WikiMechanics

Let a generic particle P be characterized by some repetitive chain of events noted as

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

where each orbital cycle is a bundle of $N$ seeds

$\sf{\Omega} = \left\{ \sf{Z}^{1} , \sf{Z}^{2} \; \ldots \; \sf{Z}^{\it{i}} \; \ldots \; \sf{Z}^{\it{N}} \right\}$

and each seed is known by its audibility $\varepsilon$ and its vis viva $\hat{K}$. The particle is characterized by a sum over these component seeds

$\begin{align} T \sf{(ºC)} \equiv \mathit{ \frac{ \mathrm{2} }{Nk_{B}} \sum_{i\sf{=1}}^{N} \varepsilon ^{\it{i}} \it{\hat{K}}^{\it{i}} } \end{align}$

where $k_{B}$ is Boltzmann's constant. Definition: the number $T \sf{(ºC)}$ is called the Celsius temperature of P. This temperature may be positive, negative or zero depending on the particle's composition and the choice of a thermometric reference sensation. To establish numerical values start with the bottom-quark

$\sf{b} \equiv \{ \sf{B}, \sf{O} \}$

for which

$T^{ \sf{b}} \, k_{B} = \it{\hat{K}} \left( \sf{B} \right) - \it{\hat{K}} \left( \sf{O} \right)$

Then, if a bottom-seed has the same vis viva as a conjugate-seed

$\it{\hat{K}} \left( \sf{B} \right) = \it{\hat{K}} \left( \sf{O} \right)$ and $T^{\sf{b}} =0 \ \sf{(ºC)}$

Consider experimental practice to obtain this consistently; for example, by using bottom quarks as a reference to calibrate the measurement of temperature. This would depend on what we mean by touching ice because this feeling was used to define freezing reference sensations and objectified to define bottom seeds. But there are many different kinds of ice and to make reliable measurements we therefore need to specify the reference sensation more precisely. So, by "touching ice" we mean; touching a slushy mix of frozen solid water and clean pure liquid water in an open container near sea level on Earth. This is an utterly conventional way of defining zero on the Celsius temperature scale. So we note such a convention by calling T(℃) the Celsius temperature and using the Celsius degree for a temperature unit.

Theorem: an ordinary quark and its associated anti-quark have the same temperature. To see this, consider the generic quarks

$\sf{z} = \left\{ \sf{Z}, \sf{O} \right\}$ and $\bar{\sf{z}} = \left\{ \sf{Z}, \overline{\sf{O}} \right\}$

By the foregoing definition, the temperature of these particles is given by

$T^{\sf{z}} \it{k_{B}} = \hat{K} \left( \sf{Z} \right) - \hat{K} \left( \sf{O} \right)$ and $T^{ \sf{\bar{z}}} \it{k_{B}} = \hat{K}\left( \sf{Z} \right) - \hat{K} \sf{(} \sf{\overline{O}} \sf{)}$

But the assumption of conjugate symmetry presumes that $\hat{K} ( {\sf{O}} ) = \hat{K} ( \overline{\sf{O}} )$. So both quarks have the same temperature and we can use the quark index $\zeta$ to refer to either quark

$T^{\sf{z}} = T^{\sf{\bar{z}}} = T^{\zeta}$

Consider that each each orbital cycle of P may also be described as 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

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

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Temperature