Equation 74 · Part 7 · No Particle Without a Cosigner

subtraction

\int_0^\infty \Omega\,(e^{2\pi c\Omega/a}-1)^{-1}\,d\Omega = a^2/(24c^2)

subtraction

What this part means

Subtract the following term or group from the preceding one. A leading minus marks a negative quantity.

Its job in the formula

Subtract the following term or group from the preceding one. A leading minus marks a negative quantity.

Full expressionsubtractionArticle meaning

The passage around this formula

Write the normalized accelerated spectrum in terms of the dimensionless ratio u $\equiv$ c $\Omega$ /a , using the exact Bose-Einstein shape derived above. Carrying out the normalization integral, $\int_0^\infty$ $\Omega$ \,( $e^{2\pi c\Omega/a}$ -1)^{-1}\,d $\Omega$ = $a^2$ /(24 $c^2$ ) , using the standard result $\int_0^\infty$ x( $e^x$ -1)^{-1}dx= $\pi^2$ /6 , gives the scale-family form

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Learn the underlying idea

Addition combines quantities; subtraction measures the signed difference between them. Parentheses show what is combined before the rest of the expression is evaluated.

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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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