Comment by Waxwing on Sleep math: red clay blue clay · 2021-03-08T22:45:33.759Z · LW · GW

Having checked this, you're totally correct. 

Here's a short Python program that checks this empirically.  

import numpy as np

n = 100

b = np.zeros(n)
r = np.ones(n)

for ib in range(n):
   for ir in range(n):
       old_b = b[ib]
       old_r = r[ir]
       new_b = new_r = np.mean([old_b,old_r])
       r[ir] = new_r
       b[ib] = new_b

print(np.mean(b))

A graph of the scaling of the max blue temperature with respect to number of divisions is here: 

https://imgur.com/a/r4hdaJD

When T is 1000 you can get the blue to 0.98 of what red was initially... so I imagine in the infinite limit you can get all the way there.

This seemed weird to me, so I looked at the final temperature distribution of all of the $n$ blue blobs, as $n$ increases.  I've plotted them here (blue lines have $n$ being small; red lines have $n$ being big). You can see that as the number of blobs becomes very big, the fraction which have temperature less than 1 falls... (i.e. knee of graph moves to the right...)

Comment by Waxwing on Sleep math: red clay blue clay · 2021-03-08T15:40:47.511Z · LW · GW

Treat the blue clay (mass 1, temperature T) as a single lump. Feed it infinitesimal lumps of red clay (mass $dX$, temperature $T_R$). After each infinitesimal feeding, the temperature of the blue clay changes by

$dT=\frac{T+T_{R}dX}{1+dX}-T=-TdX+T_{R}dX$

(Final equality comes from series expansion).

Then you can integrate: ${\int }_0^{T_{final}}\frac{dT}{T_R-T}={\int }_0^{1}dX.$

i.e. $\log (\frac{T_R}{T_R-T_{final}})=1$ 

The final solution is:

$T_{final}=T_R\cdot (1-1/e)$.

It's worth saying that this is the most efficient way to transfer energy from red to blue because each feeding step is thermodynamically reversible.