Hacker News Reader

Comment by bubblyworld | Hacker News Reader

Irradiance is a density function - it's telling you how much energy or how many photons are hitting a given area per unit of the dependent/spectral variable (you can see this by inspecting the units in the plots, for instance).

This means that changes of variable come with an additional factor - the Jacobian. You cannot simply substitute the relationship directly into the formula for the one spectral variable, like you would for a "point function" (you need to account for the change of unit in the y-axis too, if you like, not just the x-axis).

It's not explained well but this is fundamentally what causes the moving peaks.

View on HN

This subject is also a wonderful example why the maximum of a densitiy function has very little meaning. The amount of energy is an integral over the densitiy function, and if you look at the solar-spectrum, you'll notice, that the bulk of the energy is in the infrared.

Indeed, the value of a density function at any single point (maximum or not) tells you very little. What's important is how it changes over a range of values.

> Irradiance is a density function - it's telling you how much energy or how many photons are hitting a given area per unit of the [in]dependent/spectral variable (you can see this by inspecting the units in the plots, for instance).

Well, I can see that in the labels on the y-axis, but I assumed it was a mistake.

So you have a graph that tells you that, for light of wavelength 1000 nm, measured irradiance is 3.5e+18 photons per square meter per second per nanometer.

And since there "are" 1000 nanometers (?!?), this means that the actual irradiance is 3.5e+21 photons per square meter per second.

Or does the "per nanometer" really mean something less stupid than that? What's being measured? What kind of nanometers are those on the y-axis?

Not quite - the nanometres on the y-axis are "deltas" of your spectral variable. The density allows you to answer questions like "how much power am I getting per square metre in a range [A, B] of wavelengths". You would integrate your density between A and B to obtain the value.

For example, pick some small value "k" close to 0 in units of nm. Then in your example, the amount of irradiation contributed by the small window [1000-k, 1000+k] nm of wavelengths is roughly equal to k*3.5e18 photons per square metre per second (you can check that the units work out). The smaller k is, the more accurate the approximation. If you want to get an answer for a larger interval you can break it up into lots of k-sized pieces and sum the results up. Recall from calculus that this is exactly what integration is (yes, I know the truth is a little more complicated in general measure theory).

Does that help? It's a bit like a continuous probability distribution, in the sense that to get an actual probability out of it you have to integrate. Formally a mathematician would say that a density corresponds to a "measure" over the space of all possible values of your spectral values.

Reply on news.ycombinator.com