C.29.1:5.3 - Change temperature coordinates, then choose what can be forgotten

Situation and physical meaning. Consider two bodies with equal, constant heat capacity C > 0, each represented by one uniform temperature, T₁ or T₂. The bodies exchange heat only with each other. For one chosen sampling interval, stipulate a heat transfer from body 1 to body 2 of Q = αC(T₁ − T₂), with fixed 0 ≤ α ≤ 1/2. Negative Q means transfer in the opposite direction. Temperature differences use kelvins; temperature values below are expressed in degrees Celsius.

This is a supplied discrete model. Equal heat capacities and the opposite heat changes give:

T₁' = T₁ − Q/C = (1−α)T₁ + αT₂
T₂' = T₂ + Q/C = αT₁ + (1−α)T₂.

The range of α makes each new temperature a convex combination of the old temperatures, without reversing which body is warmer. The model does not prescribe the continuous temperature history inside an interval.

The working question is whether the warmer body’s modeled temperature is at most 60 °C after two intervals. Direct iteration is possible. A coordinate change can also expose which information the answer needs.

Construct the coordinates:

m = (T₁+T₂)/2
d = (T₁−T₂)/2

T₁ = m+d
T₂ = m−d.

The inverse recovers both temperatures. Thus (m,d) loses no distinction between admitted temperature pairs. Substitute the source updates and compare:

m' = (T₁'+T₂')/2 = (T₁+T₂)/2 = m
d' = (T₁'−T₂')/2 = (1−2α)(T₁−T₂)/2 = (1−2α)d.

The receiving operation preserves the mean and scales the contrast. It is a simpler expression of the same discrete update. Equal heat capacities explain the physical significance of the preserved mean: the two heat changes cancel. The warmer temperature is M = max(T₁,T₂) = m + |d|.

Relabelling and evolution. Let S(T₁,T₂) = (T₂,T₁) swap the bodies, and let U denote the update above. Equal heat capacities and the shared α give S(U(T₁,T₂)) = U(S(T₁,T₂)): both sides equal (αT₁+(1−α)T₂, (1−α)T₁+αT₂). The same update therefore applies after the swap. The warmer temperature M is invariant under S. It changes under U: for α = 1/4, (20,80) becomes (35,65), so M falls from 80 °C to 65 °C. Preservation of the mean during that update follows from the separately derived identity m′ = m.

Now consider discarding d and retaining only m. The identity update m’ = m is still exact, but the warmer temperature is no longer determined. The states (0,100) and (50,50) both have m = 50. For α = 1/4, the first becomes (25,75), then (37.5,62.5); the second stays at (50,50). They give opposite answers to the 60 °C threshold question after two intervals. Mean preservation alone is insufficient.

For α = 1/4, the contrast halves each interval. Starting from (20,80) gives this trajectory:

A bound on the contrast suffices after two intervals. Suppose the individual temperatures are unknown, but m = 50 °C and |d₀| ≤ 30 K are known. Repeatedly applying the receiving update yields dₖ = 2⁻ᵏd₀. Therefore:

50 °C ≤ Mₖ ≤ 50 °C + 2⁻ᵏ × 30 K.

Temperature differences are added to a temperature on the same scale. At k = 1 the upper bound is 65 °C, leaving the 60 °C question unresolved. At k = 2 it is 57.5 °C, so every compatible initial pair has warmer temperature at most 60 °C at that sampling instant. No choice of a representative pair is needed.

If an exact maximum is wanted, retaining (m,|d|) is sufficient under this symmetric update. The sign of d is needed only for questions distinguishing which body is warmer. Retain the distinction needed by the question, rather than restoring both coordinates automatically.

Returned result and return condition. The bound answers the stated question at the second sampling instant under the supplied model. It does not assert that the bodies were always below 60 °C: the initial pair (20,80) was not. Unequal heat capacities, external heat exchange or a changed transfer rule require a new update and a new comparison. Establishing that this discrete model predicts an actual pair of bodies requires physical and measurement work; B.5.MPC connects that work to the mathematical result.