# General Relativity and acceleration

Suppose I am riding an amusement park ride. This ride is a circular Plexiglass structure that rotates at high velocity. If I were to measure the circumference of the structure with a ruler while the ride was spinning, I would find the circumference to be very slightly longer when it was moving than when still. This is due to the he ruler being shortened by special relativity, which only deals with uniform motion. Shortening the length of the ruler causes me put the ruler end over end more, thereby making the circumference of the ride longer. This idea forms the basis for General Relativity.

My question is, if the ruler is accelerating along with me and everything else on the ride, how can it be shortened?

SUMMARY

This is the essence of Ehrenfest's "Paradox". Your ruler is of course not shortened relative to the measurements it makes in a locally co-moving inertial frame, the circumference of a circle centered on the center of rotation for the merry-go-round rider really is longer, from that observer's standpoint, than its length when the merry-go-round is not spinning. The resolution of the seemingly paradoxical situation is this: spatial geometry for an observer riding on a uniformly rotating "merry go round" is non-Euclidean: a circle's circumference is different from $2\,\pi$ times its radius and indeed the circumference to diameter ratio depends on the diameter and the position of the circle's center.

Special relativity can handle a remarkably wide range of nonuniform motions, especially through (1) the momentary comoving inertial frame and (2) Riemann Normal Co-ordinates notions; the assertion otherwise is a common misconception.

THE LEWD DETAILS

The basic problem you speak of is called Ehrenfest's "Paradox", whose analysis has a long and interesting history. Your statement "This idea forms the basis for General Relativity" is true insofar that there is some historical evidence that the pondering of Ehrenfest's paradox helped Einstein to complete his theory, but one cannot push this statement too far.

It is a myth that special relativity deals only with uniform motion. Special relativity is fine as long as the problem is considered in small enough chunks of spacetime that there are "no significant tidal effects"). More precisely, you define Riemann Normal Co-ordinates around a point $A$ and then, experimentally, if the difference between the normal co-ordinates of any nearby point $B$ and those that $A$ surveys with a straightedge and clock are negligible, then SR is good to go in a spacetime neighborhood comprising only points $B$ that this statement is true for.

So this last paragraph is true for an nonrotating observer $A$ looking at the rotating disk. Special relativistic calculations are OK from this observer's standpoint, and we can work out everything about this problem from this standpoint. However, from the standpoint of an observer $B$ on the disk, his/her momentary comoving inertial frame is one moving on a geodesic (Euclidean straight line, relative to the nonrotating observer $A$) tangential to circular paths that $A$ sees $B$ making. $B$'s Riemann normal coordinates only agree then with straightedge surveys done by $B$ in a region on the disk of small subtense angle about the center of rotation and for time intervals such $B$'s position can be thought to be roughly constant relative to $A$.

So let's see what $B$ measures doing all calculations from $A$'s frame: SR is altogether fine here. $B$ carries a ruler with him/her: both the ruler and lengths measured by $B$ must be contracted together in perfect sympathy. There is no increased "end over end" effect. We now look at this measurement from a third frame $C$, also inertial, that is momentarily comoving with $B$ with velocity $v$ tangential to $B$'s circular path. Let's assume that $B$ has a ruler of length $\delta z$, also the length that comoving $C$ measures, as long as $\delta z$ is small enough. If the ruler is laid tangential to $B$'s path relative to $A$, then from $A$'s standpoint this ruler has contracted by a factor $\gamma(R)^{-1}=\sqrt{1-\frac{\omega^2\,R^2}{c^2}}$ where $\omega$ is the angular speed of the disk, as measured by $A$ and $R$ the radial co-ordinate (from $A$'s standpoint) of $B$. This means that the angle subtended (from $A$'s standpoint) by $B$'s ruler is

$$\delta\theta=\frac{\delta z}{R}\,\sqrt{1-\frac{\omega^2\,R^2}{c^2}}\tag{1}$$

So $B$ walks this distance in the tangential direction, and repeats this action until he has walked right around the circular path back to his beginning point. He will feel Coriolis forces and also torques that counter his Wigner rotation/Thomas Precession as he does so, but this is beside our point here: we assume he can impart these forces/ torques to follow the circular path and keep his frame aligned to the tangent. The point is that as he does so, if his walking speed is much less than $v=\omega\,R$, then the distance around the circle from his standpoint is $\delta z$ times as many $\delta z$s as needed to make up a total angle of $2\pi$ (from $A$'s standpoint). But if $B$ walks around to the same point by $A$'s standpoint, $B$ must conclude that he himself has walked back to his beginning point. Now, $B$ can know that his path has been circular; if he has the right surveying instruments, he can be sure of his constant distance from the center of his merry-go-round World. If he begins at this center, and surveys a radial path, this path is orthogonal at all times to a momentary co-moving inertial frame, so this $R$ is the same as that from $A$'s standpoint.

So $B$ will conclude he lives in a non Euclidean world! Not all circles have circumference $2\,\pi\,R$. Indeed,, from (1), the angle subtended by a noninfinitessimal arc walked by $B$ is, from (1):

$$\theta =\frac{1}{R}\, \sqrt{1-\frac{\omega^2\,R^2}{c^2}}\,\int_0^{s_B(\theta)} \mathrm{d} z=\sqrt{1-\frac{\omega^2\,R^2}{c^2}}\,s_B(\theta)\tag{2}$$

where $s_B$ is the total pathlength measured by $B$. Since $\theta$ is measured from $A$'s frame, $B$ makes a complete circuit when $\theta=2\,\pi$ so therefore $B$ will measure the circumference $C$ to be $s_B(2\,\pi)$, which, from (2) is:

$$C = s_B(2\,\pi)=\frac{2\,\pi\,R}{\sqrt{1-\frac{\omega^2\,R^2}{c^2}}}\tag{3}$$

No other conclusion is possible, because, as you say in your last line, $B$'s ruler is contracted, along with all $B$'s measurements, by the factor $\gamma(R)^{-1}$.

There is no "paradox" here: it simply means that $B$'s spatial geometry will be non-Euclidean. The notion of $\pi$ would not be a simple notion in a merry-go-round world: circle circumference to diameter ratio depends on the circle's radius and also its center position.

Furthermore it is to be emphasized that accelerated motions can be perfectly analysed with SR within a small enough region of spacetime using the concept of momentary comoving inertial frame.

Another seemingly paradoxical point arises when we consider the disk before it spins. We divide a circumference like the above up into $N$ equal fixed lengths, each subtending an angle $2\pi/N$. Now we spin the disk at uniform speed. There are still $N$ equal arcs around the circumference, how can they amount to a length of more than $2\,\pi\,R$ as calculated above, if the merry go roundstays rigid?

The answer is that the merry go round cannot stay rigid between its still state and its uniformly rotating state. Five points:

1. A "Rigid Body" that transforms by Euclidean isometries as in Newtonian / Eulerian dynamics is altogether incompatible with either special or general relativity. To stay rigid, a disturbance at one end pushing the body must propagate infinitely fast to make the body move as a rigid whole;

2. In special relativity, a notion of Born Rigidity must therefore replace the Newtonian/Eulerian notion of rigid;

3. Not even Born rigidity can be upheld in transient periods whilst the merry go round is accelerating;

4. Under uniform rotation, the merry go round is again Born-rigid. But now it is in a stressed / strained state so that the equal arcs above still subtend an angle of $2\pi/N$ from $A$'s standpoint, but their lengths are stretched to $2\,\pi\,R\,{\sqrt{1-\frac{\omega^2\,R^2}{c^2}}}/N$. We cannot think of infinite elastic constants even in principle, because the stretch must happen to match the geometry. If it does not, the disk will shatter (or, if the stress becomes humungous, it will alter the stress energy tensor so that gravitational effects will begin to creep in, but that is another problem);

5. So suppose we (1) paint a short, unit length on a circumference in $A$'s frame before accelerating the merry go round,(2) cut off a unit length rod of the same length in the same frame, (3) accelerate the merry go round to its steady state and (4) accelerate the rod to the frame inertially comoving with with merry go round arc. At the instant the rod and moving arc are compared the arc will have stretched owing to the merry go round's stress/strain state. Only the rod, in an inertial frame, can give the correct, Lorentz transformed length. The length of the arc is that of a material in a different stress/ strain state from when it was at rest.

. Riemann normal co-ordinates annul the connexion co-efficients at one point (in general, only one point, unless there is no curvature). There can be curvature of the spacetime manifold but our neighborhood must be small enough that the deviation from Minkowski space is small over the neighborhood: one must limit the neighborhood size in inverse proportion to the magnitude of the curvature's components.

I think it's a fun question. The circular movement does put a spin on a classic problem.

In a nutshell, the ruler, and anything traveling with you would appear normal. We're ignoring that in your giant plexiglass carousel, the centrifugal force would turn you into silly putty, and the entire carousel would stretch and fly apart itself long before it reached any measurable relativistic speeds. Also, there would be a slight time dilation difference between your left side and your right side, assuming you were facing in the direction of movement of the ride, so as you turned the ruler in a rotating circle at relativistic speed, your left side in a slightly slower speed than your right, you might notice some very slight effect in it's shape, not unlike the fact that if you stand above a black hole, time travels slower for your feet than it does for your head, but lets assume for simplicity that all of you and what you carry is traveling at the same speed - then the ruler you carry will never appear different to you, but to a person watching you from outside the ride, it would appear squashed to them.

So, the center of the carousel's distance to you won't change because it's perpendicular.

Lets say you're traveling at .86C with a time dilation of 50%, so you live 1 second for every 2 seconds the people waiting for the ride live.

Another curious circumstance, they could advertise it as a 3 minute ride, but by your watch, it would only last 1.5 minutes. You would feel like you'd been had.

And lets say the carousel is very large with a diameter of 160,000 miles (radius 25,000 miles - I'm rounding.)

The person watching you would see you make a complete circle every second but you would see yourself making 2 complete rotations every second, because your clock runs at half the speed as his clock.

If you measure the distance you're passing as you go around, you would measure that each rotation you flew past 80,000 miles and what you passed would appear squashed to half it's distance as you passed by it, but if you measure the radius, you'd get 25,000 miles - so, as WetSavannahAnimal pointed out, 2πR no longer applies.

But if you stood up and paced around the carousel, and measured it by walking around it, you would measure the full 160,000 miles.

At least, I think that's right, but I invite correction if it's not.