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I was thinking about constraints on the stress-energy tensor of the vacuum, and came to an interesting conclusion that the vacuum itself should only be isotropic in one rest frame if there is a cosmological constant.

If we start with a vaccuum that is homogeneous and isotropic in some cartesian frame, we demand that

T = diag(rho,P,P,P)

We can demand that T always be diagonal when transformed by a boost. By setting T_01 equal to zero, and doing a boost [tex]T_{cd} = \Lambda^a{}_c \Lambda^b{}_d T_{ab}[/itex], we can find that rho = -P, as [tex]\Lambda^0{}_0 = \Lambda^1{}_1 = \beta[/tex], [tex]\Lambda^1{}_0 = \Lambda^0{}_1 = -\beta \gamma[/tex], so

[tex]T'_{01} = \gamma^2 \beta (T_{00} + T_{01}) = 0[/tex]

Thus T = diag(-P,P,P,P)

However, while T will always be diagonal, it won't be isotropic in the boosted frame. If we boost T, we get something like

diag(-(1-[itex]\beta^2[/itex])P,(1-[itex]\beta^2[/itex])P,P,P)

Thus if there is a cosmological constant, the vacuum itself should have a unique frame in which it is isotropic.

If we start with a vaccuum that is homogeneous and isotropic in some cartesian frame, we demand that

T = diag(rho,P,P,P)

We can demand that T always be diagonal when transformed by a boost. By setting T_01 equal to zero, and doing a boost [tex]T_{cd} = \Lambda^a{}_c \Lambda^b{}_d T_{ab}[/itex], we can find that rho = -P, as [tex]\Lambda^0{}_0 = \Lambda^1{}_1 = \beta[/tex], [tex]\Lambda^1{}_0 = \Lambda^0{}_1 = -\beta \gamma[/tex], so

[tex]T'_{01} = \gamma^2 \beta (T_{00} + T_{01}) = 0[/tex]

Thus T = diag(-P,P,P,P)

However, while T will always be diagonal, it won't be isotropic in the boosted frame. If we boost T, we get something like

diag(-(1-[itex]\beta^2[/itex])P,(1-[itex]\beta^2[/itex])P,P,P)

Thus if there is a cosmological constant, the vacuum itself should have a unique frame in which it is isotropic.

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