Discussion:
Equivalence principle
(too old to reply)
Luigi Fortunati
2024-06-08 17:40:15 UTC
Permalink
[Moderator's note: In the moderation process for this posting, I've cut
by accident the first lines of the message; so here's there complete
posting again. HvH]

In the video
at minute 6:56
it states that there is no measurement that can be made to distinguish
whether you’re being accelerated or whether you are sitting still on the
surface of a planet.

So, I ask: what stops us from measuring the presence (or absence) of
tidal forces? If tidal forces are there, then we are stationary on the
surface of a planet, if they are not there, we are experiencing a
non-gravitational acceleration.

Luigi Fortunati

[Moderator's remark: One has to keep in mind that the equivalence
principle is a local concept, i.e., the equivalence between the
observations in a gravitational field and in an accelerated frame of
reference in free space refers only to very small space-time regions. A
"true gravitational field" is of course never entirely equivalent to an
accelerated frame in flat Minkowski space, because according to GR the
gravitational field leads to space-time curvature, i.e., a non-vanishing
Riemann tensor, while Minkowski space is flat, which are
coordinate-independent notions, and only such notions are physically
interpretable.

Of course tidal forces are well observable, cf. the tides on Earth,
where the name "tidal force" refers to.

HvH.]
Mikko
2024-06-10 11:46:37 UTC
Permalink
Post by Luigi Fortunati
In the video http://youtu.be/R3LjJeeae68 at minute 6:56
it states that there is no measurement that can be made to distinguish
whether you’re being accelerated or whether you are sitting still on the
surface of a planet.
So, I ask: what stops us from measuring the presence (or absence) of
tidal forces? If tidal forces are there, then we are stationary on the
surface of a planet, if they are not there, we are experiencing a
non-gravitational acceleration.
Consider a situation where you are not sitting on a surface of a planet
but acclerated by a real non-gravitational interaction; and this happens
near a planet or a star: you can measure a tidal force (if your instruments
are big and sensitive enough).
--
Mikko
Hendrik van Hees
2024-06-10 12:10:37 UTC
Permalink
Or to put it simpler. In a local inertial reference frame, realized by a
point-like non-rotating body in free fall, you observe (e.g., by using
pointlike test particles) only the "true gravitational forces", i.e.,
the tidal forces.

If you sit on the surface of a planet, you are not in free fall, because
there are (electromagnetic) forces keeping you there.

That's why the accelerometer of your smart phone at rest on Earth shows
an acceleration of 9.81 m/s^2, because it measures accelerations
relative to a local inertial frame of reference! See, e.g.,

https://physlab.org/wp-content/uploads/2016/03/primer_smartphones.pdf
Post by Mikko
Post by Luigi Fortunati
In the video http://youtu.be/R3LjJeeae68 at minute 6:56
it states that there is no measurement that can be made to distinguish
whether you’re being accelerated or whether you are sitting still on the
surface of a planet.
So, I ask: what stops us from measuring the presence (or absence) of
tidal forces? If tidal forces are there, then we are stationary on the
surface of a planet, if they are not there, we are experiencing a
non-gravitational acceleration.
Consider a situation where you are not sitting on a surface of a planet
but acclerated by a real non-gravitational interaction; and this happens
near a planet or a star: you can measure a tidal force (if your instruments
are big and sensitive enough).
--
Hendrik van Hees
Goethe University (Institute for Theoretical Physics)
D-60438 Frankfurt am Main
http://itp.uni-frankfurt.de/~hees/
Luigi Fortunati
2024-06-11 07:05:55 UTC
Permalink
Or to put it simpler. In a local inertial reference frame, realized by a point-like non-rotating body in free fall, you observe (e.g., by using pointlike test particles) only the "true gravitational forces", i.e., the tidal forces.
If you sit on the surface of a planet, you are not in free fall, because there are (electromagnetic) forces keeping you there.
That's why the accelerometer of your smart phone at rest on Earth shows an acceleration of 9.81 m/s^2, because it measures accelerations relative to a local inertial frame of reference! See, e.g.,
Your reasoning is based on two preconceptions.

The first is that the accelerometer measures accelerations (and instead
it only measures forces) and the second is that free fall is an inertial
reference system despite its very evident mutual acceleration towards
the other body (also) in free fall.

Luigi Fortunati
Hendrik van Hees
2024-06-11 08:46:14 UTC
Permalink
Post by Luigi Fortunati
Or to put it simpler. In a local inertial reference frame, realized by a point-like non-rotating body in free fall, you observe (e.g., by using pointlike test particles) only the "true gravitational forces", i.e., the tidal forces.
If you sit on the surface of a planet, you are not in free fall, because there are (electromagnetic) forces keeping you there.
That's why the accelerometer of your smart phone at rest on Earth shows an acceleration of 9.81 m/s^2, because it measures accelerations relative to a local inertial frame of reference! See, e.g.,
Your reasoning is based on two preconceptions.
These are preconceptions well tested with high precision for centuries.
Physics is an empirical science, and theories are built based on precise
quantitative observations of nature.
Post by Luigi Fortunati
The first is that the accelerometer measures accelerations (and instead
it only measures forces) and the second is that free fall is an inertial
reference system despite its very evident mutual acceleration towards
the other body (also) in free fall.
I don't know, what's evident in your misconception. By definition bodies
which move without any interactions except the gravitational interaction
are by definition in free fall, and according to the equivalence
principle such bodies define a LOCAL (!!!!) inertial reference frame.
According to GR there are NO GLOBAL inertial frames as there are in
Newtonian mechanics and special-relativistic physics. Pointlike test
particles move on geodesics of spacetime, determined by the
energy-momentum-stress distributions due to the presence of other
bodies, if there are no other forces than gravity, i.e., they are not
accelerated. Geodesics here refer of course to spacetime not to
trajectories in "position space" of some arbitrary observer. E.g., the
motion of the planets around the Sun are such geodesics in spacetime for
an observer resting far away from the Sun (whose spacetime is locally
approximately described by special-relativistic, flat Minkowski
spacetime) the spatial trajectories are of course very close to Kepler
ellipses.
Post by Luigi Fortunati
Luigi Fortunati
--
Hendrik van Hees
Goethe University (Institute for Theoretical Physics)
D-60438 Frankfurt am Main
http://itp.uni-frankfurt.de/~hees/
Luigi Fortunati
2024-06-13 08:29:43 UTC
Permalink
Post by Hendrik van Hees
Post by Luigi Fortunati
The first is that the accelerometer measures accelerations (and instead
it only measures forces) and the second is that free fall is an inertial
reference system despite its very evident mutual acceleration towards
the other body (also) in free fall.
I don't know, what's evident in your misconception. By definition bodies
which move without any interactions except the gravitational interaction
are by definition in free fall, and according to the equivalence
principle such bodies define a LOCAL (!!!!) inertial reference frame.
The inertial reference frame is one where no forces act.

In free fall, tidal forces act and, therefore, you and Einstein are
wrong when you say that free fall is an inertial reference (whether
local or non-local).

Luigi Fortunati
Hendrik van Hees
2024-06-13 08:38:48 UTC
Permalink
The misconception is on your side, not Einstein's ;-).

An inertial frame of reference is operationally defined by Newton's Lex
II: A body moves uniformly (or stays at rest) if it does not interact
with anything.

The mathematical version of the equivalence principle in GR is that
spacetime is described by a torsion-free pseudo-Riemannian (Lorentzian)
manifold. An inertial frame can only be local, i.e., you can choose at
any given spacetime point a Galilean local reference frame. Physically
such a frame is realized by a point-like body in free fall, i.e., by a
body on which only gravitational forces are acting and (non-rotating,
i.e., Fermi-Walker transported) tetrads along its world line.

True gravitational fields always show up in terms of tidal forces, and
any extended test body is thus not force-free. To which extent you can
neglect these forces depends on the extension of this test body. It's
only "force-free" as long as its extensions is smaller than the
curvature radius of space time at the reference point of your
free-falling non-rotating reference frame.
Post by Luigi Fortunati
Post by Hendrik van Hees
Post by Luigi Fortunati
The first is that the accelerometer measures accelerations (and instead
it only measures forces) and the second is that free fall is an inertial
reference system despite its very evident mutual acceleration towards
the other body (also) in free fall.
I don't know, what's evident in your misconception. By definition bodies
which move without any interactions except the gravitational interaction
are by definition in free fall, and according to the equivalence
principle such bodies define a LOCAL (!!!!) inertial reference frame.
The inertial reference frame is one where no forces act.
In free fall, tidal forces act and, therefore, you and Einstein are
wrong when you say that free fall is an inertial reference (whether
local or non-local).
Luigi Fortunati
--
Hendrik van Hees
Goethe University (Institute for Theoretical Physics)
D-60438 Frankfurt am Main
http://itp.uni-frankfurt.de/~hees/
Luigi Fortunati
2024-06-14 19:42:18 UTC
Permalink
Post by Hendrik van Hees
The misconception is on your side, not Einstein's ;-).
An inertial frame of reference is operationally defined by Newton's Lex
II: A body moves uniformly (or stays at rest) if it does not interact
with anything.
Exactly, let's evaluate everything on Newton's principles and leave
aside the curvatures of spacetime which have nothing to do with Newton.

Is the space (not spacetime) of a free-falling elevator (invented by
Einstein and not me) enough for you as a *local* reference system? I
hope so.

Let's see what happens in such an elevator.

In the center, there is the one meter rod placed vertically and there
are two material points (A and B) initially next to the two ends of the
rod.

During the fall, the rod maintains its initial length because it is rigid.

Instead, the two points A and B start moving further and further away
from each other and also moving away from the ends of the rod.

Are the two material points stationary? No!

Do they move with uniform motion? No!

So where is the inertia of the elevator?

Luigi Fortunati

Luigi Fortunati
2024-06-11 07:28:24 UTC
Permalink
Post by Luigi Fortunati
[Moderator's note: In the moderation process for this posting, I've cut
by accident the first lines of the message; so here's there complete
posting again. HvH]
In the video http://youtu.be/R3LjJeeae68 at minute 6:56
it states that there is no measurement that can be made to distinguish
whether you're being accelerated or whether you are sitting still on the
surface of a planet.
So, I ask: what stops us from measuring the presence (or absence) of
tidal forces? If tidal forces are there, then we are stationary on the
surface of a planet, if they are not there, we are experiencing a
non-gravitational acceleration.
Luigi Fortunati
[Moderator's remark: One has to keep in mind that the equivalence
principle is a local concept, i.e., the equivalence between the
observations in a gravitational field and in an accelerated frame of
reference in free space refers only to very small space-time regions].
Okay, but the woman in the video is talking about a person-sized person
in an even larger box, not a very small space-time region.

So, at what small size does distinguishability cease and
indistinguishability begin?

Luigi Fortunati

[[Mod. note -- As I and others out have pointed out before, that depends
on your accuracy tolerance. Your question is sort of like asking, at what
size can you distinguish a straight line from part of some non-straight
curve?
-- jt]]
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