Geodesic Equations for the Wormhole Metric

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1 Geodesic Equations for the Wormhole Metric Dr R Herman Physics & Physical Oceanography, UNCW February 14, 2018

2 The Wormhole Metric Morris and Thorne wormhole metric: [M S Morris, K S Thorne, Wormholes in spacetime and their use for interstellar travel: A tool for teaching general relativity, Am J Phys 56, , 1988] ds 2 = c 2 dt 2 + dr 2 + (b 2 + r 2 )(dθ 2 + sin 2 θ dφ 2 )

3 Embedding ds 2 = dt 2 + dr 2 + (b 2 + r 2 )(dθ 2 + sin 2 θ dφ 2 ) Consider 2D slices (t =const, θ = π/2) Then, (c = 1) becomes dσ 2 = dr 2 + (b 2 + r 2 ) dφ 2 Compare to cylindrical coordinate line element (ρ(r), ψ, z(r)) [ (dz ) 2 ( ) ] dρ 2 ds 2 = dρ 2 + ρ 2 dψ 2 + dz 2 = + dr 2 + ρ 2 (r) dφ 2 dr dr Then, ρ 2 = r 2 + b 2 and ( ) dz 2 = b2 dr b 2 + r 2 z ψ ρ = b cosh z b Integrating, z = bsinh 1 r b, or b ρ ρ = b cosh z b

4 Lagrangian Begin with metric ds 2 = g αβ dx α dx β Then, (c = 1) τ AB = 1 0 g αβ dx α dx β Euler-Lagrange Equations Geodesic Equations ( ) d L ẋ γ L x γ = 0 where γ = 0, 1, 2, 3, ẋ γ = dx γ, and L(x γ, ẋ γ dx ) = g α αβ dx β =

5 Geodesic Equations The Key Equations d 2 x α 2 or in terms of the four-velocity: + dx β Γα βγ dx γ = 0, du α + Γα βγ uβ u γ = 0 Here the Christoffel Symbols are given by g αδ Γ δ βγ = 1 2 [ gαβ x γ + g αγ x β g ] βγ x α where Γ δ βγ = Γδ γβ

6 Wormhole Geodesics via Lagrangian Begin with the proper time, 2 = ds 2 = dt 2 dr 2 (b 2 + r 2 )(dθ 2 + sin 2 θ dφ 2 ), Write the Lagrangian, L = ( dt ) 2 ( ) dr 2 (b 2 + r 2 ) ( ( ) dθ 2 + sin 2 θ ( ) ) dφ 2, Apply the Euler-Lagrange equation for each variable: t, r, θ, φ Example - time variable t, d ṫ dt : ( ) L L ṫ t = 0

7 Time Equation Lagrangian: [ṫ2 1/2 L = ṙ 2 (b 2 + r 2 )( θ 2 + sin 2 θ φ )] 2 Geodesic Equation for t: [Recall that L d = d d ( ) d L ṫ ( ) 2 dt 2L ( ) dt L d d 2 t 2 = 0 = L t = 0 = 0 ]

8 Radial Equation Lagrangian: [ṫ2 1/2 L = ṙ 2 (b 2 + r 2 )( θ 2 + sin 2 θ φ )] 2 Geodesic Equation for r: ( ) d L = L ṙ r L d ( 1 ) dr L d 2 r 2 = r = 1 2L (2r) [ ( ) dθ 2 + sin 2 θ [ ( ) dθ 2 + sin 2 θ ( ) ] dφ 2 ( ) ] dφ 2

9 The θ-equation Lagrangian: [ṫ2 1/2 L = ṙ 2 (b 2 + r 2 )( θ 2 + sin 2 θ φ )] 2 Geodesic Equation for θ: ( ) d L θ L d ( b2 + r 2 ) dθ L = L θ = 1 2L (b2 + r 2 ) ( d (b 2 + r 2 ) dθ ) = (b 2 + r 2 ) sin θ cos θ [ 2 sin θ cos θ ( ) dφ 2 ( ) ] dφ 2

10 The φ-equation Lagrangian: [ṫ2 1/2 L = ṙ 2 (b 2 + r 2 )( θ 2 + sin 2 θ φ )] 2 Geodesic Equation for φ: ( ) d L φ L d ( b2 + r 2 sin 2 θ dφ ) L = L φ = 0 ( d (b 2 + r 2 ) sin 2 θ dφ ) = 0

11 Set of Geodesic Equations ( d (b 2 + r 2 ) dθ ) ( d (b 2 + r 2 ) sin 2 θ dφ ) d 2 t 2 = 0 d 2 r 2 = r [ ( ) dθ 2 + sin 2 θ = (b 2 + r 2 ) sin θ cos θ = 0 ( ) ] dφ 2 ( ) dφ 2 Solve for geodesics (t(τ), r(τ), θ(τ), φ(τ)) Read off Christoffel Symbols, d2 x α 2 = Γ α βγ dxβ dx γ

12 Christoffel Symbols Start with general Geodesic Equation: d 2 x α 2 = dx β dx γ Γα βγ d 2 t 2 = 0 d 2 r 2 = r Read off the coefficients: Γ t βγ = 0 Γ r θθ = r, Γr φφ = r sin2 θ [ ( ) dθ 2 + sin 2 θ ( ) ] dφ 2

13 Christoffel Symbols (cont d) d 2 x α 2 = dx β dx γ Γα βγ ( d (b 2 + r 2 ) dθ ) (b 2 + r 2 ) d 2 θ dr dθ + 2r 2 ( dφ = (b 2 + r 2 ) sin θ cos θ ( dφ = (b 2 + r 2 ) sin θ cos θ ) 2 ) 2 d 2 θ 2 = 2r dr dθ b 2 + r 2 + sin θ cos θ ( ) dφ 2 Γ θ θr = r = Γ θ b 2 +r 2 rθ, Γθ φφ = sin θ cos θ Note: Γ θ θr and Γθ rθ contribute equally, thus there is no 2

14 Christoffel Symbols (cont d) d 2 x α 2 ( d (b 2 + r 2 ) sin 2 θ dφ ) (b 2 + r 2 ) sin 2 θ d 2 φ d 2 φ = dx β dx γ Γα βγ = 0 = 2r sin 2 θ dr dφ 2(b 2 + r 2 ) sin θ cos θ dθ dφ = 2r dr dφ dθ b 2 + r 2 2 cot θ dφ Γ φ φr = r = Γ φ b 2 +r 2 rφ, Γφ φθ = cot θ

15 Christoffel Symbols from Metric The Christoffel symbols are defined by [ gαβ g αδ Γ δ βγ = 1 2 x γ For the wormhole metric, + g αγ x β g ] βγ x α ds 2 = dt 2 + dr 2 + (b 2 + r 2 )(dθ 2 + sin 2 θ dφ 2 ) g µν = b 2 + r (b 2 + r 2 ) sin 2 θ, or, g tt = 1, g rr = 1, g θθ = b 2 + r 2, g φφ = (b 2 + r 2 ) sin 2 θ

16 Christoffel Symbols Γ t βγ The nonzero metric elements are g tt = 1, g rr = 1, g θθ = b 2 + r 2, g φφ = (b 2 + r 2 ) sin 2 θ Let α = t and x α = t, then g tδ Γ δ βγ = 1 [ gtβ 2 x γ + g tγ x β g ] βγ t Since the g tµ is nonzero and constant for µ = t, g tt Γ t βγ = 1 2 g tt Γ t tt = 1 2 So, Γ t αβ = 0 for all α and β [ gtβ x γ [ gtt t + g tγ x β g βγ t ] + g tt t g tt t ] = 0 (1)

17 Christoffel Symbols Γ r αβ The metric elements are g tt = 1, g rr = 1, g θθ = b 2 + r 2, g φφ = (b 2 + r 2 ) sin 2 θ Let α = r and x α = r, then g rδ Γ δ βγ = 1 [ grβ 2 x γ + g rγ x β g ] βγ r Thus, δ = r and either β = γ = θ or β = γ = φ So, we have g rr Γ r θθ = 1 [ grθ 2 θ + g rθ θ g ] θθ r g rr Γ r φφ = 1 [ grφ 2 φ + g rφ φ g ] φφ r Therefore, since g rr = 1, Γ r θθ = r, Γr φφ = r sin2 θ

18 Christoffel Symbols Γ θ αβ The metric elements are g tt = 1, g rr = 1, g θθ = b 2 + r 2, g φφ = (b 2 + r 2 ) sin 2 θ Let α = θ and x α = θ, then g θδ Γ δ βγ = 1 2 [ gθβ x γ + g θγ x β g ] βγ θ Thus, δ = θ We take β = θ or β = φ due to symmetry So, we have g θθ Γ θ θγ = 1 [ gθθ 2 x γ + g θγ θ g ] θγ θ g θθ Γ θ φγ = 1 2 [ gθφ x γ + g θγ φ g φγ θ Nonzero terms occur for γ = r in first and γ = φ in second equation ]

19 Christoffel Symbols Γ θ αβ (cont d) Since g θθ = b 2 + r 2 and g φφ = (b 2 + r 2 ) sin 2 θ, we have g θθ Γ θ θr = 1 [ gθθ + g θr 2 r θ g ] θr θ (b 2 + r 2 )Γ θ θr = 1 g θθ = r 2 r Γ θ r θr = b 2 + r 2 = Γθ rθ and g θθ Γ θ φφ = 1 2 [ gθφ φ + g θφ φ g φφ θ ] (b 2 + r 2 )Γ θ φφ = 1 g φφ 2 θ = (b2 + r 2 ) sin θ cos θ Γ θ φφ = sin θ cos θ

20 Christoffel Symbols Γ φ αβ The metric elements are g tt = 1, g rr = 1, g θθ = b 2 + r 2, g φφ = (b 2 + r 2 ) sin 2 θ Let α = φ and x α = φ, then g φδ Γ δ βγ = 1 2 [ gφβ x γ + g φγ x β g ] βγ φ Thus, δ = φ and we take β = φ due to symmetry So, we have g φφ Γ φ φγ = 1 [ gφφ 2 x γ + g φγ φ g ] φγ φ = 1 g φφ 2 x γ Since g φφ = (b 2 + r 2 ) sin 2 θ, then γ = r or γ = θ

21 Christoffel Symbols Γ φ αβ (cont d) Since g φφ = (b 2 + r 2 ) sin 2 θ, we have Therefore, we have g φφ g φφ Γ φ φr = 1 2 r (b 2 + r 2 ) sin 2 θγ φ φr = r sin 2 θ g φφ Γ φ φθ = 1 g φφ 2 θ (b 2 + r 2 ) sin 2 θγ φ φθ = (b 2 + r 2 ) sin θ cos θ Γ φ φr = r b 2 + r 2 = Γφ rφ, Γφ φθ = cot θ = Γφ θφ

22 Wormhole Metric and Geodesic Equations ds 2 = dt 2 + dr 2 + (b 2 + r 2 )(dθ 2 + sin 2 θ dφ 2 ) ( d (b 2 + r 2 ) dθ ) ( d (b 2 + r 2 ) sin 2 θ dφ ) d 2 t 2 = 0 d 2 r 2 = r [ ( ) dθ 2 + sin 2 θ = (b 2 + r 2 ) sin θ cos θ = 0 ( ) ] dφ 2 ( ) dφ 2 Christoffel Symbols Γ r θθ = r, Γr φφ = r sin2 θ, Γ θ θr = r = Γ θ b 2 +r 2 rθ, Γ θ φφ = sin θ cos θ, Γφ φr = r = Γ φ b 2 +r 2 rφ, Γφ φθ = cot θ = Γφ θφ

23 Maple Routine