Archive for the ‘math’ Category


Cosmic background radiation

September 1, 2007

I finally finished my course on Quantum Optics; in the book we used was a question about the “number of photons per unit volume excited in a cavity at temperature T” (Loudon, Quantum Theory of Light). After deriving a formula for this, you can show that the cosmic background radiation contains about 4-5*10^5 photons per litre.
I always got stuck at the same point, but now I finally know how to solve this.


Read the attached pdf if you want to know the solution.


Calculating Pi for fun and profit

May 19, 2007

This one is for my grandchildren :P. I found a nice intuitive way of calculating Pi. The only thing you need to understand is the Pythagoras Theorem. Consider the figure below:


To calculate the area of the circle, we take small steps.

  • First we calculate the area of the grey part. It’s simple to see that this area is 0.5(1×1).
  • Then the area of the red part. This is a little more tricky. To calculate the area, we first want to know the length of S. We see that the length from A to B equals 2S. With the use of Pythagoras, we find that 2S equals the square root of 2. So we know S: 0.5*sqrt(2). If we would also know T (and hence 1-T), then we could find the red area. But T is equal to S. A different way of finding T, is using Pythagoras again; we know OA=1, and we know S, hence we know T: the square root of 0.5. Thus the area of the red part equals: 0.5*0.5*sqrt(2)*(1-sqrt(0.5))=0.104.
  • Now the green area. Well, we know (1-T) and S. So we know V. And W is 0.5*V. Now we only need to know the small side of the green triangle. This is where it gets tricky. (And messy). If we know O-fi then we’re all set. But OA=1, and we know Afi=W! Using Pythagoras (remember him?) we find filambda=1-fiO=1-sqrt(OA^2-Afi^2)=0.0439. So the area is approximately 0.0439*W=0.0439*0.29289=0.01286.

Note: wordpress doesn’t allow Greek symbols. So that explains “fi” and “lambda”
The only thing that needs to be mentioned, is that we have to multiply the grey are with 4 for a complete circle (we only considered 1 quadrant), the red area with 8, the green area with 16, etc.

Of course it’s possible to make a small script for this. I did all the fun for you, and coded it in Matlab. The code is a mess, but it’s posted here as a reference. If you understood the above (or if you came to this point without weeping), the code shouldn’t be hard to follow.

The real Pi is 3.1415926535897932384626433832795
# of iterations             found value

1                                    2.0000000000000000000000000000000
2                                    2.4142135623730949234300169337075
3                                   3.0614674589207182551770244842800
4                                   3.1214451522580524351896677430859
5                                   3.1365484905459392823695871570607
10                                 3.1415877252771597505203060598746
25                                 3.1415926535897902256560013790759
30                                 3.1415926535897951086718985911613
50                                 3.1415926535897951086718985911613

This means there’s a difference of 1.7764e-015=0.0000000000000017764, or a percentage of 0.5953e-13%=0.0000000000005953%. Not bad.

After about 30 iterations, we don’t see any change in the calculated Pi anymore. This is due to my coding skills, and the precision of the script. But I think it’s clear it does what it needs to do; the method works.