- #1

- 68

- 0

## Main Question or Discussion Point

In

Changing the symbol of the Einstein tensor and the cosmological constant to L and Ω so as not to confuse myself, I can see how with [tex]R_{\alpha\beta}-\frac{1}{2}g_{\alpha\beta}R=L_{\alpha\beta}+\Omega g_{\alpha\beta}[/tex] if I first set [tex]\Omega=0[/tex] I then get

[tex]R_{\alpha\beta}-\frac{1}{2}g_{\alpha\beta}R=L_{\alpha\beta}[/tex] which is of course nothing more than the EFE. But if instead I take [tex]R_{\alpha\beta}-\frac{1}{2}g_{\alpha\beta}R=L_{\alpha\beta}+\Omega g_{\alpha\beta}[/tex] but let [tex]\Omega\neq 0[/tex] I get [tex]L_{\alpha\beta}+\Omega g_{\alpha\beta}=0[/tex] so that [tex]L_{\alpha\beta}=-\Omega g_{\alpha\beta}[/tex] when spacetime is flat. If I then set [tex]\Omega=0[/tex] I then end up again with the Einstein tensor [tex]R_{\alpha\beta}-\frac{1}{2}g_{\alpha\beta}R=L_{\alpha\beta}[/tex] by pulling a -1 out of the coefficients since [tex]\Omega g_{\alpha\beta}=L_{\alpha\beta}+G_{\alpha\beta}[/tex].

So what I am not understanding is if Einstein required a multiple of the metric in the field equation but it caused him so much anguish to not have flat spacetime, why didn't he just go with a [tex]\Omega g_{\alpha\beta}-L_{\alpha\beta}=G_{\alpha\beta}[/tex] substitution to do so?

*Gravitation*by MTW in section 17.3, they state(I assume it was a typo putting the cosmological constant in the center portion)After much anguish, one concludes that the assumption which one might drop with the least damage to the beauty and spirit of the the theory is the assumption (1), that "G" vanish when spacetime is flat. But even dropping this assumption is painful: (1) although "G" might still be in some sense a measure of geometry, it can no longer be a measure of curvature; and (2) flat, empty spacetime will no longer be compatible with the geometrodynamic law (G≠0 in flat, empty space, whereT=0). Nevertheless, these consequences were less painful to Einstein than a dynamic universe.

The only tensor that satisfies conditions (2) and (3) [with (1) abandoned] is the Einstein tensor plus a multiple of the metric : [tex]``G_{\alpha\beta}''=R_{\alpha\beta}-\frac{1}{2}g_{\alpha\beta}R+\Lambda g_{\alpha\beta}=G_{\alpha\beta}+\Lambda g_{\alpha\beta}[/tex]

Changing the symbol of the Einstein tensor and the cosmological constant to L and Ω so as not to confuse myself, I can see how with [tex]R_{\alpha\beta}-\frac{1}{2}g_{\alpha\beta}R=L_{\alpha\beta}+\Omega g_{\alpha\beta}[/tex] if I first set [tex]\Omega=0[/tex] I then get

[tex]R_{\alpha\beta}-\frac{1}{2}g_{\alpha\beta}R=L_{\alpha\beta}[/tex] which is of course nothing more than the EFE. But if instead I take [tex]R_{\alpha\beta}-\frac{1}{2}g_{\alpha\beta}R=L_{\alpha\beta}+\Omega g_{\alpha\beta}[/tex] but let [tex]\Omega\neq 0[/tex] I get [tex]L_{\alpha\beta}+\Omega g_{\alpha\beta}=0[/tex] so that [tex]L_{\alpha\beta}=-\Omega g_{\alpha\beta}[/tex] when spacetime is flat. If I then set [tex]\Omega=0[/tex] I then end up again with the Einstein tensor [tex]R_{\alpha\beta}-\frac{1}{2}g_{\alpha\beta}R=L_{\alpha\beta}[/tex] by pulling a -1 out of the coefficients since [tex]\Omega g_{\alpha\beta}=L_{\alpha\beta}+G_{\alpha\beta}[/tex].

So what I am not understanding is if Einstein required a multiple of the metric in the field equation but it caused him so much anguish to not have flat spacetime, why didn't he just go with a [tex]\Omega g_{\alpha\beta}-L_{\alpha\beta}=G_{\alpha\beta}[/tex] substitution to do so?