- #1
Messenger
- 68
- 0
I can fairly well grasp the trace relationship between the Einstein tensor and the Ricci tensor, and see that Ricci tensor is a multiple of the metric. If the cosmological constant is included I don't get why the Einstein tensor shouldn't become a multiple of the metric (leaving out physical considerations) to still achieve a Ricci flat manifold. Any text suggestions to help me explore this appreciated.
[tex]G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}=0[/tex]
[tex]g^{\mu\nu} G_{\mu\nu}=g^{\mu\nu} R_{\mu\nu}-\frac{1}{2}R g^{\mu\nu} g_{\mu\nu}=0[/tex]
[tex]G=R-2R=0[/tex]
[tex]G=-R=0[/tex]
or
[tex]G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}=0[/tex]
[tex]G_{\mu\nu}-\Lambda g_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}=0[/tex]
[tex]\Lambda g_{\mu\nu}-\Lambda g_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}=0[/tex]
[tex]\Lambda g^{\mu\nu} g_{\mu\nu}-\Lambda g^{\mu\nu} g_{\mu\nu}=g^{\mu\nu} R_{\mu\nu}-\frac{1}{2}R g^{\mu\nu}g_{\mu\nu}=0[/tex]
[tex]4\Lambda - 4\Lambda =R- 2R=-R=0[/tex]
[tex]G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}=0[/tex]
[tex]g^{\mu\nu} G_{\mu\nu}=g^{\mu\nu} R_{\mu\nu}-\frac{1}{2}R g^{\mu\nu} g_{\mu\nu}=0[/tex]
[tex]G=R-2R=0[/tex]
[tex]G=-R=0[/tex]
or
[tex]G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}=0[/tex]
[tex]G_{\mu\nu}-\Lambda g_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}=0[/tex]
[tex]\Lambda g_{\mu\nu}-\Lambda g_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}=0[/tex]
[tex]\Lambda g^{\mu\nu} g_{\mu\nu}-\Lambda g^{\mu\nu} g_{\mu\nu}=g^{\mu\nu} R_{\mu\nu}-\frac{1}{2}R g^{\mu\nu}g_{\mu\nu}=0[/tex]
[tex]4\Lambda - 4\Lambda =R- 2R=-R=0[/tex]