8
$\begingroup$

I was asked by a student a derivation problem on 3-form Einstein euqation obtained from Palatini action. He posted some pictures on the book he have been reading.

1 2

I derive firstly the gravitational part from Platini action which reads (for simplification I consider cosmological constant as zero so the term vanishes $\int\sqrt{-g}d^4x\;2\Lambda=0$): \begin{align*} S_G&=\frac{1}{2\kappa}\int d^4x\,\sqrt{-g}R\xrightarrow{\epsilon_{IJKL}e^I\wedge e^J\wedge R^{KL}=-2|e|R\,dx^4} \frac{-1}{4\kappa}\int\epsilon_{IJKL}e^I\wedge e^J\wedge R^{KL}=\frac{-1}{4\kappa}\int \text{Tr}(e\wedge e\wedge R)\\ \delta S_G&=\frac{-1}{4\kappa}\delta\int\text{Tr}(e\wedge e\wedge R)=\frac{-1}{2\kappa}\int\text{Tr}(\delta e\wedge e\wedge R)+\frac{-1}{4\kappa}\int\text{Tr}(e\wedge e\wedge\delta R)\\ &(\text{Tr}(\delta e\wedge e\wedge R)+\text{Tr}(e\wedge\delta e\wedge R)=2\text{Tr}(\delta e\wedge e\wedge R)) \end{align*} These two parts can be calculated as \begin{align*} \int\text{Tr}(\delta e\wedge e\wedge R)&=\int\text{Tr}(e\wedge R\wedge\delta e)=\int\epsilon_{IJKL} e^I\wedge R^{JK}\wedge\delta e^L\\ \int\text{Tr}(e\wedge e\wedge\delta R)&=\int d\text{Tr}(e\wedge e\wedge\delta\omega)-2\int\text{Tr}(De\wedge e\wedge\delta\omega)=-2\int\text{Tr}(\mathscr{T}\wedge e\wedge\delta\omega)\\ &(R=d\omega+\omega\wedge\omega\;,\;De^I=de^I+\omega^I_J\wedge e^J=\mathscr{T}^I) \end{align*} Then come to matter field part: \begin{align*} \delta S_m&=\int d^4x\,\delta(\sqrt{-g}\mathcal{L}_m)\xrightarrow{-g=-\det(g_{\mu\nu})=\det(e)^2=|e|^2}\int d^4x\,\delta(|e|\mathcal{L}_m) =\int|e|d^4x\,\frac{1}{|e|}\delta(|e|\mathcal{L}_m)\\ &=\int\mathcal{V}\frac{1}{|e|}\delta(|e|\mathcal{L}_m)=\int*\left(\frac{1}{|e|}\delta(|e|\mathcal{L}_m)\right)\\ \\ \frac{1}{|e|}\delta(|e|\mathcal{L}_m)&=\frac{1}{\sqrt{-g}}\frac{\delta (\sqrt{-g}\mathcal {L}_m)}{\delta g_{\alpha\lambda}}\delta g_{\alpha\lambda}=\left(\frac{\delta\mathcal{L}_m}{\delta g_{\alpha\lambda}}-\frac{1}{2}g^{\alpha\lambda}\mathcal{L}_m\right)\delta g_{\alpha\lambda}=-\frac{1}{2}T^{\alpha\lambda}\delta g_{\alpha\lambda}\\ &=-\frac{1}{2}T^{\alpha\lambda}\eta_{NL}(e^{N}_\alpha +\delta^{\lambda}_\alpha \delta^{L}_{N} e_{\lambda}^{L})\delta e_{\lambda}^{L}=-\frac{1}{2}(T^{\alpha\lambda}e_{\alpha L}+T^{\lambda \lambda}e_{\lambda L})\delta e_{\lambda}^{L}=-T_L^\lambda\delta e_{\lambda}^{L}\\ \\ *(T_L^\lambda\delta e_{\lambda}^{L})&=\frac{|e|}{(4-0)!}T_L^\lambda\delta e_{\lambda}^{L}\epsilon_{\mu\nu\rho\lambda}dx^\mu\wedge dx^\nu\wedge dx^\rho\wedge dx^\lambda=\frac{1}{4}T_L\wedge\delta e^L \end{align*} Above derivation I have used 4-form voulme and 3-form definition of energy-momentum tensor \begin{align*} \mathcal{V}&=|e|d^4x=\frac{1}{4!}\epsilon_{IJKL}e^I_\mu e^J_\nu e^K_\rho e^L_\lambda dx^\mu\wedge dx^\nu\wedge dx^\rho\wedge dx^\lambda=\frac{1}{4!}\epsilon_{IJKL} e^I\wedge e^J\wedge e^K\wedge e^L\\ T_L&=|e|T^\lambda_L\eta_\lambda=\frac{|e|}{3!}T^\lambda_L\epsilon_{\mu\nu\rho\lambda}dx^\mu\wedge dx^\nu\wedge dx^\rho\\ \end{align*} When we condider within zero torsion spacetime ($\mathscr{T}=0$) then we will encounter the Einstein equation in 3-form: \begin{align*} \epsilon_{IJKL} e^I\wedge R^{JK}=\frac{\kappa}{2} T_L \end{align*} Which is $4\pi G$ gravitation coefficient rather then $2\pi G$ on the book.

Could anybody tell me where did I go wrong?

$\endgroup$
  • 1
    $\begingroup$ If you tell us what book this comes from it will be much easier for us to help you. $\endgroup$ – Blazej Aug 23 '16 at 19:38
  • 1
    $\begingroup$ @Blazej I have searched the key words inside the pictures and found that it should be C. Rovelli's book books.google.de/books/about/… $\endgroup$ – Tom Gao Aug 23 '16 at 19:41
  • $\begingroup$ $\uparrow$ Which page? $\endgroup$ – Qmechanic Apr 14 '17 at 15:36
  • $\begingroup$ @Qmechanic You could see it is §2.1 in that book. $\endgroup$ – Tom Gao Apr 14 '17 at 15:55
3
$\begingroup$

After some days, I suppose I probably have figured out where the problem is.

First, I found a mistake in my original derivation on the 4-form energy-momentum tnsor, lacking of other three terms, so it should be: \begin{align*} \frac{1}{|e|}\delta(|e|\mathcal{L}_m)&=-T^{\alpha\beta}\eta_{NL}(e^{N}_\alpha +\delta^{\beta}_\alpha \delta^{L}_{N} e_{\beta}^{L})\delta e_{\beta}^{L}=-T_L^\beta\delta e_{\beta}^{L}\\ *(T_L^\beta\delta e_{\beta}^{L})&=\frac{|e|}{(4-0)!}(T_L^{\underline\beta}\delta e_{\underline\beta}^{L})\epsilon_{\mu\nu\rho\lambda}dx^\underline\mu\wedge dx^\underline\nu\wedge dx^\underline\rho\wedge dx^\underline\lambda=4\cdot\frac{1}{4}T_L\wedge\delta e^L =T_L\wedge\delta e^L \end{align*} Hence the Einstein's field equation in such case should gain the form as $\epsilon_{IJKL} e^I\wedge R^{JK}=2\kappa T_L$, the coefficient is exactly the one in action $16\pi G$.

Second, We could check this by turning the 3-form back to normal tensor form: \begin{align*} \epsilon_{IJKL} e^I\wedge R^{JK}&=\frac{1}{2}\epsilon_{IJKL} e^I\wedge R^{JK}_{\;\;\;MN} e^M\wedge e^N =\frac{1}{2}\epsilon_{IJKL}\epsilon^{IMNE}R^{JK}_{\;\;\;MN}\eta_{EP}*e^P\\ &=-\frac{1}{2}(3!\delta^M_{[J}\delta_K^N\delta^E_{L]}R^{JK}_{\;\;\;MN}\eta_{EP}*e^P)=2(R^{PL}-\frac{1}{2}\eta_{PL}R)*e^P\\ &=\frac{|e|}{3!}2(R^{PL}-\frac{1}{2}\eta_{PL}R)e^P_\lambda\epsilon_{\mu\nu\rho\lambda}dx^\mu\wedge dx^\nu\wedge dx^\rho\\ &=\frac{|e|}{3!}2(R^L_\lambda-\frac{1}{2}Re_\lambda^L)\epsilon_{\mu\nu\rho\lambda}dx^\mu\wedge dx^\nu\wedge dx^\rho\\ T_L&=\frac{|e|}{3!}T^\lambda_L\epsilon_{\mu\nu\rho\lambda}dx^\mu\wedge dx^\nu\wedge dx^\rho\\ \end{align*} Then the equation could go back to the correct usual way $R^L_\lambda-\frac{1}{2}R e_\lambda^L=8\pi G\,T_\lambda^L$.

One can refer to this book Theory of Gravitational Interactions in Apendix 4, which shows the same result as I have obtained.

Hence unfortunately, Rovelli may have made a mistake in his book.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.