Equation 86 · No Particle Without a Cosigner

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e^{2\pi u}(1-\pi u)=1

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Inputs and operations1

Result or conditione^2pi u(1-pi u)

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e^{2\pi u}

Symbol e^2pi u

$e^2$ pi u is part of the quantity the equation computes from the expression on the right.

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\pi

Symbol pi

pi is part of the quantity the equation computes from the expression on the right.

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u

Symbol u

exactly twice that maximum.

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=

The expressions on both sides represent the same quantity under the stated assumptions.

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superscript

A raised number can be a power. When it is a label or bound, it selects a case or the upper limit of a sum; the formula’s structure distinguishes these uses.

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What the article says around this equation

This is the same trick used to derive Wien’s displacement law: because u\,g(u) rises from zero, passes through a single interior maximum, and falls back to zero, the integral of the absolute value of its derivative over all u is exactly twice that maximum, with no need to evaluate the integral directly. Finding the maximum of u\,g(u) = 24\, $u^2$ /( $e^{2\pi u}$ -1) means solving $e^{2\pi u}(1-\pi u)$ =1 , the same class of transcendental equation that fixes the peak of a Planck spectrum weighted by an extra power of frequency. Solved by Newton’s method to five-figure precision, the nontrivial root is $u^\ast$ $\approx$ 0.25363 , at which $u^\ast$ g( $u^\ast$ ) $\approx$ 0.3937 . The leading-order sensitivity is…

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This is the same trick used to derive Wien’s displacement law: because u\,g(u) rises from zero, passes through a single interior maximum, and falls back to zero, the integral of the absolute value of its derivative over all u is exactly twice that maximum, with no need to evaluate the integral directly. Finding the maximum of u\,g(u) = 24\, $u^2$ /( $e^{2\pi u}$ -1) means solving $e^{2\pi u}(1-\pi u)$ =1 , the same class of transcendental equation that fixes the peak of a Planck spectrum weighted by an extra power of frequency. Solved by Newton’s method to five-figure precision, the nontrivial root is $u^\ast$ $\approx$ 0.25363 , at which $u^\ast$ g( $u^\ast$ ) $\approx$ 0.3937 . The leading-order sensitivity is therefore

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Sources cited in the article section

[9] Asymptotic States of Accelerated Detectors and Universality of the Unruh Effect ↗

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