Ben Galehouse

@kevinr@masto.free-dissociation.com @lcamtuf@infosec.exchange And if you ask it to write a detailed spec based on its implementation, and then separately to write an implementation of that spec? https://www.allaboutcircuits.com/news/how-compaqs-clone-computers-skirted-ibms-patents-and-gave-rise-to-eisa/

Flatland and Minimum Action [Addendum] A few more observations about minimum action. The process of minimizing the action actually leads to the conservation of energy, at least in cases where \(V\) is only a function of position. That is, you need both endpoints, and \(L\) to perform the minimization process, and as a result of that process \(T+V\) is constant over the resulting curve. So the global minimization process leads to a quantity which is both local and constant. I find it interesting to compare this to the role of energy in quantum wave equations, e.g. the frequency of a photon is a local property, constant through a long narrow region of spacetime. This emergent conserved quantity \(T+V\) also explains why cutting up the path of a particle doesn’t really work. If you fix \(x(0), x(1), x(2)\) minimizing over the two segments \([x(0), x(1)]\) and \([x(1), x(2)]\) does not in general lead to a valid path for \([x(0), x(2)]\). In order to conserve energy at \(x(1)\), we need to choose \(x(1)\) carefully - essentially by solving the longer segment first.

Flatland and Minimum Action [2 of 2] Now let’s consider a simple but non-trivial example of how the minimum action principle works. Assume a uniform potential \(V=-kx\) and a kinetic energy \(T=\frac{1}{2} m \dot x^2\). Finally, assume that \(x(0)=x(1)=0\) Then we consider all sufficiently differential paths \(x(t)\) with those endpoints, and find one which minimizes the path integral \[\int_0^1 L =\int_0^1 T-V= \int_0^1 \frac{1}{2} m \dot x^2 + kx\] The Euler-Lagrange equation tells us that our solution will be such that \(m \ddot x = k\) and then integration plus our boundary conditions give \(x = \frac{k}{2m} (t^2 - 1)\). Some might argue that it was a choice to fix the endpoints. But a minimization process in which either endpoint moves would not give the same answer. Furthermore, the derivation of the Euler-Lagrange equation only proves that the equation holds on an interval \((a,b)\) under the assumption that \(x(a)\) and \(x(b)\) are fixed. And finally, this gives a bit of intuition about the negative sign in \(L=T-V\). During the minimization process, a potential trajectory is pulled towards a higher potential field. But then when we consider the effect on a particle over time, we see it as accelerating towards a lower potential field. Geometrically, this is because the endpoints are, ahm, nailed down. And so the action minimization principle feels a little like an artifact from outside of flatland.