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. ′ 1 Fig. 1 Reference frames and , with moving in the direction at velocity . ′ ′ Frame moves to the right in the -direction at velocity relative to the frame . In ′ applications at this point in the textbook the velocity was always set to a constant for all times. ( ) In the case here is a function of time chosen so that matches the speed of an accelerating rocket at the instant . The frame is called the “instantaneous rest frame of the rocket” as ′ shown in Fig. 2. 2 Fig. 2 Instantaneous rest frame of rocket. Gas nozzle velocity points to the left ′ 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 everywhere in ′ ′ in the usual way as discussed in the textbook in Chapters 2 and 3. To begin, let’s find the rocket’s velocity ( ) and position ( ) in the inertial frame . This is a one dimensional problem, so we suppress the transverse spatial dimensions and forego 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. Suppose that at time in the rocket’s instantaneous rest frame the rocket of ( ) ( ) ′ rest mass ejects rest mass at a nozzle velocity . The mass ejected could consist ′ ′ of gases as in a conventional rocket∆ or even light (photons) in 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 that is a constant for all time .We take to be positive by convention and note that the ejected travels in the - ′ ′ direction,as shown in the figures. The energy and momentum carried by the∆ ejected and lost from the rocket is = and = / , where we recalled the general∆ energy- ′ 2 ′ ′ 2 ∆ −∆ ∆ ∆ 3 momentum relation = ( ) and we noted that the velocity of the ejected mass is in 2 the instantaneous rest frame⁄ of the rocket. What principles of dynamics have we used∆ here? 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 , ∆in the⁄ frame∆ by making a ′ ′ ( ) boost of velocity , ∆⁄ ∆ = ( + ) = (1 + / ) = (1 + / ) ′ ′ ′ 2 2 2 ∆ = (∆ + ∆ / )=∆ ( /)( + )−=∆ ( + )/ (1) ′ ′ 2 ′ 2 2 These expressions∆ tell ∆us how the∆ energy -momentum ∆ four vector −( (∆) , ()) change s with ( ) ( ) ( ) time. The energy is related to the rocket’s velocity and rest mass ⁄ = with 2 ( ) = (1 ( ) ) . The momentum of the rocket in frame is ( ) =( ( )/ ) ( ). 2 2 −1⁄2 2 We differentiate− this⁄ expression and substitute in Eq. 1, = ( ) = + = 1 + + (2) 2 2 2 2 ∆ ∆ ∆ ∆ −∆ � � ∆ But the left-hand side is = ( + ) from Eq.1, 2 ∆ −∆ ( + ) = 1 + + (3) 2 2 2 −∆ −∆ � � ∆ We can solve for , ∆ = + + = (1 ) 2 (4) ∆ 2 2 2 ∆ −∆ − ∆ ∆ −∆ − ⁄ So, we can calculate the rate at which ( ) increases due to the ejection of mass at nozzle velocity , = (5) 2 ∆ ∆ − This equation is easy to integrate and find how ( ) varies as a function of ( ) where ( ) = = 0 , ⁄0 0 ( ) = = ln( ( ) ) (6) 2 ∫0 − ∫0 − ⁄0 4 On the left-hand side, = = = + = ln ( )( ) (7) 2 1 1 1 1+⁄ 2 2 ∫0 ∫ 1− ⁄ ∫ 1+⁄ 1−⁄ 2 ∫ �1+⁄ 1−⁄� 2 �1−⁄� Collecting our partial results, / ln( ( ) ) = ln = ln (8) 1+⁄ 1−⁄ 2 ⁄0 − 2 �1−⁄� �1+⁄� So, ( ) ( ) = ( ) (9) 1− ⁄ ⁄2 0 �1+ ⁄� which shows how ( ) decreases as ( ) increases. Eq. 9 can be inverted, ( ( ) ) ( ) = (10) ( ( ) )2⁄ 1− ⁄0 2⁄ �1+ ⁄0 � Which predicts how ( )/ increases as ( ) decreases for a given nozzle velocity . It is a scaling relation. The speed limit sets the ⁄scale0 of all velocities in the problem. You are⁄ ( ) encouraged to plot Eq. 10 for various chooses of and needed to produce velocities near the speed of light. ⁄ ⁄0 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 ( ) = 0.5 ⁄can be achieved and the final velocity of the rocket payload is 0.60 , a very impressive ⁄ 0 velocity. Unfortunately, a rocket engineer could argue quite forcefully tha 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 “relativistic ≪ rocket travel” and learn 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! 5 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 ∆at nozzle−∆ velocity . In the frame where the rocket has ( ) instantaneous velocity , = ∆ . (Positive points in the negative direction, so ( ) ′ an observer at rest in frame ∆ measures−∆ the− velocity of to be , according to Galilean mechanics.). Since = non-relativistically, we have,∆ − ( ) = ( ) ( ) +∆ = −∆( ) −+( ) (11) So, ∆ ∆ − ∆ ∆ = (12) ∆ ∆ − which integrates to, = (13) − Giving, ∫ ∫ ( ) ( ) = ln (14) − 0 which predicts how the velocity of the rocket increases as it burns its fuel and ( ) decreases. This expression can be inverted, ( ) ( ) = (15) − ⁄ 0 Does the non-relativistic analysis agree with its relativistic version when ? The crucial non-relativistic expression is Eq. 12, ≪ = (16) − The corresponding relativistic expression is Eq. 5, 6 = (17) 2 − , As long as is well approximated by unity with corrections of order 2 , and Eq. 16 2 2 is retrieved. Success!≪ � � 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, = = (18) where is the component of the particle’s relativistic momentum, = . Denote = = , a fixed acceleration. Eq. 18 is easy to solve, ⃗ ⃗ ⁄ ⁄ ( ) = (19) So, = (20) Since, = (1 ) , Eq. 20 predicts ( ), after some algebra, 2 2 −1⁄2 − ⁄ ( ) = (21.a) ( ) 2 �1+ ⁄ if we assume the initial condition = 0 at = 0. We discussed in the textbook that for small , ( ) ( ( )) , then = and we retrieve Newton’s prediction that if a particle expe≪ riences⁄ a constant ≈ acceleration⁄ ⁄ its velocity grows proportional to .We also noted that in the ( ) extreme relativistic limit, , then and the charged particle’s velocity approaches the speed limit. ≫ ⁄ → 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. 7 To begin, what is the acceleration that the

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