- #1
leo.
- 96
- 5
I'm going through the "Advanced Lectures on General Relativity" by G. Compère and got stuck with solving one set of conditions on the subject of asymptotic flatness. Let ##(M,g)## be ##4##-dimensional spacetime and ##(u,r,x^A)## be a chart such that the coordinate expression of ##g## is in Bondi gauge, so that ##x^A## are coordinates on the ##2##-sphere with ##A=2,3## and: $$g_{rr}=g_{rA}=0,\quad \partial_r\det\frac{g_{AB}}{r^2}=0.\tag{1}$$ Let ##X^\mu## be a vector field. The infinitesimal transformation generated by ##X^\mu## preserves the Bondi gauge (1) if $$L_X g_{rr}=0,\quad L_X g_{rA}=0,\quad L_X\partial_r\det\frac{g_{AB}}{r^2}=0,\tag{2}$$ where ##L_X## is the Lie derivative with respect to ##X^\mu##. Compère says in page 76 that it is possible to solve (2) exactly and write the components ##X^\mu## in terms of four functions of just ##(u,x^A)##. That's what I'm trying to do.
I've tried writing down the equations (2) explicitly using that $$L_X g_{\mu\nu}=X^\lambda \partial_\lambda g_{\mu\nu}+(\partial_\mu X^\lambda)g_{\lambda \nu}+(\partial_\nu X^\lambda)g_{\mu \lambda}\tag{3}.$$
By employing (3) and the Bondi gauge conditions (1) I was able to write the first two equations in (2) as $$2(\partial_r X^u) g_{ur} = 0,\quad (\partial_r X^u)g_{uA}+(\partial_r X^B)g_{AB}+(\partial_A X^u)g_{ur}=0.\tag{4}$$ Now there is that determinant equation. Since the determinant of ##\frac{g_{AB}}{r^2}## is just a real-valued function, I think the Lie derivative becomes just $$X^\mu \partial_\mu \partial_r \det\frac{g_{AB}}{r^2}=0.\tag{5}$$ Then I have considered expanding the determinant using the Levi-Civita symbol as $$\det \frac{g_{AB}}{r^2}=\frac{1}{r^{4}2!}\sum \varepsilon_{A_1A_2}\varepsilon_{B_1B_2}g_{A_1B_1}g_{A_2B_2}\tag{6},$$ and therefore writing the equation as $$X^\mu \partial_\mu \partial_r \frac{1}{r^{4}2!}\sum \varepsilon_{A_1A_2}\varepsilon_{B_1B_2}g_{A_1B_1}g_{A_2B_2}=0\tag{7}.$$
But now I'm stuck. Indeed the first equation in (4) can be solved assuming ##g_{ur}\neq 0## to give ##X^u = f(u,x^A)##. Now I have the other equation in (4) which doesn't seem immediate to solve. Finally there is the determinant equation, but the way I have written it I can't see how it can be solved in terms of functions of just ##(u,x^A)##.
So how can I proceed to do what Compère says and "solve (2) exactly expressing the components ##X^\mu## in terms of functions of just ##(u,x^A)##"? I don't want a full solution, I prefer some guidance on how to continue by myself.
Thanks a lot !
I've tried writing down the equations (2) explicitly using that $$L_X g_{\mu\nu}=X^\lambda \partial_\lambda g_{\mu\nu}+(\partial_\mu X^\lambda)g_{\lambda \nu}+(\partial_\nu X^\lambda)g_{\mu \lambda}\tag{3}.$$
By employing (3) and the Bondi gauge conditions (1) I was able to write the first two equations in (2) as $$2(\partial_r X^u) g_{ur} = 0,\quad (\partial_r X^u)g_{uA}+(\partial_r X^B)g_{AB}+(\partial_A X^u)g_{ur}=0.\tag{4}$$ Now there is that determinant equation. Since the determinant of ##\frac{g_{AB}}{r^2}## is just a real-valued function, I think the Lie derivative becomes just $$X^\mu \partial_\mu \partial_r \det\frac{g_{AB}}{r^2}=0.\tag{5}$$ Then I have considered expanding the determinant using the Levi-Civita symbol as $$\det \frac{g_{AB}}{r^2}=\frac{1}{r^{4}2!}\sum \varepsilon_{A_1A_2}\varepsilon_{B_1B_2}g_{A_1B_1}g_{A_2B_2}\tag{6},$$ and therefore writing the equation as $$X^\mu \partial_\mu \partial_r \frac{1}{r^{4}2!}\sum \varepsilon_{A_1A_2}\varepsilon_{B_1B_2}g_{A_1B_1}g_{A_2B_2}=0\tag{7}.$$
But now I'm stuck. Indeed the first equation in (4) can be solved assuming ##g_{ur}\neq 0## to give ##X^u = f(u,x^A)##. Now I have the other equation in (4) which doesn't seem immediate to solve. Finally there is the determinant equation, but the way I have written it I can't see how it can be solved in terms of functions of just ##(u,x^A)##.
So how can I proceed to do what Compère says and "solve (2) exactly expressing the components ##X^\mu## in terms of functions of just ##(u,x^A)##"? I don't want a full solution, I prefer some guidance on how to continue by myself.
Thanks a lot !