Translational Kinetic Energy — Question 8

Section 19-4 · Physics pages 3106–3107

Question 8 Finite samples fluctuate even at fixed equilibrium temperature For independent classical equilibrium molecules, use the supplied moments ⟨K⟩= 3 2kBT and Var(K) = 3 2(kBT)2. A temperature estimate from N independently sampled molecular energies is bT = 2K/(3kB). N σb T /T Axes and labels specify the model; the sketch is schematic. (a) Derive the relative standard deviation of bT and evaluate it for N = 600. (b) Find the minimum independent sample count needed for a relative standard deviation at most 1.00%. Explain why repeated readings of strongly correlated particles do not automatically count as independent samples.
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Question 8 - Solution Finite samples fluctuate even at fixed equilibrium temperature N σb T /T (a) Method and calculation Var(K) = Var(K) N , Var( bT) = 4 9k2 B 3(kBT)2 2N = 2T 2 3N . σb T T = r 2 3N = 1 30 ≈3.333% (N = 600). (b) Interpretation and further calculation q 2/(3N) ≤.0100 ⇒N ≥ 2 3(.0100)2 = 6666.666 . . . . Nmin = 6667. With correlations, covariance terms enter the variance of the mean; the 1/N law no longer follows from the raw reading count. Physical check This is a statistical standard deviation, not a guaranteed hard error bound.

Question 8

Finite samples fluctuate even at fixed equilibrium temperature

For independent classical equilibrium molecules, use the supplied moments ⟨K⟩=32kBT\langle K\rangle=\tfrac32k_BT and Var⁡(K)=32(kBT)2\operatorname{Var}(K)=\tfrac32(k_BT)^2. A temperature estimate from NN independently sampled molecular energies is T̂=2K¯/(3kB)\widehat T=2\overline K/(3k_B).

Diagram shown in the original SVG worksheet.

  1. Derive the relative standard deviation of T̂\widehat T and evaluate it for N=600N=600.

  2. Find the minimum independent sample count needed for a relative standard deviation at most 1.00%. Explain why repeated readings of strongly correlated particles do not automatically count as independent samples.

Question 8 - Solution

Finite samples fluctuate even at fixed equilibrium temperature

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Var⁡(K¯)=Var⁡(K)N,Var⁡(T̂)=49kB23(kBT)22N=2T23N.\operatorname{Var}(\overline K)=\frac{\operatorname{Var}(K)}N,\quad\operatorname{Var}(\widehat T)=\frac{4}{9k_B^2}\frac{3(k_BT)^2}{2N}=\frac{2T^2}{3N}.σT̂T=23N=130≈3.333%(N=600).\frac{\sigma_{\widehat T}}T=\sqrt{\frac2{3N}}=\frac1{30}\approx3.333\%\quad(N=600).

(b) Interpretation and further calculation

2/(3N)≤.0100⇒N≥23(.0100)2=6666.666….\sqrt{2/(3N)}\le.0100\ \Rightarrow\ N\ge\frac2{3(.0100)^2}=6666.666\ldots.Nmin=6667.N_{\min}=6667.With correlations, covariance terms enter the variance of the mean; the 1/N1/N law no longer follows from the raw reading count.

Physical check

This is a statistical standard deviation, not a guaranteed hard error bound.