By Charles W. Misner, Kip S. Thorne, John Archibald Wheeler, John Wheeler, Kip Thorne
Basic relativity is a gorgeous scheme for describing the gravitational fieldan dth equations it obeys. these days this concept is frequently used as a prototype for different, extra problematic structures to explain forces among common debris or different branches offundamental physics. this is because in an creation to basic relativity it's of value to split as essentially as attainable some of the materials that jointly provide form to this paradigm.
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Extra info for Introduction to General Relativity
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 .