Van Vleck determinant

2.5 Van Vleck determinant

2.5.1 Definition and properties

The van Vleck biscalar $Δ (x,x ′)$ is defined by

As we shall show below, it can also be expressed as

where $g$ is the metric determinant at $x$ and $g′$ the metric determinant at $x′$ .

Equations (63 ) and (80 ) imply that at coincidence, $[Δα ′′] = δ α′′ β β$ and $[Δ ] = 1$ . Equation (89 ), on the other hand, implies that near coincidence

so that

This last result follows from the fact that for a “small” matrix $a$ , $2 det(1 + a ) = 1 + tr(a) + 𝒪 (a )$ .

We shall prove below that the van Vleck determinant satisfies the differential equation

which can also be written as $α α (ln Δ ),ασ = 4 − σ α$ , or

in the notation introduced in Section 2.1.4. Equation (99

) reveals that the behaviour of the van Vleck determinant is governed by the expansion of the congruence of geodesics that emanate from $x′$ . If $θ∗ < 3$ , then the congruence expands less rapidly than it would in flat spacetime, and $Δ$ increases along the geodesics. If, on the other hand, $∗ θ > 3$ , then the congruence expands more rapidly than it would in flat spacetime, and $Δ$ decreases along the geodesics. Thus, $Δ > 1$ indicates that the geodesics are undergoing focusing, while $Δ < 1$ indicates that the geodesics are undergoing defocusing. The connection between the van Vleck determinant and the strong energy condition is well illustrated by Equation (97

): The sign of $Δ − 1$ near $′ x$ is determined by the sign of $α′ β′ R α′β′ σ σ$ .

2.5.2 Derivations

To show that Equation (95 ) follows from Equation (94 ) we rewrite the completeness relations at $x$ , $αβ ab α β g = η eaeb$ , in the matrix form $−1 T g = E ηE$ , where $E$ denotes the $4 × 4$ matrix whose entries correspond to $α ea$ . (In this translation we put tensor and frame indices on equal footing.) With $e$ denoting the determinant of this matrix, we have $1∕g = − e2$ , or $e = 1∕ √−-g$ . Similarly, we rewrite the completeness relations at $x′$ , $gα′β′ = ηabeα′eβ′ a b$ , in the matrix form $′−1 ′ ′T g = E ηE$ , where $′ E$ is the matrix corresponding to $α′ ea$ . With $′ e$ denoting its determinant, we have $′ ′2 1∕g = − e$ , or $′ √ ---′ e = 1 ∕ − g$ . Now, the parallel propagator is defined by $′ gαα′ = ηabg α′β′eαaeβb$ , and the matrix form of this equation is $ˆg = E ηE ′Tg′T$ . The determinant of the parallel propagator is therefore $---- ˆg = − ee′g′ = √ − g′∕√ −-g$ . So we have

and Equation (95

) follows from the fact that the matrix form of Equation (94

) is $Δ = − ˆg−1g− 1σ$ , where $σ$ is the matrix corresponding to $σ αβ′$ .

To establish Equation (98 ) we differentiate the relation $1 γ σ = 2σ σγ$ twice and obtain $γ γ σ αβ′ = σ ασ γβ′ + σ σ γαβ′$ . If we replace the last factor by $σαβ′γ$ and multiply both sides by $α′α − g$ we find

′ ′ Δ αβ′ = − gα α(σ γασ γβ′ + σγσ αβ′γ).

In this expression we make the substitution $α′ σ αβ′ = − gαα′Δ β′$ , which follows directly from Equation (94 ). This gives us

where we have used Equation (78

). At this stage we introduce an inverse $(Δ− 1)α ′′ β$ to the van Vleck bitensor, defined by $α′ −1 β′ α′ Δ β′(Δ ) γ′ = δ γ′$ . After multiplying both sides of Equation (101

) by $−1 β′ (Δ ) γ′$ we find

α′ α′ β α − 1γ′ α′ γ δ β′ = g αg β′σ β + (Δ )β′Δ γ′;γσ ,

and taking the trace of this equation yields

α −1 β′ α′ γ 4 = σ α + (Δ ) α′Δ β′;γσ .

We now recall the identity $δlndet M = tr(M −1δM )$ , which relates the variation of a determinant to the variation of the matrix elements. It implies, in particular, that $−1 β′ α′ (Δ )α′Δ β′;γ = (ln Δ ),γ$ , and we finally obtain

which is equivalent to Equation (98

) or Equation (99