# Gravity and free fall

54 views

### Luigi Fortunati

Mar 4, 2022, 3:17:02 AMMar 4
to
Why does the force of gravity disappear in the free-falling elevator and
on Earth (which is also in free-fall) does it not disappear?

### Stefan Ram

Mar 4, 2022, 5:06:04 AMMar 4
to
Luigi Fortunati <fortuna...@gmail.com> writes:
>Why does the force of gravity disappear in the free-falling elevator and
>on Earth (which is also in free-fall) does it not disappear?

What people perceive to be "the force of gravity" on the
surface of the Earth actually is a /combination/ of /two/
forces: The actual downward force of gravity pulling
everything towards the center of the Earth and an effective
force of the Earth's surface that is directed upwards and is
preventing things to actually fall further towards the
center of the Earth.

This combination of forces can be characterized by its
action on two test bodies: one body resting on the Earth's
surface and one body one meter above the first body in the
air. The combined action of the force of gravity and the
force of the Earth's surface will reduce the distance of
those two bodies in the course of time. And it is this
squeezing together of the two bodies that we perceive as the
"force of gravity". Let me call this force the "squeezing
force". People are actually shorter in the evening because
they have been squeezed all day while they were sitting,
standing, or walking vertically and they grow a bit again
when allowed to rest horizontally for some hourse. This
squeezing force is called "force of gravity" in everyday
language.

If you put the same two test bodies into an elevator that is
falling freely, one on the floor of the elevator and the
other one into the air one meter above the first one,
their distance will /not/ decrease because the force of the
Earth's surface and thus the squeezing force is missing here.
Since the squeezing force is called "force of gravity" in
everyday language, a layman would say, "There is no force of
gravity in the free-falling elevator".

From the point of view of physics, however, gravity is
always there; it is the force of the Earth's surface (and
thus the squeezing force) that is sometimes there and
sometimes not. If there would be no force of gravity on
the falling elevator, it would not be falling!

(The above has been written in the spirit of Newton's view
of gravitation as a force. Within general relativity [GR],
gravitation is not regarded to be a force anymore, but
I have ignored GR above since it is not required to answer
the question at hand. For the same reason, I have not
written anything about how exactly the force of the Earth's

### Phillip Helbig (undress to reply)

Mar 4, 2022, 4:04:36 PMMar 4
to
In article <gravity-202...@ram.dialup.fu-berlin.de>,
r...@zedat.fu-berlin.de (Stefan Ram) writes:

> If you put the same two test bodies into an elevator that is
> falling freely, one on the floor of the elevator and the
> other one into the air one meter above the first one,
> their distance will /not/ decrease because the force of the
> Earth's surface and thus the squeezing force is missing here.

Minor nitpick (the summary is otherwise excellent): in practice, the
distance between the two test bodies in an elevator will INCREASE
because gravity is slightly stronger lower down. That is an example of
a tidal force. Similarly, two bodies side-by-side in an elevator will
approach each other. At rest or in uniform motion in no gravitational
field, neither would happen. In other words, this form of the
equivalence principle is valid only in the limit of an arbitrarily small
elevator.

[[Mod. note -- To add to what Stefan and Phillip wrote:

One might ask how tidal forces are distinguished from "non-tidal" ones
like those that define the Newtonian "little g". The answer is to write
the coordinate positions of test bodies as power series in time and space;
tidal forces enter at higher order than "little g" forces.

That is, in a Newtonian reference frame (which need not be inertial),
we can write the position of a test body (in 1 dimension for simplicity)
as a power series in time (taking $t=0$ to be a nearby "reference time")
$x(t) = x_0 + x_1 t + (1/2!) x_2 t^2 + (1/3!) x_3 t^3 + ...$
Then $x_0$ and $x_1$ tell us the position & velocity of our test particle
(with respect to our coordinate system), i.e., they are specific to that
test particle. $x_2$, in contrast, tells us the Newtonian "little g"
(with respect to our coordinate system) near $t=0$,$x=0$. In Newtonian
mechanics $x_2$ is the same for all test particles near $t=0$,$x=0$; this
is called the "universality of free fall". It's because $x_2$ is the
*same* for all test particles (near $t=0$,$x=0$) that we can say that the
distance between two different test particles won't change (given suitable
initial positions/velocities).

$x_3$ and higher coefficients describe the tidal field which Phillip
referred to.

So, when we say "in the limit of an arbitrarily small elevator", what
we really mean is "ignoring $x_3$ and higher terms in the power series.
-- jt]]

### Stefan Ram

Mar 4, 2022, 4:53:23 PMMar 4
to
hel...@asclothestro.multivax.de (Phillip Helbig (undress to reply)) writes:
>In article <gravity-202...@ram.dialup.fu-berlin.de>,
>r...@zedat.fu-berlin.de (Stefan Ram) writes:
>>If you put the same two test bodies into an elevator that is
>>falling freely, one on the floor of the elevator and the
>>other one into the air one meter above the first one,
>>their distance will /not/ decrease because the force of the
>>Earth's surface and thus the squeezing force is missing here.
>Minor nitpick (the summary is otherwise excellent): in practice, the
>distance between the two test bodies in an elevator will INCREASE

Yes, but I wrote "not decrease", which includes "increase".

### Luigi Fortunati

Mar 5, 2022, 8:12:55 AMMar 5
to
Stefan Ram venerd=EC 04/03/2022 alle ore 10:12:00 ha scritto:
> From the point of view of physics, however, gravity is
> always there;

On this, of course, I absolutely agree: the force of gravity in the
free-falling elevator does not disappear.

But doesn't Einstein say quite the opposite?

(Su questo, ovviamente, sono assolutamente d'accordo: la forza di=20
gravit=E0 nell'ascensore in caduta libera non scompare. Ma Einstein non
dice proprio il contrario?)

### Stefan Ram

Mar 7, 2022, 3:54:46 PMMar 7
to
Luigi Fortunati <fortuna...@gmail.com> writes:
>Stefan Ram venerd=EC 04/03/2022 alle ore 10:12:00 ha scritto:
>>From the point of view of physics, however, gravity is
>>always there;
>On this, of course, I absolutely agree: the force of gravity in the
>free-falling elevator does not disappear.
>But doesn't Einstein say quite the opposite?

In Newton's worldview, there is an attracting force F, with

F = G m_1 m_2 / r^2 and
G = 6.67430(15) x 10^(-11) N m^2 kg^(-2),

along the (imagined) connecting line between two systems of
masses m_1 and m_2, respectively, and with distance r. This
does not change when one of the masses is placed within an
elevator, free falling or not. That force F does not depend
on the speed or acceleration of either of those two systems.

In the theory of general relativity (GR), there is never any
"force of gravity" at all. Bodies always move on what is the
generalization of a straight line in curved space-time, when
no force (a concept which does not include gravity now) is
acting on them. This theory is much more complicated and not
necessary here.

One should not mix concepts from different theories,
but needs to describe everything using the same theory.
I decided to use Newton's worldview. In Newton's worldview,
the force of gravity between two systems with masses never
disappears, it is always "G m_1 m_2 / r^2".

### Mike Fontenot

Mar 7, 2022, 4:09:29 PMMar 7
to
On 3/4/22 2:04 PM, Phillip Helbig (undress to reply) wrote:
> Minor nitpick (the summary is otherwise excellent): in practice, the
> distance between the two test bodies in an elevator will INCREASE
> because gravity is slightly stronger lower down. That is an example of
> a tidal force. Similarly, two bodies side-by-side in an elevator will
> approach each other. At rest or in uniform motion in no gravitational
> field, neither would happen. In other words, this form of the
> equivalence principle is valid only in the limit of an arbitrarily small
> elevator.

But with a UNIFORM and constant gravitational field (absolutely parallel
field lines, and no variation in field strength), the equivalence
principle says this is equivalent to two test bodies undergoing the same
constant acceleration, with no gravitational field present.

[[Mod. note -- Note that the gravitational field near the Earth (whether
in an elevator or not) isn't uniform. But over sufficiently small scales
we can approximate it as uniform, essentially neglecting higher-order
terms in the Taylor series I alluded to in a previous moderator's note.

Note also that in the context of general relativity, one needs to be
careful in invoking the equivalence principle for a *uniform* *constant*
gravitational field. The problem is that such a field is the result of
(i.e., implies the presence of) an infinte mass plane, which means that
spacetime is *not* asymptotically flat. That has a number of "interesting"
consequences...
-- jt]]

### Mike Fontenot

Mar 10, 2022, 4:01:11 AMMar 10
to
On 3/7/22 2:09 PM, Mike Fontenot wrote:

> But with a UNIFORM and constant gravitational field (absolutely parallel
> field lines, and no variation in field strength), the equivalence
> principle says this is equivalent to two test bodies undergoing the same
> constant acceleration, with no gravitational field present.
>
> [[Mod. note --
> Note also that in the context of general relativity, one needs to be
> careful in invoking the equivalence principle for a *uniform* *constant*
> gravitational field. The problem is that such a field is the result of
> (i.e., implies the presence of) an infinite mass plane,

True,

>which means that
> spacetime is *not* asymptotically flat. That has a number of "interesting"
> consequences...
> -- jt]]
>

Can you elaborate on that? Why does that imply that spacetime "is not
asymptotically flat"? And what exactly does "asymptotically flat" mean?

Mike Fontenot

### Phillip Helbig (undress to reply)

Mar 10, 2022, 4:21:30 AMMar 10
to
In article <4e3fc804-15b4-edeb...@comcast.net>, Mike
Fontenot <mlf...@comcast.net> writes:

> On 3/7/22 2:09 PM, Mike Fontenot wrote:
>
>> But with a UNIFORM and constant gravitational field (absolutely parallel
>> field lines, and no variation in field strength), the equivalence
>> principle says this is equivalent to two test bodies undergoing the same
>> constant acceleration, with no gravitational field present.
>>
>> [[Mod. note --
>> Note also that in the context of general relativity, one needs to be
>> careful in invoking the equivalence principle for a *uniform* *constant*
>> gravitational field. The problem is that such a field is the result of
>> (i.e., implies the presence of) an infinite mass plane,
>
> True,

Which also implies that it very probably doesn't exist; it certainly
doesn't exist in our Universe.

>> which means that
>> spacetime is *not* asymptotically flat. That has a number of "interesting"
>> consequences...
>> -- jt]]
>>
>
> Can you elaborate on that? Why does that imply that spacetime "is not
> asymptotically flat"? And what exactly does "asymptotically flat" mean?

Basically, it means that if you are far enough away from masses,
spacetime is Minkowski space, i.e. space itself is Euclidean and the
fourth dimension is time multiplied by the speed of light and there is
no curvature (neither of space nor of spacetime). Obviously, if there
is an infinite mass, one could never get "far enough" away from it.

### Mike Fontenot

Mar 12, 2022, 4:12:12 AMMar 12
to
On 3/10/22 2:21 AM, Phillip Helbig (undress to reply) wrote:

>>> which means that
>>> spacetime is *not* asymptotically flat. That has a number of "interesting"
>>> consequences...
>>> -- jt]]
>>>

What are the "interesting" consequences?

[[Mod. note -- "Asymptotically flat" is a property a spacetime can have
which means that (roughly speaking) there's a "place" (suitable region
of spacetime) where the graviational field is "weak", i.e., where the
spacetime is locally almost flat (here "flat" means Minkowski spacetime,
where special relativity holds). (We often call this place the "weak-field"
or "far-field" region of spacetime. Formalizing this mathematically is
a bit tricky, but it can be done in at least two ways: one relatively
easy way which assumes the existence of a coordinate system with certain
properties, and one (mathematically harder) which is coordinate-free.
There's a very nice and readable overview of asymptotic flatness in the
Wikipedia article
https://en.wikipedia.org/wiki/Asymptotically_flat

Examples of asymptotically flat spacetimes include Schwarzschild and
Kerr spacetimes (non-rotating and rotating black holes, respectively).
Examples of non-asymptotically-flat spacetimes include the Friedman-
Lemaitre-Robertson-Walker spacetime (which describes the "standard model
of cosmology"), and the spacetime containing (only) an infinite uniform
mass sheet.

Some of the "interesting" consequences of this are:

The definitions of "event horizon" and "black hole" explicitly refer
to asymptotic flatness (roughly speaking, a black hole is a region of
spacetime from which *no* future-pointing light-ray path to the weak-field
region exists; an event horizon is just the boundary of a black hole).
But in a non-asymptotically-flat spacetime there need not be any
"weak-field region". So, in a non-asymptotically-flat spacetime there's
no (or at least no standard) way to define the concepts of "event horizon"
and "black hole".

Defining the total energy (mass) or angular momentum of a spacetime is
always somewhat tricky: there are multiple (different) "reasonable"
definitions. Fortunately, the different definitions almost always agree
for "ordinary" spacetimes like Schwarzschild or Kerr black holes. But
these definitions usually require asymptotic flatness -- they're formulated
in terms of surface integrals over a "Gaussian surface" placed in the
weak-field region. So, in a spacetime which isn't asymptotically flat,
we can't easily define the total energy (mass) or angular momentum of
energy" being necessarily >= 0, and = 0 only for Minkowski spacetime.
It also means we can't make statements about total energy or total angular
momentum" being conserved. Ouch -- Doing physics without conservation
of energy is painful! :)

Seriously, though, the Usenet Physics FAQ has a nice entry "Is Energy
Conserved in General Relativity?" which discusses some of the issues
around how to define the energy of a "region of spacetime" (which might
be "the entire spacetime"), and whether this is "conserved":
http://www.edu-observatory.org/physics-faq/Relativity/GR/energy_gr.html
-- jt]]

### Stefan Ram

Mar 12, 2022, 4:12:33 AMMar 12
to
hel...@asclothestro.multivax.de (Phillip Helbig (undress to reply)) writes:
>Minor nitpick (the summary is otherwise excellent): in practice, the
>distance between the two test bodies in an elevator will INCREASE
>because gravity is slightly stronger lower down. That is an example of
>a tidal force.

By the way, recently there were news[1] about a quantum
gravity gradiometer that can detect tiny changes (like, by
one nano-g) in gravitation (ten times more accurate than
previous comparable devices). Such a device is also
described in [2], where one can read:

|Our gradiometer is based on two atom interferometers
|(gravimeters) apart in the vertical direction to measure the
|differential acceleration between two clouds of cold rubidium
|atoms in free fall.
from [2].

So, that would be an application and direct illustration
of that tidal force! In the case of [1], the two sensors
really seem to be 1 meter apart, just as in my thought
experiment.

|This places the measurement positions of the sensor at
|approximately 0.5 m from the road surface for the lower
|sensor, and 1.5 m for the upper sensor.
from [1]

Such devices allow, for example, to detect structures hidden
in the ground, such as tunnels, by their gravitational force.

1 Michael Holynski, Quantum sensing for gravity cartography,
Nature (2022). DOI: 10.1038/s41586-021-04315-3.

2 Wei Lyu et al, "Development of a compact high-resolution
absolute gravity gradiometer based on atom interferometers"

### Jonathan Thornburg [remove -color to reply]

Mar 16, 2022, 5:30:00 AMMar 16
to
In article <cb878d08-ead1-b8e1...@comcast.net>, I gave a
brief overview of the concept of asymptotic flatness in general
relativity (GR), and I mentioned some very useful concepts (event
horizons, black holes, total energy & angular momentum) which are only
defined if spacetime is asymptotically flat (i.e., which are generally
*not* defined in a non-asymptotically-flat spacetime).

Another very useful concept which is much harder to define without
asymptotic flatness is that of gravitational waves (GWs). The problem
here is that it's rather hard to answer the question "what is a GW?"
unless you have some sort of "background".

That is, our intuitive notion of a GW is that it's a "ripple of
spacetime curvature", i.e., some sort of (propagating) perturbation in
the spacetime curvature. But how do we tell which part of the spacetime
curvature is the "ripple" (perturbation) versus which is "just part of
our spacetime"? In an asymptotically flat spacetime, we can go out to
the weak-field region (recall that this is roughly the $r \to \infty$
limit),
[I'm glossing over the fact that there is more than one
$r \to \infty$ limit. E.g., there's the obvious limit
$r \to \infty$, $t \to \text{constant}$ where $t$ is a
"suitable" time coordinate. But in studying GWs it's often
more convenient to instead consider the simultaneous limits
$r \to \infty$, $t \to \infty$, and $t-r \to \text{constant}$;
this part of the far-field region is known as "null infinity".
Roughly speaking, this is "where an outgoing GW or light signal
ends up after an infinite time".]
and precisely because spacetime is almost flat (Minkowski) there (and
because of a bunch of other conditions about derivatives of the metric
being small there), there is a mathematically clean way to separate
"perturbation" from "background". Namely, in the weak-field region
there's a *unique* Minkowski spacetime singled out by the physics, up to
Lorenz boosts and rotations
[and some things called "supertranslations" which I
alas have never quite understood].
So, we can say that the difference between the actuasl metric and that
"unique" Minkowski (flat) metric, a.k.a. the "metric perturbation",
*is* the GW.

With this definition we can then make do interesting/useful calculations
(& prove theorems) about GWs, i.e., we can calculate the response of a
detector (located in the far-field region) to incident GWs, we can
calculate the GW energy flux flowing outwards through a Gaussian sphere
at $r = \infty$, and we can even prove that the time derivative of the
total energy of spacetime is equal to the negative of that energy flux.

A famous example: suppose we have a "nearly Newtonian" spacetime which
is close to Minkowski *everywhere*, and that spacetime is asymptotically
flat, and empty except for a binary star system. How much energy does
that binary star system radiate in GWs? For a nearly-Newtonian
asymptotically flat spacetime it's fairly easy to derive an excellent
(see https://en.wikipedia.org/wiki/Quadrupole_formula for more on this)
which says that the radiated energy flux is proportional to the square
of the 3rd time derivative of the binary star's mass quadrupole tensor.
Again for a nearly-Newtonian-everywhere spacetime, there's a simple
derivation of the quadrupole formulation (see any beginning GR textbook)
which is "rigorous enough for a physicist".

But now suppose our spacetime *isn't* nearly-Newtonian. E.g., what if
our orbiting "stars" are neutron stars or black holes? That simple
proof assumes that the metric is close to Minkowski *everywhere*,
including near to and inside the "stars", so that proof no long holds.
So is the quadrupole formula still valid for such a
binary-relativistic-star system? In the 1980s there was a lot of
approximation methods of uncertain mathematical validity) that in this
situation the actual radiated energy might be very different from what
the quadrupole formula would give, possibly even different in sign!
After a lot of hard work by a whole bunch of very talented mathematical
relativists, this controversy was eventually resolved (by the mid to
late 1990s, I think) and a mathematically rigorous proof of the
quadrupole formula was constructed which applies even to binary black
holes.

All (or at least a great deal) of the mathematical machinery which lets
us prove these nice theorems is only defined if the spacetime is
asymptotically flat. I suspect that there may be ways of defining these
things for *some* non-asymptotically-flat spacetimes, but we would
certainly loose a big part of the nice mathematical framework we have
for an asymptotically flat spacetime. (I suppose an analogy might be
trying to do calculus with functions which aren't Riemann-integrable.)

So, in conclusion, asymptotic flatness is a very nice property for a
spacetime to have, and if it doesn't hold then we loose a bunch of
useful and important concepts and mathematical tools.

[Moderator's note: Those interested in the history of the concept of
gravtational waves might enjoy arXiv:2111.00330:

@INPROCEEDINGS{ MDiMauroEN21a ,
AUTHOR = "Marco Di Mauro and S. Esposito and A.
TITLE = "Towards detecting gravitational waves:~a
contribution by Richard Feynman",
BOOKTITLE = "{T}he {S}ixteenth {M}arcel {G}rossmann
{M}eeting",
YEAR = "2022",
EDITOR = "Remo Ruffini and Gregory Vereshchagin",
PUBLISHER = "World Scientific"
}

(The above is preliminary publication information; probably at most the
title might not be the final one.) -P.H.]

-- "Jonathan Thornburg [remove -color to reply]" <jthor...@gmail-pink.com>
Dept of Astronomy & IUCSS, Indiana University, Bloomington, Indiana, USA
currently on the west coast of Canada
"C++ is to programming as sex is to reproduction. Better ways might
technically exist but they're not nearly as much fun." -- Nikolai Irgens
"that applies to Perl, too!" -- me

[Moderator's note: As a Fortran man, I pity your sex life. :-) -P.H.]

### Eric Flesch

Mar 20, 2022, 6:22:53 PMMar 20
to
On 4 Mar 2022 08:16:58 GMT, Luigi Fortunati
<fortuna...@gmail.com> wrote:
>Why does the force of gravity disappear in the free-falling elevator and
>on Earth (which is also in free-fall) does it not disappear?

If the Earth were in the shape of a homogeneous hollow shell, then you
would indeed free-float within it.

[[Mod. note -- I think we also need the conditionn "... a homogenous
*spherical* hollow shell. -- jt]]

### Luigi Fortunati

Mar 21, 2022, 4:04:55 AMMar 21
to
Eric Flesch domenica 20/03/2022 alle ore 09:22:48 ha scritto:
>> Why does the force of gravity disappear in the free-falling elevator and
>> on Earth (which is also in free-fall) does it not disappear?
>
> If the Earth were in the shape of a homogeneous hollow shell, then you
> would indeed free-float within it.
>
> [[Mod. note -- I think we also need the conditionn "... a homogenous
> *spherical* hollow shell. -- jt]]

In a hollow and homogeneous spherical shell I would certainly float but
the free falls of the lift and the Earth do not occur in any hollow
shell.