Waxwing

• Waxwing 8 Mar 2021 22:45 UTC

  1 point

  in reply to: Stevie Lantalia Metke’s comment on: Sleep math: red clay blue clay

  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...)

• Waxwing 8 Mar 2021 15:40 UTC

  3 points

  on: Sleep math: red clay blue clay

  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.