LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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Exponential mode sums for the periodic photon box

Statement

Let $L, β, ℏ, c > 0$ be real parameters, set $a = 2 π ℏ c / L$ , and for $n \in Z^{3} ∖ {0}$ define $ϵ_{n} = a ∣ n ∣, q_{n} = e^{- β ϵ_{n}},$ where $∣ \cdot ∣$ is the Euclidean norm. For $j \geq 1$ , the integer max-norm shell ${n \in Z^{3} : ∣ n ∣_{\infty} = j}$ has $24 j^{2} + 2$ points. Consequently $\sup_{n \neq 0} q_{n} = e^{- β a} < 1, \sum_{n \neq 0} q_{n} < \infty, \sum_{n \neq 0} ϵ_{n}^{p} q_{n} < \infty$ for every integer $p \geq 0$ . In particular, the positive mode-energy sum $\sum_{n \neq 0} \frac{ϵ_{n}}{e^{β ϵ_{n}} - 1}$ is finite. Repeating each mode twice for two polarization labels preserves all these conclusions.

Facts & Assumptions

Given: $L, β, ℏ, c > 0$ , the periodic wavevector labels $n \in Z^{3} ∖ {0}$ , and the mode energies and weights in the Statement.

[A1]

All sums are ordinary nonnegative series over the stated countable lattice; no physical interpretation of the positive parameters is required.

Proof

1.1givenA1algebra

The number of integer vectors with $∣ n ∣_{\infty} = j$ is $(2 j + 1)^{3} - (2 j - 1)^{3} = 24 j^{2} + 2$ . On this shell $j \leq ∣ n ∣ \leq \sqrt{3} j$ , so $q_{n} \leq e^{- β a j}$ and $ϵ_{n}^{p} q_{n} \leq (a \sqrt{3})^{p} j^{p} e^{- β a j}$ . The shell contributions are bounded by $C_{p} j^{p + 2} e^{- β a j}$ for a finite constant $C_{p}$ ; the ratio of consecutive terms tends to $e^{- β a} < 1$ , so their series converges.

2.1givenstep 1.1algebra

Since $∣ n ∣ \geq 1$ on the nonzero lattice and equality occurs, $\sup q_{n} = e^{- β a} < 1$ ; the $p = 0$ case of step 1.1 gives $\sum q_{n} < \infty$ , and each $p \geq 0$ gives the asserted energy-weighted sum.

3.1step 1.1step 2.1algebra∎

For every $n \neq 0$ , $ϵ_{n} / (e^{β ϵ_{n}} - 1) = ϵ_{n} q_{n} / (1 - q_{n}) \leq (1 - e^{- β a})^{- 1} ϵ_{n} q_{n}$ . Summing and applying step 1.1 with $p = 1$ proves finiteness; a second polarization multiplies each nonnegative sum by two.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Mehran Kardar, MIT OpenCourseWare 8.333 Statistical Mechanics I, Lecture 20, photon-gas section