Original by Michael Weiss 1995.

John Bell described this Special Relativity paradox in the essay "How to teach special relativity" in his collection "Speakable and Unspeakable in Quantum Mechanics".  He did not invent the puzzle, but we'll call it Bell's Spaceship Paradox.

To begin, a statement of the paradox—and if you notice some ambiguities in my formulation, that's the point!  (That's always the point in SR paradoxes.)  Bell asks us to consider two rocket ships, each accelerating at the same constant rate, one chasing the other.  The ships start out at rest in some coordinate system (the "lab frame").  Since they have the same acceleration, their speeds should be equal at all times (relative to the lab frame) and so they should stay a constant distance apart (in the lab frame).  But after a time they will acquire a large velocity, and so the distance between them should suffer Lorentz contraction.  Which is it?

I think the best approach is through spacetime diagrams.  I'll lay out one explanation in detail, then sketch two more.  The first two are pure SR, the last has the flavor of GR but not the substance.

Don't tackle the physical situation directly; instead, imagine some pictures.  Geometrical figures seem less prone to the "simultaneity confusion" that bedevils descriptions of SR paradoxes.  Once you have the geometrical picture in hand, it's a piece of cake to translate it into physical terms.  I won't try to draw any pictures (well, hardly any), but please make rough sketches—the explanation makes no sense without that.  [Not to keep you in suspense, I'll follow each geometrical statement with the physical translation in brackets.]  (Yes, I know you can do pictures on the Web.  You think I'm getting paid for this?)

If you're not familiar with spacetime diagrams, the main thing to bear in mind is how the x and t axes change when going from one frame to another frame in uniform motion with respect to the first.  (The first frame is called the "lab frame"; the second is the frame of the "co-moving observer".)

```
t-axis	t′-axis
|      /
|     /
|    /	           . x′-axis
|   /	       .
|  /       .
| /    .
|/ .
---------------------|------------------------ x-axis
.  /|
.     / |
.        /  |
.           /   |
.              /    |
/     |

```
The only real problem is drawing the x′-axis.  It should look just like the t′-axis, slanted so that the angle between the x-axis and x′-axis equals the angle between the t-axis and t′-axis.  (This is assuming we choose units so c = 1.)

The x-axis contains all events (spacetime points) that, according to the lab frame, occur at t = 0; that is, according to the lab people these events occur simultaneously.  The x′-axis contains all events that happen at t′ = 0, and so are simultaneous according to the moving observers.  This easy graphical representation of the famous "failure of simultaneity" is one of the great strengths of Minkowski's idea.

First picture: we draw cartesian coordinates in the plane, label the vertical axis the t-axis, and the horizontal axis the x-axis.  [The (t, x) system is the lab frame.]  We also draw one branch of the hyperbola x2 − t2 = 1, say the right-hand branch x = sqrt(1 + t2).  Next we draw a parallel copy of this branch, translated rightwards some fixed distance k, i.e. we draw x = k + sqrt(1 + t2).  [These two curves are the world lines of our two rocket ships.]

For each curve, dx/dt = 0 at t = 0.  [The ships are initially at rest.]  Also, exercise: d2x/dt2 = 1 at t = 0.  [Initial acceleration = 1.]  Obviously d2x/dt2 is not constant.  However, pick a point P on one of the curves, and draw Minkowski coordinates (t′, x′) with origin at P and with t′-axis tangent to the curve.  [The (t′, x′) system is the frame of the inertial instantaneously co-moving observer.]  That is, if P has coordinates (t0, x0) in the original system, then:

```     t′ = γ(t−vx) + C0   ,         v = dx/dt at t = t0
x′ = γ(x−vt) + D0   ,         γ = 1/sqrt(1−v2)
```
where C0 and D0 are chosen so that the P has coordinates (0, 0) in the primed system.  Exercise (or take my word for it): d2 x′/dt2 = 1 at P.  [The acceleration of each ship is 1 as measured by inertial instantaneously co-moving observers, at all times.]

Note a few facts about this picture.  First, any horizontal line t = t0 crosses the two curves a constant x-distance apart; the constant is the number k from a few paragraphs back.  [According to the lab frame, the ships stay a constant distance apart.]  If we pick a point P on the left-hand curve, and draw Minkowski coordinates through P as above, then the x′-axis will cross the two curves at two points whose x′-coordinates differ by more than k.  [The co-moving observer says the ships have gotten farther apart.]  If k is small compared to γ, then γk is a first-order approximation to this x′-difference.  [The lab frame distance between the ships is the co-moving distance, which is "Lorentz expanded" and then subjected to Lorentz contraction, and the expansion and contraction cancel each other so that the ships don't get closer together in the lab frame.]

[How did the ships get farther apart, if they maintained the same constant acceleration at all times?]  In the (t′, x′) coordinate system, dx′/dt′ = 0 at t′ = 0 for the left-hand curve, but dx′/dt′ > 0 at t′ = 0 for the right-hand curve.  [The co-moving observers say the pursuing ship is momentarily at rest, but the pursued ship is moving, thanks to that old relativistic standby, the failure of simultaneity.  So the pursued ship is "pulling away".]

The x′-distance between the curves is actually slightly greater than γk.  You can see this by a geometrical construction.  Remember that P is point on the left-hand curve, and the t′-axis passes through P and is tangent to that curve.  Draw a horizontal line (a line t = constant) through P; let this line cross the right-hand curve at Q.  Draw a line parallel to the t′-axis through Q (a line of the form t′ = constant).  Fact: the x′-axis crosses the two slanted lines, t′ = 0 and t′ = constant, at x′-coordinates 0 and γk, respectively.  Since the right-hand curve is tangent to t′ = constant at Q, the x′-axis will cross the right-hand curve at x′ > γk.  [Exercise: rephrase this physically!]

This first picture interprets "two ships with the equal constant accelerations" to mean "equal in the lab frame, and constant for co-moving observers".  Note that the lab frame says that the accelerations are not constant, and the co-moving observers say the accelerations are not equal!  (More precisely, any particular co-moving observer says this.  The phrase "the co-moving observers" does not refer to a single frame of reference, unlike the phrase "the lab frame".)  The lab frame says the ships maintain a constant distance from each other; the co-moving observers don't agree.

Second picture: pick the same left-hand curve as before, but pick the right-hand curve to be:

```    x = sqrt(K2 + t2),    K > 1.
```
Misner, Thorne, and Wheeler's book "Gravitation" discusses this case in more detail under the name "Fermi–Walker transport".  Here it turns out that the distance between the ships is constant according to co-moving observers.  The lab frame people measure a Lorentz-contracted distance.  The co-moving observers again say that the ships maintain constant acceleration.  The lab frame says that the pursuer accelerates at a greater rate than the pursued—yet the pursuer never catches his prey!  (Reminds me of Achilles and the Tortoise—or Keats' Grecian Urn.)

Last picture, edging just a bit towards GR: impose the following (indefinite) metric on the (t, x) plane:

```   dτ2 = e2x dt2 − dx2
```
This spacetime possesses a "uniform gravitation field".  More precisely, time-like geodesics with the initial condition dx/dτ = 0 at some point P satisfy d2x/dτ2 = −1 at P.  So if a lab-frame observer (that is, (t, x) coordinate system) lets go of an object, he'll measure it to drop with acceleration 1.

Spaceships in this universe keep stationary by setting their engines to constant thrust.

You might be tempted to think that this universe is equivalent to ordinary flat spacetime via a coordinate transform.  Not so!  Take a couple of spaceships, stationary in the lab frame.  They keep their engines blasting away with constant force.  (I'll refrain from the obvious Star Trek jokes).  If someone steps off a ship and starts falling, he becomes (for a moment) one of our co-moving observers.  (He's in free fall, and so inertial by definition!)  He falls at the universal constant acceleration of 1—or in his humble opinion, the ship is accelerating at this rate.

So if our brave new world really is flat spacetime in disguise, then the "intrinsic acceleration" of each ship—the acceleration as measured by a co-moving observer—is always 1.  So our two ships must trace out parallel hyperbolae, as in the first picture.  But then the distance between them would increase with time, as measured by the co-moving observers.  But it doesn't!  So our new universe is not flat.

You can clinch the matter by computing the curvature—you should get R = −2 (at least I did).  The 4-D variant:

```    dτ2 = e2zdt2 − dx2 − dy2 − dz2
```
is also amusing to play with.  The Einstein field equations imply a stress–energy tensor with zero density but non-zero (and non-isotropic) pressure—i.e., not physically realistic.

My guess is that the Einstein empty-space equations forbid a uniform gravitational field in the above sense.  I haven't checked this though.

Historically, Einstein had some trouble with these issues for a time.  See the discussion of Born's theory of relativistically rigid bodies in Pais's biography, "Subtle is the Lord...".  You may also want to look at The Rigid Rotating Disk in Relativity.