Comes from a Markov triple and beyond

Thermodynamic derivation of the energy density[edit]

The fact that the energy density of the box containing radiation is proportional to ${\displaystyle T^{4}}$ can be derived using thermodynamics.^[17][18] This derivation uses the relation between the radiation pressure p and the internal energy density ${\displaystyle u}$, a relation that can be shown using the form of the electromagnetic stress–energy tensor. This relation is:

${\displaystyle p={\frac {u}{3}}.}$

(x^2+y^2+z^2)/3= xyz

Which can be generalized to the general Shannon equations in N dimensions.   The equipartition theorem is simply the -iLog(i) must all partitions be within an integer. 

But then they go on, which Markov N-tuple do not. They separate kinetic along the transformed axis and provide the connection to N, total energy, which is what the stress-energy matrix does, it is a 4D. Total energy is a Hamitonian, and the 3 becomes a 3+1 after clever cancellations.  

That makes 4T be the log of something because xyz it total number of combinations, Temperature, no? But they have boundary conditions.  Then they need to scale by light steps as a wavelength, and essentially quantize x,y,z; that is, determine how many deviation counts they are allowed per coloring.

Use a Markov 4-Tuple, find a z deviation count that matches the need for euclidean precision, a z rate that divides up a complete sequence adequately to count integers. Then on your lower three counters,  Then the w,x,y counters  on the Markov 4-Tuple need to be -iLog(i)., that gives you the precision of the three axis is,w,x,y in Euclidean.   Then encode your w,x,y,z as you and sum along the z count. w,x,y,z will be a string of continued fractions. z is essentially taking half the deviation counts.

Remember, you are working in deviations in w,,x,y in the 3-tuple, continued fractions. z should count integers because you made it unconstrained from path merging.  But the Markov Nth integer, time,  still needs t be -iLog)i)I will keep repeating this to my self. The model is setting a fake N and suffering the consequences in kinetic energy on the paths in euclidean space.

When you are done, decode back to approximate original scale. Then set some standard ;like feet/sec.  You will find the general shape of the curve a result of the generalized Shannon equation for M dimensions, and M will mostly be three.

Now the three body in gravitation problem.

Start with a 4-tuple. m,x,y,z, where m,x,y,z are deviations in a surface covering the three point masses in Newton's gravity.  The problem is unresolvable, we know that. So our finite element strategy will be one of slightly being off count, relatively prime,  on the current deviation count. I gave mass the lower axis as it goes linear.  the gravity equations.    Now select your precision and find a matching 5-Tuple where the granularity of time are suitable..

Repeat the process, find the -iLog(i) for your m,x,y,z,t; then for each step of z, count the lower counters appropriately, convert all four counter back (only one or two changed).. 

Note, I am not computing the equations of motions, the transformed m,x,y,z,m are small differentials, the reals M,X,Y,Z,T  are integers with continued fractions.. The user has to move  differentials on an axis along with its finite unit of mass, and do the (X-Z)^2 and ( X-Y)^2  and M linear outside the Markov.  It is the model that takes us from Markov deviations to fractional real in another transformation. That is not a Markov thing, but the N-tuple concept is quite flexible, it can require a lot of extra counting in t, but you boil it down to a summation over a closed line.

It is simply the optimum frame reference for computing a model to fractional approximation. It works when model frame has the same equipartition as the Markov frame, the appropriate axis resolution.  Where works ,mean you gain the most precision with the fewest transactions.