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From: "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
Newsgroups: sci.physics.research
Subject: Re: The Elevator in Free Fall
Date: 21 Dec 2024 08:27:44 GMT
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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
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