My last two posts have been about 'taking the derivative under the integral' and integrating the Planck function with different limits. I previously stated that I didn't know how to integrate the complete Planck Function over all frequencies. Well, I've since looked it up, and now I do!
Here's the integral to get the total radiance as a function of temperature for a blackbody, as discussed in my previous two posts:
Since we have an (h*nu)/(k*T) in the exponential, we want that same group of terms out in front, so we multiply everything by (h^2 k^3 T^3)/(h^2 k^3 T^3) and arrange them accordingly:
Let x = (h*nu)/(k*T) so that dx/dnu = h/(k*T) and thus dnu = (k*T/h)*dx
Now we plug x and dx in and get:
Which becomes:
So, that first step reveals a good strategy for doing any tricky integral: try to simplify the integral as much as possible, in this case matching the x^3 with the e^x.
Here is where things get a little tricky. I guess the advice for the next step would be to try and get rid of that 1/(e^x-1). You have to think: geometric series.
The basic geometric series goes like:
Notice that our 1/(e^x-1) is similar in form. Let's try to manipulate it to look the same. First, let's say e^x=a for simplicity. Now we mess around with 1/(a-1) to try and get something of the geometric series form 1/(1-x):
There we go!
Now we just substitute a^-1 where x is in the geometric series:
Notice that the n=0 term in the series is equal to 1, a^-0=a^0=1, and this 1 will cancel out with the -1 at the end. We then just have the series starting at n=1. Finally, we can plug the a=e^x back in! The whole process looks like this:
Let's plug it back into the integral! [NOTE: The next sum should go from n=1 to n=infinity, NOT n=0 to n=infinity. Sorry, typo, don't want to latex it all out again :) ]
The sum can be taken out of the integral since it doesn't depend on x. We are now working with:
Does the integral look familiar? It should! It's a perfect 'derivative under the integral' integral!
In my first post, I calculated a general exponential 'derivative under the integral' case. Recall:
What do you know, it looks just like our case! replace nu with x, n with 3, and A with n, and you get:
And, if I hadn't remembered that general formula, I could have easily taken the derivative under the integral again and figured it out!
Let's plug this result back in to the full equation:
That sum we have now is actually so common that it has its own name, the Riemann zeta function.
Riemann zeta functions have the form:
These have specific solutions that you can easily look up. Our case is zeta(4), you'll notice.
Plugging that in, we can get our final answer!
There we go! You can check that we get the same answer I had gotten with Mathematica in my last post. But now I calculated it myself! Fun times.
The moral of this story, I think, is, when facing a tricky integral, to simplify, look for mathematical series, Taylor expansions, or trig identities, try to remember that things like Riemann zeta functions exist, and take the derivative under the integral!
Just for fun, we can work this number out: pi=3.14 k(Boltzmann)=1.38×10^-23 m^2*kg/s^2/K, c=3.00x10^8 m/s, and h=6.63×10^-34 m^2*kg/s:
The final line gives the physical units when T is measured in K. You'll find that W/m^2/Hz/sr is the same as kg/s^2.
So now you can find the energy per time per area per frequency per solid angle for any blackbody of temperature T!
From this equation, you can see that something with T = [1/(1.79x10^-8)]^(1/4) K will have a total radiance of 1 J/s/m^2/Hz/sr. That would be 86.45 K, or -304 degF or -187 degC.
Our sun has T=5800 K. This gives it a radiance of 2x10^7 J/s/m^s/Hz/sr. That's a lot of energy!
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