• March 17, 2018
• Physics

By Myron Fish Eddy

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Extra resources for Aeronautic radio;: A manual for operators, pilots, radio mechanics,

Example text

6) Frequently g0i = 0, in which case the metric of space simpliﬁes to 3 dl2 = −gij dxi dxj . 7) As an application consider a curve along the x1 -axis, so that dx2 = dx3 = √ 0, and dl = −g11 dx1 . The distance between x1 = a and x1 = b is l = b√ ∫a −g11 dx1 , and this explains the formula used earlier. Strong equivalence The metric gαβ is determined by the mass distribution and the choice of co-ordinates, through the ﬁeld equations that are yet to come. When other co-ordinates are used, the metric becomes diﬀerent as well, in such a way that ds2 and all other physical quantities remain invariant.

Given a 2D Riemann space with co-ordinates x, y, a metric and two vectors in the tangent space of the point (x, y): ds2 = dx2 + 4dxdy + dy 2 ; 1 4 Aα = ; Bα = y x . Write down gµν and g µν and show that all Christoﬀel symbols are zero. Compute Aν and B ν:ν . 24) for the Christoﬀel symbols; g µν = 1 3 −1 2 2 −1 ; Aµ = 9 6 ; B ν:ν = 4 . 3 The Γ’s being zero we have B ν:ν = B ν,ν . 7 Riemann tensor and curvature The metric tensor does not tell us whether a space is ﬂat, because the use of ‘strange’ co-ordinates is not prohibited.

32 2 Geometry of Riemann Spaces Eq. 3 The simplest is δ s˙ dp = 0, and this is equivalent to δ F (s) ˙ dp = 0 provided F is monotonous, F = 0. We choose δ s˙ 2 dp = 0 , or δ L dp = 0 ; L(xα , x˙ β ) = (ds/dp)2 = gαβ x˙ α x˙ β . 36) The solution is determined by the Euler-Lagrange equations (Appendix C) d ∂L = ∂xλ dp ∂L ∂ x˙ λ . 37) Now, ∂L/∂xλ = gαβ,λ x˙ α x˙ β because only gαβ depends on {xµ }. By using ˙ λ = 2gαλ x˙ α . 37): ∂ x˙ α /∂ x˙ λ = δα λ one gets ∂L/∂ x gαβ,λ x˙ α x˙ β = 2(gαλ x˙ α )˙ = 2(gαλ,β x˙ β x˙ α + gαλ x ¨α ) , or ¨α + gαλ x 1 2 (2gλα,β − gαβ,λ ) x˙ α x˙ β = 0 .