Astronet > 1.3 Postroenie sohranyayushihsya tokov i superpotencialov na proizvol'nom vspomogatel'nom fone. Problema edinstvennosti

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1.3 Postroenie sohranyayushihsya tokov i superpotencialov na proizvol'nom vspomogatel'nom fone. Problema edinstvennosti

Rassmotrim proizvol'nuyu kovariantnuyu teoriyu s lagranzhianom [20] (snachala on v neyavno kovariantnom vide): $\hat L = \hat L (A_B; A_{B,\alpha}; A_{B,\alpha\beta})$ , gde A_B -- eto nabor dinamicheskih peremennyh, proizvol'nyh tenzornyh plotnostei. Vneshnee zadannoe fonovoe prostranstvo-vremya vvedem prostoi tozhdestvennoi zamenoi: $A_{B,\tau} \equiv \overline D_\tau A_B - \overline \Gamma^\sigma_{\tau\rho} \left. A_B \right\vert _\sigma^\rho$ . Togda lagranzhian perepishetsya v yavno kovariantnoi forme:

$\begin{displaymath} \hat L = {\hat{\cal L}}_c = {{\hat{\cal L}}}_c (A_B; \overline D_\alpha A_{B}; \overline D_\beta \overline D_\alpha A_{B}). \end{displaymath}$

(1.36)

My sleduem Mickevichu [8], chtoby opredelit' proizvodnuyu Li:

$\begin{displaymath} {\pounds_\xi} A_B \equiv - \xi^\alpha \overline D_\alpha A_B... ...left.A_B\right\vert^\alpha_\beta \overline D_\alpha \xi^\beta. \end{displaymath}$

(1.37)

Poskol'ku lagranzhian (1.36) -- eto skalyarnaya plotnost', sostavlyaem dlya nego tozhdestvo Neter: ${\pounds_\xi} {\hat{\cal L}}_c +\partial_\alpha \left(\xi^\alpha {\hat{\cal L}}_c\right) \equiv 0$ , kotoroe tozhdestvenno preobrazuetsya v

	-	$\displaystyle \left[{{\delta {\hat{\cal L}}_c} \over {\delta A_B}} \overline D_... ...} \over {\delta A_B}} \left.A_B\right\vert^\beta_\alpha\right)\right]\xi^\alpha$
	+	$\displaystyle \overline D_\alpha \left[\hat u^\alpha_\sigma\xi^\sigma + \hat m^... ...\beta}_\sigma \overline D_{\beta} \overline D_{\tau}\xi^\sigma\right] \equiv 0,$	(1.38)

gde koefficienty odnoznachno opredeleny lagranzhianom sleduyushim obrazom:

$\displaystyle \hat u^\alpha_\sigma$	$\textstyle \equiv$	$\displaystyle {\hat{\cal L}}_c \delta^\alpha_\sigma + {{\delta {\hat{\cal L}}_c} \over {\delta A_B}} \left.A_B\right\vert^\alpha_\sigma$
	-	$\displaystyle \left[{{\partial {\hat{\cal L}}_c} \over {\partial (\overline D_\... ...ine D_\beta \overline D_\alpha A_{B})}}\right) \right] \overline D_\sigma A_{B}$
	-	$\displaystyle {{\partial {\hat{\cal L}}_c} \over {\partial (\overline D_\beta \... ...A_{B}- \hat n^{\alpha\tau\beta}_\lambda \overline R^\lambda_{ \tau\beta\sigma},$	(1.39)

$\displaystyle \hat m^{\alpha\tau}_\sigma$	$\textstyle \equiv$	$\displaystyle \left[{{\partial {\hat{\cal L}}_c} \over {\partial (\overline D_\... ...ta \overline D_\alpha A_{B})}}\right)\right] \left.A_{B}\right\vert^\tau_\sigma$
	-	$\displaystyle {{\partial {\hat{\cal L}}_c} \over {\partial (\overline D_\tau \o... ...rline D_\alpha A_{B})}} \overline D_\beta (\left.A_{B}\right\vert^\tau_\sigma),$	(1.40)

$\begin{displaymath} \hat n^{\alpha\tau\beta}_\sigma \equiv {\textstyle{\frac{1}{... ...D_\alpha A_{B})}} \left.A_{B}\right\vert^\beta_\sigma\right]. \end{displaymath}$

(1.41)

1.3.1 Obobshennyi tok

Podrobnyi analiz (1.38) daet drugoe sil'noe tozhdestvo:

$\begin{displaymath} {{\delta {\hat{\cal L}}_c} \over {\delta A_B}} \overline D_\... ...delta A_B}} \left.A_B\right\vert^\beta_\alpha\right) \equiv 0, \end{displaymath}$

(1.42)

ispol'zovanie kotorogo v tozhdestve (1.38) privodit ego k vidu

$\begin{displaymath} \overline D_\alpha \left[\hat u^\alpha_\sigma\xi^\sigma + \h... ...erline D_{\beta} \overline D_{\tau}\xi^\sigma\right] \equiv 0. \end{displaymath}$

(1.43)

Vyrazhenie pod divergenciei v etom tozhdestve vpolne mozhet byt' interpretirovano kak obobshennyi tok:

$\begin{displaymath} \hat i^\alpha \equiv - \left[\hat u^\alpha_\sigma\xi^\sigma ... ...sigma \overline D_{\beta} \overline D_{\tau}\xi^\sigma\right], \end{displaymath}$

(1.44)

a samo tozhdestvo Neter (1.43) perepisyvaetsya v vide:

$\begin{displaymath} \overline D_\alpha \hat i^\alpha \equiv \partial_\alpha \hat i^\alpha \equiv 0. \end{displaymath}$

(1.45)

1.3.2 Obobshennyi superpotencial

Perepishem tozhdestvo (1.43) v vide:

$\begin{displaymath} \hat{\cal O}_\sigma\xi^\sigma + \hat{\cal O}^\alpha_\sigma \... ...ta}_\sigma \overline D_{(\alpha\tau\beta)}\xi^\sigma \equiv 0. \end{displaymath}$

(1.46)

Poskol'ku komponenty $\xi^\mu$ i proizvodnye ot nih proizvol'ny v kazhdoi tochke, to koefficienty v (1.46) po otdel'nosti takzhe tozhdestvenno ravny nulyu. Eto daet nabor tozhdestv:

$\displaystyle \overline D_\alpha \hat u^\alpha_\sigma + {\textstyle{\frac{1}{2}... ...\tau\beta}_\lambda \overline D_\beta \overline R^{ \lambda}_{\sigma \tau\alpha}$	$\textstyle \equiv$	0,
$\displaystyle \hat u^\alpha_\sigma + \overline D_\lambda \hat m^{\lambda \alpha... ...r 3}} \hat n^{\lambda\tau\beta}_\sigma \overline R^{\alpha}_{ \tau\beta\lambda}$	$\textstyle \equiv$	0,
$\displaystyle \hat m^{(\alpha\tau)}_\sigma + \overline D_\lambda \hat n^{\lambda(\alpha\tau)}_\sigma$	$\textstyle \equiv$	0,
$\displaystyle \hat n^{(\alpha\tau\beta)}_\sigma$	$\textstyle \equiv$	0.	(1.47)

Sama forma tozhdestva (1.45) govorit o tom, chto obobshennyi tok dolzhen vyrazhat'sya v vide:

$\begin{displaymath} \hat i^\alpha \equiv \overline D_\tau \hat \Phi^{\alpha\tau}(\xi), \end{displaymath}$

(1.48)

gde superpotencial v pravoi chasti tozhdestvenno udovletvoryaet: $\overline D_{\alpha\tau} \hat \Phi^{\alpha\tau}(\xi) \equiv 0$ . Deistvitel'no, ispol'zuya tozhdestva (1.47) v opredelenii toka (1.44) my predstavlyaem ego v vide (1.48), gde obobshennyi superpotencial imeet vid:

$\begin{displaymath} \hat \Phi^{\alpha\tau} (\xi) \equiv \left(\hat m^{\tau\alpha... ...n^{[\alpha\tau]\lambda}_\sigma \overline D_\lambda \xi^\sigma. \end{displaymath}$

(1.49)

Uchet tret'ego iz tozhdestv (1.47) govorit, chto superpotencial (1.49) yavno antisimmetrichen.

1.3.3 Vklad ot divergencii v lagranzhiane

Prostym, no ochen' vazhnym yavlyaetsya vopros o vklade v tok i superpotencil ot divergencii v lagranzhiane. Otvet byl dan eshe Kacem [14]. Zdes' my kratko daem rezul'tat. Poskol'ku $\Delta_{(div)} \hat L = \partial_\nu \hat k^\nu$ takzhe yavlyaetsya skalyarnoi plotnost'yu, to dlya etoi dobavki otdel'no vypolnyaetsya tozhdestvo Neter: $\partial_\alpha ({\pounds_\xi} \hat k^\alpha + \xi^\alpha \partial_\nu \hat k^\nu ) \equiv 0$ , prostoi analiz kotorogo privodit k dobavkam v koefficientah (1.39) - (1.41) i v superpotenciale (1.49):

$\displaystyle \Delta_{(div)}\hat u^\alpha_\sigma$	=	$\displaystyle 2 \overline D_\tau\left(\delta^{[\alpha}_\sigma \hat k^{\tau]} \right),$
$\displaystyle \Delta_{(div)}\hat m^{\alpha\tau}_\sigma$	=	$\displaystyle 2 \left(\delta^{[\alpha}_\sigma \hat k^{\tau]} \right),$
$\displaystyle \Delta_{(div)}\hat n^{\alpha\tau\beta}_\sigma$	=	0,
$\displaystyle \Delta_{(div)}\hat \Phi^{\alpha\tau}$	=	$\displaystyle -2 \left(\xi^{[\alpha} \hat k^{\tau]} \right).$	(1.50)

Neobhodimo otmetit', chto forma etih dobavok nikak ne zavisit ot struktury vektornoi plotnosti $\hat k^{\tau}$ !

1.3.4 Neopredelennost' i edinstvennost' v opredelenii tokov i superpotencialov

Horosho izvestno, chto bez izmeneniya tozhdestva (1.45) k toku $\hat i^\alpha$ mozhet byt' dobavlena proizvol'naya velichina $\Delta \hat i^\alpha(\xi)$ , lish' by divergenciya ot nee tozhdestvenno obrashalas' v nul': $\partial_\alpha \left(\Delta \hat i^\alpha(\xi)\right) \equiv 0$ . Odnako, ,,isporchennyi'' tok tem zhe samym sposobom mozhet byt' ,,ispravlen''. Velichina $\Delta \hat i^\alpha(\xi)$ ne svyazana s lagranzhianom i proceduroi Neter. S drugoi storony, v opredelenii (1.44) vse koefficenty (1.39) - (1.41) strogo opredeleny lagranzhianom, zadannym v obshei (neyavnoi) forme. V formule (1.44) net ni odnogo chlena divergenciya ot kotorogo yavno by obrashalas' v nul'. Tozhdestvo (1.45) vypolnyaetsya kak vse vyrazhenie v celom, kak i dolzhno byt' dlya edinogo sohranyayushegosya toka Neter. Takim obrazom, mozhno sdelat' utverzhdenie:

Tok (1.44) -- $\hat i^\alpha$ -- v smysle procedury Neter opredelyaetsya edinstvennnym obrazom lagranzhianom teorii.

Podobnye vyvody spravedlivy i dlya opredeleniya superpotenciala (1.49). V principe, bez izmeneniya toka $\hat i^\alpha$ k superpotencialu v (1.48) mozhet byt' dobavlena proizvol'naya velichina $\Delta \hat \Phi^{\alpha\tau}(\xi)$ , lish' by divergenciya ot nee tozhdestvenno obrashalas' v nul': $\partial_\tau \left(\Delta \hat \Phi^{\alpha\tau}(\xi)\right) \equiv 0$ . Odnako, eta dobavka nikak ne svyazana s lagranzhianom i proceduroi Neter. Kak ee dobavili, tochno takzhe legko unichtozhit'. S drugoi storony, v opredelenii (1.49) vse chleny strogo opredeleny lagranzhianom i opisannoi proceduroi -- ih nevozmozhno otkinut'. Poetomu my delaem utverzhdenie:

Superpotencial (1.49) -- $\hat \Phi^{\alpha\tau}(\xi)$ -- v smysle procedury Neter opredelyaetsya edinstvennnym obrazom lagranzhianom teorii.

Sdelav eto utverzhdenie, my, krome togo, proillyustriruem, chto dvusmyslennost' v superpotenciale ne opasna, vo vsyakom sluchae dlya postroeniya global'nyh sohranyayushihsya velichin. Tak, tozhdestvo $\partial_\tau \left(\Delta \hat \Phi^{\alpha\tau}(\xi)\right) \equiv 0$ v svoyu ochered' govorit o tom, chto dobavka dolzhna byt' vyrazhena kak $\Delta \hat \Phi^{\alpha\beta}(\xi) \equiv \partial_\gamma \hat j^{\alpha\beta\gamma}(\xi)$ , gde $\partial_{\beta\gamma} \hat j^{\alpha\beta\gamma}(\xi) \equiv 0$ . Sochetanie trebovaniya kovariantnosti i neobhodimosti ispol'zovat' chastnye proizvodnye dlya postroeniya zakonov sohraneniya i global'nyh velichin privodit k tomu, chto velichina $\hat j^{\alpha\beta\gamma}(\xi) = \hat j^{[\alpha\beta\gamma]}(\xi)$ , to est' dolzhna byt' antisimmetrichna po vsem indeksam. Teper' perepishem zakon sohraneniya (1.48) v izmenennom vide:


		$\displaystyle \hat i^\alpha \equiv \partial_\beta \left[\hat \Phi^{\alpha\beta}(\xi)+ \partial_\gamma \hat j^{\alpha\beta\gamma}(\xi)\right],$

chto daet vozmozhnost' postroit' sohranyayushiesya velichiny analogichno (1.30):


		$\displaystyle {\cal P}(\xi) = \int_{\Sigma} \hat i^0(\xi)d^3 x = \oint_{\partia... ...[\hat \Phi^{[0n]}(\xi)+ \partial_m \left(\hat j^{[0nm]}(\xi)\right)\right]ds_n.$

V silu teremy Stoksa poslednii chlen ne daet vklada v integral i, takim obrazom, ne izmenyaet velichinu ${\cal P}(\xi)$ .

1.3.5 Edinstvennost' tokov i superpotencialov v opredelenii KBL

Vernemsya k voprosu b) v konce predydushei chasti o edinstvennosti velichin v opredelenii KBL. Prezhde vsego obsudim lagranzhian (1.27). Perepishem skalyarnuyu kriviznu v yavno ,,kovariantizovannom'' vide:


		$\displaystyle \hat R = \hat g^{\theta\sigma}\left( \overline D_\rho \Delta^\rho... ...lta^\eta_{\theta\rho}\right) + \hat g^{\theta\sigma}\overline R_{\theta\sigma},$

gde, napomnim,

$\hat k^\mu = \hat g^{\mu\rho} \Delta^\sigma_{\rho\sigma} - \hat g^{\rho\sigma} \Delta^\mu_{\rho\sigma}$ i

$\Delta^\alpha_{\mu\nu} \equiv \Gamma^\alpha_{\mu\nu} - \overline{\Gamma^\alpha_... ...ta\nu} + \overline D_{\nu} g_{\beta\mu} - \overline D_{\beta} g_{\mu\nu}\right)$ .

S uchetom togo, chto my znaem kakoi vklad dayut divergencii (sm. (1.50)), lagranzhian (1.27) yavlyaetsya kak raz togo vida, chto i lagranzhian (1.36) v etoi chasti. Sledovatel'no, lagranzhian (1.27) mozhet byt' podstavlen v obshie formuly etoi chasti. Pryamaya podstanovka pokazyvaet, chto vse formuly KBL poluchayutsya etim putem. Tak, podstanovka (1.27) v obobshennyi tok (1.44) daet KBL tok (1.32). Bolee detal'no: Podstanovka v koefficient (1.39) i uchet uravnenii Einshteina daet tenzor energii-impul'sa (1.33). Podstanovka v (1.40) daet spinovyi koefficient (1.35). Podstanovka v (1.41) daet tochno Z-chlen v (1.32). (Neobhodimo tol'ko uchityvat' raznicu znakov v formulah (1.39) - (1.41) i v opredelenii toka (1.44).) Takoe zhe sootvetstvie i mezhdu superpotencialami: (1.49) perehodit tochno v (1.31). Iz skazannogo ostaetsya sdelat' vyvod:

Rezul'taty KBL opredeleny odnoznachno v smysle procedury Neter lagranzhianom modeli (1.27).

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