Post by Luigi FortunatiPost by Richard LivingstonPost by Luigi FortunatiFor example, if I put two clocks at the maximum distance on the floor,
they remain synchronized and, therefore, horizontally there is no
acceleration (there is quiet).
If, on the other hand, I put a clock on the floor and one on the
ceiling, they don't stay synchronized and, therefore, there is
acceleration vertically (there is no quiet).
According to GR, yes.
Rich L.
In the free-falling elevator, does the clock on the floor stay
synchronized with the clock on its ceiling?
Luigi
[[Mod. note -- Yes, apart from tidal effects. -- jt]]
This is exactly the point!
Do the clocks on the floor of the free-falling elevator stay
synchronized with those on the ceiling in the presence of the tides or
not?
[[Mod. note -- The short non-mathematical answer is that the clocks in
(at rest with respect to) the free-falling elevator stay synchronized
to within a tidal-field tolerance.
A more precise answer is both longer and involves a bit of mathematics:
Consider a pair of clocks A and B. A is stationary with respect to the
elevator (whether the elevator is free-falling or not), say sitting on
the elevator floor. B starts out right next to A, synchronized with A.
Now we quickly raise B a vertical distance y, wait some fixed time interval
(much longer than the raising time), then quickly lower B back to be next
to A again. Then we measure the difference D between A and B's clock
readings, i.e., we measure the desynchronization between A and B.
[Notice that A and B are stationary next to each other
when we make this measurement, so the measurement is
unambiguous and doesn't depend on the exact time when
we make the measurement.]
Now we (gedanken) repeat this experiment for many different values of y,
and investigate how D varies with y. In particular, let's write D as a
Taylor series in y:
D(y) = D0 + D1*y + (1/2!)*D2*y^2 + (1/3!)*D3*y^3 + ...
If y=0, then we never moved clock B, so D(y=0) must be 0. Hence the
power-series coefficient D0 must be 0.
The Taylor-series coefficient D1 measures that part of the clock
de-synchronization which is linear in (i.e., proportional to) the height y.
According to general relativity, D1 is proportional to the Newtonian
"little g" in the elevator, so D1=0 if the elevator is in free-fall.
According to general relativity, the Taylor-series coefficients D2, D3, ...
are determined by the tidal fields. Unless there's something rather unusual
about the tidal fields, D2, D3, ... are all non-zero whether the elevator
is in free-fall or not.
So the more precise answer to your question is: according to general
relativity, if (and only if) the elevator is in free-fall then D1=0,
i.e., the clocks stay synchronized to 1st (linear) order in their
height difference y. Unless there's something very unusual about the
local tidal fields, the clocks are desynchronized at 2nd and higher
order in y.
(The immediately previous paragraph is precisely what we mean by the
phrase "the clocks stay synchronized apart from tidal effects". That
is, "tidal effects" means the D2, D3, ... terms in the Taylor series,
so saying that the clocks stay synchronized apart from tidal effects
means precisely that they stay synchronized in the D1 (linear) term.
-- jt]]