Geodesic Dynamics in Reissner-Nordström Geometry

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Geodesic Dynamics in Reissner-Nordström Geometry Thai Journal of Physics Vol. 38 No. 2 (2021) 69-98 Original Research Article Geodesic Dynamics in Reissner-Nordström Geometry Daniel Abramson* 9/72 Soi Kasemsan 3, Rama 1 Road, Wangmai, Pathumwan, Bangkok 10330 *Corresponding author E-mail: [email protected] Received: 28th May 2021, Revised: 22nd July 2021, Accepted: 24th July 2021 Abstract We analyze geodesics (trajectories of free-falling bodies) in the geometry created by a Reissner-Nordström black hole. Bound and unbound radial orbits are obtained first in the superextremal case, then in the extremal and subextremal cases where geodesics penetrate a horizon, inside of which an outside observer cannot see. We determine the location of geodesics in conformal (Penrose) spacetime diagrams. We also deduce that the “river model”, a useful idea to explain the behavior of Schwarzschild black holes, fails in the case where electric charge is present. Keywords: Reissner-Nordström, Black hole, Geodesic, Penrose diagram Introduction The solution of Einstein’s field equations for a spherically symmetric mass distribution was found by Karl Schwarzschild in 1916 [1], leading to the theoretical possibility of black holes. The spacetime geometry created by a static, non-rotating black hole with mass 푚 and electric charge 푞 was determined independently by Hans Reissner in 1916 [2] and Gunnar Nordström in 1918 [3].1,2 The geometry created by such a charged black hole is determined by the metric 2푚 푞2 푑푟2 푑푠2 = − (1 − + ) 푑푡2 + + 푟2푑휃2 + (푟 sin 휃)2푑휙2 . 푟 푟2 1 − 2푚⁄푟 + 푞2⁄푟2 _____________________ 1 Geometrized units where 퐺 = 푐 = 1 are used in this paper, so that mass and charge have units of length. These quantities can be converted to units of length by multiplying by the conversion factors 퐺⁄푐2 = 7.43 × 10−28 m⁄kg 4 −18 √퐺⁄4휋휖0푐 = 8.62 × 10 m⁄C 2 We investigate only the case of static (eternal), spherically symmetric, non-rotating black holes. Black holes created at a finite prior time (by stellar collapse or some other means) are not considered. 69 Thai Journal of Physics Vol. 38 No. 2 (2021) 69-98 This is not a vacuum solution since the geometry is determined by both the mass and the electric field with energy density 퐄2⁄8휋 = 푞2⁄8휋푟4. Reissner-Nordström black holes receive cursory treatments in the text literature [4,5] or are introduced as exercises [6,7]. This can be attributed to circumstances and hypotheses which suggest that the phenomena associated with charged black holes are likely to be unphysical.3 Notwithstanding this, the Reissner-Nordström solution has intrinsic interest and shares features of the Kerr solution for rotating black holes. Useful Formulas and Identities We establish several elementary results which will be used repeatedly. A. Notation 푑휏푔 denotes proper time along a geodesic. When there is no possibility of confusion, we will write 푑휏. 푑휏푟 denotes proper time for a stationary shell observer 풮푟 located at radius 푟. ̇ A dot ⬚ always denotes differentiation with respect to proper time along a geodesic: 푟̇ ≡ 푑푟⁄푑휏푔. B. Shell observer speed We consider metrics of the form 푑푟2 푑푠2 = −푓(푟) 푑푡2 + + 푟2푑휃2 + (푟 sin 휃)2푑휙2 . (1) 푓(푟) We assume 푓(∞) = 1, so the spacetime geometry is asymptotically flat. In the case of radial motion (푑휃 = 푑휙 = 0) the metric simplifies to 푑푟2 푑푠2 = −푓(푟) 푑푡2 + . 푓(푟) A shell observer 풮푟, who is stationary on the sphere of radius 푟, measures the speed of an object in (timelike) motion in his frame of reference: 푑푠| proper length 푑푡=0 푣푟 = = . proper time √−푑푠2| 푑퐱=0 _____________________ 3 Such as the cosmic censorship conjecture, mass inflation instability, and the sheer amount of excess charge needed for the inner event horizon radius to reach the Compton wavelength of an elementary particle or even the Planck length. 70 Thai Journal of Physics Vol. 38 No. 2 (2021) 69-98 Therefore 푑푟2⁄푓(푟) 푑푟2⁄푑푡2 1 푑푟 푣2 = = ⟹ 푣 = . (2) 푟 푓(푟) 푑푡2 푓(푟)2 푟 푓(푟) 푑푡 4 We choose the positive square root in (2) so that 푣푟 has the same sign as 푑푟⁄푑푡 when 푓(푟) > 0. The proper time element 푑휏푟 for a shell observer 풮푟 is 2 푑휏푟 = √−푑푠 | = √푓(푟) 푑푡 . (3) 푑퐱=0 Therefore, by (2) and (3): 2 2 1 푑푟 1 푑푟 푑휏푔 푑휏푟 푟̇√1 − 푣푟 √푓(푟) 푟̇√1 − 푣푟 푣푟 = = = = . 푓(푟) 푑푡 푓(푟) 푑휏푔 푑휏푟 푑푡 푓(푟) √푓(푟) 2 Solving for 푣푟 : 푟̇2 푣2 = . (4) 푟 푓(푟) + 푟̇2 It follows that |푣푟| < 1 if and only if 푓(푟) > 0. C. Geodesic motion Timelike geodesic trajectories in the equatorial plane 휃 = 휋⁄2 satisfy the equations of motion:5 ℓ 휙̇ = 푟2 푒 푡̇ = (5) 푓(푟) ℓ2 푟̇2 = 푒2 − (1 + ) 푓(푟) 푟2 _____________________ 4 푓(푟) < 0 corresponds to a black or white hole; see (37) and Figure 13. Along a timelike path in such a region, 푑푟 never changes sign — 푑푟 < 0 in a black hole — but 푑푡 can be positive, negative or zero. Thus “ingoing” objects (푑푟⁄푑푡 < 0) and “outgoing” objects (푑푟⁄푑푡 > 0) both move in the direction of decreasing 푟. 5 A straightforward generalization of the derivation given by Dray [8] (pp. 21-23) for Schwarzschild geometry. 71 Thai Journal of Physics Vol. 38 No. 2 (2021) 69-98 where 푒 and ℓ are constants of the motion.6 For a geodesic that starts from rest at infinity, the third equation in (5) shows that 푒2 = 푓(∞) = 1. For a radial geodesic, 휙̇ = 0 implies ℓ = 0, whence 푟̇2 = 푒2 − 푓(푟). Inserting this into (4) yields two useful relations (taking 푒 > 0): 푟̇ 푣 = (6) 푟 푒 푓(푟) 푣 = 1 − (7) 푟 푒2 From (7) we may deduce that 푒 is the ratio of total energy 퐸 to mass 푀 of a free-falling object measured by a distant, stationary observer.7 We verify this as follows. In a potential energy field 푈 the total energy is 퐸 = 훾푀 + 푈, where 훾 = (1 − 푣2)−1⁄2 and 푈 are both functions of position. Hence: 푀 2 1 2 푓(푟) 푓(푟) ( ) = ( ) = 1 − 푣2 = 1 − (1 − ) = . 퐸 − 푈 훾 푒2 푒2 Thus (퐸 − 푈)⁄푀 = 푒⁄√푓(푟). Letting 푟 → ∞ we obtain 퐸⁄푀 = 푒. Henceforth all geodesics will henceforth be assumed to be radial. Preliminaries The Reissner-Nordström metric is given by (1) with 2푚 푞2 2푚(푟 − 훽) 푓(푟) = 1 − + = 1 − (8) 푟 푟2 푟2 where 훽 ≡ 푞2⁄2푚. The radial equation of motion along a geodesic is given by (5): 2푚(푟 − 훽) 푟̇2 = 푒2 − (1 − ) . (9) 푟2 For an object that starts from rest at infinity, 푒 = 1, so its equation of motion is 2푚(푟 − 훽) 푟̇2 = . (10) 푟2 _____________________ 6 ℓ is the angular momentum per unit mass. 7 All test “objects” are assumed to be uncharged, and therefore do not interact with the electric field. 72 Thai Journal of Physics Vol. 38 No. 2 (2021) 69-98 Since the trajectory is ingoing, we conclude: √2푚(푟 − 훽) 푟̇ = − . (11) 푟 We see that the object comes to a stop at 푟 = 훽 and moves no closer to the singularity. A shell observer 풮푟 measures the object’s speed from (6): 푟̇ √2푚(푟 − 훽) 푣 = = − (12) 푟 푒 푟 Since a shell observer cannot observe an object with nonzero mass traveling with speed |푣푟| ≥ 1, it must be that 2푚(푟 − 훽) 2푚(푟 − 훽) < 1 ⟺ 0 < 1 − = 푓(푟) ⟺ 푃(푟) = 푟2 − 2푚푟 + 푞2 > 0 . 푟2 푟2 2 2 푃(푟) has discriminant 4(푚 − 푞 ), so there are three distinct scenarios: (i) If |푞| > 푚, then 푃(푟) has no real zeros and |푣푟| < 1 for all 푟. This is a superextremal singularity. 2 2 (ii) If |푞| < 푚, then 푃(푟) = 0 at 푟± = 푚 ± √푚 − 푞 . In the region 푟− < 푟 < 푟+ we find 푓(푟) < 0. In this region |푣푟| > 1 so no shell observers can exist. The radii 푟− and 푟+ are the inner and outer event horizons. This case is called subextremal. (iii) If |푞| = 푚, then 푃(푟) = 0 at 푟 = 푚. There is only a single horizon, which separates regions where 푃(푟) > 0. This case is called extremal. We consider these cases separately. Superextremal Singularities A. Geodesics originating from 풓 = ∞ In the superextremal case, 푓(푟) > 0 for all 푟, so by (4) we have |푣푟| < 1 for all 훽 ≤ 푟 < ∞. This is apparent in Figure 1, which shows 푣푟 given by (12) for various values of 푞. The curves for 푞 > 푚 = 1 correspond to superextremal singularities; the other curves will be referenced later. 73 Thai Journal of Physics Vol. 38 No. 2 (2021) 69-98 Figure 1 Shell-observer speed 푣푟 for a free-falling object starting from rest at infinity. 푞 > 0 is Reissner-Nordström geometry; 푞 = 0 is Schwarzschild geometry. 푞 > 푚 (푚 = 1) corresponds to a superextremal singularity.8 The trajectory is obtained by integrating 푑푟 √2푚(푟 − 훽) 푟̇ = = − (13) 푑휏 푟 to obtain 1 −푟푑푟 2 2 휏 = ∫ = − ( ) √푟 − 훽 (푟 + 2훽) + 퐶 . (14) √2푚(푟 − 훽) 9푚 Choosing 퐶 = 0 is equivalent to the condition 푟 = 훽 at 휏 = 0. Eqn. (14) can be inverted to obtain the trajectory 푟(휏). Free fall times. A freely falling object starting from rest at infinity falls from 푟 = 푅 to 푟 = 훽 in finite proper time: 1 2 2 ∆휏 = ( ) √푅 − 훽 (푅 + 2훽) . (15) 9푚 _____________________ 8 We choose 푚 = 1 in all figures in this paper. 74 Thai Journal of Physics Vol.
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