Relativistic Rocket Dynamics and Accelerating Reference Frames (Rindler Coordinates)

Supplemental Lecture 3

Acceleration: Relativistic Rocket Dynamics and Accelerating Reference Frames (Rindler Coordinates)

Abstract

An accelerating rocket is studied in special relativity and its equation of motion is contrasted to its non-relativistic treatment. Uniformly accelerating reference frames are formulated and the transformation between measurements in this non-inertial reference frame and an inertial frame is obtained and applied. The space-time metric inside an accelerating rocket is derived and the accelerating frame is found to be locally equivalent to the environment in a uniform gravitational field. The global properties of the accelerating frame (Rindler Wedge) are different from those of a uniform gravitational field: there is a past and future horizon of the accelerating frame but not for the uniform gravitational field. Analogies to the Schwarzschild black hole are drawn.

This lecture supplements material in the textbook: Special Relativity, Electrodynamics and General Relativity: From Newton to Einstein (ISBN: 978-0-12-813720-8). The term “textbook” in these Supplemental Lectures will refer to that work.

Keywords: Special relativity, acceleration, inertial frames of reference, accelerating grid, Rindler coordinates, Equivalence Principle, Horizons, Black Holes.

Rockets in Special Relativity

Let’s study rocket dynamics as an illustration of relativistic dynamics and an introduction to accelerating reference frames and general relativity [1,2]. Along the way we will discuss the practical challenges to interstellar travel via “conventional” rocketry. Consider inertial frames and in the usual orientation as shown in Fig. 1. ′ 𝑆𝑆 𝑆𝑆

Fig. 1 Reference frames and , with moving in the direction at velocity . ′ 𝑆𝑆 𝑆𝑆 𝑆𝑆 ′ 𝑥𝑥 𝑣𝑣 Frame moves to the right in the -direction at velocity relative to the frame . In ′ applications𝑆𝑆 at this point in the textbook𝑥𝑥 the velocity was𝑣𝑣 always set to a constant𝑆𝑆 for all times. ( ) In the case here is a function of time chosen so that𝑣𝑣 matches the speed of an accelerating rocket at the instant𝑣𝑣 . The frame is called𝑡𝑡 the “instantaneous𝑣𝑣 𝑡𝑡 rest frame of the rocket” as ′ shown in Fig. 2. 𝑡𝑡 𝑆𝑆

Fig. 2 Instantaneous rest frame of rocket. Gas nozzle velocity points to the left ′ 𝑆𝑆 𝑤𝑤 We imagine that there is a grid of clocks and rods in the accelerating reference frame and they ′ serve to “coordinatize” it. Clocks in are synchronized and rods are calibrated everywhere𝑆𝑆 in ′ ′ in the usual way as discussed in the textbook𝑆𝑆 in Chapters 2 and 3. 𝑆𝑆 To begin, let’s find the rocket’s velocity ( ) and position ( ) in the inertial frame . This

is a one dimensional problem, so we suppress𝑣𝑣 the𝑡𝑡 transverse spatial𝑥𝑥 𝑡𝑡 dimensions and forego𝑆𝑆 vector notation in most places. We will set up the dynamics in the rocket’s instantaneous rest frame at the time and then transform to frame . This strategy reduces the full problem to two ′ relatively easy steps.𝑡𝑡 Suppose that at time in the𝑆𝑆 rocket’s instantaneous rest frame the rocket of ( ) ( ) ′ rest mass ejects rest mass at𝑡𝑡 a nozzle velocity . The mass ejected could consist ′ ′ of gases as𝑀𝑀 in𝑡𝑡 a conventional rocket∆𝑀𝑀 or𝑡𝑡 even light (photons) in𝑤𝑤 a hypothetical rocket whose engine consists of annihilating matter and anti-matter. The nozzle velocity is the speed of the ejected

material in the instantaneous rest frame of the rocket. We suppose that𝑤𝑤 is a constant for all time .We take to be positive by convention and note that the ejected𝑤𝑤 travels in the - ′ ′ direction,as𝑡𝑡 shown𝑤𝑤 in the figures. The energy and momentum carried by the∆𝑀𝑀 ejected and 𝑥𝑥lost from the rocket is = and = / , where we recalled the general∆𝑀𝑀 energy- ′ 2 ′ ′ 2 ∆𝐸𝐸 −∆𝑀𝑀𝑐𝑐 ∆𝑃𝑃 ∆𝐸𝐸 𝑤𝑤 𝑐𝑐 3

momentum relation = ( ) and we noted that the velocity of the ejected mass is in 2 the instantaneous rest𝑃𝑃 frame𝐸𝐸⁄ of𝑐𝑐 the𝑣𝑣 rocket. What principles of dynamics have we used∆ here?𝑀𝑀 𝑤𝑤 Energy-momentum conservation! The energy-momentum carried away by ejected material is lost from the rocket’s energy-momentum. Now we want to view the rocket’s motion in the frame . Since ( , ) constitute the ( ) ′ ′ and components of a four vector, we can calculate 𝑆𝑆 , ∆in𝐸𝐸 the⁄𝑐𝑐 frame∆𝑃𝑃 by making a ′ ′ ( ) boost𝑡𝑡 of𝑥𝑥 velocity , ∆𝐸𝐸⁄𝑐𝑐 ∆𝑃𝑃 𝑆𝑆 𝑣𝑣= 𝑡𝑡 ( + ) = (1 + / ) = (1 + / ) ′ ′ ′ 2 2 2 ∆𝐸𝐸 = 𝛾𝛾(∆𝐸𝐸 + 𝑣𝑣∆𝑃𝑃 / )𝛾𝛾=∆𝐸𝐸 ( /𝑣𝑣𝑣𝑣)(𝑐𝑐 + )−=𝛾𝛾∆𝑀𝑀𝑐𝑐 ( +𝑣𝑣𝑣𝑣 )𝑐𝑐/ (1) ′ ′ 2 ′ 2 2 These expressions∆𝑃𝑃 tell𝛾𝛾 ∆us𝑃𝑃 how𝑣𝑣 the∆𝐸𝐸 energy𝑐𝑐 -momentum𝛾𝛾 ∆𝐸𝐸 𝑐𝑐 four𝑣𝑣 vector𝑤𝑤 −( 𝛾𝛾(∆𝑀𝑀) 𝑣𝑣, (𝑤𝑤)) change𝑐𝑐 s with ( ) ( ) ( ) time. The energy is related to the rocket’s velocity and rest mass𝐸𝐸 𝑡𝑡 ⁄𝑐𝑐 𝑃𝑃=𝑡𝑡 with 2 ( ) = (1 ( )𝐸𝐸 𝑡𝑡 ) . The momentum of the rocket in frame is𝐸𝐸 𝑡𝑡( ) =𝛾𝛾𝛾𝛾( 𝑡𝑡( 𝑐𝑐)/ ) ( ). 2 2 −1⁄2 2 We𝛾𝛾 𝑡𝑡 differentiate− 𝑣𝑣 𝑡𝑡 this⁄𝑐𝑐 expression and substitute in Eq. 1, 𝑆𝑆 𝑃𝑃 𝑡𝑡 𝐸𝐸 𝑡𝑡 𝑐𝑐 𝑣𝑣 𝑡𝑡

= ( ) = + = 1 + + (2) 2 2 𝑣𝑣𝑣𝑣 2 2 ∆𝑃𝑃𝑐𝑐 ∆ 𝐸𝐸𝐸𝐸 𝑣𝑣∆𝐸𝐸 𝐸𝐸∆𝑣𝑣 −𝛾𝛾∆𝑀𝑀𝑐𝑐 𝑣𝑣 � 𝑐𝑐 � 𝛾𝛾𝛾𝛾𝑐𝑐 ∆𝑣𝑣 But the left-hand side is = ( + ) from Eq.1, 2 ∆𝑃𝑃𝑐𝑐 −𝛾𝛾∆𝑀𝑀 𝑣𝑣 𝑤𝑤 ( + ) = 1 + + (3) 2 𝑣𝑣𝑣𝑣 2 2 −𝛾𝛾∆𝑀𝑀 𝑣𝑣 𝑤𝑤 −𝛾𝛾∆𝑀𝑀𝑐𝑐 𝑣𝑣 � 𝑐𝑐 � 𝛾𝛾𝑀𝑀𝑐𝑐 ∆𝑣𝑣 We can solve for ,

𝑀𝑀∆𝑣𝑣 = + + = (1 ) 2 (4) ∆𝑀𝑀𝑣𝑣 𝑤𝑤 2 2 2 𝑀𝑀∆𝑣𝑣 −∆𝑀𝑀𝑀𝑀 − ∆𝑀𝑀𝑀𝑀 ∆𝑀𝑀𝑀𝑀 𝑐𝑐 −∆𝑀𝑀𝑀𝑀 − 𝑣𝑣 ⁄𝑐𝑐 So, we can calculate the rate at which ( ) increases due to the ejection of mass at nozzle

velocity , 𝑣𝑣 𝑡𝑡 𝑤𝑤 = (5) 2 ∆𝑀𝑀 𝛾𝛾 ∆𝑣𝑣 − 𝑀𝑀 𝑤𝑤 This equation is easy to integrate and find how ( ) varies as a function of ( ) where ( ) = = 0 , 𝑣𝑣 𝑡𝑡 𝑀𝑀 𝑡𝑡 ⁄𝑀𝑀0 𝑀𝑀0 𝑀𝑀 𝑡𝑡 ( ) = = ln( ( ) ) (6) 𝑣𝑣 2 𝑀𝑀 𝑡𝑡 𝑑𝑑𝑑𝑑 ∫0 𝛾𝛾 𝑑𝑑𝑑𝑑 −𝑤𝑤 ∫𝑀𝑀0 𝑀𝑀 −𝑤𝑤 𝑀𝑀 𝑡𝑡 ⁄𝑀𝑀0

On the left-hand side,

= = = + = ln ( )( ) (7) 𝑣𝑣 2 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 1 1 1 𝑐𝑐 1+𝑣𝑣⁄𝑐𝑐 2 2 ∫0 𝛾𝛾 𝑑𝑑𝑑𝑑 ∫ 1−𝑣𝑣 ⁄𝑐𝑐 ∫ 1+𝑣𝑣⁄𝑐𝑐 1−𝑣𝑣⁄𝑐𝑐 2 ∫ �1+𝑣𝑣⁄𝑐𝑐 1−𝑣𝑣⁄𝑐𝑐� 𝑑𝑑𝑑𝑑 2 �1−𝑣𝑣⁄𝑐𝑐� Collecting our partial results,

/ ln( ( ) ) = ln = ln (8) 𝑐𝑐 1+𝑣𝑣⁄𝑐𝑐 1−𝑣𝑣⁄𝑐𝑐 𝑐𝑐 2𝑤𝑤 𝑀𝑀 𝑡𝑡 ⁄𝑀𝑀0 − 2𝑤𝑤 �1−𝑣𝑣⁄𝑐𝑐� �1+𝑣𝑣⁄𝑐𝑐� So,

( ) ( ) = ( ) (9) 1−𝑣𝑣 𝑡𝑡 ⁄𝑐𝑐 𝑐𝑐⁄2𝑤𝑤 𝑀𝑀 𝑡𝑡 𝑀𝑀0 �1+𝑣𝑣 𝑡𝑡 ⁄𝑐𝑐� which shows how ( ) decreases as ( ) increases. Eq. 9 can be inverted,

𝑀𝑀 𝑡𝑡 𝑣𝑣 𝑡𝑡 ( ( ) ) ( ) = (10) ( ( ) )2𝑤𝑤⁄𝑐𝑐 1− 𝑀𝑀 𝑡𝑡 ⁄𝑀𝑀0 2𝑤𝑤⁄𝑐𝑐 𝑣𝑣 𝑡𝑡 𝑐𝑐 �1+ 𝑀𝑀 𝑡𝑡 ⁄𝑀𝑀0 � Which predicts how ( )/ increases as ( ) decreases for a given nozzle velocity . It

is a scaling relation. The𝑣𝑣 𝑡𝑡 speed𝑐𝑐 limit sets𝑀𝑀 the𝑡𝑡 ⁄scale𝑀𝑀0 of all velocities in the problem. You 𝑤𝑤are⁄𝑐𝑐 ( ) encouraged to plot Eq. 10 for various𝑐𝑐 chooses of and needed to produce velocities near the speed of light. 𝑤𝑤⁄𝑐𝑐 𝑀𝑀 𝑡𝑡 ⁄𝑀𝑀0 An interesting special case of Eq. 10 is the “photon rocket”, already solved in the textbook in a problem set. In that case the motor of the rocket consists of separate matter and anti-matter containers, “magnetic bottles” presumably, and the matter and anti-matter are allowed to annihilate in a controlled fashion into light which is directed through the nozzle. In this case 2 = 2. If half of the rocket consists of annihilating matter and anti-matter, then ( ) =

0𝑤𝑤.5 ⁄can𝑐𝑐 be achieved and the final velocity of the rocket payload is 0.60 , a very impressive𝑀𝑀 𝑡𝑡 ⁄𝑀𝑀 0 velocity. Unfortunately, a rocket engineer could argue quite forcefully tha𝑐𝑐 t our naïve exercise is not a realistic model of a practical rocket. For a chemical rocket and the resulting ( ) is

much less impressive. The reader is encouraged to Google “relativistic𝑤𝑤 ≪ 𝑐𝑐 rocket travel” and𝑣𝑣 learn𝑡𝑡 about the limitations and hazards of extreme relativistic velocities and interstellar travel. One consideration which we do not face here is the fact that interstellar space is hardly empty: there are cosmic rays, interstellar gases, background radiation, etc. These phenomena become daunting perils for space travel at hypothetical velocities approaching the speed of light!

Back to our discussion of ideal rocketry dynamics. It is interesting to contrast the relativistic rocket to its non-relativistic predecessor. In the non-relativistic case, energy-momentum conservation becomes separate momentum and mass conservation laws. In addition, Lorentz transformations are replaced by Galilean transformations, just the addition of velocities. In the instantaneous rest frame of the rocket = is the change of the rocket’s momentum ′ when it ejects a differential mass ∆at𝑃𝑃 nozzle−∆ velocity𝑀𝑀𝑀𝑀 . In the frame where the rocket has ( ) instantaneous velocity , = ∆𝑀𝑀 . (Positive 𝑤𝑤 points in the negative𝑆𝑆 direction, so ( ) ′ an observer at rest in frame𝑣𝑣 ∆𝑃𝑃 measures−∆𝑀𝑀 𝑤𝑤 the− 𝑣𝑣velocity of 𝑤𝑤 to be , according𝑥𝑥 to Galilean mechanics.). Since = 𝑆𝑆non-relativistically, we have,∆𝑀𝑀 𝑤𝑤 − 𝑣𝑣

𝑃𝑃 𝑀𝑀𝑀𝑀 ( ) = ( )

( ) +∆ 𝑀𝑀𝑀𝑀 = −∆(𝑀𝑀 𝑤𝑤) −+𝑣𝑣( ) (11)

So, ∆𝑀𝑀 𝑣𝑣 𝑀𝑀∆𝑣𝑣 − ∆𝑀𝑀 𝑤𝑤 ∆𝑀𝑀 𝑣𝑣

= (12) ∆𝑀𝑀 ∆𝑣𝑣 −𝑤𝑤 𝑀𝑀 which integrates to,

= (13) 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 −𝑤𝑤 𝑀𝑀 Giving, ∫ ∫

( ) ( ) = ln (14) 𝑀𝑀 𝑡𝑡 𝑣𝑣 𝑡𝑡 −𝑤𝑤 𝑀𝑀0 which predicts how the velocity of the rocket increases as it burns its fuel and ( ) decreases.

This expression can be inverted, 𝑀𝑀 𝑡𝑡

( ) = ( ) (15) 𝑀𝑀 𝑡𝑡 −𝑣𝑣 𝑡𝑡 ⁄𝑤𝑤 𝑀𝑀0 𝑒𝑒 Does the non-relativistic analysis agree with its relativistic version when ? The crucial

non-relativistic expression is Eq. 12, 𝑣𝑣 ≪ 𝑐𝑐

= (16) 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 −𝑤𝑤 𝑀𝑀 The corresponding relativistic expression is Eq. 5,

= (17) 2 𝑑𝑑𝑑𝑑 𝛾𝛾 𝑑𝑑𝑑𝑑 −𝑤𝑤 𝑀𝑀 , As long as is well approximated by unity with corrections of order 2 , and Eq. 16 2 𝑣𝑣 2 is retrieved.𝑣𝑣 Success!≪ 𝑐𝑐 𝛾𝛾 𝑂𝑂 �𝑐𝑐 �

Uniform Proper Acceleration and Non-Inertial Reference Frames

In Chapter 6 of the textbook we solved the problem of a relativistic charged particle in a uniform electric field, as would be the case in a large capacitor, which points in the + direction. The

equation of motion reads, 𝑥𝑥

= = (18) 𝑑𝑑 𝑑𝑑𝑑𝑑 𝑝𝑝𝑥𝑥 𝐹𝐹𝑥𝑥 𝑞𝑞𝐸𝐸𝑥𝑥 where is the component of the particle’s relativistic momentum, = . Denote =

𝑝𝑝𝑥𝑥= , a fixed𝑥𝑥 acceleration. Eq. 18 is easy to solve, 𝑝𝑝⃗ 𝛾𝛾𝛾𝛾𝑣𝑣⃗ 𝐹𝐹𝑥𝑥⁄𝑚𝑚 𝑞𝑞𝑞𝑞𝑥𝑥⁄𝑚𝑚 𝛼𝛼 ( ) = (19) 𝑑𝑑 𝑑𝑑𝑑𝑑 𝛾𝛾𝛾𝛾 𝛼𝛼 So,

= (20)

Since, = (1 ) , Eq. 20 𝛾𝛾predicts𝛾𝛾 𝛼𝛼𝛼𝛼 ( ), after some algebra, 2 2 −1⁄2 𝛾𝛾 − 𝑣𝑣 ⁄𝑐𝑐 ( ) = 𝑣𝑣 𝑡𝑡 (21.a) ( ) 𝑐𝑐 2 𝑣𝑣 𝑡𝑡 �1+ 𝑐𝑐⁄𝛼𝛼𝛼𝛼 if we assume the initial condition = 0 at = 0. We discussed in the textbook that for small , ( ) ( ( )) , then 𝑣𝑣= and𝑡𝑡 we retrieve Newton’s prediction that if a particle𝑡𝑡 expe𝑡𝑡 ≪ 𝑐𝑐riences⁄𝛼𝛼 a constant𝑣𝑣 𝑡𝑡 ≈ acceleration𝑐𝑐⁄ 𝑐𝑐⁄𝛼𝛼𝛼𝛼 its𝛼𝛼 velocity𝛼𝛼 grows proportional to .We also noted that in the ( ) extreme relativistic limit, , then and the charged particle’s𝑡𝑡 velocity approaches the speed limit. 𝑡𝑡 ≫ 𝑐𝑐⁄𝛼𝛼 𝑣𝑣 𝑡𝑡 → 𝑐𝑐 Let’s use this problem to solve an interesting problem in general relativity: What is the space-time metric of a uniformly accelerating reference system? This development will take several steps.

To begin, what is the acceleration that the charged particle experiences in its instantaneous rest frame of velocity ( )? We need to know in given in , the inertial lab frame. In ′ ′ ′ fact, = 𝑆𝑆 . We can understand𝑣𝑣 𝑡𝑡 this from several𝐸𝐸𝑥𝑥 perspectives.𝑆𝑆 𝐸𝐸 𝑥𝑥First,𝑆𝑆 in Chapter 8 of the ′ textbook𝐸𝐸𝑥𝑥 we𝐸𝐸𝑥𝑥 derived the transformation law for , , electric and magnetic fields, and derived = . In more elementary fashion we can consider�𝐸𝐸�⃗ 𝐵𝐵�⃗� the electric field along the axis of a large ′ 𝐸𝐸capacitor.𝑥𝑥 𝐸𝐸𝑥𝑥 Making boosts in the direction of the axis leaves the charge/area on a capacitor plate invariant and by Gauss’ law this leaves the electric field unchanged.

We come to the conclusion that is the proper acceleration of the particle. So, is the acceleration that an observer who is 𝛼𝛼in the rest frame of the accelerating particle 𝛼𝛼experiences ′ for all . 𝑆𝑆

Next𝑡𝑡 let’s find the trajectory of the accelerating particle in the frame , ( , ). Eq. 21.a can be written in the form, 𝑆𝑆 𝑐𝑐𝑐𝑐 𝑥𝑥

( ) = (21.b) ( ) 𝛼𝛼𝛼𝛼 2 𝑣𝑣 𝑡𝑡 �1+ 𝛼𝛼𝛼𝛼⁄𝑐𝑐 which can be integrated to find,

( ) = = 1 + ( ) (22) ( ) 2 𝛼𝛼𝛼𝛼𝛼𝛼𝛼𝛼 𝑐𝑐 2 2 𝑥𝑥 𝑡𝑡 ∫ �1+ 𝛼𝛼𝛼𝛼⁄𝑐𝑐 𝛼𝛼 � 𝛼𝛼𝛼𝛼⁄𝑐𝑐 where we chose ( = 0) = . A more insightful way to write this result is to square it and ( ) 2 treat and 𝑥𝑥 on𝑡𝑡 an equal 𝑐𝑐footing,⁄𝛼𝛼

𝑥𝑥 𝑡𝑡 𝑐𝑐𝑐𝑐 = (23) 2 2 2 4 2 which we recognize as invariant hyperbolas𝑥𝑥 − 𝑐𝑐 [1,2]𝑡𝑡 ! In𝑐𝑐 ⁄a 𝛼𝛼boosted frame, ( , ) ( , ), Eq. 23 becomes, 𝑐𝑐𝑐𝑐 𝑥𝑥 → 𝑐𝑐𝑡𝑡̃ 𝑥𝑥�

= (24) 2 2 2 4 2 These results are shown in the Minkowski𝑥𝑥� − diagram𝑐𝑐 𝑡𝑡̃ 𝑐𝑐 Fig.⁄𝛼𝛼 3.

Fig. 3 Hyperbola of constant proper acceleration and line of constant .

The invariant hyperbola Eq. 23 has a simple interpretation. Imagine the rocket beginning𝑡𝑡̃ its flight at = 0 from an initial position on the axis of = . At the same time launch a 2 light ray 𝑡𝑡toward the rocket but starting at the origin.𝑥𝑥 Will𝑥𝑥0 the𝑐𝑐 light⁄𝛼𝛼 ray catch and overtake the rocket? The light ray travels at the speed limit from the origin, but the rocket accelerates from a starting position at = . (Imagine that is a “typical” acceleration like the acceleration at 2 2 the surface of the earth𝑥𝑥0 , 9𝑐𝑐.80⁄𝛼𝛼 m/s , and verify 𝛼𝛼that is a huge distance.) The outcome of the race is clear from the Minkowski diagram: the light𝑥𝑥 0ray never catches the rocket! We are interested in the distance between the light ray and the rocket, ( ) , and this can be read off

the hyperbola, 𝑥𝑥 𝑡𝑡 − 𝑐𝑐𝑐𝑐

= ( )( + ) = (25) 2 2 2 4 2 Using Eq. 22 for ( ) we have,𝑥𝑥 for− 𝑐𝑐 >𝑡𝑡 0, 𝑥𝑥 − 𝑐𝑐𝑐𝑐 𝑥𝑥 𝑐𝑐𝑐𝑐 𝑐𝑐 ⁄𝛼𝛼

𝑥𝑥 𝑡𝑡 𝑡𝑡 = = ~ (26) 4 2 (4 2 ) 2 𝑐𝑐 ⁄𝛼𝛼 𝑐𝑐 ⁄𝛼𝛼 1 𝑐𝑐 𝑐𝑐 2 𝑥𝑥 − 𝑐𝑐𝑐𝑐 𝑥𝑥+𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐��1+ 𝑐𝑐⁄𝛼𝛼𝛼𝛼 +1� 2 𝛼𝛼 �𝛼𝛼𝛼𝛼�

where we took 1 in the last expression to calculate that asymptotically ( )~ : so −1 the light ray approaches𝛼𝛼𝛼𝛼⁄𝑐𝑐 ≫ but never catches the rocket. Note that the light ray in Fig.𝑥𝑥 − 3 𝑐𝑐is𝑐𝑐 the𝑡𝑡 asymptote of the hyperbola. The hyperbola approaches the light ray in the limit . In this sense light propagation occurs with a “divergent proper acceleration”. Another way𝛼𝛼 → to∞ express this result is to note that boosts translate points on the hyperbola. So, the point in Fig. 3 can be boosted to point and we learn that in the instantaneous rest frame of the rocket,𝐵𝐵 the invariant distance-squared𝐴𝐴 to the light ray is for all . 4 2 𝑐𝑐 ⁄𝛼𝛼 𝑡𝑡 An Accelerating Coordinate System: The Rindler Wedge

Now let’s get more ambitious and use what we have learned to make an accelerating grid, an accelerating reference frame where we could run experiments and make space-time measurements. Eq, 23 suggests that we consider the class of hyperbolas, = ( + ) (27) 2 2 2 2 2 for > 0 as shown in Fig. 4. The axis𝑥𝑥 −lies𝑐𝑐 on𝑡𝑡 the 𝑥𝑥� axis𝑐𝑐 at⁄𝛼𝛼 = 0. 𝑥𝑥� 𝑥𝑥� 𝑥𝑥 𝑡𝑡

Fig. 4 Coordinates of accelerating grid.

Instead of thinking of each hyperbola as the space-time path of a particular rocket, we can think of as measuring the distance off the floor of a huge “mother” ship with markings of increasing

as𝑥𝑥� shown in Fig. 5. 𝑥𝑥�

Fig. 5 Spatial coordinates along the length of the accelerating rocket.

𝑥𝑥� The mother ship could be taken as long as one likes. It is also convenient to put synchronized clocks (synchronized in the mother ship’s instantaneous rest frame) at each marker telling time

. 𝑥𝑥� ( ) 𝑡𝑡 ̃ To better understand the accelerating grid , let’s differentiate Eq. 27 with respect to for fixed and , 𝑐𝑐𝑡𝑡̃ 𝑥𝑥� 𝑡𝑡 𝑥𝑥� 𝛼𝛼 2 2 = 0 (28.a) 𝑑𝑑𝑑𝑑 2 Or, identifying = , 𝑥𝑥 𝑑𝑑𝑑𝑑 − 𝑐𝑐 𝑡𝑡

𝑑𝑑𝑑𝑑⁄𝑑𝑑𝑑𝑑 𝑣𝑣 = (28.b) 𝑣𝑣 2 But the Lorentz transformation to the rocket𝑡𝑡 having𝑐𝑐 𝑥𝑥 instantaneous velocity ( ) reads, ( ) = 𝑣𝑣 𝑡𝑡 𝑥𝑥�= 𝛾𝛾 𝑥𝑥 − 𝑣𝑣𝑣𝑣 𝑣𝑣 𝑡𝑡̃ 𝛾𝛾 �𝑡𝑡 − 2 𝑥𝑥� 11 𝑐𝑐

So, the line = , Eq. 28.b,is a line of constant . These are rays from the origin, the axes, at 𝑣𝑣 2 velocity . We𝑡𝑡 recognize𝑐𝑐 𝑥𝑥 this as an example of the𝑡𝑡 Relativitỹ of Simultaneity emphasized𝑥𝑥� in

Chapter 2𝑣𝑣 and 3 of the textbook. A line of constant is shown in Fig. 6 where we also show that this is a line of constant = , the tangents to𝑡𝑡 thẽ hyperbolas. This shows that all the points along the mother ship 𝑑𝑑have𝑑𝑑⁄𝑑𝑑 the𝑑𝑑 same𝑣𝑣 velocity at any given time in the mother ship’s rest 𝑥𝑥frame.� So, the mother ship accelerates without internal stresses and𝑡𝑡� it provides a useful coordinate grid for the accelerating, non-inertial frame [1,2]. In other words, there is a common velocity ( ) of the entire ship, or, equivalently, the accelerating coordinate grid.

𝑣𝑣 𝑡𝑡

Fig. 6 Rindler coordinates showing lines of constant ( ) and constant .

𝑣𝑣 𝑡𝑡 𝑡𝑡̃ 12

The grid is static in the variable - the points of fixed maintain the same distance between them that they had at = 0 when𝑡𝑡� the entire rocket was 𝑥𝑥�simultaneously at rest in the inertial frame . Note that the𝑡𝑡 top of the mother ship, points of large , accelerate less than points near its bottom,𝑆𝑆 small. From Eq. 27 we could define, 𝑥𝑥�

𝑥𝑥� = + (2 ) 2 𝑐𝑐 𝑐𝑐 > 0 (0) = 𝑥𝑥� where and . This guarantees𝛼𝛼 𝑥𝑥� that the L𝛼𝛼orentz contraction of the ship is given by ( ) the global𝑥𝑥� transformation𝛼𝛼 law,𝛼𝛼 = where is the proper length of the ship and = ( ) is its instantaneous velocity:∆𝑥𝑥 since𝑙𝑙0⁄ the𝛾𝛾 𝑣𝑣 ship contracts𝑙𝑙0 as measured in the frame , its top𝑣𝑣 𝑣𝑣must𝑡𝑡 accelerate less than its bottom to accommodate a constant velocity. 𝑆𝑆 Next we need the explicit transformation from ( , ) to ( , ). First, let’s consider the relation between and at = 0, the base of the rocket.𝑐𝑐𝑐𝑐 𝑥𝑥 A clock𝑐𝑐𝑡𝑡̃ 𝑥𝑥 �at = 0 in the rocket measures . When the rocket𝑡𝑡̃ has 𝑡𝑡velocity𝑥𝑥� at time , must be related to 𝑥𝑥 �by time dilation, = . 𝑡𝑡Integrating̃ this relation, 𝑣𝑣 𝑡𝑡 𝑑𝑑𝑡𝑡̃ 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝛾𝛾𝛾𝛾𝑡𝑡̃ = = = sinh ( ) (29) ( ) 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑐𝑐 −1 2 where we use Eq. 21 to𝑡𝑡 ̃ write∫ 𝛾𝛾 in ∫terms�1+ 𝛼𝛼of𝛼𝛼⁄ 𝑐𝑐 ( ) and𝛼𝛼 we used𝛼𝛼𝛼𝛼 ⁄the𝑐𝑐 indefinite integral,

1 + = sinh . Eq.𝛾𝛾 29 can be inverted,𝑣𝑣 𝑡𝑡 2 −1 𝑑𝑑𝑑𝑑⁄� 𝑦𝑦 𝑦𝑦 = sinh( ) ∫ 2 (30) 𝑐𝑐 which applies at the base of the𝑐𝑐𝑐𝑐 rocket𝛼𝛼 = 𝛼𝛼0.𝑡𝑡 ̃⁄Here𝑐𝑐 sinh is the hyperbolic function, sinh = ( ). Finally, using Eq.𝑥𝑥� 22, we can write𝑥𝑥 ( ) in terms of for = 0, 1 𝑥𝑥 −𝑥𝑥 𝑥𝑥 2 𝑒𝑒 −(𝑒𝑒) = 1 + ( ) = 1 + sinh ( 𝑥𝑥 𝑡𝑡 ) = cosh(𝑡𝑡̃ 𝑥𝑥)� 2 2 2 (31) 𝑐𝑐 2 𝑐𝑐 2 𝑐𝑐 where we used 𝑥𝑥the𝑡𝑡 hyperbolic𝛼𝛼 � identity,𝛼𝛼𝛼𝛼⁄𝑐𝑐 cosh𝛼𝛼 � = 1 + sinh𝛼𝛼𝑡𝑡̃⁄𝑐𝑐, and cosh𝛼𝛼 =𝛼𝛼𝑡𝑡̃⁄(𝑐𝑐 + ). 2 2 1 𝑥𝑥 −𝑥𝑥 As discussed in Eq. 27, the generalizations𝑥𝑥 of Eq. 30 and 𝑥𝑥31 to non-zero𝑥𝑥 2>𝑒𝑒0 should𝑒𝑒 be [3],

= + sinh( ) 𝑥𝑥� 2 𝑐𝑐 𝑐𝑐𝑐𝑐 �𝑥𝑥� � 𝛼𝛼𝑡𝑡̃⁄𝑐𝑐 𝛼𝛼 = + cosh( ) 2 (32) 𝑐𝑐 𝑥𝑥 �𝑥𝑥� 𝛼𝛼 � 𝛼𝛼𝑡𝑡̃⁄𝑐𝑐 These expressions reproduce the class of hyperbolas, = ( + ) discussed in Eq. 2 2 2 2 2 27. 𝑥𝑥 − 𝑐𝑐 𝑡𝑡 𝑥𝑥� 𝑐𝑐 ⁄𝛼𝛼

An Introduction to the Equivalence Principle of General Relativity

Let’s end this discussion with two points of interest in general relativity. First, what is the metric in the non-inertial, accelerating coordinate grid ( , )? This is a uniformly accelerating environment so the Equivalence Principle of general𝑐𝑐𝑡𝑡̃ 𝑥𝑥� relativity states that that its local properties should be the same as those of a uniform gravitational field with a gravitational acceleration = ( ) ( ) and a gravitational potential = = for weak fields, 1. Second, the 𝑔𝑔 2 𝛼𝛼space, called the Rindler Wedge𝑉𝑉 after𝑥𝑥� its𝑔𝑔 𝑥𝑥�inventor,𝛼𝛼𝑥𝑥� W.Rindler, has𝑉𝑉 past𝑥𝑥� ⁄ and𝑐𝑐 ≪future horizons which make it a toy model of black holes!

To begin let’s calculate the metric in these variables [3]. This is easy. We just change variables using Eq. 32. In the ( , ) variables we just have the Minkowski metric so,

𝑐𝑐𝑐𝑐 𝑥𝑥= = (33) 2 2 2 2 𝜇𝜇 𝜈𝜈 𝜇𝜇𝜇𝜇 𝜇𝜇𝜇𝜇 To calculate we simply change𝑑𝑑𝑑𝑑 variables𝑐𝑐 𝑑𝑑𝑑𝑑 − following𝑑𝑑𝑑𝑑 ∑ Eq.𝑔𝑔 32𝑑𝑑 and𝑥𝑥� 𝑑𝑑 take𝑥𝑥� differentials,

𝑔𝑔𝜇𝜇𝜇𝜇 = sinh( ) + + cosh( ) 2 𝑐𝑐 𝛼𝛼 𝑐𝑐𝑐𝑐𝑐𝑐 𝛼𝛼𝑡𝑡̃⁄𝑐𝑐 𝑑𝑑𝑥𝑥� �𝑥𝑥� � 2 𝛼𝛼𝑡𝑡̃⁄𝑐𝑐 𝑐𝑐𝑐𝑐𝑡𝑡̃ 𝛼𝛼 𝑐𝑐 = sinh( ) + 1 + cosh( ) (34) 𝛼𝛼𝑥𝑥� 2 𝑐𝑐𝑐𝑐𝑐𝑐 𝛼𝛼𝑡𝑡̃⁄𝑐𝑐 𝑑𝑑𝑥𝑥� � 𝑐𝑐 � 𝛼𝛼𝑡𝑡̃⁄𝑐𝑐 𝑐𝑐𝑐𝑐𝑡𝑡̃ And similarly,

= cosh( ) + 1 + sinh( ) (35) 𝛼𝛼𝑥𝑥� 2 𝑑𝑑𝑑𝑑 𝛼𝛼𝑡𝑡̃⁄𝑐𝑐 𝑑𝑑𝑥𝑥� �� 𝑐𝑐 � 𝛼𝛼𝑡𝑡̃⁄𝑐𝑐 𝑐𝑐𝑐𝑐𝑡𝑡̃ The invariant interval is then,

= = 1 + (36) 2 2 2 2 2 𝛼𝛼𝑥𝑥� 2 2 2 2 𝑑𝑑𝑑𝑑 𝑐𝑐 𝑑𝑑𝑑𝑑 − 𝑑𝑑𝑑𝑑 � 𝑐𝑐 � 𝑐𝑐 𝑑𝑑𝑡𝑡̃ − 𝑑𝑑𝑥𝑥� after some algebra that uses the hyperbolic identities stated above. We learn from this that,

= 1 + = 1 + 2 + (37) 𝛼𝛼𝑥𝑥� 2 𝛼𝛼𝑥𝑥� 𝛼𝛼𝑥𝑥� 2 2 2 2 𝑔𝑔00 � 𝑐𝑐 � 𝑐𝑐 �𝑐𝑐 � describes the metric of the Rindler Wedge!

In Chapter 11 of the textbook we considered weak, static gravitational fields and found ( ) 1 + 2 ( )/ . So, we can identify here, that for an accelerating reference frame with 2 𝑔𝑔proper00 𝑥𝑥� acceleration≈ 𝑉𝑉 𝑥𝑥� , 𝑐𝑐the equivalent problem in gravitation has the potential,

𝛼𝛼 ( ) = (38)

This was our expectation. But this discussion𝑉𝑉 𝑥𝑥� also𝛼𝛼𝑥𝑥� applies to strong gravity where the inequality ( ) 1 does not apply. In that case, from Chapter 11.4.4, 2 𝑉𝑉 𝑥𝑥� ⁄𝑐𝑐 ≪ = 1 + = ( ) (39) 2 2 𝛼𝛼𝑥𝑥� 2𝑉𝑉 𝑥𝑥� ⁄𝑐𝑐 2 𝑔𝑔00 � 𝑐𝑐 � 𝑒𝑒 So,

( ) = ln 1 + (40) 2 𝛼𝛼𝑥𝑥� 2 𝑉𝑉 𝑥𝑥� ⁄𝑐𝑐 � 𝑐𝑐 � This entire exercise in accelerating reference frames is particularly interesting since it provides a soluble example of the Equivalence Principle. Equally important, however, is that it shows the limitations of the principle as well. Recall from the textbook that the Equivalence Principle applies only to local measurements that one can make in accelerating frames and uniform gravitational fields. It is obvious that the physics in the two sorts of frames are not equivalent beyond that: if an experimenter was in an accelerating elevator but could look through a window at his surroundings, he would certainly know that he is not at rest on a massive planet: the long range propagation of signals in the two environments are clearly distinct. We can see this in the Minkowski diagram Fig. 7 which shows the inertial and accelerating frames in Wedge I.

Fig. 7 Rindler space showing horizons, particles passing through horizons and accelerating grid.

Suppose that a particle meanders in the accelerating frame into region II, passing through the light cone = on the𝑃𝑃 way. Now the particle has no way to communicate with an accelerating𝑥𝑥 observer𝑐𝑐𝑐𝑐 because that would require𝑃𝑃 a signal that could travel in excess of the 𝑂𝑂 16

speed of light! itself can never pass back through the light cone and reach observer . Thus the

accelerated frame,𝑃𝑃 the Wedge I, has a simple future horizon and can be described as a 𝑂𝑂model of a black hole. The figure also shows a past horizon where particles can enter Wedge I and be detected but cannot leave the Wedge until > 0 and do so through the future horizon.

We can illustrate the existence of the future𝑡𝑡 horizon in the Rindler Wedge by doing a simple experiment. Consider a planet at rest in the inertial frame. Suppose it is a distance to the right

of a rocket which blasts off from position = 0 at = 0 . At time the rocket’s position𝐷𝐷 is given by Eq. 22 adjusted for these initial conditions,𝑥𝑥 𝑡𝑡 𝑡𝑡

( ) = 1 + ( ) 1 2 (41) 𝑐𝑐 2 𝑥𝑥 𝑡𝑡 𝑔𝑔 �� 𝑔𝑔𝑔𝑔⁄𝑐𝑐 − � So, an observer on the rocket measures the distance to the planet as ( ) / ( ) where we

have accounted for Lorentz contraction. Recall that ( ) = 1 + ( �𝐷𝐷 −) 𝑥𝑥 . So,𝑡𝑡 � 𝛾𝛾 𝑡𝑡 2 𝛾𝛾 𝑡𝑡 � 𝑔𝑔𝑔𝑔⁄𝑐𝑐 1 + ( ) 1 + ( ) 2 2 = 𝑐𝑐 2 = ( ) �𝐷𝐷 − ��1 + ( 𝑔𝑔𝑔𝑔⁄)𝑐𝑐 − �� 1 + ( 𝑐𝑐 ) 2 �𝐷𝐷 − 𝑥𝑥 𝑡𝑡 � 𝑔𝑔 �𝐷𝐷 𝑔𝑔 � 𝑐𝑐 2 2 − 𝛾𝛾 𝑡𝑡 𝑔𝑔 which approaches as becomes� large.𝑔𝑔𝑔𝑔 The⁄𝑐𝑐 negative sign� simply𝑔𝑔 𝑔𝑔means⁄𝑐𝑐 that the planet is 2 now behind the rocket,− 𝑐𝑐 ⁄but𝑔𝑔 the𝑡𝑡 surprise is that the same distance is found for all ! In 2 other words, there is a “horizon” at : from the point of view− 𝑐𝑐 of⁄ the𝑔𝑔 rocket, which 𝐷𝐷would 2 be at rest in the accelerating coordinate− 𝑐𝑐 grid⁄𝑔𝑔 of Rindler, all planets eventually stop receding and accumulate a distance behind the rocket. 2 Another way of describing− 𝑐𝑐 ⁄𝑔𝑔 this effect is to note that from the perspective of an observer at rest in the inertial frame , the rocket is accelerating to the right and its velocity is approaching the

speed limit. Therefore,𝑆𝑆 signals sent towards it at a later time can never reach it if is too large. The rocket’s location is given by Eq. 41. Suppose that a burst𝑡𝑡0 of light is sent toward𝑡𝑡0 the rocket ( ) at = 0 from position . Its position at later times is = + .𝐿𝐿 So, 𝑡𝑡 𝑥𝑥0 𝑥𝑥𝐿𝐿 𝑡𝑡 𝑐𝑐𝑐𝑐 𝑥𝑥0 ( ) ( ) = 1 + ( ) 1 ~ + 2 2 2 3 𝑐𝑐 2 𝑐𝑐 𝑐𝑐 𝑥𝑥 𝑡𝑡 − 𝑥𝑥𝐿𝐿 𝑡𝑡 �� 𝑔𝑔𝑔𝑔⁄𝑐𝑐 − � − 𝑐𝑐𝑐𝑐 − 𝑥𝑥0 − − 𝑥𝑥0 2 So, once 2 , the observer𝑔𝑔 on the rocket cannot detect the axis𝑔𝑔 to the left𝑔𝑔 of𝑡𝑡 . 2 𝑡𝑡 ≫ 𝑐𝑐⁄ 𝑔𝑔 𝑥𝑥 − 𝑐𝑐 ⁄𝑔𝑔 17

The hyperbolic coordinates of the Rindler Wedge provide a coordinate system for Minkowski space that is well equipped to describe accelerated reference systems. But the underlying space- time is just flat Minkowski space. The metric in Rindler coordinates reads,

= 1 + 2 2 𝛼𝛼𝑥𝑥� 2 2 2 𝑑𝑑𝑑𝑑 � 2 � 𝑐𝑐 𝑑𝑑𝑡𝑡̃ − 𝑑𝑑𝑥𝑥� This is just a change of coordinates from an inertial𝑐𝑐 coordinate system ( , ) where

= 𝑐𝑐𝑐𝑐 𝑥𝑥 2 2 2 2 These are the same space-times. Geometric𝑑𝑑𝑑𝑑 quantities,𝑐𝑐 𝑑𝑑𝑑𝑑 − such𝑑𝑑𝑑𝑑 as the Ricci scalar must be the same

in either description. = 0 is clearly the case for = . Is =𝑅𝑅 0 for the Rindler 2 2 2 2 coordinate system? Let’s𝑅𝑅 answer the question physically𝑑𝑑𝑑𝑑 𝑐𝑐. 𝑑𝑑The𝑑𝑑 − Rindler𝑑𝑑𝑑𝑑 Wedge𝑅𝑅 experiences a uniform gravitational acceleration: Two balls at identical heights but having any transverse separation fall to the floor of the mother ship at the some time while maintaining their transverse separation. In other words,, there are no tidal forces in the Rindler𝑡𝑡̃ Wedge. But vanishing tidal forces means vanishing curvature, as we learned in chapter 12 of the textbook. We must consider non-uniform gravitational fields to find curved space-times.

We can make a similar observation in the context of classical differential geometry. Recall the equation for the Gaussian curvature from the textbook,

1 1 1 = + 𝜕𝜕 𝜕𝜕√𝐺𝐺 𝜕𝜕 𝜕𝜕√𝐸𝐸 𝐾𝐾 − � � � � �� where the metric for the surface embedded√𝐸𝐸𝐸𝐸 𝜕𝜕𝜕𝜕 in√ 𝐸𝐸three𝜕𝜕𝜕𝜕 dimensional𝜕𝜕𝜕𝜕 √𝐺𝐺 Euclidean𝜕𝜕𝜕𝜕 space reads,

= + 2 2 2 Imagine that we have a space with =𝑑𝑑𝑑𝑑1 and𝐸𝐸 𝑑𝑑𝑑𝑑= (1𝐺𝐺+𝑑𝑑𝑑𝑑) . Then the second term in the 2 (1 + ) expression for the Gaussian curvature𝐺𝐺 is proportional𝐸𝐸 to 𝑣𝑣2 which vanishes identically 𝜕𝜕 2 and this implies = 0. 𝜕𝜕𝑣𝑣 𝑣𝑣

𝐾𝐾 References

1. W. Rindler, Introduction to Special Relativity, Oxford University Press, Oxford, 1991.

2. W. Rindler, Essential Relativity, Springer-Verlag, Berlin, 1971.

3. Gerard, ‘t Hooft, Introduction to General Relativity, Rinton Press, Inc., Princeton, New Jersey, 2001.