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[Phys-l] feeling gravity, or not



On 10/27/2006 04:31 PM, Edmiston, Mike wrote:?

I say there is distortion and I say I feel gravity. I
realize if you put me in free fall the sensation is different. So what?

I'll tell you what. Thereby hangs the entire tale. See below.

If you stand me on earth and turn off gravity that will feel different
also. That is, if there were a gravity switch, and you strapped me to a
chair bolted to earth, and then you turn gravity on and off, I could
sense the difference between off and on. And I wouldn't only sense the
difference in my butt. I would sense it all over.

That claim is implausible and unsubstantiated. In other words, PbBA.

Is there any physics principle or equation that supports this claim?

I realize you can still call this a particular viewpoint.

Is there any reason to prefer this viewpoint over the other viewpoint,
-- emphasized by John M. -- that what people feel are the _non-uniformities_
in the field?

In particular, consider the following greatly-simplified situation.
We simplify the point-to-point variation in the earth's gravitational
field by considering only one component (i.e. the component that is
diagrammed as left/right in the diagrams below). Also we diagram
only the /sign/ of the field, so that the force is either "rightward"
or "leftward". We also neglect the mass of the observer.

The the forces on the various parts of the earth are like this:


_.-""""-._
.rrrrrrlllll`.
/rrrrrrrlllllll\
|rrrrrrrrllllllll|
|rrrrrrrrllllllll|
|rrrrrrrrllllllll|
\rrrrrrrlllllll/
`.rrrrrlllll.'
`-....-'

At every point on the left half of the diagram gravity exerts a
rightward force ("r") while at every point on the right half of
the diagram, gravity exerts a leftward force ("l").

We need not quantify the magnitude of the field; by symmetry,
there exist "l/r" pairs, such that every "l" has an "r" of equal
magnitude. So diagramming just the sign ("l" or "r") tells us
what we need to know.

If we consider just the observer ("o") and "his half" of the earth,

_.-""
.rrrrrr
/rrrrrrr
|rrrrrrrr
o |rrrrrrrr
|rrrrrrrr
\rrrrrrr
`.rrrrr
`-..


then physics says that the observer will experience weightlessness,
because the observer and the half-earth will be comoving and equally
accelerated. I claim the observe will not perceive gravity, and
will not be able to measure gravity with any instrument. This is
a version of the Einstein elevator argument (although a similar
argument was presented by Galileo in 1638).

When we consider both halves of the earth, neither half is freely
falling w.r.t its local field. That's because each half is
acted upon by mechanical forces (pressure gradients and the like)
transmitted through the earth.

I claim that when the observer feels any force at all in this
situation, it is because a leftward force is inhibiting the
observer's rightward free-fall. That is, there is a leftward
force pushing on the observer's feet. Without this leftward
force, there would be nothing the observer could observe, as
previously discussed.

The punch line is that this crucial leftward force is entirely
attributable to the far side of the earth.

The only reason you feel gravity is because of nonuniformities
in the gravitational field. Einstein's equivalence principle
asserts that a uniform field is unfeelable.