Temperature — Question 10

Question 10

A local temperature requires separation of time scales

For a driven material, local microscopic equilibration takes $\tau_{\mathrm{loc}}=.002\,\mathrm s$. A sensor averages over time $\Delta t$, while the macroscopic state changes appreciably in $\tau_{\mathrm{drive}}=.50\,\mathrm s$. Adopt the explicit criteria $\Delta t\ge10\tau_{\mathrm{loc}}$ and $\Delta t\le\tau_{\mathrm{drive}}/10$ for a meaningful time-resolved local-temperature report.

1. Find the allowed averaging-time interval. Assess choices $.005$, $.030$ and $.20\,\mathrm s$.

2. If the driving time falls to $.10\,\mathrm s$, is any averaging time admissible? Derive the general existence condition.

Question 10 - Solution

A local temperature requires separation of time scales

(a) Method and calculation

$$10\tau_{\mathrm{loc}}=.020\,\mathrm s,\qquad\tau_{\mathrm{drive}}/10=.050\,\mathrm s.$$$$\boxed{.020\le\Delta t\le.050\,\mathrm s}.$$Only $.030$ s passes: $.005$ s is too short for the specified equilibration criterion and $.20$ s blurs the changing state.

(b) Interpretation and further calculation

A nonempty interval requires$$10\tau_{\mathrm{loc}}\le\tau_{\mathrm{drive}}/10\quad\Longleftrightarrow\quad\tau_{\mathrm{drive}}\ge100\tau_{\mathrm{loc}}.$$$$.10\,\mathrm s<100(.002\,\mathrm s)=.20\,\mathrm s.$$No averaging choice works after the change. A numerical sensor output still exists but lacks the stipulated local-equilibrium interpretation.

Physical check

The factor of 100 follows from the two chosen factors of ten, not from a universal thermodynamic constant.