Physics 554: HW#1 Solutions

Physics 554: HW#1 Solutions Katrin Schenk 7 Feb 2001 Problem 1.Properties of linearized Riemann tensor: Part a: We want to show that the expression for the linearized Riemann tensor, given by R αβγδ = [h αδ,βγ h αγ,βδ + h βγ,αδ h βδ,αγ ] /2 (1) has all the symmetries of the Riemann tensor. To check the first symmetry R (αβ)γδ = 0, we get For the pair exchange symmetries we find, R βαγδ = h βδ,αγ h βγ,αδ + h αγ,βδ h αδ,βγ = R αβγδ. (2) R γδαβ = h γβ,δα h γα,δβ + h δα,γβ h δβ,γα = h αδ,βγ h αγ,βδ + h βγ,αδ h βδ,αγ = R αβγδ, (3) where we have used the symmetry of h αβ, and the fact that partial derivatives commute. To prove the last symmetry, R α[βγδ] = 0, we note that this can be written as R α{βγδ} = 0, where the curly brackets indicate a sum over cyclic permutations. The two pieces of Eq. (1), which can be written as h αδ,βγ h αγ,βδ = h αδ,βγ h αγ,δβ h βγ,αδ h βδ,αγ = h βγ,δα h δβ,γα (4) must cancel in pairs when the indices δβγ are permuted cyclicly, so we get R αβγδ + R αγδβ + R αδβγ = 0. (5)

2 Part b: Let h αβ h αβ α ξ β β ξ α = h αβ 2ξ (β,α) (6) with ξ = ξ(x γ ). Substituting this into Eq. (1) we get R αβγδ R αβγδ 2 [ξ δ,αβγ ξ γ,αβδ + ξ γ,βαδ ξ δ,βαγ ] = R αβγδ. (7) Problem 2. Part a Beginning with the Minkowski metric in cylindrical coordinates ds 2 = dt 2 + dρ 2 + ρ 2 dφ 2 + dz 2, (8) we define a rotating coordinate system by φ = φ Ωt. Direct substitution of this transformation into the metric (8) yields, ds 2 = dt 2 + dρ 2 + ρ 2 d φ 2 + dz 2 + 2Ωρ 2 d φdt + O(Ω 2 ). (9) The metric (9) can be written as g αβ = ˆη αβ + h αβ + O(Ω 2 ), (10) where ˆη αβ is the flat metric given by ˆη αβ dx α dx β = dt 2 + dρ 2 + ρ 2 d φ 2 + dz 2, (11) and h αβ is given by h αβ dx α dx β = 2Ωρ 2 d φdt. (12) Note that the flat metric ˆη αβ is not the physical metric that would be measured by rods and clocks (which is g αβ ). However, this fact does not prevent us from using our mathematical

3 formalism for linearized perturbations of Minkowski spacetime on the metric (10). Note that Eq. (12) is equivalent to h t φ = h φt = Ωρ 2, (13) with all other h αβ = 0. The gravitomagnetic potential is given by A (g) i = 1 4 h 0i = 1 4 h 0i, (14) which yields The vector obtained from this one-form is A (g) α dx α = 1 4 h 0idx i = 1 4 Ωρ2 d φ. (15) A (g) = 1 4 Ωρ e ˆφ, (16) where e ˆφ = 1 ρ φ is the unit vector in the φ direction. The gravitomagnetic field is given by B (g) = 4 A (g) = Ωρ e ˆφ = 1 eẑ ρ ρ ρ(ωρ) = 2Ω e ẑ. (17) Thus a = v B (g) = 2v Ω = 2Ω v, (18) which is exactly the Coriolas acceleration. Part b. We use the formalism of linearized gravity: h αβ = 16πT αβ, (19)

4 where h αβ, and T αβ are tensors on Minkowski spacetime. As such, we can use either Cartesian coordinates (t, x, y, z) to describe h αβ and T αβ, or cylindrical polar coordinates (t, ρ, φ, z) in which the Minkowskip metric is η αβ dx α dx β = dt 2 + dρ 2 + ρ 2 dφ 2 + dz 2. (20) The corresponding orthonormal basis is eˆt =, t eˆρ =, e ρ ˆφ = 1, e ρ φ ẑ =. z In Cartesian coordinates we have (at leading order) T ti = ρ (m) v i, (21) where ρ (m) is the mass density and v i is the 3 velocity. This formula is still true if we transform to cylindrical polars, since the coordinate transformation involves only the spatial coordinates, (x, y, z) (ρ, φ, z). The mass density is ρ (m) (ρ, φ, z) = σδ(ρ R). (22) The 3 velocity is v = ω r, where ω = ω eẑ. This gives v = ωr e ˆφ. (23) Combining (21), (22), and (23) yields T 0i dx i = σωrρδ(ρ R)dφ = σωr 2 δ(ρ R)dφ, (24) where we have used the fact that the one form obtained by lowering the index of the vector e ˆφ is ρdφ. Solving for the Gravitomagnetic field: We have A (g) = 4πj, (25)

5 where A (g) i = 1 4 h 0i and j i = T 0i. Since the source is time-independent we have 2 A (g) = 4πj. (26) Taking the curl of both sides of (26) we get 2 A (g) = 4π j, (27) or 2 B (g) = 16π j, (28) since B (g) = 4 A (g) and the curl and Laplacian operators commute in the flat background. We solve Eq. (28) directly to obtain B (g), rather than first computing A (g). Let s first compute the RHS. In cylindrical polar coordinates ( j) α = ɛ αβγ β j γ 1 = g ˆɛαβγ β j γ, (29) where ˆɛ αβγ takes on values +1, 1 or 0. This gives, using g = ρ, ( j) z = 1 ρ ( ρj φ φ j ρ ) = 1 ρ ρj φ (ρ) = 1 ρ σωr2 δ (ρ R) (30) where δ (ρ R) is the derivative of the delta function with respect to ρ and we have used Eq. (24). Inserting (28) into (30), we see that the only non-vanishing component of B (g) is in the z direction, and that 2 B z = 16πσωR 2 1 ρ δ (ρ R) (31) or 1 ρ ρ(ρb z,ρ) = 16πσωR 2 1 ρ δ (ρ R) (32)

6 which immediately yields, ρb z,ρ = 16πσωR 2 δ(ρ R) + k 0, (33) where k 0 is a constant of integration. This gives B z,ρ = 16πσωRδ(ρ R) + k 0 /ρ, (34) and we can solve for B z to get B z (ρ) = 16πσωRΘ(ρ R) + k 0 ln ρ + k 1. (35) Here k 1 is another constant and Θ is the step function given by 1 x 0 Θ(x) = 0 x < 0. (36) [Note that the general solution to 2 B z = 0 that is independent of φ and z is B z = k 1 + k 0 ln(ρ)]. Inside the cylinder, B z (ρ) should be finite as ρ 0, and so k 0 = 0. Outside the cylinder, the gravitomagnetic force should go to zero as ρ. So k 1 = 16πσωR. Then our solution (35) for B z becomes B z (ρ) = 16πσωR [1 Θ(ρ R)] = 16πσωRΘ(R ρ). (37) Thus inside the cylinder, B (g) = 2Ω eẑ, where Ω = 8πσωR, (38) while outside the cylinder B (g) = 0. Part c. No. Locally the two situations have the same metric, and hence no local experiment can distinguish between the 2 possibilities. Problem 3.Isotropic coordinates for spherical stars:

7 Part a (Solution by Paul Wiggins.) By inspection of the two metrics, the first in Schwarzschild coordinates ds 2 = e 2Φ(r) dt 2 + e 2Λ(r) dr 2 + r 2 dω 2, (39) and the second in isotropic coordinates ds 2 = e 2Φ( r) dt 2 + e 2µ( r) d r 2 + e 2µ( r) r 2 dω 2, (40) we can write down relations between the two coordinate systems: e Λ(r) dr = e µ( r) d r r = e µ( r) r. (41) Combining the above two equations yields a differential equation for r in terms of r and Λ(r), whose solution is dr r eλ(r) = d r r (42) [ r r(r) = C exp dr exp ] Λ(r ), (43) where C is a constant. Then from Eq. ( 41) the potential µ( r) is given by r e µ( r) = r( r) r. (44) Part b: (Solution by Paul Wiggins.) Starting with the form ( 40) of the metric in isotropic coordinates, we expand to obtain ds 2 = [ 1 + 2Φ + O(Φ 2 ) ] dt 2 + [ 1 + 2µ + O(Φ 2 ) ] [ d x 2 + dȳ 2 + d z 2]. (45) Comparing with the results from MTW box 16.2 we get Φ = Φ newtonian = µ. (46)

8 Beginning with the Eqs. ( 41) we can write d ( re µ) = dre µ (1 µ r) = d r, (47) so e Λ = e µ d r dr = (1 µ r), (48) where µ dµ dr Λ µ r = Φ r. (49)