By G. terHooft

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**Example text**

In any case, a residual gravitational field emanating from the vacuum, if it exists at all, must be extraordinarily weak. For a long time, it was presumed that the cosmological constant Λ = 0 . Only very recently, strong indications were reported for a tiny, positive value of Λ. Whether or not the term exists, it is very mysterious why Λ should be so close to zero. In modern field theories it is difficult to understand why the energy and momentum density of the vacuum state (which just happens to be the state with lowest energy content) are tuned to zero.

Exercise: show that this is twice what one would expect if a light ray could be regarded as a nonrelativistic particle in a hyperbolic orbit around the star. 27) in ordinary units reads as δϕ = 6πGN m , a(1 − ε2 ) c2 where a is the major axis of the orbit, ε its excentricity and c the velocity of light. 40) 13. GENERALIZATIONS OF THE SCHWARZSCHILD SOLUTION. a). The Reissner-Nordstrøm solution. Spherical symmetry can still be used as a starting point for the construction of a solution of the combined Einstein-Maxwell equations for the fields surrounding a “planet” with electric charge Q and mass m.

The expansion continues forever if Λ > 0 and k = −1. If Λ < 0, the expansion always comes to a halt, at which point the universe begins to shrink (amax in Fig. ) a) V 0 a max a b) c) Fig. 6. 20) for the cases a) k = 0, Λ < 0 , b) k = −1, Λ = 0 and c) k = 0, Λ > 0. In the case (a), there is a turning point at a = amax . 17) when there are other relations between the pressure and the density. For instance in a radiation-filled universe, we have p = ρ/3, and since we may assume that the radiation is thermal, and the number of photons is conserved, we may conclude that ρ = ρ0 /a4 instead of Eq.