Geodesic Equations for the Wormhole Metric

Dr. R. Herman

Physics & Physical Oceanography, UNCW

February 14, 2018 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, 395-412, 1988.]

ds2 = −c2dt2 + dr 2 + (b2 + r 2)(dθ2 + sin2 θ dφ2) Embedding ds2 = −dt2 + dr 2 + (b2 + r 2)(dθ2 + sin2 θ dφ2) Consider 2D slices (t =const, θ = π/2). Then, (c = 1) becomes

dΣ2 = dr 2 + (b2 + r 2) dφ2.

Compare to cylindrical coordinate line element (ρ(r), ψ, z(r)) " # dz 2 dρ2 dS2 = dρ2 + ρ2 dψ2 + dz2 = + dr 2 + ρ2(r) dφ2. dr dr

Then, ρ2 = r 2 + b2 and z ψ dz 2 b2 = . z dr b2 + r 2 ρ = b cosh b Integrating, z = bsinh−1 r , or ρ b b z ρ = b cosh . b Lagrangian 2 α β Begin with metric ds = gαβ dx dx . Then, (c = 1) r Z 1 dxα dxβ τAB = −gαβ dλ. 0 dλ dλ Euler-Lagrange Equations ⇒ Geodesic Equations

d ∂L ∂L − = 0. dλ ∂x˙ γ ∂xγ

dxγ where γ = 0, 1, 2, 3, x˙ γ = , and dλ r dxα dxβ dτ L(xγ, x˙ γ) = −g = . αβ dλ dλ dλ Geodesic Equations

The Key Equations

d2xα dxβ dxγ + Γα = 0, dτ 2 βγ dτ dτ or in terms of the four-velocity: duα + Γα uβuγ = 0. dτ βγ

Here the Christoﬀel Symbols are given by

1 ∂g ∂g ∂g g Γδ = αβ + αγ − βγ αδ βγ 2 ∂xγ ∂xβ ∂xα

δ δ where Γβγ = Γγβ. Wormhole Geodesics via Lagrangian Begin with the proper time,

dτ 2 = −ds2 = dt2 − dr 2 − (b2 + r 2)(dθ2 + sin2 θ dφ2),

Write the Lagrangian, v u 2 2 2 2! u dt dr dθ dφ L = t − − (b2 + r 2) + sin2 θ , dλ dλ dλ dλ

Apply the Euler-Lagrange equation for each variable: t, r, θ, φ. ˙ dt Example - time variable t, t ≡ dλ :

d ∂L ∂L − = 0. dλ ∂t˙ ∂t Time Equation Lagrangian:

h i1/2 L = t˙2 − r˙2 − (b2 + r 2)(θ˙2 + sin2 θφ˙2)

d d Geodesic Equation for t: [Recall that L dτ = dλ ] d ∂L ∂L = dλ ∂t˙ ∂t d 2¡ dt = 0 dλ 2¡L dλ d dt L = 0 dτ dτ

d2t = 0. dτ 2 Radial Equation Lagrangian:

h i1/2 L = t˙2 − r˙2 − (b2 + r 2)(θ˙2 + sin2 θφ˙2)

Geodesic Equation for r:

d ∂L ∂L = dλ ∂r˙ ∂r " # d 1 dr 1 dθ 2 dφ2 L − = − (2r) + sin2 θ dτ L dλ 2L dλ dλ

" # d2r dθ 2 dφ2 = r + sin2 θ . dτ 2 dτ dτ The θ-Equation Lagrangian:

h i1/2 L = t˙2 − r˙2 − (b2 + r 2)(θ˙2 + sin2 θφ˙2)

Geodesic Equation for θ:

d ∂L ∂L = dλ ∂θ˙ ∂θ " # d b2 + r 2 dθ 1 dφ2 L − = − (b2 + r 2) 2 sin θ cos θ dτ L dλ 2L dλ

d dθ dφ2 (b2 + r 2) = (b2 + r 2) sin θ cos θ . dτ dτ dτ The φ-Equation Lagrangian:

h i1/2 L = t˙2 − r˙2 − (b2 + r 2)(θ˙2 + sin2 θφ˙2)

Geodesic Equation for φ:

d ∂L ∂L = dλ ∂φ˙ ∂φ d b2 + r 2 dφ L − sin2 θ = 0 dτ L dλ

d dφ (b2 + r 2) sin2 θ = 0. dτ dτ Set of Geodesic Equations

d2t = 0 dτ 2 " # d2r dθ 2 dφ2 = r + sin2 θ dτ 2 dτ dτ d dθ dφ2 (b2 + r 2) = (b2 + r 2) sin θ cos θ dτ dτ dτ d dφ (b2 + r 2) sin2 θ = 0. dτ dτ

I Solve for geodesics (t(τ), r(τ), θ(τ), φ(τ)). d2xα α dxβ dxγ I Read off Christoffel Symbols, dτ 2 = −Γβγ dτ dτ Christoffel Symbols Start with general Geodesic Equation:

d2xα dxβ dxγ = −Γα dτ 2 βγ dτ dτ

d2t = 0 dτ 2 " # d2r dθ 2 dφ2 = r + sin2 θ dτ 2 dτ dτ

Read off the coefficients: t I Γβγ = 0. r r 2 I Γθθ = −r, Γφφ = −r sin θ. Christoffel Symbols (cont’d)

d2xα dxβ dxγ = −Γα dτ 2 βγ dτ dτ

d dθ dφ2 (b2 + r 2) = (b2 + r 2) sin θ cos θ dτ dτ dλ d2θ dr dθ dφ2 (b2 + r 2) + 2r = (b2 + r 2) sin θ cos θ dτ 2 dτ dτ dλ d2θ 2r dr dθ dφ2 = − + sin θ cos θ . dτ 2 b2 + r 2 dτ dτ dλ

θ r θ θ I Γθr = b2+r 2 = Γrθ, Γφφ = − sin θ cos θ. θ θ I Note: Γθr and Γrθ contribute equally, thus there is no 2. Christoﬀel Symbols (cont’d)

d2xα dxβ dxγ = −Γα dτ 2 βγ dτ dτ

d dφ (b2 + r 2) sin2 θ = 0. dτ dτ d2φ dr dφ (b2 + r 2) sin2 θ = −2r sin2 θ dτ dτ dτ dθ dφ −2(b2 + r 2) sin θ cos θ . dτ dτ d2φ 2r dr dφ dθ dφ = − − 2 cot θ . dτ b2 + r 2 dτ dτ dτ dτ

φ r φ φ I Γφr = b2+r 2 = Γrφ, Γφθ = cot θ. Christoffel Symbols from Metric The Christoffel symbols are defined by

1 ∂g ∂g ∂g g Γδ = αβ + αγ − βγ . αδ βγ 2 ∂xγ ∂xβ ∂xα

For the wormhole metric,

ds2 = −dt2 + dr 2 + (b2 + r 2)(dθ2 + sin2 θ dφ2).

 −1 0 0 0   0 1 0 0  gµν =   ,  0 0 b2 + r 2 0  0 0 0 (b2 + r 2) sin2 θ

2 2 2 2 2 or, gtt = −1, grr = 1, gθθ = b + r , gφφ = (b + r ) sin θ. t Christoﬀel Symbols Γβγ The nonzero metric elements are 2 2 2 2 2 gtt = −1, grr = 1, gθθ = b + r , gφφ = (b + r ) sin θ. Let α = t and xα = t, then

1 ∂g ∂g ∂g g Γδ = tβ + tγ − βγ . tδ βγ 2 ∂xγ ∂xβ ∂t

Since the gtµ is nonzero and constant for µ = t,

1 ∂g ∂g ∂g g Γt = tβ + tγ − βγ tt βγ 2 ∂xγ ∂xβ ∂t 1 ∂g ∂g ∂g g Γt = tt + tt − tt = 0. (1) tt tt 2 ∂t ∂t ∂t

t So, Γαβ = 0 for all α and β. r Christoﬀel Symbols Γαβ The metric elements are 2 2 2 2 2 gtt = −1, grr = 1, gθθ = b + r , gφφ = (b + r ) sin θ. Let α = r and xα = r, then

1 ∂g ∂g ∂g g Γδ = rβ + rγ − βγ . rδ βγ 2 ∂xγ ∂xβ ∂r

Thus, δ = r and either β = γ = θ or β = γ = φ. So, we have

1 ∂g ∂g ∂g g Γr = rθ + rθ − θθ . rr θθ 2 ∂θ ∂θ ∂r 1 ∂g ∂g ∂g g Γr = rφ + rφ − φφ . rr φφ 2 ∂φ ∂φ ∂r

Therefore, since grr = 1,

r r 2 Γθθ = −r, Γφφ = −r sin θ. θ Christoﬀel Symbols Γαβ The metric elements are 2 2 2 2 2 gtt = −1, grr = 1, gθθ = b + r , gφφ = (b + r ) sin θ. Let α = θ and xα = θ, then

1 ∂g ∂g ∂g g Γδ = θβ + θγ − βγ . θδ βγ 2 ∂xγ ∂xβ ∂θ

Thus, δ = θ. We take β = θ or β = φ due to symmetry. So, we have 1 ∂g ∂g ∂g g Γθ = θθ + θγ − θγ . θθ θγ 2 ∂xγ ∂θ ∂θ 1 ∂g ∂g ∂g g Γθ = θφ + θγ − φγ . θθ φγ 2 ∂xγ ∂φ ∂θ Nonzero terms occur for γ = r in ﬁrst and γ = φ in second equation. θ Christoﬀel Symbols Γαβ (cont’d) 2 2 2 2 2 Since gθθ = b + r and gφφ = (b + r ) sin θ, we have

1 ∂g ∂g ∂g g Γθ = θθ + θr − θr . θθ θr 2 ∂r ∂θ ∂θ 1 ∂g (b2 + r 2)Γθ = θθ = r θr 2 ∂r r Γθ = = Γθ . θr b2 + r 2 rθ and 1 ∂g ∂g ∂g g Γθ = θφ + θφ − φφ . θθ φφ 2 ∂φ ∂φ ∂θ 1 ∂g (b2 + r 2)Γθ = − φφ = −(b2 + r 2) sin θ cos θ φφ 2 ∂θ θ Γφφ = − sin θ cos θ. φ Christoﬀel Symbols Γαβ The metric elements are 2 2 2 2 2 gtt = −1, grr = 1, gθθ = b + r , gφφ = (b + r ) sin θ. Let α = φ and xα = φ, then

1 ∂g ∂g ∂g g Γδ = φβ + φγ − βγ . φδ βγ 2 ∂xγ ∂xβ ∂φ

Thus, δ = φ and we take β = φ due to symmetry. So, we have

1 ∂g ∂g ∂g g Γφ = φφ + φγ − φγ φφ φγ 2 ∂xγ ∂φ ∂φ 1 ∂g = φφ . 2 ∂xγ

2 2 2 Since gφφ = (b + r ) sin θ, then γ = r or γ = θ. φ Christoﬀel Symbols Γαβ (cont’d) 2 2 2 Since gφφ = (b + r ) sin θ, we have

1 ∂g g Γφ = φφ . φφ φr 2 ∂r 2 2 2 φ 2 (b + r ) sin θΓφr = r sin θ 1 ∂g g Γφ = φφ . φφ φθ 2 ∂θ 2 2 2 φ 2 2 (b + r ) sin θΓφθ = (b + r ) sin θ cos θ Therefore, we have r Γφ = = Γφ , Γφ = cot θ = Γφ . φr b2 + r 2 rφ φθ θφ Wormhole Metric and Geodesic Equations

ds2 = −dt2 + dr 2 + (b2 + r 2)(dθ2 + sin2 θ dφ2).

d2t = 0 dτ 2 " # d2r dθ 2 dφ2 = r + sin2 θ dτ 2 dτ dτ d dθ dφ2 (b2 + r 2) = (b2 + r 2) sin θ cos θ dτ dτ dτ d dφ (b2 + r 2) sin2 θ = 0. dτ dτ

r r 2 θ r θ Christoﬀel Symbols Γθθ = −r, Γφφ = −r sin θ, Γθr = b2+r 2 = Γrθ, θ φ r φ φ φ Γφφ = − sin θ cos θ, Γφr = b2+r 2 = Γrφ, Γφθ = cot θ = Γθφ. Maple Routine