Thread by @MarkusDeserno

“The sun’s black body Planck spectrum peaks in the visible range.” Have you heard that before? Is it true? Want to follow me into a rabbit hole about densities and transformation laws? Then buckle up for another colorful physics/math thread!

1/24

Let’s make sure we’re all on the same page: the sun is hot, and that’s why it emits light. Not the same amount, though, for each color. How much light is emitted at what color is called a "spectrum". For the sun, it’s mostly the famous “black body spectrum”.

2/24

So why don’t we just plot it to answer our question? Sure—here we go: the attached image shows the sun’s Planck spectrum (more precisely: the spectral energy density) as a function of wavelength. I indicated the visible range by a rainbow-band.

3/24

Uh, well, aren’t we done here? It obviously peaks in the visible range, decaying both towards lower wavelengths (“ultraviolet”) and higher ones (“infrared”). Why the fuss?

Bear with me, this is more interesting than you might think!

4/24

A good place to start is to check the units of the spectral energy density: J/m³/μm. Or, in other words: energy per length to the fourth power. This looks odd. Why the fourth power? Why is the energy density not just energy per volume?

5/24

Ah, because it’s a *spectrally resolved* energy density! To get something of the units “energy per volume”, we need to *integrate* this function over a range of wavelengths, and what we find is the total energy density over that range. This will become important.

6/24

Let’s now open the Pandora box and ask what this spectrally resolved energy density will look like if we instead plot it as a function of the light’s frequency. Well, here it is:

7/24

Well—whoopsie. This function does definitely NOT peak in the visible; its maximum is in the infrared, and the center of the visible range is almost a factor 2 off in terms of frequency. Why did this happen? Or better: How COULD this even happen???

8/24

Students tend to be justifiably surprised (sometimes upset) when this conundrum first dawns upon them. The obvious expectation is that if we transform from wavelength λ to frequency f=c/λ, the λ-maximum of the spectrum gets mapped to the f-maximum. Well, not so!

9/24

Why is that so? Well, this is related to the fact that both these functions are *spectral densities*. You only get an “ordinary” energy density after integrating the spectral density either over a wavelength range or a frequency range.

10/24

It’s now important to appreciate that it’s that *integrated* energy density that is physically important, and which we must make sure we “preserve” upon transforming. The common way to indicate this is to talk about the energy density within a tiny spectral range.

11/24

Let us call the wavelength resolved spectral function uλ (think of the “λ” as a subscript), and it depends on the wavelength as its argument: uλ(λ). (Why two lambdas? Because the name of a function and the symbol for its argument are not the same thing!)

12/24

The frequency resolved function is then uf(f). Naively, we might think that since λ=c/f, we could just transform via uf(f)=uλ(c/f). If that were true, the maximum of uλ would indeed map to the maximum of uf, as expected, but this transformation is wrong!

13/24

It is wrong because it doesn’t shove the right amount of energy density contained between λ and λ+dλ into the transformed region between f and f+df. To insist on that, we can write down an equation that does this:

uf(f) df = uλ(λ) dλ.

14/24

This equation, by construction, takes the energy density in a “λ-bin” and transports it into an “f-bin”. We can easily solve it for the frequency resolved spectral energy density:

uf(f) = uλ(c/f) dλ/df

but behold: WE GET A JACOBIAN!

15/24

Actually, since the sign of that derivative doesn’t matter, we typically put the Jacobian under a modulus:

uf(f) = uλ(c/f) |dλ/df|

This makes a HUGE difference! The Jacobian is a nontrivial function of f, multiplying our naively transformed spectrum.

16/24

Specifically, we get uf(f) = uλ(c/f) × (c/f²), and that extra factor c/f² of course messes with the location of the peak.

So the mathematics is clear. But can we still get a bit more intuition, maybe? Please?

17/24

OK, I’ll try! The following picture illustrates how the λ-spectrum gets “bounced off” the nonlinear transformation from λ→f to give you the f-spectrum (plotted wonkily on the lower right). Crucially, you can see how the width of the λ- and f-bins is itself λ-dependent!

18/24

This means that as we increase λ from small to large values, the corresponding f-spectrum doesn’t just inherit this info. It also notices that fixed λ-bins turn into increasingly smaller f-bins, and so we must additionally increase the height of uf to preserve the area!

19/24

These two effects combine into a shift of the new maximum in uf(f) away from where the old function uλ(λ) had its maximum.

20/24

And for those of you who love animations, I’ve also tried my best to do that here—moving the λ-value around, keeping the λ-bins at a fixed size, and then see what happens to the f-bins. How gut-feeling “convincing” this is of course depends on your take.

21/24

So does the sun’s spectrum peak in the visible range? Well, it depends on how you plot it! But the maybe even better answer is that this question doesn’t make much sense in isolation. I can shift the maximum ANYWHERE by suitably devious plotting!

22/24

Let me close by saying that these kinds of shenanigans ALWAYS happen when you transform densities. A rather prominent example are probability densities that you wish to express in new variables. And forgetting the Jacobian is a VERY common mistake!

23/24

OK, we’re done here! Thanks for sticking to the end! And as always, if you found this lesson interesting or valuable (or, ideally, both!), I’d be grateful for a re-tweet, so that others can share in the fun! 🙏

Bye, until next time! 👋

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