Discussion:
Inertial frame
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Luigi Fortunati
2022-05-07 10:38:37 UTC
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A free-falling brick is an inertial frame?

Are a pair of free-falling half-bricks an inertial frame or are they=20
two distinct inertial frame?
Julio Di Egidio
2022-05-07 13:00:52 UTC
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Post by Luigi Fortunati
A free-falling brick is an inertial frame?
A free-falling brick is *in* "an" inertial frame. Which more precisely
means that we can find an inertial frame in which the brick is in
uniform motion, then just a special case is an inertial frame in
which the brick is at rest, and and even more special case is the
frame in which the brick is at rest at the origin of space. And we
might call that last one "the brick's own frame", because there is
indeed something "privileged" about it as far as that brick is
concerned: OTOH, though, notice that the fact that the brick is in
*free-fall* requires no frame to state or verify at all, it's altogether
a *true* physical property that can be verified *locally*.
Post by Luigi Fortunati
Are a pair of free-falling half-bricks an inertial frame or are they
two distinct inertial frame?
It should now be clear that that is simply upside down: if two
bricks are in free fall, of course one can find an inertial frame in
which both are in uniform motion, in fact infinitely many of them.

Julio
Luigi Fortunati
2022-05-08 07:41:36 UTC
Permalink
Post by Julio Di Egidio
A free-falling brick is an inertial frame?=20
A free-falling brick is *in* "an" inertial frame. Which more precisely
means that we can find an inertial frame in which the brick is in
uniform motion, then just a special case is an inertial frame in=20
which the brick is at rest, and and even more special case is the
frame in which the brick is at rest at the origin of space. And we
might call that last one "the brick's own frame", because there is
indeed something "privileged" about it as far as that brick is
concerned: OTOH, though, notice that the fact that the brick is in
*free-fall* requires no frame to state or verify at all, it's altogethe=
r
Post by Julio Di Egidio
a *true* physical property that can be verified *locally*.
How small must this be "locally"?

As small as a brick? A half brick? A tenth of a brick?
Julio Di Egidio
2022-05-08 14:21:21 UTC
Permalink
Post by Luigi Fortunati
Post by Luigi Fortunati
A free-falling brick is an inertial frame?
A free-falling brick is *in* "an" inertial frame. Which more precisely
means that we can find an inertial frame in which the brick is in
uniform motion, then just a special case is an inertial frame in
which the brick is at rest, and and even more special case is the
frame in which the brick is at rest at the origin of space. And we
might call that last one "the brick's own frame", because there is
indeed something "privileged" about it as far as that brick is
concerned: OTOH, though, notice that the fact that the brick is in
*free-fall* requires no frame to state or verify at all, it's altogether
a *true* physical property that can be verified *locally*.
How small must this be "locally"?
As small as a brick? A half brick? A tenth of a brick?
[Apologies for my first reply, not very constructive: please disregard,
I am trying an actual answer.]

It's literally the *point*, which is an ideal condition of course, but one
of those that can be approximated arbitrarily well: keeping in mind
that this is classical physics, not quantum, so we don't need further
cautions.

In practice, it's just a little bit more articulated than that, it goes this
way: *you* start moving until your motion is in sync with that of the
brick, which you can check by having a rigid rod guarantee that your
distance and orientation relative to the brick does not change (and
then would you ask why and how that works?), then, by *transitivity*
of the condition of being inertial, you just check that *you* are
inertial: and that is as local as it needs be, modulo approximations
that are simply structural to doing experiments.

And I am sure most here could rephrase that in better terms, but I'd
rather invite you to take a step back and this point and reconsider
how the very progression goes: what it even means for a property
to be a *true physical property* vs e.g. an artefact of the coordinate
system.

Julio
Tom Roberts
2022-05-15 10:10:42 UTC
Permalink
How small must this be "locally"? As small as a brick? A half
brick? A tenth of a brick?
[... verbiage that does not answer the question]
You both miss the key concept about locally inertial frames: they are
only APPROXIMATELY inertial, and the maximum size they can be is
determined by your measurement resolution; better resolution puts a
smaller limit on the size of the region in which they can be considered
inertial.

Here's a simple example.

Suppose you start with a 4-meter-wide elevator in freefall near the
surface of the earth, and you release two small ball bearings 3 meters
apart horizontally, at rest relative to the inside of the elevator. They
will APPROXIMATELY remain at rest relative to the elevator, but we know
that they are each falling toward the center of the earth, so they will
slowly approach each other as the elevator continues to fall.

Imagine you can measure the distance between them with a resolution of 1
millimeter. Since the earth radius is 6.371E6 meters, the elevator can
fall 1,460 meters until they are 2.999 meters apart, which will take
17.2 seconds.

Suppose, instead, that you can measure the distance between them with a
resolution of 1 micron. The elevator can fall 1.46 meters until they are
2.999999 meters apart, which will take 0.54 seconds.

Clearly the limited region of spacetime over which this locally inertial
frame is approximately inertial depends on how well you can measure.

Other types of measurements will put different constraints on the size
of the region. For instance, recent measurements of optical clocks at
NIST will put a limit of a few centimeters tall before such clocks at
rest in the elevator at its ceiling and floor will cease to remain in
sync.
[I] invite you to [...] reconsider [...] what it even means for a
property to be a *true physical property* vs e.g. an artefact of the
coordinate system.
Modern physics has a simple and very general requirement: "true physical
properties" must be invariant under changes of coordinates. This is why
modern physical theories are all expressed in terms of tensors, which
are completely independent of coordinates, and therefore invariant under
changes of coordinates.

Tom Roberts
Lou
2022-05-20 12:44:11 UTC
Permalink
Post by Tom Roberts
How small must this be "locally"? As small as a brick? A half
brick? A tenth of a brick?
[... verbiage that does not answer the question]
You both miss the key concept about locally inertial frames: they are
only APPROXIMATELY inertial, and the maximum size they can be is
determined by your measurement resolution; better resolution puts a
smaller limit on the size of the region in which they can be considered
inertial.
Here's a simple example.
Suppose you start with a 4-meter-wide elevator in freefall near the
surface of the earth, and you release two small ball bearings 3 meters
apart horizontally, at rest relative to the inside of the elevator. They
will APPROXIMATELY remain at rest relative to the elevator, but we know
that they are each falling toward the center of the earth, so they will
slowly approach each other as the elevator continues to fall.
Imagine you can measure the distance between them with a resolution of 1
millimeter. Since the earth radius is 6.371E6 meters, the elevator can
fall 1,460 meters until they are 2.999 meters apart, which will take
17.2 seconds.
Suppose, instead, that you can measure the distance between them with a
resolution of 1 micron. The elevator can fall 1.46 meters until they are
2.999999 meters apart, which will take 0.54 seconds.
Clearly the limited region of spacetime over which this locally inertial
frame is approximately inertial depends on how well you can measure.
An interesting point you make. It follows that there is no such
thing as an inertial frame because no matter how small you make two
points in space and how close together they are they will always be
seperate “inertial” frames. Being pulled towards the Center of gravity
from slightly different angles. So an inertial frame cannot actually
exist in 3D space.
Post by Tom Roberts
Other types of measurements will put different constraints on the size
of the region. For instance, recent measurements of optical clocks at
NIST will put a limit of a few centimeters tall before such clocks at
rest in the elevator at its ceiling and floor will cease to remain in
sync.
[I] invite you to [...] reconsider [...] what it even means for a
property to be a *true physical property* vs e.g. an artefact of the
coordinate system.
Modern physics has a simple and very general requirement: "true physical
properties" must be invariant under changes of coordinates. This is why
modern physical theories are all expressed in terms of tensors, which
are completely independent of coordinates, and therefore invariant under
changes of coordinates.
Tom Roberts
Tom Roberts
2022-05-20 15:15:32 UTC
Permalink
Post by Lou
the key concept [is that] locally inertial frames [...] are
only APPROXIMATELY inertial [...]
An interesting point you make.
It's not my point, this has been known for over a century, since
Einstein began thinking about gravity.
Post by Lou
It follows that there is no such
thing as an inertial frame
Hmmm. Physics is not math. Yes, in the world we inhabit, and in any
manifold of GR with any nonzero energy density, there are no EXACTLY
inertial frames. But in physics, where measurements are always inexact,
approximately inertial frames can be very useful, because in a locally
inertial frame one can use SR rather than GR, which GREATLY simplifies
the analysis. The point is to make sure the approximation involved
affects answers less than measurement resolutions.

[For example, at the LHC no individual event lasts more
than 1 microsecond. A truly inertial frame that is at
rest in the lab at the start of the event will fall at
most a few picometers during the event. As their
detectors have resolutions at least a million times
larger than that, the difference between the lab frame
and that inertial frame is completely negligible.]
Post by Lou
So an inertial frame cannot actually
exist in 3D space.
Approximately-inertial frames are aspects of (4D) spacetime, not (3D)
space.

Tom Roberts

Julio Di Egidio
2022-05-09 00:38:46 UTC
Permalink
Post by Luigi Fortunati
Post by Luigi Fortunati
A free-falling brick is an inertial frame?
A free-falling brick is *in* "an" inertial frame. Which more precisely
means that we can find an inertial frame in which the brick is in
uniform motion, then just a special case is an inertial frame in
which the brick is at rest, and and even more special case is the
frame in which the brick is at rest at the origin of space. And we
might call that last one "the brick's own frame", because there is
indeed something "privileged" about it as far as that brick is
concerned: OTOH, though, notice that the fact that the brick is in
*free-fall* requires no frame to state or verify at all, it's altogether
a *true* physical property that can be verified *locally*.
How small must this be "locally"?
Can I assume the first part is clear now?
Post by Luigi Fortunati
As small as a brick? A half brick? A tenth of a brick?
*Local* has a technical definition (do you need me to tell it to
you?), just like everything else here, so either you have some
serious question starting from that definition, or you should
simply get a good introductory course to the matter. Otherwise
I too don't see why you are even allowed to keep posting here.

Julio
Rock Brentwood
2022-05-17 06:45:31 UTC
Permalink
[[Mod. note -- I have rewrapped overly-long lines. -- jt]]
Post by Luigi Fortunati
A free-falling brick is an inertial frame?
If it's not rotating. Otherwise, the free-falling object is in a
rotating frame, if the frame is attached to and associated with the
object, itself. If it's not rotating, then it is locally inertial.
The distinction between the two rests on Newton's bucket thought
experiment.
Post by Luigi Fortunati
Are a pair of free-falling half-bricks an inertial frame or are they
two distinct inertial frame?
The curvature of space-time is precisely the warping of a field of
locally inertial free-fall frames whereby those that are initially
at rest with respect to one another start to accelerate with respect
to each other. If you display them as worldlines in a 4-dimensional
graph (or in a 3-dimensional graph, where one of the spatial
dimensions is suppressed, for the benefit of the unlucky few who
are visually impaired to see in 4 dimensions) then you'll see the
worldlines for locally inertial free-fall frames - initially parallel
in a time-like direction - starting to curve into one another -
hence the "falling" action associated with gravity. In this sense,
the gravity one feels and experiences is actually a warping in time,
first and foremost, rather than a warping in space. The actual
contraction is quantified and accounted for in the Raychaudhuri
equation, which is closely related to the "geodesic deviation
equation".

In a flat space-time spatially separated locally inertial frames,
initially at rest with respect to one another, remain at rest; and
so can be said to comprise the different locations of a global
inertial frame.

All of the foregoing applies independent of paradigm - to *both*
relativistic *and* non-relativistic theory; so it is neither a
construct nor innovation of "general relativity", but rather one
which first fully emerged at the onset of general relativity and
so has been (falsely) associated with it as a characteristic feature
of it. It is a general feature of any theory of gravity that respects
the Equivalence Principle.

In fact, both Newtonian gravity and Einsteinian gravity (specifically:
the Schwarzschild solution) can be unified as a one-parameter family
of geometries, that are warped versions of the 5-dimensional Bargmann
Geometry, via the line element + constraint:

dx^2 + dy^2 + dz^2 + 2 dt du + a du^2 - 2 V dt^2 + 2aV/(1 + 2aV) dr^2 = 0

where r = root(x^2 + y^2 + z^2), dr = (x dx + y dy + z dz)/r and
V = -GM/r is the potential of a gravitating body of mass M located
at r = 0.

The extra, u, coordinate is the non-relativistic limit of (s - t)
c^2, as c goes to infinity, where s is proper time. This has meaning
... which also (by the way) shows that such things as "time dilation"
and "twin paradox" are *also* rooted in non-relativistic theory in
disguised form as u, and are not features specific to Relativity!
The u coordinate shows up, physically, as negative the action per
unit mass for an inertial particle.

When a = 0, this is Newtonian gravity, and it can be generalized
by having V be the total potential for all gravitating bodies,
rather than just for one. The geodesics for this geometry are the
orbits of Newtonian gravity. Since V is a function of the coordinates
and velocities of individual bodies, rather than a bona fide field
quantity, it is very tempting to try and quantize this geometry
directly in quantum *mechanics*. But for the fact that you still
have the self-energy and self-force problems to deal with (in starker
form, in fact) you'd almost have a full-fledged *geometric* quantum
theory for Newtonian gravity - one in which space-time itself is
quantized. But the whole "quantizing field theory as mechanics"
strategy has these same issues roadblocks, here, as did Feynmann
and Wheeler's attempt to do the same with electromagnetic theory
in the 1940's.

The case where a = 0 and V = 0 is the Bargmann geometry, which is
the natural geometric arena for non-relativistic theory.

When a > 0, this is the Schwarzschild solution in which the proper
time is given as s = t + a u, and in which c = root(1/a) is an
invariant speed (i.e. "light speed").

The term "dx^2 + dy^2 + dz^2" is the legacy of Euclidean geometry;
while "2 dt du" is the legacy of *Galileo*'s principle of Relativity,
which is where space and time *actually* became unified into the
chrono-geometry of spacetime. The secret eloping of the two, however,
went largely unnoticed until it was fully consummated by the addition
of the Poincare' term "a du^2", which turns this into a geometry
for Minkowski space. The warping of time associated with Newtonian
gravity is in the "-2V dt^2" term, while the warping of space,
itself, associated with General Relativity is limited to the
substantially smaller "2aV/(1 + 2aV) dr^2" term.

The only effects of paradigm, here, are those limited to the "Special
Relativity" term a du^2 and the "General Relativity" term
2aV/(1 + 2aV) dr^2 (so that whatever "relativistic corrections" there
are, which are made to trajectories have to come from these terms).
The main thrust of gravity - and the essential background behind your
query - resides with the "Newtonian Gravity" term -2 V dt^2.
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