Tag Archives: numerical values

Switching Units

(i) Homunculus sitting at a desk: "Continuously switching from atomic to SI to Plank units devastated my brain." -- The doorbell rings. (ii) "Fresh & Frozen Ltd. Are you the idiot who ordered 9.11E31 kg vanilla ice cream?" -- Homunculus: "Shit, I knew it."

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(i) Homunculi sitting in the audience waiting. One saying: "That takes rather long." (ii) The aromatic Thio-Homunculus got up and starts speaking: "Since no professor appears to come, I'll hold a lecture instead. I will discuss why f and f(x) are not at all the same, how to label axes on a plot properly, the concept of numbers with finite precision, the fundamental meaning of a physical quantity, the correct usage of Wikipedia and... Hey, why are you all leaving?"

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0.000000125 are Not a Coincidence

Recently, I had to use some ab initio results for a kinetic computation. I was interested in a transition dipole moment of the molecule but none was given. However, there were some “infrared intensities” given in — crazy enougth — km mol-1. I was rather confused about that because, by all convention, this is no unit of an intensity.

[Klippenstein:1996] gives a conversion factor that claims basically:

A = 1.25E-7 I(in km mol-1) ν2(in cm-1)

where A is the Einstein coefficient for spontaneous emission.

I generally do not trust such formulas. Furthermore I think they are an unnecessary attack on healthy physical thinking. But how to do better? A colleague found a tutorial of Dunbar where he presented essentially the same formula.

Finally [Neugebauer:2002] could give me deeper insight. The values given as “infrared intensities” seem to be what Neugebauer calls integrated absorption coefficients. Using the definitions given in [Neugebauer:2002] it is a simple and straight forward exercise to calculate the Einstein coefficient as

A = I 8 π c NA-1 (ν/c)2

In this formula I let I be the symbol for the integrated molar absorption coefficient for the sake of continuity.

If one investigates the numerical value of 8 π c NA-1 he finds it to be about 1.25115E-14. The difference of a factor of E7 compared with Klippenstein’s formula is explained by recalling that he is suggesting to use km and cm-2 instead of SI units.


[Klippenstein:1996] Klippenstein et. al. Ion–Molecule Radiative Association Kinetics. J. Chem. Phys., 104(12), 1996.

[Neugebauer:2002] Neugebauer et. al. Raman and IR Spectra for Buckminsterfullerene. J. Comp. Chem., 23(9), 2002.

A Tribut to C++ Homework

For my very first homework in C++ (about half a year ago) I was asked to write a program that computes the exponential function. Up till now I was never fully satisfied by my former solution. So here is the ultimate version. A little late, but anyhow.

WeiredExponent is a program that combines a maximum of system incompatibility and user frustration potential with a minimum of calculation speed and reliability. It is a little C++ program that computes the exponential function from a number passed to it as a command line argument. However, none of the calculation steps is actually preformed by C++ code. Instead, it forks children that use Java, Fortran, Pascal and Python to carry out small pieces of work. — Each in a very inefficient way. Nevertheless, the program gets the correct results and there is no code that isn’t used. (Such like a += 0;.)

To run it, you’ll need:

WeiredExponent is of course Free Software. This means, that if you open the window and — as loud as you can — shout out: “Hello world!” then you can freely use it for any purpose you want. It is not very recommended to use WeiredExponent in important exams or as dog food. The author of this software is not responsible for bad marks, neither he is for sick dogs.

Download the C++ source code

The Magic Tool

Just quoting a recent Physics exercise:

Question: A man with a mass of 80 kg jumps horizontally from a wagon (on friction free horizontal rails) with a mass of 800 kg. At the liftoff, his speed relative to the wagon is 3 m/s. How much energy did the jump cost the man?
Answer: E = 327.3 J

The other day, I broke my window pane and had to purchase a new glass. Hence I had to know the dimensions of my window. Unfortunately, I didn’t have an appropriate measuring tape and one has to know it very exactly when ordering glass. No problem: I expected the window to be almost quadratic and I guessed that the area would be about one and a half square meter. Now I pulled out my good old pocket calculator, hit 1.5 + sqrt + enter and – surprise – got exactly 1.2247448713915890491 m as the side length. I think most scientific laboratories would – even using sophisticated equipment – have had serious troubles to meet that precision.

P.S. The inconvenient fact that the new pane did not fit, I can only explain by incredible malpractice of the manufacturer.