Essence of the paradox
Imagine a disk of radius R rotating with constant angular velocity ω. Let us fix the reference frame to the stationary center of the disk. Then the magnitude of the relative velocity of any point in the disk circumference is ωR. So the circumference will undergo Lorenz contraction by a factor of (1- (ω R)^2/C^2)^0,5. However the radius, being perpendicular to the direction of motion, will not undergo any contraction. So we have circumference/diameter = (2πR (1- (ω R)^2/C^2)^0,5) / 2R = π (1- (ω R)^2/C^2)^0,5. This is paradoxical, since Euclidean geometry tells us it should be exactly equal to π. Ehrenfest considered an ideally rigid cylinder that is made to rotate.
Assuming that the cylinder does not expand or contract, its radius stays the same. But measuring rods laid out along the circumference 2πR should be Lorentz-contracted to a smaller value than at rest, by the usual factor γ. This leads to the paradox that the rigid measuring rods would have to separate from one another due to Lorentz contraction; the discrepancy noted by Ehrenfest seems to suggest that a rotated rigid disk should shatter.Paradox solution
Even to this day, there are conflicting explanations for this 'paradox'. The simplest way to look at it is from a perspective of simultaneity. There is no way to define simultaneity for the spinning disk as a whole. In simpler words, if we synchronize a clock sitting at the center of the disk with a clock at the perimeter of the stationary disk and then spin the disk, the two clocks will go out of synchronization, just like the clocks and calendars of the twins in the twin paradox did and when two observers cannot agree on the time, they will not agree on the measured lengths of moving objects.
Interpretation of paradox solution in MT.
The matter exposed further corroborates considerations of MT about effective physical consistence of space time distortions, both in SR and in GR. However these are spatial contractions and temporal dilatations measured from different (inertial or not inertial) reference systems. In the local reference system is not appreciated any de synchronization and then there isn't any lenght contraction.
Sequentially doesn't exist any curvature gradient in the disk frame and therefore it doesn't shatter.
This involves that assignement of the freefall motion and, sequentially, of the bodies weight to the space time curvature remains an arbitrary postulate.
Stefano Gusman
Mario ludovico wrote :
RispondiEliminaStrange presentation of a paradox that doesn't appear to be so.
(1)If the reference frame is the laboratory, then the radius of the rotating disk is continuously changing its motion direction. Furthermore, in no way the disk can be considered as an inertial system in uniform motion with respect to the laboratory frame; which makes the thought experiment inappropriate from the special relativity view point.
(2) Otherwise,if one thinks of an immensely large rigid rotating disk, fixing the reference frame origin in the disk's center, then in a roughly approximate way the rotation of the disk might be considered, for an infinitesimal instant, as that of a system of bodies in linear uniform motion with respect to the frame. In such an artificial theoretical case, however (while forgetting the internal tension affecting the disk's material), there is no need to account for questions of simultaneity. The uniform rotation is observed from the origin of the reference system, so that relativistic contractions may be allowed for - with respect to that frame - concerning each of the infinite contiguous circumferences (linear bodies)of which the disk consists. To mean - the rotation period T being constant - that each circumference length may be thought of as undergoing the respective relativistic contraction. This depends on the relative speed, which in turn depends on the distance from the reference frame origin. If p=2(PI)r is any of the infinite number of circumferences of the disk, then its particular relativistic contraction is approximately expressed by
dp/dr = 2(PI)/(1 - v^2/c^2)^(-3/2),
where v = 2(PI)r/T.
This implies that the contraction of each circumference must be associated with the contraction of the respective radius. To conclude that, in the particular case considered, the whole rotating disk contracts (if neglecting the tension due to centrifugal forces, obviously).
Marius answered :
(1)I agree. The right example would be an observer solidal with a reference system rotating with constant angular speed ω around the disk fixed to a standing reference system. The observer could say , for the equivalence principle, to stand into a gravitational field as the one of the Earth. He would see the circumference of the disk rotating in the opposite way. However the things wouldn’t change. Infact in infinitesimal space – time regions of a not inertial system stand valid RS rules, since during an infinitesimal interval of time, radium is perpendicular to the observer solidal with laboratory, therefore only circumference lenght is contracted and due radium unchanging the “film moving” appears to the observer as a distorted frame (note that “appears” not “is”).
(2) Not exactly. Also in this case, equivalence principle allows us to consider every point of the disk standing in a not inertial system. For this reason we must apply the RG rules, therefore the clock standing in the centre is faster then every other clock on the disk frame. Otherwise every clock on the disk slowsdown respect to the one in the centre, and sequentially the observer in the centre sees the lenghts undergoing Lorenz's contractions (time dilatations always are tied with lenghts contractions).
Marius,
RispondiEliminaho letto un tuo commento qui, a proposito di Entanglement. E vorrei passassi su questo post dove ho scritto un articolo In memoria di Marcello.