**Maksvell taqsimoti** - statistik muvozanat holatida turgan fizik sistema tarkibidagi klassik mexanika qonunlariga boʻysunuvchi zarralar (molekula, atom, elektron va boshqalar) ning tezlik boʻyicha taqsimlanishi qonuni. 1859-yilda J. K. Maksvell aniqlagan. Maksvell taqsimotiga, asosan, statistik muvozanat holatidagi gazda ixtiyoriy molekula tezligi *v* ning koordinatalardagi proyeksiyalarining mos ravishda erishishi mumkin boʻlgan istalgan *vx vy,vz* kattaliklarga toʻgʻri keluvchi birlik intervaldagi qiymatlardan birortasiga teng boʻlish ehtimolligini ifodalaydi. Maksvell taqsimoti birinchi marta 1920-yilda nemis fizigi O. Shtern tomonidan qizdirish natijasida kumush atomlarining bugʻlanib chikib turishidan foydalanib oʻtkazgan tajribalarida toʻla tasdiqlangan

**Gaz molekulalarining tezliklari bo`yicha taqsimlanishi**. Maksvell taqsimoti. Muvozanat holatda turgan gaz molekulalari agar gazga hech qanday tashqi kuchlar maydoni ta’sir etmayotgan bo`lsa, o`zaro to`qnashib turadi. Har bir to`qnashish jarayonida, energiya almashinuvi tufayli, molekula o`z tezligini ham miqdori bo`yicha, ham yo`nalishi bo`yicha o`zgartiradi. Maksvell extimollik nazariyasidan foydalanib 1859 yilda gaz molekulalarining tezlikka qarab taqsimlanish qonunini aniqladi, uning fikricha: 1. Тezliklar ichida extimolligi eng katta bo`lgan shunday e tezlik mavjudki, ko`pchilik molekulalar unga yaqin bo`lgan tezliklarda harakatlanadi. Тezligi e dan juda katta va juda kichik bo`lgan molekulalar oz miqdorni tashkil etadi. 2. Harakat tartibsiz bo`lgani uchun aniq bir tezlikda harakatlanayotgan molekulalar sonini hisoblab bo`lmaydi. Lekin ma’lum , + d oraliqdagi tezlikda harakatlanayotgan molekulalar sonini hisoblash mumkin. Buning uchun Maksvell nisbiy tezlikdan foydalanadi.

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{\displaystyle f(v)~d^{3}v={\biggl [}{\frac {m}{2\pi kT}}{\biggr ]}^{\frac {3}{2}}\,\exp \left(-{\frac {mv^{2}}{2kT}}\right)~d^{3}v,.}

{\displaystyle f(v)~d^{3}v={\biggl [}{\frac {m}{2\pi kT}}{\biggr ]}^{\frac {3}{2}}\,\exp \left(-{\frac {mv^{2}}{2kT}}\right)~d^{3}v,.}
 298,15 haroratda bir nechta asil gazlar tezligining tezlik ehtimoli zichligi funktsiyalari K (25 °C). *Y* - o'qi s / m ga teng, shuning uchun egri chiziqning istalgan kesimi ostidagi maydon (bu tezlikning ushbu diapazonda bo'lish ehtimolini ifodalaydi) o'lchamsizdir.

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{\displaystyle f(v\_{x})~dv\_{x}={\sqrt {\frac {m}{2\pi kT}}}\,\exp \left(-{\frac {mv\_{x}^{2}}{2kT}}\right)~dv\_{x},.}

{\displaystyle f(v_{x})~dv_{x}={\sqrt {\frac {m}{2\pi kT}}}\,\exp \left(-{\frac {mv_{x}^{2}}{2kT}}\right)~dv_{x},.}

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{\displaystyle f(v)={\biggl [}{\frac {m}{2\pi kT}}{\biggr ]}^{\frac {3}{2}}\,4\pi v^{2}\exp \left(-{\frac {mv^{2}}{2kT}}\right)..}

{\displaystyle f(v)={\biggl [}{\frac {m}{2\pi kT}}{\biggr ]}^{\frac {3}{2}}\,4\pi v^{2}\exp \left(-{\frac {mv^{2}}{2kT}}\right)..}



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{\displaystyle {\begin{aligned}&0=kTvf'(v)+f(v)(mv^{2}-2kT),\\[4pt]&f(1)={\sqrt {\frac {2}{\pi }}}\,{\biggl [}{\frac {m}{kT}}{\biggr ]}^{\frac {3}{2}}\exp \left(-{\frac {m}{2kT}}\right);\end{aligned}}.}

{\displaystyle {\begin{aligned}&0=kTvf'(v)+f(v)(mv^{2}-2kT),\\[4pt]&f(1)={\sqrt {\frac {2}{\pi }}}\,{\biggl [}{\frac {m}{kT}}{\biggr ]}^{\frac {3}{2}}\exp \left(-{\frac {m}{2kT}}\right);\end{aligned}}.}



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{\displaystyle {\begin{aligned}&0=a^{2}xf'(x)+\left(x^{2}-2a^{2}\right)f(x),\\[4pt]&f(1)={\frac {1}{a^{3}}}{\sqrt {\frac {2}{\pi }}}\exp \left(-{\frac {1}{2a^{2}}}\right).\end{aligned}}.}

{\displaystyle {\begin{aligned}&0=a^{2}xf'(x)+\left(x^{2}-2a^{2}\right)f(x),\\[4pt]&f(1)={\frac {1}{a^{3}}}{\sqrt {\frac {2}{\pi }}}\exp \left(-{\frac {1}{2a^{2}}}\right).\end{aligned}}.}
 Maksvell-Boltzman tezligini taqsimlash tomon bo'shashgan 2D gaz simulyatsiyasi

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{\displaystyle P(s<|{\vec {v}}|<s+ds)={\frac {ms}{kT}}\exp \left(-{\frac {ms^{2}}{2kT}}\right)ds.}

{\displaystyle P(s<|{\vec {v}}|<s+ds)={\frac {ms}{kT}}\exp \left(-{\frac {ms^{2}}{2kT}}\right)ds.}
Solar Atmosphere Maxwell–Boltzmann Distribution. Quyosh atmosferasiga mos keladigan Maksvell-Boltzman taqsimoti. Zarrachalar massasi bitta proton massasi, *m*p = 6973167000000000000♠1.67×10−27 kg ≈ 7000100000000000000♠1 Da, harorat esa quyosh fotosferasining samarali harorati, *T* = 5800 K {\displaystyle {\tilde {V}}}, {\displaystyle {\bar {V}}}, va *V*rms mos ravishda eng ehtimoliy, o'rtacha va ildiz o'rtacha kvadrat tezliklarni belgilaydi. Ularning qadriyatlari {\displaystyle {\tilde {V}}} ≈ 7003979000000000000♠9.79 km/s, {\displaystyle {\bar {V}}} ≈ 7004110500000000000♠11.05 km/s, va *V*rms ≈ 7004120000000000000♠12.00 km/s .


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{\displaystyle v\_{\text{p}}\approx {\sqrt {\frac {2\cdot 8.31\ {\text{J}}\cdot {\text{mol}}^{-1}{\text{K}}^{-1}\ 300\ {\text{K}}}{0.028\ {\text{kg}}\cdot {\text{mol}}^{-1}}}}\approx 422\ {\text{m/s}}..}

{\displaystyle v_{\text{p}}\approx {\sqrt {\frac {2\cdot 8.31\ {\text{J}}\cdot {\text{mol}}^{-1}{\text{K}}^{-1}\ 300\ {\text{K}}}{0.028\ {\text{kg}}\cdot {\text{mol}}^{-1}}}}\approx 422\ {\text{m/s}}..}



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{\displaystyle {\begin{aligned}v\_{\mathrm {rms} }&={\sqrt {\langle v^{2}\rangle }}=\left[\int \_{0}^{\infty }v^{2}\,f(v)\,dv\right]^{\frac {1}{2}}\\[2pt]&=\left[4\pi \left({\frac {b}{\pi }}\right)^{\frac {3}{2}}\int \_{0}^{\infty }v^{4}e^{-bv^{2}}dv\right]^{\frac {1}{2}}\\[2pt]&=\left[4\pi \left({\frac {b}{\pi }}\right)^{\frac {3}{2}}{\frac {3}{8}}\left({\frac {\pi }{b^{5}}}\right)^{\frac {1}{2}}\right]^{\frac {1}{2}}={\sqrt {\frac {3}{2b}}}\\[2pt]&={\sqrt {\frac {3kT}{m}}}={\sqrt {\frac {3RT}{M}}}={\sqrt {\frac {3}{2}}}v\_{\text{p}}\end{aligned}}.}

{\displaystyle {\begin{aligned}v_{\mathrm {rms} }&={\sqrt {\langle v^{2}\rangle }}=\left[\int _{0}^{\infty }v^{2}\,f(v)\,dv\right]^{\frac {1}{2}}\\[2pt]&=\left[4\pi \left({\frac {b}{\pi }}\right)^{\frac {3}{2}}\int _{0}^{\infty }v^{4}e^{-bv^{2}}dv\right]^{\frac {1}{2}}\\[2pt]&=\left[4\pi \left({\frac {b}{\pi }}\right)^{\frac {3}{2}}{\frac {3}{8}}\left({\frac {\pi }{b^{5}}}\right)^{\frac {1}{2}}\right]^{\frac {1}{2}}={\sqrt {\frac {3}{2b}}}\\[2pt]&={\sqrt {\frac {3kT}{m}}}={\sqrt {\frac {3RT}{M}}}={\sqrt {\frac {3}{2}}}v_{\text{p}}\end{aligned}}.}

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{\displaystyle v\_{\text{p}}\approx 88.6\%\ \langle v\rangle <\langle v\rangle <108.5\%\ \langle v\rangle \approx v\_{\mathrm {rms} }..}

{\displaystyle v_{\text{p}}\approx 88.6\%\ \langle v\rangle <\langle v\rangle <108.5\%\ \langle v\rangle \approx v_{\mathrm {rms} }..}

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{\displaystyle c={\sqrt {\frac {\gamma }{3}}}\ v\_{\mathrm {rms} }={\sqrt {\frac {f+2}{3f}}}\ v\_{\mathrm {rms} }={\sqrt {\frac {f+2}{2f}}}\ v\_{\text{p}},.}

{\displaystyle c={\sqrt {\frac {\gamma }{3}}}\ v_{\mathrm {rms} }={\sqrt {\frac {f+2}{3f}}}\ v_{\mathrm {rms} }={\sqrt {\frac {f+2}{2f}}}\ v_{\text{p}},.}



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{\displaystyle {\begin{aligned}v\_{\rm {rel}}\equiv \langle |{\vec {v}}\_{1}-{\vec {v}}\_{2}|\rangle &=\int \!d^{3}v\_{1}\,d^{3}v\_{2}\left|{\vec {v}}\_{1}-{\vec {v}}\_{2}\right|f({\vec {v}}\_{1})f({\vec {v}}\_{2})\\[2pt]&={\frac {4}{\sqrt {\pi }}}{\sqrt {\frac {kT}{m}}}={\sqrt {2}}\langle v\rangle \end{aligned}}.}

{\displaystyle {\begin{aligned}v_{\rm {rel}}\equiv \langle |{\vec {v}}_{1}-{\vec {v}}_{2}|\rangle &=\int \!d^{3}v_{1}\,d^{3}v_{2}\left|{\vec {v}}_{1}-{\vec {v}}_{2}\right|f({\vec {v}}_{1})f({\vec {v}}_{2})\\[2pt]&={\frac {4}{\sqrt {\pi }}}{\sqrt {\frac {kT}{m}}}={\sqrt {2}}\langle v\rangle \end{aligned}}.}

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{\displaystyle f({\vec {v}})\equiv \left[{\frac {2\pi kT}{m}}\right]^{-{\frac {3}{2}}}\exp \left(-{\frac {1}{2}}{\frac {m{\vec {v}}^{2}}{kT}}\right)..}

{\displaystyle f({\vec {v}})\equiv \left[{\frac {2\pi kT}{m}}\right]^{-{\frac {3}{2}}}\exp \left(-{\frac {1}{2}}{\frac {m{\vec {v}}^{2}}{kT}}\right)..}



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{\displaystyle {\begin{aligned}\int \_{0}^{+\infty }v^{a}\exp \left(-{\frac {mv^{2}}{2kT}}\right)dv&=\left[{\frac {2kT}{m}}\right]^{\frac {a+1}{2}}\int \_{0}^{+\infty }e^{-x}x^{a/2}\,dx^{1/2}\\[2pt]&=\left[{\frac {2kT}{m}}\right]^{\frac {a+1}{2}}\int \_{0}^{+\infty }e^{-x}x^{a/2}{\frac {x^{-1/2}}{2}}\,dx\\[2pt]&=\left[{\frac {2kT}{m}}\right]^{\frac {a+1}{2}}{\frac {\Gamma \left({\frac {a+1}{2}}\right)}{2}}\end{aligned}}.}

{\displaystyle {\begin{aligned}\int _{0}^{+\infty }v^{a}\exp \left(-{\frac {mv^{2}}{2kT}}\right)dv&=\left[{\frac {2kT}{m}}\right]^{\frac {a+1}{2}}\int _{0}^{+\infty }e^{-x}x^{a/2}\,dx^{1/2}\\[2pt]&=\left[{\frac {2kT}{m}}\right]^{\frac {a+1}{2}}\int _{0}^{+\infty }e^{-x}x^{a/2}{\frac {x^{-1/2}}{2}}\,dx\\[2pt]&=\left[{\frac {2kT}{m}}\right]^{\frac {a+1}{2}}{\frac {\Gamma \left({\frac {a+1}{2}}\right)}{2}}\end{aligned}}.}


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{\displaystyle {\begin{aligned}\langle v\rangle &={\frac {\displaystyle \int \_{0}^{+\infty }v\cdot v^{n-1}\exp \left(-{\tfrac {mv^{2}}{2kT}}\right)\,dv}{\displaystyle \int \_{0}^{+\infty }v^{n-1}\exp \left(-{\tfrac {mv^{2}}{2kT}}\right)\,dv}}\\[4pt]&={\sqrt {\frac {2kT}{m}}}\ {\frac {\Gamma \left({\frac {n+1}{2}}\right)}{\Gamma \left({\frac {n}{2}}\right)}}\end{aligned}}.}

{\displaystyle {\begin{aligned}\langle v\rangle &={\frac {\displaystyle \int _{0}^{+\infty }v\cdot v^{n-1}\exp \left(-{\tfrac {mv^{2}}{2kT}}\right)\,dv}{\displaystyle \int _{0}^{+\infty }v^{n-1}\exp \left(-{\tfrac {mv^{2}}{2kT}}\right)\,dv}}\\[4pt]&={\sqrt {\frac {2kT}{m}}}\ {\frac {\Gamma \left({\frac {n+1}{2}}\right)}{\Gamma \left({\frac {n}{2}}\right)}}\end{aligned}}.}


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{\displaystyle {\begin{aligned}\langle v^{2}\rangle &={\frac {\displaystyle \int \_{0}^{+\infty }v^{2}\cdot v^{n-1}\exp \left(-{\tfrac {mv^{2}}{2kT}}\right)\,dv}{\displaystyle \int \_{0}^{+\infty }v^{n-1}\exp \left(-{\tfrac {mv^{2}}{2kT}}\right)\,dv}}\\[2pt]&=\left[{\frac {2kT}{m}}\right]{\frac {\Gamma ({\frac {n+2}{2}})}{\Gamma ({\frac {n}{2}})}}\\[2pt]&=\left[{\frac {2kT}{m}}\right]{\frac {n}{2}}={\frac {nkT}{m}}\end{aligned}}.}

{\displaystyle {\begin{aligned}\langle v^{2}\rangle &={\frac {\displaystyle \int _{0}^{+\infty }v^{2}\cdot v^{n-1}\exp \left(-{\tfrac {mv^{2}}{2kT}}\right)\,dv}{\displaystyle \int _{0}^{+\infty }v^{n-1}\exp \left(-{\tfrac {mv^{2}}{2kT}}\right)\,dv}}\\[2pt]&=\left[{\frac {2kT}{m}}\right]{\frac {\Gamma ({\frac {n+2}{2}})}{\Gamma ({\frac {n}{2}})}}\\[2pt]&=\left[{\frac {2kT}{m}}\right]{\frac {n}{2}}={\frac {nkT}{m}}\end{aligned}}.}
* Boltsman kvant tenglamasi
* Maksvell-Boltzman statistikasi
* Maksvell-Jyuttner taqsimoti
* Boltsman taqsimoti
* Rayleigh taqsimoti
* Gazlarning kinetik nazariyasi