Blackbody Radiation

Anything with a temperature above absolute 0 (which, considering that the background temperature of space is 2.7 K, is pretty much everything) emits lights with varying degrees of efficiency. Stars, for instance, are some of the most efficient emitters. Ideal emitters (an object that absorbs all light energy that hits it and reradiates that energy in a certain spectrum) reflect no light, and are thus known as blackbodies, and the reradiated energy is called blackbody radiation. Stars are not actually ideal emitters, but they are close enough  to roughly approximate one. Some example blackbody spectra are shown on this graph:

As you can see from this graph, the blackbody spectrum varies with temperature. If you want to see an example of some of the things you can do with this fact, keep reading.




Suppose you have a system with a Sun-like star and a Y Dwarf. This is a kind of Brown Dwarf, which is like a star, only too small to sustain Hydrogen fusion at its core. Consequently, they are much cooler than stars, and Y Dwarfs are the coolest of them all, with temperatures generally less than 500 K (We'll say this one has a temperature of 350 K). To understand how cool this is, consider that the room you're in now is probably 293-298 K.

Once you've wrapped your mind around this, consider this question: what is the best wavelength of light in which to observe this particular Y Dwarf? Well, if we treat this object as a blackbody emitter, then we can relate it's temperature to its wavelength output. If you look at the graph above, which plots intensity vs wavelength, you can probably guess that the best wavelength for seeing the object is the wavelength at which the intensity peaks.

The relation between temperature and this maximum wavelength is given by Wien's Displacement Law:



Well, that's handy. For T = 350 K, we have a wavelength of about 8.3 microns, which falls in the near-infrared electromagnetic spectrum.

Okay, so now assume that the Y Dwarf has a radius equal to that of Jupiter, and that the nearby star is once again about the same as the Sun. Can we figure out how many photons emitted from the Y Dwarf at the maximum wavelength per second, per unit area, will reach an observer 30 lightyears away?

To do this, we'll need this equation:



This is the formula for the specific intensity of the object, which is the energy emitted per time, per area, per frequency, per solid angle. Solid angle is an element of a sphere's surface.

Now, suppose the observer can see half of the sphere. Thus, we'll assume that that half of the sphere's photons make it to the observer. This corresponds to a solid angle of 2 times pi.

With that out of the way, we need to change units from energy to photons. These are related by this equation:



We can use this to reduce the specific intensity equation to this:

\

This is in units of photons per time per area per wavelength per solid angle. Multiplying by 2 times pi will remove the solid angle component, and we have the wavelength necessary, so we essentially have the relation we want. But we still have that complicated exponential in the denominator. Is there any way to eliminate or simplify that?

Well, the Taylor expansion of e^x to the first order is 1 + x, where x is the expression e is raised to. If we do this, we get the much friendlier



Which reduces to



Which is entirely in terms of things we know (k is Boltzmann's constant). Multiply this by 2*pi, and we get



Worked out with David and Mee.

This was the first two parts of 2.d on the Blackbody worksheet. I will add the next two parts soon.