Logarithmic Temperature Scale

To begin, you should probably read the Wikipedia article on Absolute Zero, if you haven’t already.
The two important takeaways from it are:

Now that you’ve caught up, we can get started with this new temperature scale.

The basic idea is that, because absolute zero can’t be reached, it should be at

- \infty

. A well-known function that can achieve this is the logarithm.

Let’s use my base-less logarithm first and just plug in an absolute temperature

𝑇

, for example

𝑇 = 100 𝐾

\log (𝑇) = \log (100) + \log (K) .

Let’s try another one,

𝑇 = 0 𝐾

\log (𝑇) = \log (0) + \log (K) = - \infty + \log (K) .

And no, you can’t correct me by saying “

\log (0)

is undefined.” This is my blog, where I get to make the definitions, and I define

\ln : [0, \infty] \to [- \infty, \infty]

with

\ln (0) = - \infty

and

\ln (\infty) = \infty

.⁴

Anyway, the problem is now that we wanted

0 K

to be

- \infty

, not

- \infty + \log (𝐾)

. So we have to subtract

\log (K)

, or, really,

\log (𝑥 K)

for any

𝑥

, since then we’ll get

\log (0 K) - \log (𝑥 K) = - \infty + \log (K) - (\log (𝑥) + \log (K)) = - \infty - \log (𝑥) = - \infty .

Simplifying the left side, we also get

\log (𝑇) - \log (𝑥 K) = \log (\frac{𝑇}{𝑥 K})

and this form is very interesting. Notice how we can set

𝑇 = 𝑥 K

and get

\log (\frac{𝑇}{𝑥 K}) = \log (\frac{𝑥 K}{𝑥 K}) = \log (1) = 0 .

Thus,

𝑇_{0} ≔ 𝑥 K

is the temperature that is assigned the value 0 in the new scale.

For convenience and practicality, I’ll just copy the familiar Celsius scale for this and set

𝑥 = 273.15

We’re still not quite done since the quantity

\log (\frac{𝑇}{273.15 K})

isn’t a number yet. It’s a dimensioned quantity with dimension “logarithm”, as described in my previous post.

Since this means we can choose any unit for this, let’s just copy the Celsius scale again and choose the boiling point of water (

373.15 K

) as the second reference point.

Instead of dividing it into 100 tho, we’ll let SI prefixes handle that and set the boiling point of water to just 1.

This means our unit will be

\log (\frac{373.15}{273.15}) = \log (\frac{7463}{5463}) .

Since the two reference points are copied from the Celsius scale, this new scale shall be called the Logius scale, with unit symbol “Lo”.

The full conversion formula is then

𝑥 K = \frac{\log (\frac{𝑥}{273.15})}{\log (\frac{7463}{5463})} Lo,

or, written as a logarithm with a base,

𝑥 K = \log_{\frac{7463}{5463}} (\frac{𝑥}{273.15}) Lo .

Then, as I said above, the “actually useful” unit will be the centi-Logius, i.e.

1 cLo = 0.01 Lo .

In particular, we have

\begin{matrix} - 273.15 ° C & = & 0 K & = & - \infty Lo & = & - \infty cLo, \\ 0 ° C & = & 273.15 K & = & 0 Lo & = & 0 cLo, \\ 100 ° C & = & 373.15 K & = & 1 Lo & = & 100 cLo . \end{matrix}

To prove that the Centilogius is actually a useful unit for everyday life, look at the graph of it and the Celsius scale for everyday temperatures:

The maximum deviation in this range is at the left end, at a numeric difference of about

4.4

, tho there is also a local maximum at about

320.6 K

with a numeric difference of about

3.9

These are obviously not a lot, so someone used to the Celsius scale (i.e. most of the world’s population) can just carry over their intuitions for how hot or cold a temperature value is.

¹ or should it be “Part 1 Cour 2”? Or maybe “Part 1.3 Generation 2”? ↩︎
² I’m deliberately ignoring some negative temperature shenanigans, which are only kind of below absolute zero. ↩︎
³ Probably. Maybe. Thermodynamics says so, at least. But who knows. ↩︎
⁴ I’m using $\ln$ here instead of $\log$ since I don’t really have a good formal notation for the domain and range of my base-less logarithm. Maybe I should change that in a part 3. We’ll see. ↩︎