B.5.TU:5 - Archetypal Grounding

B.5.TU:5.1 - Discover whether one input is enough

A practitioner wants to use two available operations, f: X -> Y and g: X -> Z, on one input. The letters name distinct atomic types. The question is whether the chosen composition rules can produce both results from that input.

In a cartesian account, a copying operation Delta_X: X -> X x X supplies the pair of inputs. First copy, then apply the two operations: (f x g) after Delta_X sends x to (f(x), g(x)). The input of the parallel operation is now supplied.

Compare a resource-sensitive account generated only by f, g, identities, serial composition, tensoring and exchange of factors. Tensoring f with g takes X tensor X to Y tensor Z: it requires two inputs. Each given generator preserves the number of atomic factors, and serial composition, tensoring and exchange preserve that property. These rules therefore cannot construct X -> Y tensor Z. Supplying another input or adding an admissible copying operation changes what can be built. This is a result about the stated generated account.

Now apply the distinction. For a reusable data value interpreted through pure functions, both uses can receive that value. For a physical specimen consumed by an assay, obtain a physical way to supply what each assay needs. Dividing a specimen may supply those inputs if the assays admit the resulting portions. The word “copy” would leave that physical contribution unexplained.

The common method discovers the needed theory operation and returns its consequence to use. The cartesian construction and its contrast with tensor composition come from Baez and Stay, §2.3; the input-count argument makes the stated restricted case inspectable.

B.5.TU:5.2 - Use variational mechanics to answer a motion question

An engineer wants to understand what a variational account predicts for an ideal carriage coasting along a straight horizontal track. Adopt a one-dimensional nonrelativistic free-particle model: positive mass m, negligible resistance and no applied driving force during the interval. Its Lagrangian is L(q, v) = m*v*v/2. For this case, the action is the time integral of that quantity along a proposed path. The physical model supplies this choice of Lagrangian.

The mechanical rule selects paths of stationary action: for every small path variation that leaves the endpoint positions fixed, the first-order change of action must vanish. The calculation below finds that path and shows that, for this free particle, it minimizes the action. Reproducing the calculation uses differentiation and definite integration; the displayed action identity can also be obtained as a supplied mathematical result.

Let the carriage pass q=0 at time 0 and q=d at positive elapsed time T. Recover the operations hidden by “calculate the action”: choose a position function q(t), differentiate it to obtain velocity, evaluate L on those values, then integrate over time. This is the constructive use developed in SICM, §§1.3–1.4. The following one-dimensional case uses it.

Start with q_0(t)=d*t/T, and try paths q_a(t)=d*t/T + a*(t/T)*(1-t/T), where a is a length. Every path has the specified endpoint positions. Its velocity is v_a(t)=(d+a*(1-2*t/T))/T. Substitution and integration give S_a = m*(d*d+a*a/3)/(2*T). For m=1 kg, d=2 m and T=1 s, the straight path has action 2 J s; a=1 m gives 13/6 J s.

This comparison favors the straight path within the chosen family. The general argument is also short. For any continuously differentiable added displacement eta(t) that is zero at both endpoints, the cross term integrates to (m*d/T)*(eta(T)-eta(0))=0. Thus the change in action is (m/2)*integral(eta'(t)^2 dt), which is nonnegative. The straight path minimizes this action among those paths. For a nonzero eta, write this positive change as K. Along the paths q_0+c*eta, the action is S[q_0]+c*c*K, whose derivative at c=1 is 2*K>0. Thus a nonzero displacement from q_0 cannot be stationary. The selected motion has constant velocity d/T.

The application now has an interpreted answer: under the adopted model, passing those endpoint positions in that time entails constant velocity. A computational implementation must evaluate the path derivative in the velocity argument before integrating. It can reproduce the numerical action comparison; the displayed argument supplies the wider conclusion.

Suppose the same two-metre trip must instead begin and end at rest. The constant-velocity answer fails that condition. The useful return is that the undriven free-particle account cannot supply this trip: the design needs acceleration and deceleration, and an account of the forces producing them. The next contribution is that driven-motion model. Increasing the resolution of the same free-particle calculation would retain the missing physical contribution.

B.5.TU:5.3 - Keep a useful bound when the optimizer cannot be enacted

A planner learning optimization must select whole jobs for four available hours. At most one A-job is available, taking three hours for stipulated value 5; at most two B-jobs are available, each taking two hours for value 3. Durations and values add in this constructed problem.

A linear relaxation permits real counts 0 <= x <= 1, 0 <= y <= 2 with 3*x+2*y <= 4. It returns x=1, y=0.5, value 6.5. Recover the argument: the time condition gives y <= (4-3*x)/2, so 5*x+3*y <= 6+0.5*x <= 6.5; the returned pair attains that limit.

For the actual whole-job choice, enumerate the alternatives. With A selected, no B fits and value is 5. Without A, two B-jobs fit and value is 6. Select two B-jobs. The fractional optimizer has still supplied a useful bound: no whole-job arrangement can attain value 7 because every such arrangement is also admitted by the relaxation.

The use of the theory changes with the question. Selecting work needs an attainable arrangement; excluding a target can use a bound. The practitioner needs to recover that relation before deciding whether a further optimization step is useful. B.5:5.4 also works the changed five-hour case.