equilibrium, the surprise in cosmology is that the temperature has small variations in different directions, which give us information about the inflationary process that gave birth to the light initially.

share|cite|improve this answer

We think that the Big Bang model of the universe creation explains why the microwave background radiation is a remnant of it. If you go to the paragraph of "timeline of the big bang"

A few minutes into the expansion, when the temperature was about a billion (one thousand million; 109; SI prefix giga-) kelvin and the density was about that of air, neutrons combined with protons to form the Universe's deuterium and helium nuclei in a process called Big Bang nucleosynthesis.[20] Most protons remained uncombined as hydrogen nuclei. As the Universe cooled, the rest mass energy density of matter came to gravitationally dominate that of the photon radiation. After about 379,000 years the electrons and nuclei combined into atoms (mostly hydrogen); hence the radiation decoupled from matter and continued through space largely unimpeded. This relic radiation is known as the cosmic microwave background radiation.

At its decoupling it was black body radiation at quite high temperatures but with the cosmic expansion it has cooled down to microwaves:

This component is redshifted photons that have freely streamed from an epoch when the universe became transparent for the first time to radiation. Its discovery and detail observations of its properties are considered one of the major confirmations of the Big Bang.

share|cite|improve this answer

Black body radiation is a bit misleading name, as it may also refer to the equilibrium radiation. The two have different conditions, but they follow the same frequency distribution, i.e. Planck's law.

Black-body radiation is the type of electromagnetic radiation within or surrounding a body in thermodynamic equilibrium with its environment, or emitted by a black body (an opaque and non-reflective body) held at constant, uniform temperature. ---Wikipedia

Since the photon gas was in thermal equilibrium with matter at the start of last scattering, they followed the Planck's law.

share|cite|improve this answer
What is misleading about this? The two concepts are the same. It's light at thermal equilibrium. Do you mean that the light is propagating through space away from the black body? This is not relevant to the question, since space is filled with the blackbody radiation going in all directions. – Ron Maimon Jun 20 '12 at 19:29
@RonMaimon: Because black body radiation has connotations that only black body radiates them. – Siyuan Ren Jun 20 '12 at 19:57
Karsus Ren: The statement that only black bodies radiate blackbody radiation is true for the radiation away from black bodies, since shiny bodies only radiate to the extent that they absorb, so it isn't misleading. If you have blackbody radiation inside perfect mirrors, the mirrors neither absorb or emit anything. This is just saying that the time to make thermal equilibrium inside the mirror box is infinite, since the mirrors will preserve the initial spectrum of the photons inside for all time, no matter how nonthermal. – Ron Maimon Jun 21 '12 at 15:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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