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Example text

Eq. 17) contains the mass density and Eq. 18) the pressure. Let us assume that the pressure vanishes. 19) can be solved exactly. In terms of F (t) we have 2F F + F 2 + k = 0. 23) = −D/2F 2 . 20): F Write Eq. 28) t(ϕ) = √ (ϕ − 12 sin 2ϕ) , k k D (1 − cos2ϕ) . 29) F (ϕ) = 2k These are the equations for a cycloid. Note that 3(F˙ 2 + k) 3D = 3 . 30) G00 = 2 F F Therefore D is something like the mass density of the universe, hence positive. 31) k = 0 → ϕ inﬁnitesimal . See Fig. 5. All solutions start with a “big bang” at t = 0.

The region x > 0, y > 0 will be identiﬁed with the “ordinary world” extending far from our source. The second universe, the region of space-time with x < 0 and y < 0 has the same metric as the ﬁrst one. It is connected to the ﬁrst one by something one could call a “wormhole”. The physical signiﬁcance of this extended region however is very limited, because: 1) “ordinary” stars and planets contain matter (Tµν = 0) within a certain radius r > 2M , so that for them the validity of the Schwarzschild solution stops there.

Now we wish to compute the trajectory of a light ray. It is also a geodesic. Now however ds = 0. 4), but now we set ds/dτ = 0 , so that Eq. 5) becomes 0 = 1− 2M 2M ˙2 t − 1− r r −1 r˙ 2 − r 2 θ˙ 2 + sin2 θ ϕ˙ 2 . 28) Since now the parameter τ is determined up to an arbitrary multiplicative constant, only the ratio J/E will be relevant. Call this j. Then Eq. 15) becomes u ϕ = ϕ0 + u0 du j −2 − u2 + 2M u3 1 −2 . 29) As the left hand side of Eq. 13) must now be replaced by zero, Eq. 18) becomes u + u = 3M u2 .

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