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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.
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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.
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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-
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momentum relation =(
) and we noted that the velocity of the ejected mass is in
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 + /)
=(+/)=(/)(+)=(+)/ (1)
These expressions tell us how the energy-momentum four vector ((),()) changes with
time. The energy () is related to the rocket’s velocity and rest mass ()=() with
()=(1()
)
. The momentum of the rocket in frame is ()=(()/)().
We differentiate this expression and substitute in Eq. 1,
=()=+=1 +
+ (2)
But the left-hand side is =(+) from Eq.1,
(+)=1 +
+ (3)
We can solve for ,
=++
=(1
) (4)
So, we can calculate the rate at which () increases due to the ejection of mass at nozzle
velocity ,
=
(5)
This equation is easy to integrate and find how () varies as a function of ()
where
=(= 0),
=
()
=ln(()
) (6)
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On the left-hand side,
=
=
(
)(
)=
+
=
ln
(7)
Collecting our partial results,
ln(()
)=
ln
=ln
/ (8)
So,
()=()
()
(9)
which shows how () decreases as () increases. Eq. 9 can be inverted,
()=(()
)
(()
)
(10)
Which predicts how ()/ increases as ()
decreases for a given nozzle velocity
. It
is a scaling relation. The speed limit sets the scale 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.
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
velocity. Unfortunately, a rocket engineer could argue quite forcefully that 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!
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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)
which predicts how the velocity of the rocket increases as it burns its fuel and () decreases.
This expression can be inverted,
()
=()
(15)
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,
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=
(17)
As long as , is well approximated by unity with corrections of order
, and Eq. 16
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,
()=
(
) (21.a)
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
experiences 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.
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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,
=. We can understand this from several perspectives. First, in Chapter 8 of the
textbook we derived the transformation law for
,
, electric and magnetic fields, and derived
=. In more elementary fashion we can consider the electric field along the axis of a large
capacitor. 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 in the rest frame of the accelerating particle experiences
for all .
Next let’s find the trajectory of the accelerating particle in the frame , (,). Eq. 21.a can
be written in the form,
()=
(
) (21.b)
which can be integrated to find,
()=
(
)=
1 + (
) (22)
where we chose (= 0)=
. A more insightful way to write this result is to square it and
treat () and on an equal footing,
=
(23)
which we recognize as invariant hyperbolas [1,2]! In a boosted frame, (,)(,), Eq. 23
becomes,
=
(24)
These results are shown in the Minkowski diagram Fig. 3.
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Fig. 3 Hyperbola of constant proper acceleration and line of constant .
The invariant hyperbola Eq. 23 has a simple interpretation. Imagine the rocket beginning its
flight at = 0 from an initial position on the axis of =
. At the same time launch a
light ray toward the rocket but starting at the origin. Will the light 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
the surface of the earth, 9.80 m/s2, and verify that is a huge distance.) The outcome of the
race is clear from the Minkowski diagram: the light ray 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,
=()(+)=
(25)
Using Eq. 22 for () we have, for > 0,
=
=
(
)~
(26)
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where we took
1 in the last expression to calculate that asymptotically ()~: so
the light ray approaches but never catches the rocket. Note that the light ray in Fig. 3 is the
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 way 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, the invariant
distance-squared to the light ray is
for all .
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)
for > 0 as shown in Fig. 4. The axis lies on the axis at = 0.
Fig. 4 Coordinates of accelerating grid.
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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
as shown in Fig. 5.
Fig. 5 Spatial coordinates along the length of the accelerating rocket.
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
.
To better understand the accelerating grid (,) let’s differentiate Eq. 27 with respect to
for fixed and ,
2
2= 0 (28.a)
Or, identifying
=,
=
(28.b)
But the Lorentz transformation to the rocket having instantaneous velocity () reads,
=()
=
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So, the line =
, Eq. 28.b,is a line of constant . These are rays from the origin, the axes, at
velocity . We recognize this as an example of the Relativity of Simultaneity emphasized in
Chapter 2 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 to the hyperbolas. This shows that all the points
along the mother ship have the same velocity at any given time
in the mother ship’s rest
frame. So, the mother ship accelerates without internal stresses and 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.
Fig. 6 Rindler coordinates showing lines of constant () and constant .
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The grid is static in the variable
- the points of fixed maintain the same distance between
them that they had at = 0 when the entire rocket was simultaneously at rest in the inertial
frame . Note that the top of the mother ship, points of large , accelerate less than points near
its bottom, small. From Eq. 27 we could define,
()=+
where > 0 and (0)=. This guarantees that the Lorentz contraction of the ship is given by
the global transformation law, =() where is the proper length of the ship and =
() is its instantaneous velocity: since the ship contracts as measured in the frame , its top
must accelerate less than its bottom to accommodate a constant velocity.
Next we need the explicit transformation from (,) to (,). First, let’s consider the
relation between and at = 0, the base of the rocket. A clock at = 0 in the rocket measures
. When the rocket has velocity at time , must be related to by time dilation, =.
Integrating this relation,
=
=
(
)=
sinh(
) (29)
where we use Eq. 21 to write in terms of () and we used the indefinite integral,
1 +
=sinh. Eq. 29 can be inverted,
=
sinh(
) (30)
which applies at the base of the rocket = 0. Here sinh is the hyperbolic function,
sinh =
(). Finally, using Eq. 22, we can write () in terms of for = 0,
()=
1 + (
)=
1 + sinh(
)=
cosh(
) (31)
where we used the hyperbolic identity, cosh= 1 + sinh, and cosh =
(+).
As discussed in Eq. 27, the generalizations of Eq. 30 and 31 to non-zero > 0 should be [3],
=+
sinh(
)
=+
cosh(
) (32)
These expressions reproduce the class of hyperbolas, =(+
) discussed in Eq.
27.
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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 general 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
space, called the Rindler Wedge after its inventor, W.Rindler, has past and 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,
==
(33)
To calculate we simply change variables following Eq. 32 and take differentials,
= sinh(
)++
cosh(
)
=sinh(
)+1 +
cosh(
) (34)
And similarly,
=cosh(
)+1
+
sinh(
) (35)
The invariant interval is then,
==1 +
(36)
after some algebra that uses the hyperbolic identities stated above. We learn from this that,
=1 +
= 1 + 2
+
(37)
describes the metric of the Rindler Wedge!
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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
proper acceleration , the equivalent problem in gravitation has the potential,
()= (38)
This was our expectation. But this discussion also applies to strong gravity where the inequality
()
1 does not apply. In that case, from Chapter 11.4.4,
=1 +
=(
)
(39)
So,
()
=ln 1 +
(40)
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.
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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 the way. Now the particle has no way to communicate with an
accelerating observer because that would require a signal that could travel in excess of the
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speed of light! itself can never pass back through the light cone and reach observer . Thus the
accelerated frame, the Wedge I, has a simple future horizon and can be described as a model 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 future 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 position is
given by Eq. 22 adjusted for these initial conditions,
()=
1 + (
)1 (41)
So, an observer on the rocket measures the distance to the planet as ()/() where we
have accounted for Lorentz contraction. Recall that ()=1 + (
) . So,
()
()=
1 + (
)1
1 + (
)=+
1 + (
)
which approaches
as becomes large. The negative sign simply means that the planet is
now behind the rocket, but the surprise is that the same distance
is found for all ! In
other words, there is a “horizon” at
: from the point of view of the rocket, which would
be at rest in the accelerating coordinate grid of Rindler, all planets eventually stop receding and
accumulate a distance
behind the rocket.
Another way of describing 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, 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 burst of light is sent toward the rocket
at = 0 from position . Its position at later times is ()=+. So,
()()=
1 + (
)1~
+
2
So, once 2
, the observer on the rocket cannot detect the axis to the left of
.
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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 +
This is just a change of coordinates from an inertial coordinate system (,) where
=
These are the same space-times. Geometric quantities, such as the Ricci scalar must be the same
in either description. = 0 is clearly the case for =. Is = 0 for the Rindler
coordinate system? Let’s answer the question physically. The Rindler Wedge 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 Rindler 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
where the metric for the surface embedded in three dimensional Euclidean space reads,
=+
Imagine that we have a space with = 1 and =(1 + ). Then the second term in the
expression for the Gaussian curvature is proportional to
(1 + ) which vanishes identically
and this implies = 0.
References
1. W. Rindler, Introduction to Special Relativity, Oxford University Press, Oxford, 1991.
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2. W. Rindler, Essential Relativity, Springer-Verlag, Berlin, 1971.
3. Gerard, ‘t Hooft, Introduction to General Relativity, Rinton Press, Inc., Princeton, New
Jersey, 2001.