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1.8 Teorema viriala

Predpolozhim, chto uravnenie sostoyaniya stepennoe: $P=K\rho^\gamma$ . Togda udel'naya teplovaya energiya $E={K\over {\gamma-1}}\rho^{\gamma-1}={K\over {\gamma-1}}v^{-( \gamma-1)}$ . My znaem, chto v ravnovesii $\delta {\cal{E}}=0$ pri proizvol'noi $\delta r(m)$ . Pust' $\delta r=(1+\alpha) r \;(\vert\alpha\vert \ll 1)$ . Takoe vozmushenie opisyvaet podobnoe (gomologicheskoe) rasshirenie ili szhatie zvezdy. Togda $v'=(1+ 3\alpha)v, \;\delta U=-\alpha U, \;\delta Q=-3(\gamma-1)\alpha Q$ . Sledovatel'no,

$\displaystyle \delta {\cal{E}}=-3(\gamma-1)\alpha Q-\alpha U=0,$

otkuda

$\displaystyle Q=-{1\over {3(\gamma-1)}}U$

(eto sootnoshenie i nazyvayut teoremoi viriala). Dlya odnoatomnogo gaza s $\gamma= {5\over 3}$ imeem $Q=-{U\over 2}, \;{\cal{E}}={U\over 2}=-Q$ .

$\begin{wrapfigure}{l}{0.5\textwidth} \epsfxsize =0.45\textwidth \hbox to0.5\textwidth{\hss\epsfbox{fig/f11.ai}\hss} \end{wrapfigure}$

Ris. 11.

Teper' polozhim $r'=\beta r$ , prichem ne budem schitat' $\vert\beta-1\vert$ maloi velichinoi, a ishodnoe sostoyanie -- ravnovesnym. Oboznachim cherez i sootvetstvuyushie velichiny energii ishodnoi modeli. Togda posle preobrazovaniya $U={1\over \beta}U_1, \;Q={1\over {\beta^{3(\gamma-1)}}}Q_1$ (dlya stepennogo uravneniya sostoyaniya). Esli $\gamma= {5\over 3}$ , to $Q={1\over \beta^2}Q_1$ . Kak vyglyadit v etom sluchae krivaya ${\cal{E}} (\beta)$ ? Pri $\beta \to \infty$ asimptotika opredelyaetsya velichinoi , pri $\beta \to 0$ -- velichinoi (ris. 11). Poluchaem, chto pri $\gamma=5/3$ krivaya imeet odin i tol'ko odin minimum, t.e. ravnovesie ustoichivo.

Poluchim teoremu viriala drugim sposobom iz uravneniya ravnovesiya, prichem zavisimost' $P(\rho)$ mozhet byt' proizvol'noi. (Vyshe pri vyvode teoremy viriala iz variacionnogo principa zavisimost' $P=K\rho^\gamma$ byla sushestvenna). Umnozhim uravnenie ravnovesiya (1.6) na :

$\displaystyle -{r\over \rho} \,{dP\over {dr}}={Gm\over r}$

i prointegriruem po

$\begin{displaymath} \begin{array}{ll} \int {Gmdm\over r}&=-\int {r\over \rho} \,... ...ight\vert _0^R+3\int P \,4\pi r^2 dr=3\int PdV, \cr \end{array}\end{displaymath}$

t.e.

$\displaystyle \int {Gmdm\over r}=3\int PdV$

teorema viriala pri proizvol'nom $P(\rho)$ .

Pri stepennom uravnenii sostoyaniya, ispol'zuya $P=(\gamma-1)E \,\rho$ , imeem uzhe izvestnoe sootnoshenie $U=-3(\gamma-1)Q$ .

V deistvitel'nosti uravnenie sostoyaniya ne stepennoe, no dlya mnogih ocenok polezno znat' svoistva zvezd s takim uravneniem sostoyaniya. Dlya stepennogo uravneniya sostoyaniya imeetsya podobie, t.e. dostatochno reshit' zadachu pri dannom $\gamma$ dlya odnogo znacheniya $\rho_c$ , chtoby naiti funkcional'nuyu zavisimost' $M(\rho_c)$ i $R(\rho_c)$ . V sistemu uravnenii

$\displaystyle -{1\over \rho} \,{dP\over {dr}}={Gm\over r^2},\quad {dm\over dr}=4\pi r^2 \,\rho,\quad P=K\rho^\gamma$

vhodyat razmernye konstanty

$\displaystyle [G]={\mbox{sm}^3\over \mbox{g} \cdot \mbox{s}^2}, \;[K]={\mbox{sm... ...\over \mbox{sm}^3}\right)^{-\gamma+1}, \;[\rho_c]= {\mbox{g}\over\mbox{sm}^3}.$

Poetomu, kombiniruya $G, \;K, \;\rho_c$ v razlichnyh stepenyah, mozhno poluchit' massu, radius i drugie harakteristiki zvezdy. Etu zadachu mozhno reshit' formal'no, sostavlyaya sistemu uravnenii tipa

$\displaystyle [R]=$ sm $\displaystyle ^1=[G]^x \,[K]^y \,[\rho_c]^z=$ sm $\displaystyle ^\alpha \,$ g $\displaystyle ^\beta \,$ s $\displaystyle ^\delta\,,$

$\displaystyle \alpha=1, \;\beta=0, \;\delta=0, \;$ t.e. $\displaystyle$

$\displaystyle 3x+(3\gamma-1)y-3z=1,$

$\displaystyle -x+(1-\gamma)y+z=0,$

$\displaystyle -2x-2y=0;$

otkuda $x=-{1\over 2}, \;y={1\over 2}, \;z={{\gamma-2}\over 2}$ , t.e. $R\sim ( K/G)^{1\over 2}\rho_c^{{\gamma-2}\over 2}$ .

Bolee naglyadno eta svyaz' poluchaetsya s pomosh'yu poryadkovyh ocenok:

$\displaystyle P_c \simeq {GM^2\over R^4}, \quad\rho_c \simeq {M\over R^3}, \quad P_c=K\rho_c^\gamma.$

Smysl pervogo sootnosheniya legko ponyat', esli vspomnit', chto sila prityazheniya mezhdu dvumya polovinkami zvezdy $\sim {GM^2\over R^2}$ , a davlenie (sila na edinicu ploshadi, proporcional'noi

) $\sim {GM^2\over R^4}$ . Isklyuchaya iz etih vyrazhenii

, imeem vyrazhenie dlya

, a isklyuchaya

, nahodim

$\displaystyle {P_c\over \rho_c^{4/3}} \sim GM^{2/3} \sim K\rho_c^{\gamma-4/3}, \quad \rho_c \sim \left({GM^{2/3}\over K}\right)^{1\over {\gamma-4/3}}.$

Podcherknem, chto vid krivyh $M(\rho_c)$ i $R(\rho_c)$ zavisit ot bezrazmernoi velichiny $\gamma$ , t.e. krivye dlya raznyh $\gamma$ ne podobny.

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