The gelatin sphere

[Luigi Fortunati]

A gelatin sphere in remote space, far from any gravitational field,
maintains its spherical shape because it is in an inertial reference
frame where there are no forces that can deform it.

But if the gelatin sphere approaches a planet or a star and falls free,
it gets longer, it ovalizes.

Is this deformation of the gelatin sphere in free fall in a
gravitational field due to the action of a force or not?

[Oliver Jennrich]

Post by Luigi Fortunati
A gelatin sphere in remote space, far from any gravitational field,
maintains its spherical shape because it is in an inertial reference
frame where there are no forces that can deform it.

Actually, it maintains its spherical shape because of surface tension
or other elastic internal forces. Without it, brownian motion would
cause diffusion, as can be observed by any gas cloud.

Even in the case of surface tension and finite temperature, the
spherical form would only adhered to 'on average', as the same
brownian motion would excite the normal modes of the sphere: the
sphere would wobble.

Post by Luigi Fortunati
But if the gelatin sphere approaches a planet or a star and falls free,
it gets longer, it ovalizes.

Given that there is a mechanism to exert internal forces (such as
said surface tension) on the sphere, most parts of the now ovoid
are not in free fall anymore but deviate from the geodesic path due
to those internal forces.

If there is no mechanism to exert such internal forces, as would
be in the case of e.g. an ideal gas, all the parts of the cloud
would fall freely. But the shape of the cloud would not even be
close to an ovoid. At least not for long.

Post by Luigi Fortunati
Is this deformation of the gelatin sphere in free fall in a
gravitational field due to the action of a force or not?

The shape of a body that is held together by internal forces is
determined by the balance of the external (tidal) force and the
internal forces.

[Luigi Fortunati]

Post by Oliver Jennrich
The shape of a body that is held together by internal forces is
determined by the balance of the external (tidal) force and the
internal forces.

Yes, there are the INTERNAL forces that are always present and that do
not have a privileged direction, so they tend to keep the initial
spherical shape of the jelly ball (on average).

And there is the EXTERNAL (tidal) force that acts more on the "south"
part of the sphere than the "north" part which is farther from the
center of gravity, thus causing an elongation at the poles and a
crushing at the equator.


[[Mod. note -- Yes. As Oliver Jennrich said, the body's shape
is determined by the balance of these two forces.
-- jt]]

[Luigi Fortunati]

Luigi Fortunati a écrit

Post by Luigi Fortunati

Post by Oliver Jennrich
The shape of a body that is held together by internal forces is
determined by the balance of the external (tidal) force and the
internal forces.

Yes, there are the INTERNAL forces that are always present and that do
not have a privileged direction, so they tend to keep the initial
spherical shape of the jelly ball (on average).
And there is the EXTERNAL (tidal) force that acts more on the "south"
part of the sphere than the "north" part which is farther from the
center of gravity, thus causing an elongation at the poles and a
crushing at the equator.
[[Mod. note -- Yes. As Oliver Jennrich said, the body's shape
is determined by the balance of these two forces.
-- jt]]

It follows that the presence of an EXTERNAL force during the free fall
in a gravitational field, contrasts with the inertiality described in
the first principle.

So I explicitly ask: is the free fall in a gravitational field inertial
or accelerated?

[Phillip Helbig (undress to reply)]

Luigi Fortunati a écrit

Post by Luigi Fortunati

Post by Oliver Jennrich
The shape of a body that is held together by internal forces is
determined by the balance of the external (tidal) force and the
internal forces.

Yes, there are the INTERNAL forces that are always present and that do
not have a privileged direction, so they tend to keep the initial
spherical shape of the jelly ball (on average).
And there is the EXTERNAL (tidal) force that acts more on the "south"
part of the sphere than the "north" part which is farther from the
center of gravity, thus causing an elongation at the poles and a
crushing at the equator.
[[Mod. note -- Yes. As Oliver Jennrich said, the body's shape
is determined by the balance of these two forces.
-- jt]]

It follows that the presence of an EXTERNAL force during the free fall
in a gravitational field, contrasts with the inertiality described in
the first principle.
So I explicitly ask: is the free fall in a gravitational field inertial
or accelerated?

I'm not sure what your question is, but if you are thinking of the idea
that "a man falling from a roof feels no force", then this is true
actually only in the limit if infinitely small size. An object of
finite size will feel tidal forces (and, of course, any
non-gravitational forces present such as surface tension).

[Luigi Fortunati]

Phillip Helbigundress to reply a écrit

Post by Phillip Helbig (undress to reply)

Post by Luigi Fortunati
It follows that the presence of an EXTERNAL force during the free fall
in a gravitational field, contrasts with the inertiality described in
the first principle.
So I explicitly ask: is the free fall in a gravitational field inertial
or accelerated?

I'm not sure what your question is, but if you are thinking of the idea
that "a man falling from a roof feels no force", then this is true
actually only in the limit if infinitely small size. An object of
finite size will feel tidal forces (and, of course, any
non-gravitational forces present such as surface tension).

Is the object of finite size in free fall (in a gravitational field)
inertial (that is, is it not undergoing any force) or is it accelerated
(because it is subject to the EXTERNAL tidal forces)?

Second question: do the tidal EXTERNAL forces act only in a given
reference system or do they always act?

[Steven Carlip]

On 7/2/19 4:36 AM, Luigi Fortunati wrote:

[...]

Post by Luigi Fortunati
Is the object of finite size in free fall (in a gravitational field)
inertial (that is, is it not undergoing any force) or is it accelerated
(because it is subject to the EXTERNAL tidal forces)?

If the object is held together by internal forces, then most bits of
the object aren't inertial -- they're subject to the internal forces.
If you work hard enough, you can define a center of mass position
that moves inertially. That seems obvious, and I can't believe
that's what you're really trying to ask.

So let's take an example that avoids this, a ball of "dust," particles
that aren't bound to each other. If the ball starts out spherical,
it will be distorted in roughly the same way your gelatin sphere is
(not exactly the same way, of course, because there are no longer
any internal forces). In this case, each particle is inertial; the
shape changes because inertial motion is different at different
locations. The distortion is called "geodesic deviation" -- look
it up if you want details.

Post by Luigi Fortunati
Second question: do the tidal EXTERNAL forces act only in a given
reference system or do they always act?

Tidal "forces" in general relativity are proportional to the
curvature tensor. If they are nonzero in any reference frame,
they're nonzero in all frames.

Steve Carlip

[Oliver Jennrich]

Phillip Helbigundress to reply a écrit

Post by Phillip Helbig (undress to reply)

Post by Luigi Fortunati
It follows that the presence of an EXTERNAL force during the free fall
in a gravitational field, contrasts with the inertiality described in
the first principle.
So I explicitly ask: is the free fall in a gravitational field inertial
or accelerated?

I'm not sure what your question is, but if you are thinking of the idea
that "a man falling from a roof feels no force", then this is true
actually only in the limit if infinitely small size. An object of
finite size will feel tidal forces (and, of course, any
non-gravitational forces present such as surface tension).

Is the object of finite size in free fall (in a gravitational field)
inertial (that is, is it not undergoing any force) or is it accelerated
(because it is subject to the EXTERNAL tidal forces)?

That is a pointless question. If the object is of finite size, it
doesn't make sense to ask if the *object* ist in free fall. Most parts
of the extended object experience an acceleration as the internal forces
(e.g. tension) prevent the respective volume element to move on a
geodesic.

Now, in practical terms, there are bodies (cf the test masses of LISA
Pathfinder) that are so small compared to the typical length scale of
the curvature that the tidal forces and the ensuing deformation of the
object is unmeasurable small. But fundamentally, almost all parts of an
extended body feel an acceleration as long as there is a mechanism to
exert internal forces.

Second question: do the tidal EXTERNAL forces act only in a given
reference system or do they always act?

They always act. You cannot "transform away" curvature.

[Sylvia Else]

Post by Luigi Fortunati
So I explicitly ask: is the free fall in a gravitational field inertial
or accelerated?

It is inertial, but the only part of your gelatin sphere that is in free
fall is the part at its centre of gravity.

If we put an observer inside the sphere, but in such a way that he was
not affected by it, then, except when he's not at the centre of gravity
of the sphere, he'd find himself accelerating relative to the gelatin
surrounding him.

Sylvia.

[Luigi Fortunati]

Post by Sylvia Else

Post by Luigi Fortunati
So I explicitly ask: is the free fall in a gravitational field inertial
or accelerated?

It is inertial, but the only part of your gelatin sphere that is in free
fall is the part at its centre of gravity.

The gelatin sphere is ALL in free fall!

Luigi

[Oliver Jennrich]

Post by Luigi Fortunati

Post by Sylvia Else

Post by Luigi Fortunati
So I explicitly ask: is the free fall in a gravitational field inertial
or accelerated?

It is inertial, but the only part of your gelatin sphere that is in free
fall is the part at its centre of gravity.

The gelatin sphere is ALL in free fall!

No, it isn't.

Yo cannot prepare an extended body with internal forces that is fully in
free fall in a scenario where you have non-homogeneous gravity.

If you manage to create a gravitational field that is homogeneous (akin
the electrical field between the two electrodes of an infinitely large
capacitor) you can put any extended body fully into free fall.

In your scenario (sphere orbiting a star), you won't be able to prepare
such a situation.

[Luigi Fortunati]

Oliver Jennrich a écrit

Post by Oliver Jennrich

Post by Luigi Fortunati
The gelatin sphere is ALL in free fall!

No, it isn't.
Yo cannot prepare an extended body with internal forces that is fully in
free fall in a scenario where you have non-homogeneous gravity.

I prepare the extended body in spherical form while I am FAR away from
any gravity which, by its nature, is NEVER homogeneous as it ALWAYS
decreases with distance.

Post by Oliver Jennrich
In your scenario (sphere orbiting a star), you won't be able to prepare
such a situation.

In fact I would do it by staying VERY FAR from every star and from every
planet.

And if then I approached a neutron star, the sphere would make me
understand that there is an external (tidal) force because it EXTEND!

[Sylvia Else]

Post by Luigi Fortunati

Post by Sylvia Else

Post by Luigi Fortunati
So I explicitly ask: is the free fall in a gravitational field inertial
or accelerated?

It is inertial, but the only part of your gelatin sphere that is in free
fall is the part at its centre of gravity.

The gelatin sphere is ALL in free fall!
Luigi

If the parts of the gelatin sphere were in free fall, then those parts
further from the the body creating the gravitational field would take
longer to orbit than those parts that are closer. But all the parts of
the sphere actually take the same time to orbit, so they cannot all be
in free fall.

The parts that are not at the centre of gravity of the sphere are being
subject to forces that make them follow orbits that differ from the ones
they'd follow in free fall. Those forces come from the distortion of the
gelatin.

If the required forces become too great for the gelatin to bear, then
the sphere will break. This is analogous to the situation of a
gravitationally bound object at the Roche limit.

Sylvia.

[Lawrence Crowell]

Post by Sylvia Else

Post by Luigi Fortunati

Post by Sylvia Else

Post by Luigi Fortunati
So I explicitly ask: is the free fall in a gravitational field inertial
or accelerated?

It is inertial, but the only part of your gelatin sphere that is in free
fall is the part at its centre of gravity.

The gelatin sphere is ALL in free fall!
Luigi

If the parts of the gelatin sphere were in free fall, then those parts
further from the the body creating the gravitational field would take
longer to orbit than those parts that are closer. But all the parts of
the sphere actually take the same time to orbit, so they cannot all be
in free fall.
The parts that are not at the centre of gravity of the sphere are being
subject to forces that make them follow orbits that differ from the ones
they'd follow in free fall. Those forces come from the distortion of the
gelatin.
If the required forces become too great for the gelatin to bear, then
the sphere will break. This is analogous to the situation of a
gravitationally bound object at the Roche limit.
Sylvia.

Yes, this is the case, a tidal locked body has these internal
forces. Digressing to a Newtonian situation the antipodal points close
and far from the main gravitation body both have the same orbital
period. Yet Newton or Kepler's 3rd law tells us this is not what happens
with orbits. There are then internal forces on this body. This of course
has no influence on the orbit of the center of mass. Remember from most
elementary mechanics, say Halliday & Resnick level physics, that forces
internal to a body generate no net acceleration.

LC

[Lawrence Crowell]

Post by Luigi Fortunati
A gelatin sphere in remote space, far from any gravitational field,
maintains its spherical shape because it is in an inertial reference
frame where there are no forces that can deform it.
But if the gelatin sphere approaches a planet or a star and falls free,
it gets longer, it ovalizes.
Is this deformation of the gelatin sphere in free fall in a
gravitational field due to the action of a force or not?

There are two ways of looking at this. The Newtonian view is there is a
difference if the gravitational force across the sphere of
material. There is then a difference of force between the antipodal
points given be F - F' = -GMm(1/r^2 - 1/r'^2). Now let r' = r + d, for d
the diameter of the sphere and so using an approximation we get

F - F' = -2GMmd/r^3.

This is treated as a force.

In general relativity we have a somewhat different perspective. The geodesic equation

d^2r/ds^2 + G^r_{ab}U^aU^b = 0 ---> G^r_{ab} Christoffel symbol.

In a weak gravity field has the proper time s ~= t and U^t ~= 1 while
all other spatial U^i ~= 0. So this leads to

d^2r/ds^2 + GM/r^2 = 0,

for the Christoffel symbol of the Schwarzschild metric. This however is
considered a bit of an illusion, for the geodesic equation is a
connection coefficient based equation that is not covariant. It turns
out to give the Newtonian result in a weak spherically symmetric
case. To get the difference in force across the sphere we need to
consider the geodesic deviation equation

dU^a/ds + R^a_{bcd}U^bV^cU^c = 0.

I have to cut across stuff, for this editor is not convenient. The V^c
is the vector separating two test masses, which we take to be the
diameter, and the velocity vectors U^a have only U^t ~= 1 and we then
recover

dU^a/ds + 2GMmd/r^3 = 0.

This is the same math result, but from the interpretation of there being
a separation of neighboring geodesic flows.

The real force involved is the material resistance to this tidal
acceleration. In general relativity gravitation is not a real force. A
spring folding two masses will be distended by the different geodesic
flows. We call this the tidal acceleration or force, but really any
force involved is a resistance to this geodesic separation.