Why does the force of gravity disappear in the free-falling elevator and
on Earth (which is also in free-fall) does it not disappear?
If you put the same two test bodies into an elevator that is
falling freely, one on the floor of the elevator and the
other one into the air one meter above the first one,
their distance will /not/ decrease because the force of the
Earth's surface and thus the squeezing force is missing here.
In article <gravity-20220304100436@ram.dialup.fu-berlin.de>, >ram@zedat.fu-berlin.de (Stefan Ram) writes:
If you put the same two test bodies into an elevator that isMinor nitpick (the summary is otherwise excellent): in practice, the
falling freely, one on the floor of the elevator and the
other one into the air one meter above the first one,
their distance will /not/ decrease because the force of the
Earth's surface and thus the squeezing force is missing here.
distance between the two test bodies in an elevator will INCREASE
From the point of view of physics, however, gravity is
always there;
Stefan Ram venerd=EC 04/03/2022 alle ore 10:12:00 ha scritto:
From the point of view of physics, however, gravity isOn this, of course, I absolutely agree: the force of gravity in the >free-falling elevator does not disappear.
always there;
But doesn't Einstein say quite the opposite?
Minor nitpick (the summary is otherwise excellent): in practice, the
distance between the two test bodies in an elevator will INCREASE
because gravity is slightly stronger lower down. That is an example of
a tidal force. Similarly, two bodies side-by-side in an elevator will approach each other. At rest or in uniform motion in no gravitational
field, neither would happen. In other words, this form of the
equivalence principle is valid only in the limit of an arbitrarily small elevator.
But with a UNIFORM and constant gravitational field (absolutely parallel field lines, and no variation in field strength), the equivalence
principle says this is equivalent to two test bodies undergoing the same constant acceleration, with no gravitational field present.
[[Mod. note --
Note also that in the context of general relativity, one needs to be
careful in invoking the equivalence principle for a *uniform* *constant* gravitational field. The problem is that such a field is the result of (i.e., implies the presence of) an infinite mass plane,
which means that
spacetime is *not* asymptotically flat. That has a number of "interesting" consequences...
-- jt]]
On 3/7/22 2:09 PM, Mike Fontenot wrote:
But with a UNIFORM and constant gravitational field (absolutely parallel
field lines, and no variation in field strength), the equivalence
principle says this is equivalent to two test bodies undergoing the same
constant acceleration, with no gravitational field present.
[[Mod. note --
Note also that in the context of general relativity, one needs to be
careful in invoking the equivalence principle for a *uniform* *constant*
gravitational field. The problem is that such a field is the result of
(i.e., implies the presence of) an infinite mass plane,
True,
which means that
spacetime is *not* asymptotically flat. That has a number of "interesting" >> consequences...
-- jt]]
Can you elaborate on that? Why does that imply that spacetime "is not asymptotically flat"? And what exactly does "asymptotically flat" mean?
which means that
spacetime is *not* asymptotically flat. That has a number of "interesting" >>> consequences...
-- jt]]
distance between the two test bodies in an elevator will INCREASE
because gravity is slightly stronger lower down. That is an example of
a tidal force.
Why does the force of gravity disappear in the free-falling elevator and
on Earth (which is also in free-fall) does it not disappear?
Why does the force of gravity disappear in the free-falling elevator and
on Earth (which is also in free-fall) does it not disappear?
If the Earth were in the shape of a homogeneous hollow shell, then you
would indeed free-float within it.
[[Mod. note -- I think we also need the conditionn "... a homogenous *spherical* hollow shell. -- jt]]
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