In article <vk0k8g$2p4uk$1@dont-email.me>, Luigi Fortunati discussed,
in Newtonian mechanics and in general relativity (GR), the behavior of
an elevator which is (a) initially suspended stationary from an elevator
cable or cables, and then (b) free-falling downwards after the cable(s)
break. Luigi then went on to (c) ask some questions about the motion
of bodies placed above/below the center of gravity of a free-falling
elevator.

In this article I'll try to analyze the same system Luigi described,
and clarify a few of the tricky parts of how this system is modelled
in GR. In this article I'll focus on (a) and (b) above; I'll discuss
(c) in a following article.



Let's start with the Newtonian perspective, where gravity is a force,
and where we measure acceleration with respect to (i.e., relative to)
an inertial reference frame (IRF). To a very good approximation, we can
treat the surrounding building housing the elevator shaft as an IRF.

BEFORE the cable-break, i.e., when the elevator is suspended from
the cables, stationary with respect to the surrounding-building IRF,
there are 2 (vertical) forces acting on the elevator:
* elevator weight (F=mg, pointing down), where m is the elevator's
mass and g is the local gravitational acceleration (about 9.8 m/s^2
near the Earth's surface)
* cable tension (F=T, pointing up, i.e., the cable(s) are pulling up
on the elevator)
The net (vertical) force acting on the elevator is thus T - mg in the
upward direction.

Since we observe that the elevator is stationary with respect to the surrounding-building IRF, i.e., the elevator's acceleration with respect
to that IRF is a=0, we use Newton's 2nd law to infer that the net force
F_net acting on the elevator must also be zero:
F_net = ma = 0
and hence
T - mg = 0
and hence the cable tension must be
T = mg

Luigi wrote

The cables break and the elevator goes into free fall.
Newton told us that the elevator accelerates and, therefore, there is a
force that makes it accelerate.

That's correct.

AFTER the cable-break, when the elevator is in free-fall downwards
(let's neglect air resistance for simplicity), the only force acting
on the elevator is the elevator's weight (F=mg, pointing down), so the
net force acting on the elevator is F_net = mg (pointing down) and the elevator's acceleration with respect to the surrounding-building IRF
is a = F_net/m = g (again pointing down).



Now let's look at the same system from a GR perspective, i.e., from a perspective that gravity isn't a force, but rather a manifestation of
spacetime curvature. In this perspective it's most natural to measure accelerations relative to *free-fall*, or more precisely with respect
to a *freely-falling local inertial reference frame* (FFLIRF). An
FFLIRF is just a Newtonian IRF in which a fixed coordinate position
(e.g., x=y=z=0) is freely falling.

Like Newtonian IRFs, there are infinitely many FFLIRFs at a given
position, with differing relative positions, velocities, and
orientations, but all these FFLIRFs have zero acceleration and
rotational velocity with respect to each other. If all we care about
is acceleration, we often ignore the freedom to choose different
relative positions, velocities, and orientations, and refer to "the"
FFLIRF.

Any FFLIRF is (by definition) freely-falling, so it's accelerating
*downwards* at an acceleration of g relative to the surrounding-building Newtonian IRF.

Since the g vector points (approximately) towards the center of the
Earth, we see that the FFLIRF *changes* if you go to a different place
near the Earth's surface. For example, my FFLIRF differs from (i.e.,
has a nonzero relative acceleration with respect to) the FFLIRF of
someone 1000 km away on the surface of the Earth, or even of someone
at my latitude/longitude but 1000 km above me. This why we have the
word "local" in the phrase "freely-falling *local* inertial reference
frame". This is related to Luigi's questions (c); I'll elaborate on
this in a following article.

In GR it's easiest to first consider the situation AFTER the cable-break,
when the elevator is freely falling (we're neglecting air resistance).

Luigi wrote:

Then Einstein came along and told us that this is not true and that
there is no force that accelerates the elevator in free fall.

That's correct. There are no forces acting on the elevator (remember
we're not considering gravity to be a "force"), so the net force acting
on the elevator is zero, so Newton's 2nd law
a = F_net/m
says that a=0, i.e., the elevator has zero acceleration
*with respect to (i.e., relative to) a FFLIRF*.

Since we've already established that a FFLIRF is accelerating *downwards*
at an acceleration of g relative to the surrounding-building IRF, we
conclude that the elevator is accelerating *downwards* at an acceleration
of g relative to the surrounding-building IRF.

Luigi wrote:

But if there is no force that accelerates the elevator, it means that
the elevator does not accelerate.
And if it does not accelerate, then it moves with uniform speed.

This two sentences both leave out a key qualification, namely "with
respect to a FFLIRF". That is, a more accurate statement is that if
there is no force that accelerates the elevator, it means that the
elevator does not accelerate *with respect to a FFLIRF*, and hence it
moves with uniform speed *with respect to a FFLIRF*. The qualification
"with respect to a FFLIRF" is essential here -- without it the statement
is ambiguous (acceleration with respect to what?).

Luigi wrote:

But speed is not absolute: it is relative.
And so I ask: is there any reference system with respect to which its
speed is uniform?

Yes, the elevator's speed is uniform with respect to any FFLIRF. Since
a FFLIRF is accelerating (downwards) with respect to the surrounding
buildint's IRF, the elevator's speed is NOT uniform with respect to the surrounding building's IRF.

Now let's consider the situation BEFORE the cable-break from a GR
perspective. Now there *is* an external force acting on the elevator,
namely the cable tension (F=T pulling up on the elevator). In the
Newtonian perspective we found that T = mg, and this turns out to still
be true in GR.
[Aside: What I just wrote is true for weak gravitational
fields like the Earth's. If we were in a very strong
gravitational field (e.g., close to a neutron star or
black hole) then we might have to be more careful with
many of the statements I'm making.]
So, the net force acting on the elevator is F_net = mg, pointing up.

Newton's 2nd law then says
a = F_net/m
= mg / m
= g
i.e., the elevator (which is stationary relative to the surrounding
building) must be accelerating *up* at an acceleration of g with respect
to (i.e., relative to) any FFLIRF.

This seems a bit counterintuitive, but in fact it's correct: Since a
FFLIRF is accelerating *down* at an acceleration of g with respect to the surrounding-building IRF, the (stationary) surrounding building (and the elevator, which is stationary with respect to the building) must be accelerating *up* at an acceleration of g with respect to the FFLIRF.

[Aside: It's instructive to compare the previous paragraph
with what we'd think about a different physical system:
suppose that the building and elevator were in space far from
any other masses, and the building's foundation were replaced
by a huge rocket that's accelerating the whole building (and
the elevator suspended inside the building from cables which
haven't yet broken) upwards at an acceleration of g relative
to a Newtonian IRF.

Given our assumption of "in space far from any other masses",
a Newtonian IRF is a FFLIRF, and vice versa. So, this
"rocket-accelerated elevator in space" would have the same
upward acceleration with respect to a FFLIRF as our ordinary
elevator here on Earth (again, BEFORE the cable-break) in the
GR perspective.

This is an example of the "equivalence principle" (EP) which,
in its simplest form, says (roughly) that a uniform gravitational
field has the same local effects as a steady acceleration.
In Newtonian mechanics it's not apparent why the EP should
be true; GR sort of assumes the EP as a postulate. In fact,
assuming the EP can take you most of the way to deriving GR,
and this was roughly the route that Einstein took in originally
obtaining GR. (I'm glossing over lots of technical details here.)]


To summarize, then, in GR *free-fall* plays a similar role to that which *uniform motion* plays in Newtonian mechanics. Newton's 2nd law
a = F_net/m
is formally the same in GR and in Newtonian mechanics, but a and F_net
are interpreted somewhat differently:
* In Newtonian mechanics, "a" is interpreted as acceleration with respect
to (relative to) an IRF, and gravity is viewed as a force contributing
to F_net.
* In GR, "a" is interpreted as acceleration with respect to (relative to)
a FFLIRF, and gravity is *not* viewed as a force and does *not* contribute
to F_net. As I'll explain in a following article, gravity actually shows
up as spacetime curvature, evidenced by the relative acceleration of
FFLIRFs at different places (e.g., the relative acceleration of my
FFLIRF and the FFLIRF of someone 1000 km away).

--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
on the west coast of Canada
"the stock market can remain irrational a lot longer than you can
remain solvent" or (probably the correct original wording) "markets
can remain irrational a lot longer than you and I can remain solvent"
-- A. Gary Shilling (often misattributed to John Maynard Keynes)

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On 22/12/2024 09:57, Luigi Fortunati wrote:

Jonathan Thornburg [remove -color to reply] il 21/12/2024 09:27:44 ha scritto:

...
Now let's look at the same system from a GR perspective, i.e., from a
perspective that gravity isn't a force, but rather a manifestation of
spacetime curvature. In this perspective it's most natural to measure
accelerations relative to *free-fall*, or more precisely with respect
to a *freely-falling local inertial reference frame* (FFLIRF). An
FFLIRF is just a Newtonian IRF in which a fixed coordinate position
(e.g., x=y=z=0) is freely falling.

Can we define the interior space of the elevator as "local" or is it
too big?

If it is too big, how big must it be to be considered "local"?

If it is shown that there are real forces inside the free-falling
elevator, can we still consider this reference system inertial?

Are tidal forces real?

Do we mean by "freely falling bodies" only those that fall in the very
weak gravitational field of the Earth or also those that fall in any
other gravitational field, such as that of Jupiter or a black hole?

Luigi Fortunati.

This problems in understanding GR is, in my opinion, due to too much
emphasis on the geometrical point of view. Of course, geometry is the theoretical foundation of all of modern physics, i.e., a full
theoretical understanding of physics is most elegantly achieved by
taking the geometric point of view of the underlying mathematical
models. However, there's also a need for a more physical, i.e.,
instrumental formulation of its contents.

Now indeed, from an instrumental point of view, the gravitational
interaction is distinguished from the other interactions by the validity
of the equivalence principle, i.e., "locally" you cannot distinguish
between a gravitational force on a test body due to the presence of a gravitational field due to some body. In our example we can take as a
test body a "point mass" inside the elevator, with the elevator walls
defining a local spatial reference frame. The corresponding time is
defined by a clock at rest relative to this frame at the origin of the
frame (say, one of the edges of the elevator). Now, the equivalence
principle says that it is impossible for you to distinguish by any
physics experiment or measurement inside the elevator, whether you are
in a gavitational field (in our case due to the Earth), which can be
considered homogeneous (!!!), for all relevant (small!) distances and
times around the origin of our elevator reference frame or whether the
elevator is accelerating in empty space. A consequence is also that if
you let the elevator freely fall in the gravitational field of the
Earth, you don't find any homogeneous gravitational field, i.e., free
bodies move like free particles locally, and thus the free-falling
elevator defines a local inertial frame of reference.

Translated to the "geometrical point of view" that means that you
describe space and time in general relativity as a differentiable
spacetime manifold. The equivalence principle means that at any
space-time point you can define a local inertial frame, where the
pseudometric of Minkowski space (special relativity) defines a
Lorentzian spacetime geometry.

If you now look at larger-scale physics around the origin of the freely-falling-elevator restframe, where the inhomogeneity of the
Earth's gravitational field become important, there are "true forces"
due to gravity. In the local inertial frame these are pure tidal forces,
named because they are responsible for the tides on the Earth-moon
system freely falling in the gravitational field of the Sun.

So it's important to keep in mind that the equivalence between
gravitational fields and accelerated reference frames in Minkowski space
holds only locally, i.e., in small space-time regions around the origin
of your coordinate system, in which external gravitational fields can be considered as homogeneous (and static). T

he physically interpretible geometrical quantities are tensor (fields),
and the general-relativistic spacetime at the presence of relevant true gravitational fields due to the presence of bodies (e.g., the Sun in the
solar system) is distinguished from Minkowski space by the non-vanishing curvature tensor, and this is a property independent of the choice of
reference frames and (local) coordinates, i.e., you can distinguish from
being in an accelerated reference frame in Minkowski space (no
gravitational field present) and being under the influence of a true gravitational field due to some "heavy bodies" around you, by measuring
whether there are tidal forces, i.e., whether the curvature tensor of
the spacetime vanishes (no gravitational interaction at work, i.e.,
spacetime is described as a Minkowski spacetime) or not (gravitational interaction with other bodies present, and you have to describe the
spacetime by some other pseudo-Riemannian spacetime manifold, which you
can figure out by solving Einstein's field equations, given the energy-momentum-stress tensor of the matter causing this gravitational
field, e.g., the Schwarzschild solution for a spherically symmetric mass distribution).

--
Hendrik van Hees
Goethe University (Institute for Theoretical Physics)
D-60438 Frankfurt am Main
http://itp.uni-frankfurt.de/~hees/

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On 12/20/24 12:51 AM, Luigi Fortunati wrote:

The cables break and the elevator goes into free fall.
Newton told us that the elevator accelerates and, therefore, there is a
force that makes it accelerate.
Then Einstein came along and told us that this is not true

This is one of your problems. Physics is not about "true or false"; we
make MODELS of the world we inhabit -- Newtonian mechanics and GR are
DIFFERENT MODELS. The problem is that you intermix nomenclature
willy-nilly between them.

and that
there is no force that accelerates the elevator in free fall.
But if there is no force that accelerates the elevator, it means that
the elevator does not accelerate.
And if it does not accelerate, then it moves with uniform speed.

This is another of your problems. You did not specify the coordinates
relative to which "speed" is measured. Relative to an elevator-fixed
LOCALLY inertial frame, it moves with uniform speed.

But speed is not absolute: it is relative.

This is another of your problems. Your words are HIGHLY ambiguous.

And so I ask: is there any reference system with respect to which its
speed is uniform?

Yes. Any elevator-fixed locally inertial frame.

This is for Newton's second law: force that accelerates mass.
Instead, for the first law, Einstein says that a body in the elevator
in free fall is at rest with respect to the elevator itself.
So, why does a body placed below the center of gravity of a
free-falling elevator accelerate downwards, and if it is above the
center of gravity, it accelerates upwards?

For this last to happen the elevator does not meet the criteria of a
locally inertial frame (see my recent post in this newsgroup). It is of
course due to tidal forces inside the elevator, the acceleration of
which exceeds one's measurement accuracy. Objects separated horizontally
will also accelerate towards each other (due to tidal forces that exceed measurement accuracy).

I repeat: you need more precision in thought and words.

Tom Roberts

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On 12/22/24 2:57 AM, Luigi Fortunati wrote:

Can we define the interior space of the elevator as "local" or is it
too big?

This depends on: a) the curvature of spacetime where the elevator is
located, b) the size of the elevator (including duration), and c) one's measurement accuracy. For an ordinary-sized elevator near earth falling
for ten seconds, and a measurement accuracy of microns and microseconds
(or larger), its interior can be considered a LOCALLY inertial frame.

If it is shown that there are real forces inside the free-falling
elevator, can we still consider this reference system inertial?

Depends on the details (and the meanings of words). For internal forces
that are small enough to not significantly distort the steel elevator,
it can be considered LOCALLY inertial, as long as it meets the criteria
above.

Are tidal forces real?

This depends on the meanings of words, and is therefore ambiguous and uninteresting to me.

Do we mean by "freely falling bodies" only those that fall in the very
weak gravitational field of the Earth or also those that fall in any
other gravitational field, such as that of Jupiter or a black hole?

"freely falling" means not subject to any external forces. This is
independent of the size of nearby bodies. Note that an object with size comparable to curvature cannot be considered freely-falling. (Here
gravity is not a force.)

The size of a locally-inertial frame depends on the criteria of my first paragraph above. The inside of the elevator above but near Jupiter can
be considered a locally-inertial frame. For a one-solar-mass black hole
just outside its horizon, such an elevator is too big. For a
billion-solar-mass black hole just outside its horizon, it can be
considered a LOCALLY inertial frame. (More massive black holes have
smaller spacetime curvatures at their horizon.)

Tom Roberts

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In article <vk0k8g$2p4uk$1@dont-email.me>, Luigi Fortunati wrote:

The cables break and the elevator goes into free fall.

[[...]]

for the first law, Einstein says that a body in the elevator
in free fall is at rest with respect to the elevator itself.

Not quite: a body *sufficiently close to the elevator's center of mass*
is *unaccelerated* with respect to the elevator itself. "Unaccelerated"
means constant velocity, but that velocity need not be zero.

So, why does a body placed below the center of gravity of a
free-falling elevator accelerate downwards, and if it is above the
center of gravity, it accelerates upwards?

Assuming this elevator is near the Earth's surface, the answer to
your question is "tidal forces". In more detail...

[For the following, I'll use the sign convention that the Newtonian
"little g" vector point down, and has *positive* magnitude.]

We observe that the Newtonian "little g vector" (as measured by observers stationary with respect to (wrt) the Earth's surface; let's call this g_wrt_Earth) varies with position. In this case (measuring near to,
and above, the Earth's surface), g_wrt_Earth always points roughly
down, but decreases in magnitude as we go up in altitude away from the
Earth's surface.

If we now go to a free-falling local inertial reference frame (FFLIRF),
say one accelerating down with acceleration g_elevator wrt the Earth's
surface -- we'll measure a little-g vector (as measured wrt the FFLIRF
-- let's call this g_wrt_FFLIRF) of
g_wrt_FFLIRF(position) = g_wrt_Earth(position) - g_elevator (1)
where I've explicitly shown which terms are position-dependent.

Since g_elevator is (by the definition of FFLIRF) precisely g_wrt_Earth
at the elevator's center-of-mass position, by equation (1), g_wrt_FFLIRF
must be zero at this position. But since g_wrt_Earth varies with position, g_wrt_FFLIRF must also vary with position. That is, g_wrt_FFLIRF will in general be nonzero for positions away from the elevator center-of-mass.

In this case (elevator near to, and above, the Earth's surface),
g_wrt_Earth points down everywhere in the elevator, but at the TOP of the elevator
|g_wrt_Earth(top of elevator)|
< |g_wrt_Earth(elevator center of mass)| (2)
where I'm using | | to denote the magnitude of a vector.
By equation (1), this means that g_wrt_FFLIRF(top of elevator) points UP. Similarly, at the BOTTOM of the elevator,
|g_wrt_Earth(bottom of elevator)|
> |g_wrt_Earth(elevator center of mass)| (3)
so equation (1) tells us that g_wrt_FFLIRF(bottom of elevator) points
DOWN.

This means that test masses placed in different parts of the elevator
have a "tidal" acceleration wrt each other. Alternatively, if we hold
on to test masses in different parts of the elevator (i.e., we hold them
at fixed positions wrt the elevator, so they are NOT in free-fall), then
we'll feel "tidal" forces acting on the masses. Tidal forces/accelerations represent a deviation of our supposed FFLIRF from actually being inertial.


In article <vk8fh7$h93j$1@dont-email.me>, Luigi Fortunati wrote:

Can we define the interior space of the elevator as "local" or is it
too big?

If it is too big, how big must it be to be considered "local"?

If it is shown that there are real forces inside the free-falling
elevator, can we still consider this reference system inertial?

Tha answer to all of these questions is the same, namely, "It depends
on your accuracy tolerance for measurements".

That is, it's easy to see that |g_wrt_FFLIRF(top of elevator)| and |g_wrt_FFLIRF(bottom_of_elevator)| both vary with the size of the
elevator. In particular, if you make the elevator 10 times smaller,
both of these numbers will also be about 10 times smaller. So, if you
make the elevator small enough, then both of these numbers will be tiny
enough that we can approximate them as zero. That is, if the elevator
is small enough, then tidal accelerations/forces are negligible, i.e.,
the elevator is (to within our accuracy tolerance) an inertial reference
frame.

Similarly, the effect of any given nonzero g_wrt_FFLIRF compounds over
time, i.e., if we say that our experiments are only going to last for
some finite duration, then the shorter that time, the less the effect of
any given nonzero g_wrt_FFLIRF. Thus if our duration is short enough,
then we can neglect the effects of g_wrt_FFLIRF being zero, and the
elevator is (to within our accuracy tolerance) an inertial reference
frame.

More precisely, for any fixed accuracy tolerance, if we make the elevator
small enough and/or make our measurements for a short enough duration,
then the tidal accelerations/forces across the elevator will be small
enough to be negligible.

As Tom Roberts explained earlier in this thread, just how small the
elevator must be and/or just how short our measurement duration must be, depends on how fast the local little-g field varies with position. For example, if instead of our elevator being near to (and having a little-g
vector dominated by) the Earth, our elevator is instead near to the
event horizon of a billion-solar-mass black hole, then little-g varies
much less strongly with position than it does near the Earth. So, in
this case (near a huge black hole) the elevator can be larger, and/or
we can make measurements for a longer duration, and still stay within
the same fixed accuracy tolerance.

Are tidal forces real?

They can do work, so they must be real.

To see this, fix a stick running from floor to ceiling in the elevator,
and place two beads on it, which can slide up and down with a (small)
bit of friction. If the friction is small enough (but still nonzero),
then the beads will move under the influence of the tidal forces, and
the friction will cause the beads and the stick to heat up. This
heating shows that the tidal forces are doing work, and hence that
tidal forces must be real.

[Digression: the same question arose in general relativity,
when well into the 1950s many researchers were unsure whether
or not gravitational waves were real. The two-beads-on-a-stick
gedanken experiment for gravitational waves was proposed by
Feynman and analyzed in detail by Bondi in 1957, proving that
gravitational waves could do work and must thus be real.]

See
https://en.wikipedia.org/wiki/Tidal_power
for some real-world examples of work done by tidal forces, with (in some
cases) power output measured in the hundreds of megawatts. These systems
all ultimately exploit the fact that g_wrt_Earth is NOT constant from one
edge of the tidal power basin to the other edge, i.e., that the (entire)
tidal power basin as NOT a local inertial reference frame.

--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
currently on the west coast of Canada
"[I'm] Sick of people calling everything in crypto a Ponzi scheme.
Some crypto projects are pump and dump schemes, while others are pyramid
schemes. Others are just standard issue fraud. Others are just middlemen
skimming off the top. Stop glossing over the diversity in the industry."
-- Pat Dennis, 2022-04-25

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In article <vk0k8g$2p4uk$1@dont-email.me>, Luigi Fortunati asked

Are tidal forces real?

In article <lt1tqbFapkrU1@mid.dfncis.de>, I replied

They can do work, so they must be real.

[[...]]

See
https://en.wikipedia.org/wiki/Tidal_power
for some real-world examples of work done by tidal forces, with (in some cases) power output measured in the hundreds of megawatts. These systems
all ultimately exploit the fact that g_wrt_Earth is NOT constant from one edge of the tidal power basin to the other edge, i.e., that the (entire) tidal power basin as NOT a local inertial reference frame.

Oops, that last sentence of mine was quite wrong. :(

The variation in g_wrt_Earth over the tidal power basin is generally
negligible for a tidal power system. I should have written this instead:

These systems
all ultimately exploit the fact that g_wrt_Earth varies with time
AND varies with position on the Earth's surface. The combination
of these variations causes ocean tides (which certainly do mechanical
work).

I should also point out this GREAT animation of worldwide tides at one particularly-powerful frequency:

https://en.wikipedia.org/wiki/File:Global_surface_elevation_of_M2_ocean_tide.webm

--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
currently on the west coast of Canada
"[I'm] Sick of people calling everything in crypto a Ponzi scheme.
Some crypto projects are pump and dump schemes, while others are pyramid
schemes. Others are just standard issue fraud. Others are just middlemen
skimming off the top. Stop glossing over the diversity in the industry."
-- Pat Dennis, 2022-04-25

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On 12/29/24 2:34 PM, Luigi Fortunati wrote:

[...] *real* gravitational force

You are basing your entire argument on the meaning of one word. That is
NOT physics; at best it is linguistics.

So, it is the presence or absence of *real* gravitational forces in
the "local" frame of the elevator that establishes which of the two
models is correct and which is not.

Complete nonsense.

First, you give no method to determine whether a given force is "real"
or not. You seem to assume that this distinction is obvious, but it most definitely is NOT, and depends on the meaning one gives to that word.
The validity of physical models doe NOT depend on the meanings of words
-- the models themselves are mathematical, not linguistic. IOW: the word
"real" does not appear in any physical theory.

Second, the ONLY way to determine whether a given model (theory) is
valid is to perform experiments and make measurements on them, then
compare to the predictions of the model for the same measurements of the
same experiments.

Bottom line: if you perform experiments near the surface of the earth,
with measurement accuracies of microns and microseconds over distances
<~ 100 meters and durations <~ 100 seconds, you'll find that both
Newtonian gravity and General relativity are valid. But for certain astronomical measurements (such as the advance of mercury's perihelion
with arc-second/century accuracy) only GR is valid.

A contestation that I expected and that no one has made is this: if
General Relativity is wrong, why are its results more consistent
with observed phenomena than those of Newton's gravitation?

This is more nonsense. See above. GR is valid over a LARGER DOMAIN than Newtonian gravity. As I keep saying, "right and wrong" (also "true and
false") are not applicable to physical theories; theories are valid over
a given domain with a given measurement accuracy, and in this case those domains are different for NG and GR.

I repeat: you REALLY need to learn what physics actually is -- you have
wildly incorrect notions about it. Until you do, you will keep confusing
and mystifying yourself.

[This is getting overly repetitive and boring. Don't expect
me to continue.]

Tom Roberts

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