What is blackbody radiation and why does it matter?
In: Physics
In Physics, a “blackbody” is defined as something that absorbs 100% of the light that hits it. In other words, 0% of the light that hits it. Things that look black to us absorb a lot of light—that is why they are black. What our eyes perceive as being black is really just a lack of light being reflected back into our eyes. So, that’s why “blackbodies” are referred to as “black”.
The idea of a blackbody is an idealized construct, meaning that we can pretend they exist when we’re doing the math, even if they don’t *technically* exist in the real universe. Nothing in the universe is perfect, so really a “perfect” blackbody that perfectly absorbs 100% of the light that hits it doesn’t really exist. But we can make *approximations* and say that something is *almost* a perfect blackbody. Because it makes the math easier.
Now, to get to what you asked, which was about blackbody *radiation*. According to Planck’s Law, a blackbody at *thermal equilibrium* (simply meaning that the body is at a constant temperature and there is no net flow of energy between it and its surroundings) will emit electromagnetic radiation (fancy word for light). Light emitted from a blackbody is called (logically) blackbody radiation.
The reason that a blackbody emits light is a little more complicated, and has to do with some quantum mechanics. In order for a bit of light to be absorbed by some bit of matter, the things that are actually doing the absorbing are the electrons in “orbit” around the atoms within the mater. Electrons can exist only at specific energy levels around the atom, and when an electron absorbs a photon, it gains that photon’s energy, and so it moves up a few levels, according to how much energy it absorbed. If it absorbed more energy, it jumps up higher. If it absorbed less, it might only jump up one level. It works the other way, too: if an excited electron (one in a higher energy level) falls down to a lower level, it will *emit* a photon. That photon will carry *exactly* the amount of energy lost by the electron when it jumped down to a lower energy level.
Going back to the specific case of an ideal black body: I mentioned that a blackbody is at thermal equilibrium, meaning that there is no net flow of energy between it and its surroundings. In other words, it *emits* exactly as much energy as it *absorbs*. A blackbody also absorbs 100% of the light that hits it, so it must perfectly absorb light at *every* frequency. Visible light, infrared, ultraviolet, microwaves, radio waves, gamma rays—it doesn’t matter, it *all* gets absorbed by a perfect blackbody.
This is important because typically, when we look at the light emitted from some source—say, the glowing gas inside a neon sign—we see distinct *lines* of emission at very specific wavelengths (or colors) of light depending on the unique set of electron energy levels that the atoms possess. So you get something that looks like [this.](https://upload.wikimedia.org/wikipedia/commons/3/3d/Boron_emission_spectrum.png) A sort of “bar code” of colors that tells us what energy levels exist inside that material and, therefore, the types of atoms—whether they be hydrogen or carbon or boron or whatever.
Going back to blackbodies. Since a blackbody absorbs *every* wavelength of light, it must also *emit* every wavelength. So it’s spectrum looks more like [this](https://upload.wikimedia.org/wikipedia/commons/4/4e/Spectral_lines_continuous.png) (just a continuous rainbow).
BUT here comes an important distinction: Planck’s Law says that a blackbody will emit at *every* wavelength, but that *doesn’t* mean it emits the same *amount* of light at every wavelength. In fact, a blackbody will always emit more light at certain wavelengths than at others. There will be a peak somewhere—the wavelength where the most light is being emitted.
Here’s the important part: where that peak happens depends entirely on the temperature of the blackbody. A hotter blackbody’s peak will happen more towards the “blue” end of the spectrum than a cooler body, whose peak will happen more towards the “red” end. This is because blue light carries more energy than red light, and so a hotter blackbody (one with more energy) will emit higher-energy light.
Not only is a hotter blackbody *bluer* than a cooler one, but it will also be *brighter*; it will emit more light at *all* wavelengths than a cooler blackbody.
This idea is best visualized in [a plot like this one](https://upload.wikimedia.org/wikipedia/commons/1/19/Black_body.svg). The horizontal axis represents the wavelength (color) of light, with bluer to the left and redder to the right. The range of wavelengths that we can actually see is also represented on the graph. To the left of that range is ultraviolet light, to the right is infrared light, both of which are invisible to human eyes. The vertical axis in the plot represents the intensity of light at that wavelength—you can think of it as the relative number of photons being emitted at that specific color.
In the plot, notice that the peak of the 5000 Kelvin (about 4700 ºC or 8500 ºF) body is further left (bluer) than the 4000 K body, which is further left than the 3000 K body. Hotter=bluer.
Notice, also, that the peaks are each *higher* than the last. In fact, each of the curves fits completely *underneath* the next. Hotter=brighter *at all wavelengths*
So, that’s blackbody radiation in a nutshell. To get to why it’s important, there are a lot of applications in many fields of physics. But I will speak specifically to my area of expertise, which is astronomy.
As it turns out, stars are a pretty darn good approximation of an ideal blackbody. They aren’t perfect, but they’re close enough. This is useful for astronomers because we can look at a star and measure its color, and therefore the wavelengths of light it is emitting at. If we know that, we can sort of work Planck’s Law backward and figure out the star’s temperature. The temperature and brightness of a star can help us figure out a lot about the star: its size, mass, and even its age and stage of life.
Blackbody radiation is also used as a bit of pretty substantial evidence for the big bang theory. There’s this thing called the “Cosmic Microwave Background” (CMB for short). [It looks like this.](https://upload.wikimedia.org/wikipedia/commons/2/2d/WMAP_2010.png) The CMB is basically a map of the very first light formed near the beginning of the universe. This image alone is a *huge* piece of evidence for the big bang theory; it is evidence that the universe expanded outward from an initial hot, dense singularity.
This evidence is further supported by the fact that the CMB is a *nearly perfect blackbody*. There is *no* other object in nature that gets as close to being this perfect. The big bang theory proposes that the universe underwent a period of very rapid inflation at the very beginning, and then inflation slowed down. That period of rapid inflation would have “smoothed out” the universe, leading to an early universe that looked the same everywhere (the scientific word is *isotropic*). That “smoothness” could not have happened without that period of rapid inflation.
Because the CMB is so “smooth” (isotropic), and because it is about as close to being a perfect blackbody as you can get, it is some pretty substantial evidence for that theorized inflationary period and for the big bang theory in general.
I know that was kind of a long-winded explanation with a lot of complicated ideas, so let me know if you have any more questions! I’ll be happy to answer.
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