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Let us consider the following timelike congruences, which maps congruence parameters t,r,theta and z to points in a minkowskii space (t', x', y', z'). g is a "free parameter" (which could even be set to 1 to simplify the problem)

(t,r,theta, z) -> t' = (1/g+z) sinh(g t) , x'=r cos (w t + theta), y' = r sin(w t + theta),

z' = (1/g + z) cosh(g t)

The point (r=0, z=0), which one might describe as "the center of the disk" is undergoing hyperbolic motion in the z direction, i.e. it's accelerating at a constant rate, as can be seen by comparing it to the equations in MTW, pg 173, for hyperbolic motion. The parameter "g" represents the rate of said acceleration, and can be set to 1 if desired.

The points at r>0, theta = constant are "rotating around the center of the disk" with an angular velocity w.

The point of writing out the congruence is to make the notion of a "disk" mathematically exact and allow us to calculate things about it. I can't think of any clearer way to define a disk precisely than to write down the congruence of worldlines which make up the disk.

Now consider the case z=0 (a "thin" disk). Then we have:

(t, r, theta) ->t' = (1/g) sinh(gt), x' = r cos(w t + theta), y' = r sin(wt + theta) z' = (1/g)cosh(gt)

The question is - is this 3-parameter set of congruences Born rigid, in the appropriate reduced dimensional manner? (It's only got 1 time and 2 space dimensions).

Looking at Wald, the first step to confirm this would be to express the congruences as a function of proper time. However, the congruence parameter t is proper time for the particle at "the center of the disk", but not elsewhere.

My feeling is that this congruence "should be" Born rigid, because if we look at an instantaneously co-moving inertial frame around the point (r=0, z=0) it describes a rotating plane.

The 4-velocity of every point in the plane is the same as the 4-velocity of a rotating disk.

Further, the rotation rate (measured by clocks in the co-moving inertial frame above) doesn't change with time.

I'm hoping someone will have some insight as to how to compute the expansion tensor to confirm this idea that 2-space 1-time congruence is Born rigid WITHOUT re-parameterizing everything in terms of proper time.

The inuitive point of this (assuming the math works out) is that the thin disks for z=constant are each Born rigid, but rotate with respect to each other, so that the 3-space + 1 time congruence is not Born rigid.