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Tangent87
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I am doing this question on page 75 here (http://www.maths.cam.ac.uk/undergrad/pastpapers/2006/Part_2/list_II.pdf ).
Here is my solution so far:
[tex]\nabla_a\nabla_b\phi=\partial_a\partial_b\phi=\partial_b\partial_a\phi=\nabla_b\nabla_a\phi[/tex]
Then for the energy-momentum tensor I just wrote down this from my notes:
[tex]T_{ab}=\frac{c^4}{8{\pi}G}(R_{ab}-\frac{1}{2}Rg_{ab}+{\Lambda}g_{ab})[/tex] where [tex]R=g^{cd}R_{cd}[/tex].
Now, I believe this is the contracted Bianchi identity: [tex]\nabla^{b}(R_{ab}-\frac{1}{2}g_{ab}R)=0[/tex] which using their expression for [tex]R_{ab}[/tex] gives:
[tex]\nabla^{b}(\partial_a\phi\partial_b\phi)-\frac{1}{2}g_{ab}\nabla^{b}(g^{cd}R_{cd})=0
\Rightarrow \nabla^{b}(\partial_a\phi\partial_b\phi)=\frac{1}{2}g_{ab}g^{cd}\nabla^{b}(\partial_c\phi\partial_d\phi)[/tex].
Now I'm stuck cos I don't see what's going to cancel to give me their (*), do I have to use the energy-momentum tensor somehow?
Thanks for any help.
Here is my solution so far:
[tex]\nabla_a\nabla_b\phi=\partial_a\partial_b\phi=\partial_b\partial_a\phi=\nabla_b\nabla_a\phi[/tex]
Then for the energy-momentum tensor I just wrote down this from my notes:
[tex]T_{ab}=\frac{c^4}{8{\pi}G}(R_{ab}-\frac{1}{2}Rg_{ab}+{\Lambda}g_{ab})[/tex] where [tex]R=g^{cd}R_{cd}[/tex].
Now, I believe this is the contracted Bianchi identity: [tex]\nabla^{b}(R_{ab}-\frac{1}{2}g_{ab}R)=0[/tex] which using their expression for [tex]R_{ab}[/tex] gives:
[tex]\nabla^{b}(\partial_a\phi\partial_b\phi)-\frac{1}{2}g_{ab}\nabla^{b}(g^{cd}R_{cd})=0
\Rightarrow \nabla^{b}(\partial_a\phi\partial_b\phi)=\frac{1}{2}g_{ab}g^{cd}\nabla^{b}(\partial_c\phi\partial_d\phi)[/tex].
Now I'm stuck cos I don't see what's going to cancel to give me their (*), do I have to use the energy-momentum tensor somehow?
Thanks for any help.
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