Astronet > 4.4 Metod Neter-Belinfante i polevoi podhod k OTO

po tekstam po klyuchevym slovam v glossarii po saitam perevod po katalogu

<< 4.3 Problemy polevoi formulirovki OTO | Oglavlenie | Literatura k Lekcii 4 >>

Razdely

4.4 Metod Neter-Belinfante i polevoi podhod k OTO

Kombinaciya etih dvuh metodov pozvolyaet razreshit' opisannye v predydushei chasti problemy. Rezul'taty etoi chasti byli dolozheny na seminare [24].

4.4.1 Tozhdestvo Kaca-Bichaka-Linden-Bella

Napomnim kak byl poluchen zakon sohraneniya KBL [25]. Dlya lagranzhiana

$\begin{displaymath} {\hat{\cal L}}_G = -{1\over 2\kappa} \left(\hat R - \overline {\hat R} + \partial_\mu \hat k^\mu \right) \end{displaymath}$

(4.33)

c $\hat k^\mu \equiv \hat g^{\mu\rho} \Delta^\sigma_{\rho\sigma} - \hat g^{\rho\sigma} \Delta^\mu_{\rho\sigma}$ i $\Delta^\mu_{\rho\sigma} = \Gamma^\mu_{\rho\sigma}- \overline{ \Gamma^\mu_{\rho\sigma}}$ bylo vypisano tozhdestvo Neter

$\begin{displaymath} {\pounds_\xi} {\hat{\cal L}}_G + \partial_\mu \left(\xi^\mu {\hat{\cal L}}_G\right) \equiv 0, \end{displaymath}$

(4.34)

kotoroe zatem preobrazovano v zakon sohraneniya (sm. (1.29) v lekcii 1):

$\begin{displaymath} J^\mu(\xi) = \partial_\nu \hat J^{\mu\nu} (\xi). \end{displaymath}$

(4.35)

Esli v processe preobrazovanii (4.34) ne ispol'zovat' uravnenii Einshteina i ne vezde yavno predstavlyat' ${\hat{\cal L}}_G$ , to vmesto (4.35) my poluchim tozhdestvo

$\displaystyle \left[ {{\partial {\hat{\cal L}}_G} \over {\partial \left(\bar D_... ... \overline D_\nu g_{\rho\sigma} - {\hat{\cal L}}_G \delta^\mu_\nu\right]\xi^\nu$	+	$\displaystyle \hat\sigma^{\mu\rho\sigma}\partial_{[\rho}\xi_{\sigma]}$
$\displaystyle - {{\delta {{\hat{\cal L}}_G}} \over {\delta g_{\rho\sigma}}} \le... ... g_{\rho\sigma}}} \left.{ \overline g_{\rho\sigma}} \right\vert^\mu_\nu \xi^\nu$	+	$\displaystyle \hat Z^\mu_{(1)} \equiv \partial_\nu \hat J^{\mu\nu},$	(4.36)

gde superpotencial kak i prezhde (sm. (31) v lekcii 1) imeet vid:

$\begin{displaymath} {\hat J^{\mu\nu}} ={1\over \kappa}\big({D^{[\mu}\hat \xi^{\n... ...hat \xi^{\nu]}}\big)+ {1\over \kappa}\hat \xi^{[\mu} k^{\nu]}, \end{displaymath}$

(4.37)

a tok v levoi chasti (4.36) imeet bolee obshii vid, chem v (4.35).

4.4.2 Tozhdestvo Neter dlya polevogo lagranzhiana

Gravitacionnyi lagranzhian v polevoi formulirovke (sm. (4.26)) est'

$\begin{displaymath} -{1\over 2\kappa}\hat L^{g} =-{1\over 2\kappa} \left[\hat R... ...\mu\nu} - \overline{\hat R} + \partial_\mu{\hat k}^\mu\right]. \end{displaymath}$

(4.38)

Uchtem svyaz' ${\hat l^{\mu\nu} \equiv \hat g^{\mu\nu} - \overline {\hat g^{\mu\nu}}}$ i oboznachim $-{1\over 2\kappa}\hat L^{g} \equiv {\hat{\cal L}}^{(2)}$ . Posle etogo vidno, chto polevoi lagranzhian (4.38) i lagranzhian KBL (4.33) svyazany sootnosheniem:

$\begin{displaymath} {\hat{\cal L}}^{(2)} \equiv {\hat{\cal L}}_G + {1\over 2\kap... ...^{\mu\nu} - \overline {\hat g^{\mu\nu}}) \overline R_{\mu\nu}, \end{displaymath}$

(4.39)

yavnaya forma kotorogo

$\begin{displaymath} {\hat{\cal L}}^{(2)} \equiv {1\over 2\kappa}\left[{{\hat g}^... ...-\Delta^\rho_{\mu\sigma}\Delta^\sigma_{\rho\nu}\right)\right]. \end{displaymath}$

(4.40)

Teper' ispol'zuem tozhdestvo Neter ${\pounds_\xi} {\hat{\cal L}}^{(2)} + \partial_\mu \left(\xi^\mu {\hat{\cal L}}^{(2)}\right) \equiv 0$ , chtoby preobrazovat' ego k vidu:

$\displaystyle \left[{{\partial {\hat{\cal L}}^{(2)}} \over {\partial \left(\bar... ...erline D_\nu g_{\rho\sigma} - {\hat{\cal L}}^{(2)}\delta^\mu_\nu\right] \xi^\nu$	+	$\displaystyle \hat\sigma^{\mu\rho\sigma}\partial_{[\rho}\xi_{\sigma]}$
$\displaystyle - {{\delta {\hat{\cal L}}^{(2)}} \over {\delta g_{\rho\sigma}}} \... ...lta \bar g_{\rho\sigma}}} \left.\bar g_{\rho\sigma} \right\vert^\mu_\nu \xi^\nu$	+	$\displaystyle {\hat Z}^{�\mu}_{(2)} \equiv - \partial_\nu \left({\hat M}^{�\mu\nu}_{(2)\lambda} \xi^\lambda \right),$	(4.41)

gde

$\begin{displaymath} {\hat M}^{�\mu\nu}_{(2)\lambda} = -{\hat M}^{�\nu\mu}_{(2)\l... ...\right)}} \left.{\bar g_{\rho\sigma}} \right\vert^\nu_\lambda. \end{displaymath}$

(4.42)

Vychtem iz tozhdestva (4.36) tozhdestvo (4.41), uchtem ravenstvo


		$\displaystyle {{\delta {\hat{\cal L}}^{(2)}} \over {\delta \bar g_{\rho\sigma}}... ..._{\rho\sigma} \right\vert^\mu_\nu \equiv {1 \over {\kappa}}\hat G^{L\mu}_{\nu},$

svyaz' (4.39), opredelenie (4.42) i poluchim novoe tozhdestvo:

$\begin{displaymath} {1 \over {\kappa}} \hat G^{L\mu}_{\nu}\xi^\nu + {1 \over \ka... ...mbda \right) \equiv \partial_\nu \left(*\hat J^{\mu\nu}\right) \end{displaymath}$

(4.43)

predstavlennoe v terminah polevogo podhoda.

4.4.3 Superpotencial v polevoi formulirovke OTO

Eshe raz zapishem obshee uravnenie polevoi vormulirovki:

$\begin{displaymath} \hat G^L_{\mu\nu} + \hat \Phi^L_{\mu\nu} = \kappa\left({\hat t}^g_{\mu\nu} + {\hat t}^m_{\mu\nu}\right), \end{displaymath}$

(4.44)

kotoroe nuzhno ispol'zovat', chtoby prevratit' sil'noe tozhdestvo (4.43) v slabyi zakon sohranenniya. Odnako, chto delat' s velichinoi $\hat \Phi^L_{\mu\nu}$ , kotoraya otsutstvuet v (4.43)? Okazyvaetsya posle ispol'zovaniya opredelenii (4.25), (4.29) i (4.30) uravnenie (4.44) mozhno perepisat' v vide:

$\begin{displaymath} \hat G^L_{\mu\nu} = \kappa\left({\hat t}^g_{\mu\nu} + {\hat t}^M_{\mu\nu} - \overline{ {\hat t}^M_{\mu\nu}}\right), \end{displaymath}$

(4.45)

gde


		$\displaystyle {\hat t}^M_{\mu\nu} \equiv 2 {{\delta{\hat{\cal L}}^M \left(\over... ...iv 2 \overline{{{\delta{\hat{\cal L}}^M}\over {\delta \overline{g^{\mu\nu}}}}}.$

Teper', konechno, istochnik v (4.45) ne yavlyaetsya simmetrichnym tenzorom energii-impul'sa, sootvetstvuyushim dinamicheskomu lagranzhianu (4.25): ${\hat t}^m_{\mu\nu} \neq {\hat t}^M_{\mu\nu} - \overline{ {\hat t}^M_{\mu\nu}} \equiv \delta \hat t^M_{\mu\nu}$ . Posle podstanovki uravneniya (4.45) v tozhdestvo (4.43) poluchim:

$\begin{displaymath} \left(\delta {\hat t}_{\nu}^{M\mu} + {\hat t}_{\nu}^{g\mu} ... ...+ *\hat Z^\mu = {\partial_\nu} \left( *\hat J^{\mu\nu}\right). \end{displaymath}$

(4.46)

Obsudim eto uravnenie. Prezhde vsego, eto est' analog uravneniya KBL (4.35), zdes' v pravoi chasti takzhe stoit divergenciya teper' ot novogo superpotenciala $*\hat J^{\mu\nu}$ . Sledovatel'no, levaya chast' est' sohranyayushiisya tok. V zakone sohraneniya uchastvuyut proizvol'nye vektory $\xi^\alpha$ i ono spravedlivo na proizvol'no iskrivlennyh fonah. Takim obrazom mozhno zaklyuchit', chto postroenie (4.46) reshaet srazu pervye tri problemy polevogo podhoda, otmechennye v predydushei chasti.

Pochemu etot uspeh ne byl dostignut ran'she? Okazyvaetsya bylo oshibochnym predpolozhenie (dlya postroeniya sohranyayushegosya toka na proizvol'nom fone) ispol'zovat' istochnik ${\hat t}^m_{\mu\nu}$ v pravoi chasti (4.44), opredelennyi standartnym obrazom v (4.30). Okazyvaetsya neobhodimo ispol'zovat' vozmusheniya ${\hat t}^M_{\mu\nu} - \overline{ {\hat t}^M_{\mu\nu}} \equiv \delta \hat t^M_{\mu\nu}$ . Otmet'te takzhe, chto v levoi chasti (4.46) soderzhitsya chlen vzaimodeistviya s fonom, prodeklarirovannyi kachestvenno nami ran'she [13]. Z-chlen v (4.46), kak i vezde, obrashaetsya v nul' dlya vektorov Killinga fona.

4.4.4 Sravnenie rezul'tatov metoda Neter-Belinfante i polevogo podhoda

Glavnym rezul'tatom lekcii 2 bylo postroenie zakona sohraneniya (sm. (24) v lekcii 2):

$\begin{displaymath} \hat {\cal T}^\mu_\nu\xi^\nu + \hat {\cal Z}^\mu = \partial_\nu\hat {\cal I}^{\mu\nu}, \end{displaymath}$

(4.47)

gde obobshennyi tenzor energii-impul'sa imeet vid

$\begin{displaymath} \hat {\cal T}^{\mu\nu} = (\hat T^{(\mu}_\rho\overline g^{\nu... ... {1\over \kappa} {\hat l}^{\lambda[\mu} \bar R^{\nu]}_\lambda, \end{displaymath}$

(4.48)

a superpotencial predstavlyaet soboi summu KBL superpotenciala i popravki Belinfante:

$\begin{displaymath} {\cal I}^{\mu\nu} \equiv \hat J^{\mu\nu}+ \hat{S}^{\mu\nu\rho}\xi_\rho. \end{displaymath}$

(4.49)

Perepishem uravnenie (4.46) v kompaktnom vide:

$\begin{displaymath} *\hat \Pi^{\mu}_\nu \xi^\nu + *\hat Z^\mu = {\partial_\nu} \left( *\hat J^{\mu\nu}\right), \end{displaymath}$

(4.50)

gde

$\displaystyle *\hat \Pi_{\mu\nu}$	$\textstyle \equiv$	$\displaystyle {\hat t}_{\mu\nu}^{M} - \overline {{\hat t}_{\mu\nu}^{M}} + {\hat... ...\mu\nu}^{g} + {1 \over \kappa} \hat l_{\mu}^{\lambda} \overline R_{\lambda\nu},$
$\displaystyle {*\hat J}^{\mu\nu}$	$\textstyle \equiv$	$\displaystyle \hat J^{\mu\nu} + {\hat M}^{�\nu\mu}_{(2)\lambda} \xi^\lambda,$
$\displaystyle {\hat t}^M_{\mu\nu}$	=	$\displaystyle \hat T_{\mu\nu} -{\textstyle{\frac{1}{2}}} g_{\mu\nu}\hat T - {\t... ...eft(\hat T_{\rho\sigma} -{\textstyle{\frac{1}{2}}} g_{\rho\sigma}\hat T\right),$
$\displaystyle \overline{{\hat t}^M_{\mu\nu}}$	=	$\displaystyle \overline{\hat T}_{\mu\nu}.$	(4.51)

Okazyvaetsya, chto popravka Belinfante, opredelennaya v (23) v lekcii 2 i spinovyi chlen (4.42) sovpadayut: $\hat{S}^{\mu\nu\rho} = {\hat M}^{\nu\mu\rho}_{(2)}$ . Uchityvaya etot fakt v (4.49) i v superpotenciale v (4.51), zaklyuchaem

$\displaystyle *\hat J^{\mu\nu} ={\cal I}^{\mu\nu}$	=	$\displaystyle {1 \over \kappa} \hat l^{\rho[\mu}\overline D_\rho\xi^{\nu]} + \hat {\cal P}^{\mu\nu}{_\lambda} \xi^\lambda$
	=	$\displaystyle {1 \over \kappa} \left(\hat l^{\rho[\mu}\overline D_\rho\xi^{\nu]... ...igma \hat l^{\nu]\sigma}-\bar D^{[\mu}\hat l^{\nu ]}_\sigma \xi^\sigma \right),$

i $*\hat Z^\mu = \hat{\cal Z}^\mu$ (poslednee ravenstvo proveryaetsya takzhe pryamym vychisleniem). Poetomu dlya (4.47) i (4.50) sleduet:

$\begin{displaymath} *\hat \Pi^{\mu\nu} = \hat {\cal T}^{\mu\nu}. \end{displaymath}$

(4.52)

Takim obrazom vse svoistva $\hat {\cal T}^{\mu\nu}$ issledovannye v lekcii 2 v polnoi mere otnosyatsya k $*\hat \Pi^{\mu\nu}$ . Podvedem itog:

Zakon sohraneniya (4.50) poluchen v ramkah polevogo podhoda i yavlyaetsya tochno uravneniem (4.47), kotoroe est' sledstvie metoda Neter-Belin- fante. Dva raznyh podhoda na urovne samyh obobshennyh zakonov sohraneniya dayut edinyi otvet.

Sravnivaya tenzory energii-impul'sa (4.48) i v (4.51) nahodim, chto tenzory energii-impul'sa gravitacionnogo polya ne sovpadayut: ${\hat t}^{\mu\nu}_{g} \neq \hat \tau^{\mu\nu}$ . (Tol'ko dlya sluchaya, kogda $\overline R_{\mu\nu} = 0$ i $R_{\mu\nu} =0$ my poluchaem, chto dolzhno byt' ${\hat t}^{\mu\nu}_{g} = \hat \tau^{\mu\nu}$ .) Tem ne menee, s ispol'zovaniem fonovyh i dinamicheskih uravnenii Einshteina ravenstvo (4.52) podtverzhdaetsya pryamymi raschetami. Takaya situaciya govorit o slozhnosti v opredelenii, naprimer, energii gravitacionnyh voln na dostatochno slozhnyh fonah. Na etu problemu v chastnoi besede ukazal Kopeikin [26].

4.4.5 Razreshenie neopredelennosti Boul'vara-Dezera

Teper' rassmotrim poslednyuyu: chetvertuyu problemu polevogo podhoda -- eto neopredelennost' v vybore razbienii (4.31). Esli my vyberem proizvol'noe iz razbienii (4.31):

$\begin{displaymath} g^A = \bar g^A + h^A, \end{displaymath}$

(4.53)

to shag za shagom sleduya sposobu postroeniya polevoi formulirovki v rabote [15] (poslednii metod izlozhennyi v etoi lekcii) my poluchim uravneniya tipa (4.10), zatem tipa (4.45), i, nakonec, tipa (4.50):

$\begin{displaymath} *\hat \Pi^{\mu}_\nu(\hat l_{(A)}) \xi^\nu + *\hat Z^\mu(\ha... ...= {\partial_\nu} \left( *\hat J^{\mu\nu}(\hat l_{(A)})\right). \end{displaymath}$

(4.54)

Raznica lish' v tom, chto argument v etom uravnenii zavisit ot vybora razbieniya (4.53) i vyrazhaetsya cherez nego kak

$\begin{displaymath} \hat l^{\mu\nu}_{(A)} \equiv h^A {{ \partial \overline {\hat g^{\mu\nu}}} \over {\partial \overline {g^A}}}. \end{displaymath}$

(4.55)

Napomnim, chto superpotencial $*\hat J^{\mu\nu}(\hat l_{(A)})$ lineen po svoim argumentam $\hat l^{\mu\nu}_{(A)}$ . No esli $h_A^1 \neq h_A^2$ , togda $\hat l^{\mu\nu}_{1(A)} = \hat l^{\mu\nu}_{2(A)} +O(l^2) + ...$ . Eto oznachaet, chto neopredelennost' Boul'vara-Dezera v superpotenciale $*\hat J^{\mu\nu}(\hat l_{1(A)}) \neq *\hat J^{\mu\nu}(\hat l_{2(A)}) \neq ....$ imeet mesto nachinaya so vtorogo poryadka.

Vernemsya k metodu Neter-Belinfante: ne bylo ogranichenii v vybore peremennyh, my mogli vybrat' lyubuyu iz nih: $g_{\mu\nu},� g^{\mu\nu},�\hat g^{\mu\nu},...$ ! V rezul'tate, konechnym i lineinym okazalsya edinstvenno KBL superpotencial, $\hat J^{\mu\nu}(\hat l^{\mu\nu})$ , a zatem novyi superpotencial, $\hat {\cal I}^{\mu\nu}(\hat l^{\mu\nu})$ . Tol'ko dlya vybora (4.31), ili v (4.55), edinstvenno $h^A = \hat l^{\mu\nu}$ superpotencial $*\hat J^{\mu\nu}(\hat l_{(A)})$ v (4.54) i $\hat {\cal I}^{\mu\nu}(\hat l^{\mu\nu})$ sovpadayut. Etot fakt svidetel'stvuet o preimushestve vybora (4.22). Sushestvuyut i drugie pokazaniya v pol'zu etogo vybora i razbieniya, sootvetvuyushego emu. Superpotencial Abbotta-Dezera [22] yavlyaetsya odnim iz nabora (4.54), a imenno: $*\hat J^{\mu\nu}_{AD} =*\hat J^{\mu\nu}(h^A = h_{\mu\nu})$ . Odnako my [27] pokazali, chto takoi superpotencial ne daet pravil'nogo Bondi-Saksa impul'sa na nulevoi beskonechnosti, to est' ne udovletvoryaet odnomu iz osnovnyh estestvennyh testov, v to vremya kak $*\hat J^{\mu\nu}(\hat l^{\mu\nu})$ daet nuzhnyi rezul'tat.

<< 4.3 Problemy polevoi formulirovki OTO | Oglavlenie | Literatura k Lekcii 4 >>

Publikacii s klyuchevymi slovami: zakony sohraneniya - Obshaya teoriya otnositel'nosti - gravitaciya Publikacii so slovami: zakony sohraneniya - Obshaya teoriya otnositel'nosti - gravitaciya
Sm. takzhe: Padenie shara v Solnechnoi sisteme Gravitacionnye anomalii na Merkurii Galaktika Arp 94 Sputniki GRAIL sostavlyayut kartu lunnoi gravitacii 70 let Stivenu Hokingu Molotok protiv peryshka na Lune Sputnik Gravity Probe B podtverdil nalichie gravimagnetizma Vse publikacii na tu zhe temu >>

Obsudit' etu publikaciyu

Versiya dlya pechati

Astrometriya - Astronomicheskie instrumenty - Astronomicheskoe obrazovanie - Astrofizika - Istoriya astronomii - Kosmonavtika, issledovanie kosmosa - Lyubitel'skaya astronomiya - Planety i Solnechnaya sistema - Solnce