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

Quark Temperatures |

ζ |
z z | T (℃) |

1 | u u | -815 |

2 | d d | -1,034 |

3 | e e | 676 |

4 | g g | -1,185 |

5 | m m | -6,401 |

6 | a a | 6,529 |

7 | t t | 222 |

8 | b b | 0 |

9 | s s | -252 |

10 | c c | 100 |

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

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 $T^{\sf{b}} =0 \ \sf{(ºC)}$

*T*the

*Celsius*temperature and using the Celsius degree (℃) for a temperature unit.

^{1}of nuclear particles. 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 hypothesis 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

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

## Temperature of Compound Quarks

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

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

Related WikiMechanics articles.

Lifetime | This article defines thermodynamic temperature and gives an interpretation of negative temperatures. |

Summary |

Adjective | Definition | |

Celsius Temperature | $\begin{align} T \sf{(ºC)} \equiv \mathit{ \frac{ \mathrm{2} }{Nk_{B}} \sum_{i\sf{=1}}^{N} \varepsilon ^{\it{i}} \it{\hat{K}}^{\it{i}} } \end{align}$ | 4-8 |