Update - May 14, 2016

It's been a really long time since I've made a post here. I just wanted to say that I'm working on the next post regarding the AB magnitude system and I should be able to post it soon.

Magnitudes, or, Stupid Astronomical Systems

For this post I'm going to talk a little about the astronomical magnitude system. First, let me say that I think this system is really stupid and if I could I would eliminate it from astronomy. The only reason it exists is because it has been used for hundreds of years, but it is backwards and ridiculous.

Anyways, a star emits light. This is measured as luminosity L coming from the star, in units of energy per time, or W. Any point in space is going to receive a flux F of this light, in units of energy per time per area, or W/m^2.

For example, the Sun's luminosity L_sun is 3.839×10^26 W. The Earth is at a radius of 1 AU, or 1.5x10^11 m. This means the sphere at the radius of the Earth is 4*pi*(1 AU)^2 = 2.83x10^23 m^2. So, 3.84x10^26 W is reaching an area of 2.83x10^23 m^2, giving a flux of (3.84x10^26 W)/(2.83x10^23 m^2) = 1357 W/m^2.

This brings me to the magnitude scale, which basically tells you how bright something is compared to a reference 0 magnitude. The general equation is this:

'm' is known as the apparent magnitude of an object. This can be simplified if object 2 is considered the reference object (m_2 = 0):

A common reference is the star Vega, which has a flux in the V (visible) band of 3.55x10^-9 erg/s/cm^2/Angstrom or 3.55x10^-12 W/m^2/Angstrom. The problem is that through better measurements it has turned out that Vega actually has a magnitude of 0.035, so the actual zeropoint flux is 3.67x10^-12 W/m^2/Angstrom.

Now we can calculate the apparent magnitude of, say, the Sun. Though we calculated the Sun's flux to be 1357 W/m^2 above, that is the total flux from the Sun. The flux per wavelength in the V band for the sun is 0.184 W/m^2/Angstrom. So:

The apparent magnitude of the Sun is -26.75. The more negative the magnitude, the brighter the object. Each 5 magnitudes corresponds to a flux that is 100 times brighter, so 25 magnitudes would be 100*100*100*100*100=10 billion times brighter.

In addition to apparent magnitude there is also absolute magnitude, which is defined as the magnitude of an object if it were 10 pc away. This is an absolute measure of every object's brightness. We can calculate the absolute magnitude M of an object if we know its apparent magnitude and its distance. The relationship, known as the 'distance modulus,' can be easily derived by making m_2 in the magnitude equation the absolute magnitude, and remembering that F=L/area:

Now, for example, we can calculate the Sun's absolute magnitude. Its distance is 1 AU = 1.5x10^11 m = 4.85x10^-6 pc.

The absolute magnitude of the Sun is 4.82. Clearly, the apparent magnitude is so bright purely because we are so close to it.

Next post I'll cover AB magnitudes, so look forward to that ;)

Full Integration of the Planck Function, or, Learning Something New

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!

Further Fun With the Planck Function, or, When All Else Fails Use Mathematica

In my last post I showed an integration trick by using the Planck function:

To do so, I took the limit where h*nu >> k*T. This condition is valid for high frequencies (or probably less common, low temperatures). Then I could approximate it as this:

I forgot to mention that this is known as the Wien approximation. And after integrating I got the result:

Remember, my approximation was for high frequencies. Does it make sense to integrate over all frequencies? Maybe as an approximation, but it is important to keep valid limits in mind.

Using Mathematica, I can integrate the exact Planck function:
Mathematica> Integrate[2 h \[Nu]^3/c^2/(Exp[h \[Nu]/k/T] - 1), {\[Nu], 0, Infinity}]
This gives the solution:

Notice that my factor was 12, while the correct factor should be 2*pi^4/15=12.9879.
My approximation was close! But it was a little lower than the exact answer. I'm just trying to illustrate the importance of always thinking about the validity of your answers.

I'm curious how one would analytically integrate the complete Planck function. I'm not sure myself.

Now, what about the other possible limit, h*nu << k*T? Then we can do a Taylor expansion:

 Around a=0 using the exponential term:

We can discard everything on the order of x^2 or higher since x is small. The whole equation becomes:

This is known as the Rayleigh–Jeans law. Notice how my integral over all frequencies would now be infinite? This was known as the 'ultraviolet catastrophe.' These high-frequency and low-frequency approximations were what Planck used to ultimately formulate his function, valid at all frequencies.

Taking the Derivative Under the Integral, or, First Post

Well, this is my first post. I'm just making this blog to record thoughts I have about science, school, my research, and anything else. I don't really expect anyone to read it, except maybe future me.

The first thing I'm going to post is a math trick that I really like. To me it's called 'taking a derivative under the integral.' I'm going to show how it works in the context of integrating the Planck function. The Planck function is the 'spectral radiance' as a function of frequency (or wavelength), and it describes the radiation a blackbody at a certain temperature will emit. 'Spectral radiance' is a measure of energy, in units of W/m^2/Hz/sr (watts/meters^2/hertz/steradians, or, energy per time per area per frequency per solid angle). Here is the function:

Integrating this function over all frequencies will give the total power per area per solid angle emitted by a blackbody at a certain temperature. h*nu and k*T are both energies, and I'm going to assume that h*nu is much greater than k*T, so that (h*nu)/(k*T) is a large number, so that exp(h*nu/k/T) is a very large number, so that exp(h*nu/k/T) - 1 ~ exp(h*nu/k/T). Now we have:

Since we don't care about the constants until the end, let's arbitrarily have 2*h/c^2 = N and h/k/T = M. Now we have:

So, the integral we are concerned with is this:

Since N is a constant, we can take it out of the integral, and just add it back in at the end. Now we have:

So, the idea is to add a variable in that we will make equal to 1 at the end, and take the derivative with respect to that variable so that the integral is simplified to something we can easily do. I will add the variable alpha into the exponential, and keep in mind it will ultimately be equal to 1 so I'm not really changing anything:

Now, we want to take the derivative of this with respect to alpha so that it can replace what we currently have in the integral. Here are the steps:

So, we're getting close to having what is in the integral. To fine tune it:

(Remember, alpha=1) So there we have it. Now we can substitute this into the integral:

We can take the -1/M^3 out of the integral and put it back in later. We can also take the derivative out of the integral since it doesn't depend on nu. Now we have:

This integral is much simpler:

Now, we do the derivative:

So, I will put the N and the -1/M^3 back in:

Now, we set alpha=1 and plug in N=2*h/c^2 and M=h/k/T, and get our final result!

There you have it. This seems like an overly long process when I explain it all like this, but once you get the hang of it you can do it all pretty quickly. You could also have done 'integration by parts,' but personally I find this method easier and more straightforward.

It also works for arbitrary exponentials:

I'll leave it as an exercise to work it out yourself (heh ... it's fun, trust me!), but the solution is:

This is just one case, too. I bet there are tons of situations where it would be possible to 'differentiate under the integral.' It's just nice to keep in mind if you have a tough integral in front of you.

That's all for my first post :)