Physics 445 Solution for homework 5 Fall 2004 Cornell University 41 points) Steve Drasco 1 NOTE From here on, unless otherwise indicated we will use the same conventions as in the last two solutions: four-vectors v are denoted by an arrow, three-vectors v will be in bold, and we use units where c = 1. I. HARTLE CHAPTER 8, PROBLEM 2 8 POINTS) The line element for the surface of a two-sphere with radius a is ds 2 = a 2 dθ 2 + a 2 sin 2 θ dφ 2. 1.1) Extrema of the functional integral S[x µ λ)] = dλ g αβ [x µ λ)]ẋ α ẋ β = dλ Lx α, ẋ β ), 1.2) where here an overdot represents d/dλ, must satisfy the geodesic equation see 3 on problem set 4) ẍ α = Γ α βγẋ β ẋ γ, 1.3) and of course, the Euler-Lagrange equation Using the line element 1.1) we have L,α = ) 2s 2 φ2 sin θ cos θ, 0, L,α = d ) L dτ ẋ α. 1.4) ) d L dτ ẋ α = 2a 2 θ, φ sin 2 θ + 2 φ θ ) sin θ cos θ. 1.5) Substituting into the Euler-Lagrange equation 1.4) gives the equations of motion θ = sin θ cos θ φ 2, φ = 2 cot θ θ φ. 1.6) Comparing these with the geodesic equation 1.3) gives Γ θ φφ = sin θ cos θ, Γ φ θφ = Γφ φθ = cot θ, Γθ θφ = Γ θ φθ = Γ θ θθ = Γ φ θθ = Γφ φφ = 0. 1.7) By symmetry, the coordinates can always be chosen so that any great circle runs along the equator and is described by θ = π/2, θ = 0, and φ = constant. This satisfies the equations of motion 1.6), which means that any great circle is a geodesic. II. HARTLE CHAPTER 8, PROBLEM 4 10 POINTS) For this problem, we will use the line element ds 2 = [1 Ω 2 x 2 + y 2 )]dt 2 + 2Ωydx xdy)dt + dx 2 + dy 2 + dz 2. 2.1)
2 Introduce new coordinates x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ. 2.2) Then by comparison to the usual Cartesian and spherical coordinates, we have and by direct substitution Now compute the differentials to see that the off diagonal term in the line element is dx 2 + dy 2 + dz 2 = dr 2 + r 2 dθ 2 + r 2 sin 2 θ dφ 2, 2.3) Ω 2 x 2 + y 2 ) dt 2 = Ω 2 r 2 sin 2 θ dt 2. 2.4) dx = sin θ cos φ dr + r cos θ cos φ dθ r sin θ sin φ dφ 2.5) dy = sin θ sin φ dr + r cos θ sin φ dθ + r sin θ cos φ dφ, 2.6) Substitution of Eqs. 2.3), 2.4), and 2.7) into the line element 2.1) gives 2Ωydx xdy)dt = 2Ωr 2 sin 2 θ dφ dt. 2.7) ds 2 = dt 2 + dr 2 + r 2 dθ 2 + r 2 sin 2 θ dφ 2 + Ω 2 r 2 sin 2 θ dt 2 2Ωr 2 sin 2 θ dφ dt. 2.8) If we define ϕ = φ Ωt, we can rewrite this as the usual flat metric in spherical coordinates with φ ϕ ds 2 = dt 2 + dr 2 + r 2 dθ 2 + r 2 sin 2 θ dϕ 2. 2.9) Let the coordinates x α be Cartesian coordinates constructed from t, r, θ, ϕ) t = t, x = r sin θ cos ϕ, y = r sin θ sin ϕ, z = r cos θ = z. 2.10) The result 2.9) from A means that ds 2 = dt ) 2 + dx ) 2 + dy ) 2 + dz ) 2, 2.11) and that the x coordinates are related to the x a coordinates where a and a can be either 1 or 2) by where R b is a rotation matrix x = R b x b, 2.12) ) cos Ωt sin Ωt Rb =, 2.13) sin Ωt cos Ωt corresponding to a rotation angle of Ωt in φ about the z-axis. The geodesic equations in these coordinates must be ẍ α = 0, where an overdot represents d/dτ. Using Eq. 2.12) we can write this in terms of the rotating coordinates x α. The t and z components are unchanged ẗ = 0, z = 0, which also implies that ṫ is a constant. The x b components are changed to Now contract with the inverse/transpose of the rotation matrix to find Rb ẍ b + 2Ṙ b ẋ b + R b x b = 0. 2.14) ẍ a = 2Ra a Ṙ b ẋ b Ra a R b x b. 2.15)
Since ṫ is constant, we can easily convert the proper time derivatives of the rotation matrices into coordinate time derivatives ẍ a = 2ṫRa a drb dt Note that the rotation matrices have the properties ẋ b ṫ 2 Ra a d 2 Rb dt 2 xb. 2.16) 3 where ɛ a b is the Levi-Chivita matrix Ra a drb dt = Ωɛ a b, Ra a d 2 Rb dt 2 = Ω 2 δb a, 2.17) ɛ a b = ) 0 1 1 0 2.18) So the geodesic equations for the x α coordinates are ẗ = 0, ẍ = Ω 2 xṫ 2 + 2Ωẏṫ, ÿ = Ω 2 yṫ 2 2Ωẋṫ, z = 0. 2.19) In the non-relativistic limit ṫ 1, and the geodesic equations for x and y become C. ẗ = 0, ẍ = Ω 2 x + 2Ωẏ, ÿ = Ω 2 y 2Ωẋ, z = 0. 2.20) where an overdot represents d/dt = d/dτ, in this limit. The Coriolis acceleration is given by ẍ cor = 2Ω ẋ see your favorite classical Mechanics text). For rotation about the z-axis ẍ cor = 2Ωẏ, ÿ cor = 2Ωẋ. 2.21) The centrifugal acceleration is given by ẍ cen = Ω Ω x. For rotation about the z-axis, this gives ẍ cen = Ω 2 y, ÿ cen = Ω 2 x. 2.22) So we can rewrite the non-relativistic limit of the geodesic equation in the rotating x α frame 2.20) as ẗ = 0, ẍ = ẍ cor + ẍ cen, ÿ = ÿ cor + ÿ cen, z = 0. 2.23) III. HARTLE CHAPTER 8, PROBLEM 14 8 POINTS) In a material with index of refraction nx i ), light propagates at a speed ds/dt = 1/n, where ds 2 = dx 2 + dy 2 + dz 2. So the time T that it takes for light to travel along some path x i s) is given by T [x i s)] = dt = ds n[x i s)]. 3.1) Fermat s principle says that light follows a path which is an extremum of T [x i s)]. If we define a geometry with line element ds 2 Fermat = n2 ds 2, then these light paths will be geodesics since the geodesics are the extrema of dλ g Fermat ij dx i dx j dλ dλ = gij Fermat dx i dx j = ds Fermat = ds n[x i s)] = T [x i s)]. 3.2)
4 The metric for the line element ds 2 Fermat is g ij = n 2 δ ij, g ij = n 2 δ ij. 3.3) Recall, from II in the solutions to homework 4, that we can write the geodesic equation as ẍ i = 1 2 gij g kl,j 2g jk,l )ẋ k ẋ l = n 1 δ ij n,j δ kl 2n,l δ jk )ẋ k ẋ l, 3.4) We can rewrite this in usual 3-vector notation as ẍ = 1 n 2ẋx n) ]. n 3.5) IV. HARTLE CHAPTER 9, PROBLEM 1 5 POINTS) Using Schwarzschild coordinates, the line element is ds 2 = 1 2M r ) dt 2 + 1 2M r ) 1 dr 2 + r 2 dθ 2 + r 2 sin 2 θdφ 2. 4.1) Since the angular portion of the line element is the same as the line element for the two-sphere 1.1) with radius r, the area of any surface of constant r is the usual 4πr 2. Therefore the inner and outer surfaces of the shell, with areas 144πM 2 and 400πM 2, are located at radius r 1 = 6M and r 2 = 10M, respectively. The radial thickness of the shell is r2 r 1 dr 1 2M r ) 1/2 5 = 2M du 3 1 1 u) 1/2 4.64M. 4.2) V. RINDLER SPACETIME 10 POINTS) The line element in the Rindler spacetime is Solution prepared by Katrin Schenk ds 2 = x 2 dt 2 + dx 2 5.1) Let z α λ) be a geodesic, where λ is some affine parameter 1, and let v α = dz α /dλ = ṫ, ẋ), be a vector which is tangent to the geodesic an overdot will again represent d/dλ). We will make use of use the conserved quantity E associated with the Killing vector t, and also velocity normalization v 2 = V, where V is a constant. Note that we will not assume that V = ±1; for a general geodesic it can be arbitrary. [For spacelike or timelike geodesics it is conventional to rescale the affine parameter λ by redefining λ kλ, where k is a constant, in order to make d/dλ) 2 = ±1, but it is not necessary to do so.] The conserved quantities give As suggested, we will require the geodesic to obey the boundary conditions ṫ = Ex 2, 5.2) ẋ 2 = E 2 x 2 + V. 5.3) z α 0) = [t0), x0)] = 0, 1), 5.4) v α 0) = [ṫ0), ẋ0)] = α, β). 5.5) 1 This means that dλ/dτ is constant, for timelike geodesics, or that dλ/ds is constant for spacelike geodesics. These conditions can be derived by requiring that the geodesic equation has the same form when written in terms of λ, as it does when written in terms of τ or s.
5 Evaluating Eqs. 5.2) and 5.3) at λ = 0, then gives The solution to Eq. 5.3) which obeys the boundary conditions is given by E = α, V = β 2 α 2 5.6) xλ) = 1 + 2βλ + Vλ 2. 5.7) Using this in Eq. 5.2) and solving for t gives ) αλ tλ) = tanh 1. 5.8) 1 + βλ So at the point Q where λ = 1, we have x = 1 + 2β + V, ) α t = tanh 1. 1 + β 5.9) Using V = β 2 α 2 we can solve for α and β: α = x sinh t, β = x cosh t 1. 5.10) The validity of Eqs. 5.10) can be checked by substituting back into Eqs. 5.9). We now treat α, β as new coordinates defined by Eqs. 5.10), and we compute the metric in these new coordinates. Computing the differentials gives which, from Eq. 5.1), give dα = dx sinh t + xdt cosh t, dβ = dx cosh t + xdt sinh t, 5.11) ds 2 = dα 2 + dβ 2. 5.12)