Is angular velocity (the horizontal rotation of a massive disc falling through the air) sensitive to time dilation or invariant to it? Will its angular velocity (speed of rotation) increase, decrease or stay invariant when reaching lower altitudes?
4
-
$\begingroup$ anything that is rotating or revolving will show clocks attached to its surface to run at different speeds, as observed by someone looking at it from a non rotating lab. But the angular momentum in the lab's reference frame will be conserved if no torque acts on the disk in such frame. $\endgroup$– Pato GalmariniCommented Apr 17 at 4:12
-
$\begingroup$ So if i understand correctly, a pendulum will increase its ticking when lowered to the earth's surface from an altitude, a regular electromagnetic clock will decrease its ticking when lowered, however a rotating disc will remain a consistent clock? $\endgroup$– ApsteronaldoCommented Apr 17 at 7:52
-
$\begingroup$ I think what Pato said is the answer you are looking for however I think one should add that this really depends on the exact assumptions you make as e.g. in general spacetimes the notion of angular momentum of an object (or even its energy) may not be well defined. $\endgroup$– Thomas TappeinerCommented Apr 17 at 9:45
-
1$\begingroup$ Added a bit to my answer about the pendulum velocity. $\endgroup$– KDPCommented Apr 24 at 15:29
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
8
The angular velocity of the falling disk will slow down as measured by a stationary (non inertial) observer standing on the surface of the Earth, for two reasons:
As it falls, its vertical velocity increases and it will experience time dilation proportional to the velocity time dilation factor $\sqrt{1-v^2/c^2}$.
As it falls it moves from a higher gravitational potential to a lower gravitational potential and is subject to gravitational time dilation proportional to $\sqrt{1-\frac{2GM}{Rc^2}}$ where M is the mass of the Earth or whatever the main gravitational body is.
Total slow down of the angular velocity measured by a ground observer is proportional to: $$\sqrt{\left(1-\frac{v^2}{c^2}\right) \left(1-\frac{2GM}{Rc^2}\right)}$$
The clock of a free falling observer comoving with the disk, will also slow down by the same factor, so a co-falling inertial observer will see no change in the angular velocity of the disk (but the ground based non inertial observer will). Basically, the disk is a form of clock and is subject to velocity and gravitational time dilation like any other ideal clock. (A pendulum is not an ideal clock as it requires gravity to operate).
As a side note, what will happen to the pendulum as it falls? If the pendulum is at the top of its swing when it is released, it will appear to stop instantly the moment it is in free fall, according to a co-falling inertial observer. If it is not at the top of its swing, it continues in circles around its pivot with the tangential velocity it had at the instant it was released, rather than oscillate to and fro.
EDIT I have answered the main question. My answer to the subsidiary question of a pendulum, has been migrated to this related question.
-
1$\begingroup$ Amazing answer, thanks a lot. Question: how can the frequency of the pendulum approach infinity, if its arc swing velocity is limited by light speed c? $\endgroup$ Commented Apr 22 at 20:33
-
1$\begingroup$ You have a good point and I am working on fixing that apparent anomaly. One thing to consider is the extremely close to the event horizon is that the tidal effect becomes extreme and the force at the bottom of the swing arm is much greater than at the top of the swing arm so the assumption of a constant force working on the weight of the pendulum is false unless we have a swing arm with length approaching zero and infinite frequency with a swing arm of zero length is possible but probably undefined. Also consider that from the point of view of an observer at infinity ... $\endgroup$– KDPCommented Apr 22 at 21:16
-
$\begingroup$ Right, I guess I see your point. Somehow I would like to know from the viewpoint of the strong equivalence principle, the view from inside the local reference frame in which the light speed and laws of nature are identical to every other local reference frame. So following the laws of nature inside that particular reference frame, as if I were falling inside the black hole with the pendulum bob, and thus the truth of the local limit of light speed c. Independent of external observers. That would be my first move to get an answer to this question, I think. Wonder what you can do! Happy to hear $\endgroup$ Commented Apr 22 at 21:23
-
1$\begingroup$ Bear in mind that the equivalence principle applies to small enough area in a gravitational field that it can be considered flat space, but as the curvature of space becomes extreme close to the event horizon, then the region the equivalence principle is valid in approaches infinitesimal ,so it is difficult to obtain sensible results for a pendulum with an arm of length that is greater than infinitesimal in that extreme region. All the reasonable approximations that are valid in a weak field come back to bite us $\endgroup$– KDPCommented Apr 22 at 21:29
-
$\begingroup$ "So following the laws of nature inside that particular reference frame, as if I were falling inside the black hole with the pendulum bob, and thus the truth of the local limit of light speed c." As I mentioned in my answer, if the pendulum is free falling, a co-falling observer will see it performing complete circles with constant angular velocity whatever the height, except for at the central singularity, The pendulum will only oscillate if the pivot is at a fixed height and it is not possible for anything to maintain a constant radial coordinate at or below the event horizon. $\endgroup$– KDPCommented Apr 24 at 16:10