# Can acceleration be measured with a watch?

39 views

### Luigi Fortunati

May 23, 2022, 4:46:23 AMMay 23
to
For 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).

### Richard Livingston

May 24, 2022, 12:05:07 PMMay 24
to
According to GR, yes.

Rich L.

### Luigi Fortunati

May 26, 2022, 12:57:01 PMMay 26
to
Richard Livingston martedě 24/05/2022 alle ore 18:05:03 ha scritto:
>> For 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]]

### Luigi Fortunati

May 26, 2022, 8:47:51 PMMay 26
to
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]]

### J. J. Lodder

May 27, 2022, 3:44:19 AMMay 27
to
Luigi Fortunati <fortuna...@gmail.com> wrote:
You may note that with today's clock accuracies
these 'tidal effects' will be observable in an elevator-sized box,
such as the ISS for example.
(if it would be practical to bring those clocks up there)

So No, if you measure accurately enough,
(but still not a measurement of acceleration)

Jan

### Luigi Fortunati

May 28, 2022, 9:50:17 PMMay 28
to
Luigi Fortunati gioved=EC 26/05/2022 alle ore 12:47:45 ha scritto:
>> Richard Livingston marted=EC 24/05/2022 alle ore 18:05:03 ha scritto:
>>>> For 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.

If we accept the criterion of "tolerance", then also the clock at the
top of the tower remains synchronized with the one at the base (within
the limits of the tolerance) but this is not the case because, after a
sufficiently long time, due to the tidal force (which there is also
here) the clock at the base of the tower is further behind than the one
at the top.

So the tower (like the lift stopped at the floor) is not an inertial
reference frame and, therefore, it is an accelerated reference frame.

And if the Tower were to plummet downward like the free-falling
elevator, would its clock at the top remain synchronized with the one
at the base even for a long enough time, or would one of the two
continue to lag behind the other?

[[Mod. note --
Your questions are all answered by the Taylor series explained in a
previous moderator's note.
-- jt]]

### J. J. Lodder

Jun 8, 2022, 4:50:44 PMJun 8
to
Luigi Fortunati <fortuna...@gmail.com> wrote:

> For 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).

You misunderstand the situation.
The rate difference between the clocks, aka the gravitational red shift,
is given by their Newtonian potential difference \Delta\Phi.
(the Newtonian approximation to GTR applies, because \Delta\Phi << c^2)
Not quite correct first derivation by Einstein, 1907,
modern ones in for example MT&W and other GTR textbooks.

It is given by: \nu_1 / \nu_2 = (1 + (\Phi_1 - \Phi_2) / c^2)

Remember that the Newtonian potential is negative:
a clock that stands higher is at a higher (less negative) potential,
and it runs faster.

The acceleration that the clock would get, if released,
is given by the Newtonian force,
which is given by the gradient of the potential.

The new experimental fact that clocks with a stability of order 10^-18
have become available opens up a new window of experimental possibility.
(the rate difference is order 10^-16 per meter altitude)
This can be used to establish 'relativistic geodsy',
aka 'chronometric levelling', with centimeter accuracy.
Work is in progress to establish a European levelling network,
comparing optical clock rates over long distances (~1000 km achieved)
over optical fibre connections.

To see why this is important: the acceleration, the local g,
determined by the gradient, is a local quantity.
The potential otoh is global.
It must be obtained by solving Poisson's equation,
taking all masses on the whole Earth into account.

The distinction is of importance when you want for example
to predict the sea level rise from melting of Greenland ice.
(which changes the mass distribution, and hence the shape of the geoid)
The sea level rise will not be distributed equally over the whole Earth.

Jan