Talk:Bell's spaceship paradox/CH's essay

To do

 * copyedit (at one point we changed notation t,x,y,z -> T,X,Y,Z, and hopefully got everything changed)
 * correct numerator cosh to sinh in third expression for T at the top of the article (occurs twice)

Advanced analysis
We will use a Cartesian chart for Minkowski spacetime with the standard line element
 * $$ ds^2 = -dT^2 + dX^2 + dY^2 + dZ^2, \; \; -\infty < T, X, Y, Z < \infty $$

The proper time parameterized world line of the first observer is given by
 * $$ T = \begin{cases} s & s \leq 0 \\ \frac{\sinh(k \, s)}{k} & 0 < s < \sigma \\ \frac{\cosh(k \, \sigma)-1}{k} + s \, \cosh(k \sigma) & s \geq \sigma \end{cases} $$
 * $$ X = \begin{cases} \alpha_1 & s \leq 0 \\ \alpha_1 + \frac{\cosh(k \, s) - 1}{k} & 0 < s < \sigma \\ \alpha_1 + \frac{\cosh(k \, \sigma) - 1}{k} + s \, \sinh(k \, \sigma) & s \geq \sigma \end{cases} $$



In the figure, the first spaceship begins to accelerate at event A, which has coordinates T = 0, X = α1, and stops accelerating at event A&prime;, after time σ has elapsed by an ideal clock carried on board. During the acceleration phase, the path curvature (that is, the covariant derivative $$\nabla_{\vec{U}} \vec{U}$$ where $$\vec{U}$$ is the unit tangent vector to the world line) has constant magnitude k, as measured on board the spaceship.

Likewise, the proper time parameterized world line of the second observer is given by
 * $$ T = \begin{cases} s & s \leq 0 \\ \frac{\sinh(k \, s)}{k} & 0 < s < \sigma \\ \frac{\cosh(k \, \sigma)-1}{k} + s \, \cosh(k \sigma) & s \geq \sigma \end{cases} $$
 * $$ X = \begin{cases} \alpha_2 & s \leq 0 \\ \alpha_2 + \frac{\cosh(k \, s) - 1}{k} & 0 < s < \sigma \\ \alpha_2 + \frac{\cosh(k \, \sigma) - 1}{k} + s \, \sinh(k \, \sigma) & s \geq \sigma \end{cases} $$

In the figure, the second spaceship begins to accelerate at event B, which has coordinates T = 0, X = α2, and stops accelerating at event B&prime;, after time σ has elapsed by an ideal clock carried on board. After event B&Prime;, both spaceships can once again be treated as comoving inertial observers by the usual methods of elementary special relativity. We imagine that a string has been stretched between the two spaceships before the acceleration phase, and we wish to compare its length before and after the acceleration.

During the period $$T \leq 0$$, the two observers are separated by the constant distance $$\alpha_2-\alpha_1$$. After the acceleration phase, their world lines are once again portions of straight lines (timelike geodesics) and we can easily compute the new distance using elementary geometry of Minkowski space.

Consider the line through event A&prime; which is orthogonal (in Minkowski geometry, not euclidean geometry!) to the tangent vector to the world line there. In the language of elementary special relativity, this line represents a "space of simultaneity". It is now a simple exercise in analytic geometry to find the intersection of this line at event B&Prime; with the world line of the second observer. The new distance between the spaceships turns out to be
 * $$ (\alpha_2 - \alpha_1 ) \, \cosh(k \sigma) = (\alpha_2 - \alpha_1 ) \, \left( 1 + \frac{k^2 \sigma^2}{2} + O(\sigma^3) \right) $$

As Bell pointed out, unless the product k σ is very small, this increase in the distance between the two spaceships will break the string stretched between the two spaceships. See the figure above, where we have taken k = σ = 1, so that the length of the string has increased by about 1.5 (a factor which is more than sufficient to break most strings!).

Some might object that this conclusion would be unchanged if we replaced the hyperbolic arcs by any other timelike curve segments which have the appropriate endpoints (and the appropriate tangent vectors at these endpoints). In this case we cannot avoid analyzing the acceleration phase.

Bell observers
To assuage any remaining misgivings, we ought to verify that the string is indeed stretched continuously during the acceleration phase. This will be somewhat tricky, since it turns out (see Rindler coordinates) that even in flat spacetime, there are various distinct but operationally meaningful notions of distance which can be employed by accelerating observers. Fortunately, these all agree for very small distances. This suggests interpolating infinitely many new world lines between the two already considered. That is, we should introduce a family of ideal observers whose world lines form a timelike congruence. Then we can verify that during the acceleration phase, the distance between nearby bits of string increases continuously, without having to worry about competing notions of distance on larger scales.

We must find a unit timelike vector field whose integral curves are the world lines of form
 * $$ T = \frac{1}{k} \, \sinh(k s), \; \; X = \alpha + \frac{1}{k} \; \left( \cosh(k s) - 1 \right) $$

Differentiating with respect to proper time, we obtain
 * $$ \dot{T} = \cosh(k s), \; \; \dot{X} = \sinh (k s) $$

Eliminating s yields
 * $$ \dot{T} = \sqrt{1+k^2 \, T^2}, \; \; \dot{X} = k \, T $$

This gives a system of first order linear ordinary differential equations. Solving (or "integrating") this system gives back our family of hyperbolic curves. These are the integral curves of the vector field (first order linear partial differential operator)
 * $$ \vec{U} = \sqrt{1+k^2 \, T^2} \, \partial_T + k \, T \, \partial_X$$

At each event, this vector field is timelike and has unit length (using the line element given above). Therefore, this is the desired unit timelike vector field of tangent vectors to the world lines of our observers. We will call them Bell observers.

We can now obtain a frame field assigning an infinitesimal Lorentz frame to each event by augmenting this unit timelike vector field with three unit spacelike vector fields, such that all four vector fields are mutually orthogonal at each event. This is easily accomplished; we may take our frame field to be
 * $$ \vec{e}_0 = \begin{cases} \partial_T & T \leq 0 \\ \sqrt{1+k^2 \, T^2} \, \partial_T + k \, T \, \partial_X & T > 0 \end{cases} $$
 * $$ \vec{e}_1 = \begin{cases} \partial_X & T \leq 0 \\ k \, T \, \partial_T + \sqrt{1 + k^2 \, T^2} \, \partial_X & T > 0 \end{cases} $$
 * $$\vec{e}_2 = \partial_Y, \; \; \vec{e}_3 = \partial_Z $$

(For simplicity, in this section we have let $$\sigma \rightarrow \infty$$, but this is inessential to the discussion.)

Here $$\vec{e}_0$$ is a timelike unit vector field, the tangent vector field to a congruence of timelike curves, while $$\vec{e}_1, \; \vec{e}_2, \; \vec{e}_3$$ are spacelike unit vector fields, with all four being mutually orthogonal at each event. The congruence of timelike curves consists of parallel hyperbolas which represent the world lines of the Bell observers. In other words, this frame field assigns a infinitesimal Lorentz frame to the tangent space at each event, namely the frame representing the instantaneous state of motion of the Bell observer passing through that event.

We can now apply a standard method from the differential geometry of curves in Lorentzian manifolds, the so-called kinematic decomposition of a timelike congruence. A short computation shows that
 * $$ \nabla_{\vec{e}_0} \vec{e}_0 = \begin{cases} 0 & T \leq 0 \\ k \, \vec{e}_1 & T >0 \end{cases} $$

confirming that each Bell observer feels a constant force after time zero, as measured in his own frame, and with all the Bell observers experiencing a force of the same magnitude. The components of the expansion tensor, evaluated with respect to this frame, are
 * $$ \theta_{\hat{a}\hat{b}} = \begin{cases} 0 & T \leq 0 \\ \frac{k^2 T}{\sqrt{1+k^2 \, T^2}} \; \operatorname{diag} (1,0,0) & T>0 \end{cases} $$

The vorticity tensor vanishes, so the congruence of world lines of the Bell observers is hypersurface orthogonal. The orthogonal hyperslices are in fact locally isometric to ordinary euclidean space E3 even after time zero. These hyperslices have the form (two dimensions suppressed)
 * $$ X = c + \frac{\sqrt{1+ k^2\, T^2} - \operatorname{arctanh} \left( \frac{1}{\sqrt{1+k^2 \, T^2}} \right)}{k} $$

The nonzero component of the expansion tensor,
 * $$ \frac{k^2 T}{\sqrt{1+k^2 \, T^2}} $$

shows that--- contrary to the initial expectation of many students--- after time zero, the distance between nearby Bell observers is increasing. This component approaches k T as T decreases to zero from above, and approaches k as T grows without bound (see figure at right). That is, the expansion rate initially grows linearly and then approaches a constant value. Here T is the coordinate time of our Cartesian chart, but this can be easily re-expressed in terms of the proper time of one of our observers. In any case, using the dependence upon coordinate time rather than proper time does not affect the point of this computation: the expansion rate in the direction of acceleration approaches a positive constant. We have verified that the length of the string does grow continuously (until it breaks), and since the distance between each bit of matter in the string cannot grow without bound, we have confirmed that as Bell claimed the string must eventually break.

Rindler observers
Many readers will still be made uneasy by this claim. After all, in Newtonian mechanics, two bodies experiencing the same constant acceleration (same direction and magnitude) maintain a constant distance. Does our result imply that in relativistic physics, any accelerated object must eventually be torn apart? That would hardly make sense!

To study this issue, it is helpful to introduce another set of observers, called Rindler observers. Each of these observers also experience constant acceleration, but the value of this constant varies between the observers in just the right way to ensure that the Rindler observers do maintain constant distance to their nearest neighbors. (Actually, we get more than this: even distant Rindler observers maintain constant distance, according to any of several distinct but operationally significant notions of distance.)

The physical experience of Rindler observers may be described in terms of the frame field
 * $$ \vec{f}_0 = \frac{1}{\sqrt{X^2-T^2}} \left( X \, \partial_T + T \, \partial_X \right) $$
 * $$ \vec{f}_1 = \frac{1}{\sqrt{X^2-T^2}} \left( T \, \partial_T + X \, \partial_X \right) $$
 * $$ \vec{f}_2 = \partial_Y, \; \; \vec{f}_3 = \partial_Z $$

Notice that this frame is only defined on the region $$X>0, \; -X < T < X$$, which is often called the Rindler wedge.

Computing the kinematic decomposition of the new congruence, we find that the acceleration vector of a Rindler observer is given by
 * $$ \nabla_{\vec{f}_0} \vec{f}_0 = \frac{1}{\sqrt{X^2-T^2}} \vec{f}_1$$

Since each Rindler observer has a world line of form $$X^2-T^2 = 1/a^2$$ for some $$a>0$$, each Rindler observer in fact has a constant acceleration (but different observers in the congruence might have different constant values associated with the magnitude of their acceleration). We stress that this is the acceleration as measured by the Rindler observers themselves, not static observers. Moreover, the expansion and vorticity tensors both vanish. Therefore the congruence of the world lines of Rindler observers is the closest we can come in relativistic physics to a rigidly accelerating congruence.

To better understand these new observers, it is helpful to change coordinates to a new chart, the Rindler chart, in which the world lines of the Rindler observers are represented as vertical lines; see Rindler coordinates. Note that frame fields, like their constituent vector fields, are geometric objects which can be represented in any coordinate chart covering at least part of their domain of definition. The methods of differential geometry yield information (such as the kinematic decomposition) which is largely independent of coordinate chart, but choosing a well adapted chart may still be useful. For example, in the Rindler chart, the acceleration of each observer is manifestly constant.

However, it is easier to see in our Cartesian chart that the world lines of the Rindler observers (with two dimensions suppressed) form a certain family of nested hyperbolas which are exactly analogous to a family of concentric circles in the euclidean plane E2. Indeed, the fact that the world line of observer who experiences a constant acceleration is a hyperbola is simply the Minkowksi analog of the familiar fact that in euclidean geometry, a curve having constant path curvature along the curve is a circle. Furthermore, when we say that trailing Rindler observers have to acclerate harder than leading Rindler observers, we are saying that the world lines of trailing observers bend faster, per unit of arc length on their world line (that is, per unit of proper time) than do the world lines of leading observers, and this is analogous to the fact that in a family of concentric circles in the euclidean plane, inner circles must bend faster, per unit of arc length, than outer circles.

Once we see this, comparing the Rindler and Bell observers, it is easy to see why the expansion tensor of the congruence of world lines of Bell observers must be nonzero. (The euclidean analog of the Bell congruence consists of semicircles which all have the same radius and are all orthogonal at their midpoints to some line.)

If a rod is stretched between two such Rindler observers, "towed" by the leading observer, it will of course be under tension, and moreover after relaxation the rod should exhibit a nonzero tension gradient along its length. However, a detailed analysis of this phenomenon would require a material model. In the analysis above, note that if we replace "string: with "rod", we have tacitly assumed that a body force is applied to each element of the rod beginning at T=0. In particular, we have ignored that fact that if we intended rather to apply a force at just one point of the rod, the rest of the rod will take some time to respond.  (The paper by Nikolić cited below offers some discussion of this issue.)

This raises a new issue: the simplest material model is suggested by Hooke's law. But is this compatible with relativistic kinematics? The answer would seem to be that strictly speaking it is not, since a spring modeled by Hooke's law responds instantaneously all along its length to a tug on one endpoint.