The Temperature of Compound Quarks
 Baby Collar (detail), Dong People. China, Yunnan province. Twentieth century, 33 x 15 cm. From the collection of Tan Tim Qing, Kunming. Photograph by D Dunlop.

Consider a particle P described by some repetitive chain of events

$\Psi ^{\sf{P}} = \left( \sf{\Omega}_{1} , \sf{\Omega}_{2} , \sf{\Omega}_{3} \ \ldots \ \right)$

where each orbital cycle is a bundle of N quarks

$\sf{\Omega} = \left\{ \sf{q}_{1}, \sf{q}_{2} \ \ldots \ \sf{q}_{\it{i}} \ \ldots \ \sf{q}_{\it{N}} \right\}$

By definition each quark is composed from a pair of seeds

$\sf{q} = \left\{ \sf{Z} , \sf{Z}^{\prime} \right\}$

and characterized by its temperature

\begin{align} T ^{ \sf{q}} = \frac{ \varepsilon \hat{K} + \varepsilon^{\prime} \hat{K}^{\prime} }{ k_{B} } \end{align}

where $\hat{K}$ is the vis viva and $\varepsilon$ is the audibility of each seed. Then by the definition of temperature

\begin{align} \it{T} ^{\, \sf{P}} &\equiv \frac{1}{N k_{B}} \sum_{i=1}^{N} \varepsilon_{\it{i}} \it{\hat{K}}_{\it{i}} + \varepsilon^{\prime} _{\it{i}} \it{\hat{K}^{\prime}_{\it{i}}} \\ &= \frac{1}{N} \sum_{i=1} ^{N} T_{\it{i}} ^{\sf{q}} \end{align}

So the temperature of a compound quark is just an average over its component quarks. Moreover, by the hypothesis of conjugate symmetry an ordinary quark and its anti-quark have the same temperature. Then swapping ordinary quarks with anti-quarks cannot change the average, and so any particle and its corresponding anti-particle have the same temperature

$\it{T} \left( \sf{P} \right) = \it{T} \left( \overline{\sf{P}} \right)$

 Next step: quarks are indestructible.
page revision: 104, last edited: 06 Mar 2016 19:03