Exercises — Zeroth law — transitivity of thermal equilibrium

Statement. If $A\sim C$ and $B\sim C$, then $A\sim B$. Property supplied. Transitivity — the "chaining" property. Reflexivity ($A\sim A$) and symmetry ($A\sim B\Rightarrow B\sim A$) are obvious/free; transitivity is the physical content that had to be asserted as a law.

Same mark on $C$ both times means $A\sim C$ and $B\sim C$ with $T_A=T_C=T_B$. By transitivity, $A\sim B$: pour them together and no net heat flows. This is the whole practical payoff — comparison through a reference $C$, never directly.

False. Direction of flow (hot → cold) is the Second law's business. The Zeroth law describes only zero-flow situations (equilibrium) and their transitivity. It is silent on direction.

(a) $T_A=T_B\Rightarrow A\sim B$ (Zeroth law, via equal readings on the calibrated ladder): no net heat flow. (b) $T_D=7\ne 5=T_A$, so not equal readings $\Rightarrow A\nsim D$ (Zeroth law tells us only that they are not in equilibrium — heat will flow, but the law is silent on which way). Which way does it flow? From the hotter $D$ into the cooler $A$. ⚠️ This directional claim is the Second law, not the Zeroth. The Zeroth law alone stops at "$A\nsim D$, therefore some flow occurs." Once flow ends both read a common value between 5 and 7 — and by then, again by the Zeroth law, $A\sim D$.

Equilibrium: yes — equal temperature means $A\sim B$; joining them produces no net heat flow. Same energy: no — internal energy is extensive (scales with amount of stuff); the pool holds vastly more. Temperature is intensive. Equilibrium is about equal $T$, not equal energy.

Blank is "$=$": $T_A=T_B$. Chain: $A\sim C\Rightarrow T_A=T_C$; $B\sim C\Rightarrow T_B=T_C$; equality is transitive so $T_A=T_B$, hence $A\sim B$. Property: transitivity (Zeroth law), mirrored by transitivity of numerical equality.

Let $T_A=T_B=90^{\circ }\text{C}$ and $T_C=20^{\circ }\text{C}$. Then $A\nsim C$ and $B\nsim C$ (both disagree with $C$), yet $T_A=T_B$ so $A\sim B$. The Zeroth law only fires on the positive case ($A\sim C$ and $B\sim C$). "Not in equilibrium" is not transitive.

To attach one number $T$ to a whole equivalence class, membership must not depend on ordering: if $A$ shares the class with $B$, then $B$ shares it with $A$. If $\sim$ were not symmetric, "$A$ in $B$'s class but $B$ not in $A$'s" would let two different labels coexist for one relationship — the map body $\to$ number would be ill-defined. Once heat flow stops, there is no preferred direction, so $\sim$ is symmetric, and the label is consistent.

A thermometer reports a body's state as one number and compares bodies through itself ($C$). If $A$ and $B$ both match $C$'s reading yet do not match each other, then "same reading" no longer implies "same hotness." The reading carries no reliable meaning about $A$ vs $B$ — the whole device is useless. Transitivity is precisely the property that makes a mediated comparison valid.

Read the figure below: the two green double-arrows ($A-C$ and $B-C$) are closed loops — those equilibria hold. The red dashed double-arrow between $A$ and $B$ is drawn broken: in this broken universe it refuses to close, so no consistent label can be pinned to $A$ and $B$ together. Our real universe forbids this picture — that is exactly the content of the Zeroth law.

(a) Solve $L=L_0(1+\alpha \theta )$ for $\theta$: $\theta =\frac{L/L_0-1}{\alpha }=\frac{1.000247/1.000-1}{1.9\times 10^{-5}}=\frac{2.47\times 10^{-4}}{1.9\times 10^{-5}}=13.0^{\circ }\text{C}.$ Both cups give the same length, so ${\theta }_A={\theta }_B=13.0^{\circ }\text{C}$. (b) Equal thermometer reading $\Rightarrow A\sim C$ and $B\sim C$ at the same $T$ $\Rightarrow$ by the Zeroth law $A\sim B$: mixing produces no net heat flow. (c) Why length works — the monotonicity argument. Look at the graph: $L(\theta )=L_0(1+\alpha \theta )$ has slope $L_0\alpha >0$ everywhere, so it is strictly increasing — hotter always means longer, cooler always means shorter, with no repeats. That is what "monotonic" means and it is exactly what a temperature scale needs: (i) each length corresponds to one temperature (the line is never flat, so no two temperatures share a length), and (ii) the order of lengths matches the order of hotness (so "same length" honestly means "same hotness" — the "$\Leftarrow$" arrow from the notation box). Any state property with a strictly-rising graph vs. hotness — mercury height, electrical resistance, gas pressure — would serve equally well. If the graph ever flattened or turned back down, two different hotnesses would share one reading and the scale would lie.

Chain 1: $A\sim C_1$ (both "level 40") and $W\sim C_1$ (both "level 40") $\Rightarrow A\sim W$. Chain 2: $A\sim W$ and $C_2\sim W$ (rod calibrated on $W$ at $13.0^{\circ }\text{C}$) $\Rightarrow A\sim C_2$. So the rod, placed in $A$, must reach equilibrium at the same reading it had on $W$: $\boxed{13.0^{\circ }\text{C}}$. Two different thermometers agree because transitivity lets $W$ mediate between them.

Take two classes $[X]$ and $[Y]$ and suppose they overlap: some $Z$ has $Z\sim X$ and $Z\sim Y$.

• Pick any $W\in [X]$, so $W\sim X$. From $Z\sim X$ and symmetry $X\sim Z$; transitivity gives $W\sim Z$ (via $W\sim X\sim Z$). Then $W\sim Z$ and $Z\sim Y$ give $W\sim Y$, so $W\in [Y]$. Hence $[X]\subseteq [Y]$.
• By the same argument with $X,Y$ swapped, $[Y]\subseteq [X]$. So $[X]=[Y]$. Therefore any two classes that share even one member are equal; otherwise they are disjoint. Reflexivity ($X\sim X$) guarantees $X\in [X]$, so no body is left out. Each body belongs to exactly one class $\Rightarrow$ attach one number $T$ to each class. That number is temperature. ∎

(a) Reflexivity: $A\sim A$ trivially — the class always contains its own member; this is what lets $[A]$ be non-empty and the partition to cover everything.

(b) The $ϵ\to 0$ limit — why continuously, not by decree. When two bodies at temperatures differing by $ϵ$ touch, the net heat that flows is driven by that difference: by the Second law the flow rate vanishes when the difference vanishes (no gradient, no drive). Concretely, the amount of heat exchanged before settling scales with the gap $ϵ$ (small gap ⇒ small exchange). As $ϵ\to 0$ the exchanged heat $\to 0$ continuously — there is no sudden jump, because the driving cause shrinks smoothly to zero. So "just barely different" grades smoothly into "exactly equal": $T_A=T_B$ exactly and $A\sim B$ at the boundary. The hidden assumption is only that heat flow is a smooth response to the temperature difference (no threshold, no jump) — true for ordinary bodies.

(c) Finite heat capacity — measurement disturbs. A real thermometer $C$ has non-zero heat capacity, so reaching $C\sim A$ takes some heat from (or gives some to) $A$, nudging $A$'s temperature toward $C$'s. See the figure: the final common reading lands between $A$'s original value and $C$'s starting value, weighted by their heat capacities. The fix in practice: make $C$ tiny (small heat capacity) compared with $A$, so the nudge is negligible — the reading $\to$ $A$'s true temperature. The ideal thermometer of L5.2's "textbook limit" is the $C$-heat-capacity-$\to 0$ end of this same picture.

Take reference $C$ and two bodies $A,B$. Bring $A$ to equilibrium with $C$ (reading stops), record it; separately bring $B$ to equilibrium with the same $C$ at the same reading. Now — crucially without letting either re-equilibrate with the room — join $A$ and $B$ directly with sensitive heat-flow detection. Prediction (Zeroth law): zero net heat flow. Falsification: a reproducible, non-zero net heat flow between $A$ and $B$ despite both having matched $C$. No such result has ever been observed — which is why "$T$ exists" is safe to build all of thermometry on.

Which level tested that "$\nsim$ is not transitive"? ::: Level 3 (L3.1). What temperature did both cups share in L4.1? ::: $13.0^{\circ }\text{C}$. What is the output of the equivalence-class partition proof? ::: A single number (temperature) per class. Which law actually says heat flows hot→cold in L2.1? ::: The Second law — the Zeroth law only says "not equilibrated, so some flow occurs." What property of the rod's $L(\theta )$ graph makes it a valid thermometer? ::: Strict monotonicity (always rising) — one length per temperature, order preserved.