Quantum Black Hole and Information. Lecture (1): Acceleration, Horizon, Black Hole

Transcription

1 Quantum Black Hole and Information Soo-Jong copyright Lecture (1): Acceleration, Horizon, Black Hole [Convention: c = 1. This can always be reinstated from dimensional analysis.] Today, we shall demonstrate a striking effect of non-inertial frame. Often, non-inertial frame alias gravity gives rise to an event horizon beyond which no knowledge or information can be accessed. The key idea is that acceleration was an absolute notion in Newtonian mechanics as well as in special relativity, but no longer so in general relativity. 1 Accelerated motion 1.1 relativistic acceleration The simplest situation of non-inertial frame is a uniform acceleration. Actually, the motion can be studied entirely within special relativity. Let s study this situation in detail. In Newton s mechanics, a uniform external force exerts a uniform acceleration. The problem of this Newtonian definition of uniform acceleration is that du dt = constant where u = dx dt 1 (1.1) is incompatible with special relativity since u needs to take arbitrary value to maintain constant acceleration. We thus define relativistic uniform acceleration as follows: uniform acceleration: In special relativity, an acceleration is uniform if it takes the same value in any instantaneous comoving frame. The instantaneous comoving frame is an inertial frame that travels, at each instance, with the same velocity as the body. 1

2 1.2 hyperbolic motion We now show that relativistic motion of a body under uniform acceleration executes hyperbolic motion. Let the particle has the velocity u = (dx(t)/dt) in an inertial frame S(x,t). It suffices to restrict to R 1,1. In an instantaneous comoving inertial frame S (x,t ), the particle has zero velocity u = 0 but a constant acceleration: S : u = 0 and du dt = a. (1.2) By Lorentz transformation from S back to S, it follows that du dt = ( 1 u 2 (t) ) 3/2 a. (1.3) Solving the differential equation, we find the velocity in the inertial frame S as That is, u(t) 2 (1 u(t) 2 ) = a2 t 2. (1.4) u(t) := dx dt = So, the motion in the inertial frame S is given by at (1.5) 1+a2 t2. S : u(t) = at (1+a 2 t 2 ) 1/2 and du(t) dt = a (1+a 2 t 2 ) 3/2. (1.6) This quite satisfactory. Initially, the particle was at rest (or constant velocity). For t (a/c), the particle velocity increases linearly, the same as Newtonian mechanics. If the speed of light were infinite, c, this is the exact solution. At finite c, we expect that late time behavior is modified considerably. Indeed, for t (a/c), we see that the particle velocity approaches the speed of light. Integrating once more, x(t) = x(0)+ 1 ( ) 1+a2 t a 2 1. (1.7) 2

3 Suppose we put the particle initially at x(0) = 1. Then, the trajectory can be rewritten a in the form x(t) 2 t 2 = 1 a2. (1.8) We see that the particle follows a hyperbola in spacetime. The left and right trajectories are simply related by parity transformation, x(t) x(t). The acceleration a > 0 is oriented to increasing x(t). Both trajectories follows non-relativistic trajectory initially t 0 but then asymptotes to speed of light at late time. This transition is imperative for the accelerated motion to be compatible with basic principles of special relativity. 1.3 Rindler coordinates Consider now total set of the hyperbolic trajectories: {t,x(t)} ρ (ρ := 1 a ) (1.9) ρ You see that they cover the L and R spacelike quadrant of (1+1)-dimensional Minkowski spacetime. Either one, L or R, is called the Rindler wedge. This Rindler wedge can be conveniently parametrized: x(ρ,θ) = ρcoshθ, t(ρ,θ) = ρsinhθ (0 ρ <, θ + ). (1.10) These are called Rindler coordinates: they cover the Rindler wedge, 1/4 of the Minkowski spacetime. 1.4 event horizon Even horizon refers to the boundary of the region that can be communicated by light if waited forever. It is observer-dependent concept. Consider (1+1)-dimensional Minkowski space. By Poincare transformation, an inertial observer remains inertial observer. Take first an inertial observer A sitting at x = 0. What 3

4 events can be observed by A? If an event emanates light-flash, it will propagate maximally on light-cone. If that light-ray crosses the trajectory of A, the event is observed. You easily see that A can observe all events there is no event horizon for A. Take next a non-inertial observer A executing the hyperbolic motion. Being hyperbolic, the trajectory has two families one which is restricted to the region R containing positive x-axis and one which is restricted to the region L containing negative x-axis. The regions cannot communicate each other. This is because light signals events in region R emit / absorb cannot be absorbed / emitted events in region L and vice versa. We see that the two light-cones serve as event horizon of accelerating particles. For completeness, let us analyze the region F containing positive t-axis and the region P containing negative t-axis. Signals emanating from an event in region P can reach both regions R and L. In other words, accelerating body can receive information from region B. Of course, the body cannot send a signal to this region and get back a bounced signal. Signals emanating from a body in regions R and L can reach all events in region F. Of course, the body cannot receive a signal sent by that event. Qualitatively same conclusion follows even if the acceleration were non-uniform, since at each instance one can approximate the acceleration to be approximately uniform. This peculiarity arose entirely by the fact that these particles are under acceleration, so we conclude that freely falling body in non-inertial frame will have an event horizon as well. 4

5 t x Figure: A particle under uniform force executes hyperbolic motion. The hyperbolic motion solved above corresponds to the right curves. These particles will not be able to see signals emanating from events at upper left half. The asymptote ct = x is called the event horizon to these particles. Notice that the horizon is independent of value of acceleration and is also light-like. 1.5 derivation of (1.3) Here, we derive Eq.(1.3). Suppose that two inertial frames S(x,t) and S (x,t ) are in instantaneous relative velocity v. Then, a particle velocity u(t) and u (t ) as measured in these two inertial frames are related each other by Lorentz transformation: u 1 = u 1 +v 1+u 1 v u 2 = 1 v 2 u 2 1+u 1 v u 3 = 1 v 2 u 3 1+u 1v. (1.11) 5

6 From this, we can extract relation between du and du. Also, by Lorentz transformation, t = 1 1 v 2 (t +x v), so dt = 1 1 v 2 (1+u 1 v)dt. (1.12) Dividing du by dt, we find that du 1 dt = (1 v2 ) 3 2 du 2 dt = (1 v2 ) 3 2 du 3 dt = (1 v2 ) du 1 (1+u 1v) 3 dt [ 1 du 2 (1+u 1v) 2 dt [ 1 du 3 (1+u 1 v)2 dt u 2 v du 1 (1+u 1v) 3 dt u 3 v du 1 (1+u 1 v)3 dt ] ] (1.13) In our situation, since S is a comoving frame, we will set v to u 1 at each instance and u = 0. Then, we see that Eq.(1.3) follows for all three components of the accelerations. 2 particle horizon versus event horizon There are actually two different sorts of horizons: particle horizon versus event horizon. Consider the worldline of an observer O moving on a timelike trajectory in spacetime M. Suppose past infinity I of M is spacelike. Then, at any point P of O, the past lightcone at P is the set of events in spacetime which can be observed by O at that time. The division of events into those seen by O at P and those not seen by O at P gives rise to the particle horizon of O at P. It represents the history of those events lying at the limits of O s vision. Suppose past infinity I of M is null. Then, all events are seen by O at P. Suppose both past infinity I and future infinity I + are spacelike. If the whole history of the observer O is considered, then past light-cone of O at P on I + defines the future event horizon of O. Events outside this horizon will never be seen by O. Suppose I + is null. If O moves on a timelike trajectory, O does not possess an event horizon. However, if O moves with uniform acceleration. Then, the speed of the observer 6

7 approaches 1 asymptotically, so the trajectory ends up on I +. Then, O possesses a future event horizon. Notice that all the above event horizons are observer-dependent. This is to be contrasted with the event horizons of black holes. There, the event horizons are absolute event horizons because they are observer-independent. 7

8 Homework 1: (1) Lorentz geometry of the (1 + 1)-dimensional Minkowski spacetime is described by the line element ds 2 = dt 2 dx 2. (2.1) Show that the Rindler edge is described in terms of the Rindler coordinates by the line element ds 2 = ρ 2 dθ 2 dρ 2. (2.2) (2) Draw ρ = constant or θ = constant contours of the Rindler coordinates. (3) Describe the uniform acceleration motion in terms of the Rindler coordinates. (4) Describe light-rays in terms of the Rindler coordinates. (5) Suppose you make analytic continuation of the Rindler wedge by θ iθ. Show that (ρ,θ) describes the entire R 2. Discuss whether this is contradictory to the statement that the Rindler wedge covers only 1/4 of the R 1,1. Stated differently, identify where and how the 3/4 of the R 1,1 had been mapped to. Homework 2: Consider 2 rockets attached by a rigid-body stick of length l. Suppose they move with same proper acceleration a. What happens to the rope when the rockets are seen by (1) inertial observer, and (2) non-inertial observer who accelerates together with the rockets? [Hint: Analyze the motion using the Rindler coordinates] Homework 3: (1) Consider the spacetime described by the following line element ds 2 = 1 ρ 2[(dt)2 (dx) 2 (dρ) 2 ] (0 ρ < ) (2.3) Find whether there is an event horizon and, if there is, where it is. [Hint: Light-rays are defined by the property that the follow the trajectories ds = 0] 8

9 (2) Answer how your answers change if the line element is deformed to ds 2 = where L is an arbitrary nonzero constant. Homework 4: 1 ρ 2 +L 2[(dt)2 (dx) 2 (dρ) 2 ] (0 ρ < ) (2.4) (1) Consider the spacetime R d,1 described by the following line element: ( ds 2 = 1 GA ) ( dt 2 1 GA ) 1 dr 2 r 2 dω d 1. (2.5) r r where r = (x 1 ) 2 + +(x d ) 2 is spatial radial coordinate. Find whether there is an event horizon and, if there is, where it is. (2) Suppose you explore narrow region r = GA(1+ǫ) where 0 < ǫ 1. Show that it is reduced to product of a scaled version of the Rindler wedge (2.2) and a (d 1)-dimensional sphere. Identify the scale factor of the Rindler wedge and radius of the sphere. Problem 5: A particle moves from rest at the origin of a frame S along the x-axis, with constant acceleration a as measured in an instantaneous rest frame. (1) Show that the equation of motion is ax 2 +2c 2 x ac 2 t 2 = 0. (2.6) (2) Prove that a signal emitted after time t = c/a at the origin will never reach the receding particle. (3) A standard clock carried along with the particle is set to read zero at the beginning of the motion and reads τ at time t in S. Assuming ideal clock unaffected by acceleration (so-called clock hypothesis), prove the following relations: u c = tanh aτ c, at c = sinh aτ c, (1 u2 /c 2 ) 1 = cosh aτ c x = c2 a (cosh aτ c 1 ). (2.7) 9

10 (4) Show that, during an elapsed time T( c/a) in the inertial frame, the particle clock will record approximately the time T(1 a 2 T 2 /6c 2 ). Problem 6: Devise a (theoretical or experimental) question that tests whether a particle faster than speed of light is physically acceptable. 10