On Thu, Feb 13, 2014 at 2:32 PM, Edgar L. Owen <[email protected]> wrote:

> Jesse,
>
> Let me think about this, but it is NOT the observer in "free fall in a
> gravitational field" that is equivalent to acceleration. It is an observer
> RESISTING free fall (e.g. standing on the surface of the earth) that is
> equivalent to acceleration.
>

Suppose the observer who's not moving on a geodesic path (call her Alice)
passes through the small spacetime neighborhood where the observer who IS
moving on a geodesic path (call him Bob) is defining his  "local inertial
frame" (Bob's geodesic path can either by a free-fall path through curved
spacetime, or inertial motion in flat spacetime, since both qualify as
geodesics in their respective spacetimes). As Alice passes through this
region, she performs some experiment and notes the physical result.
Whatever physical elements are involved in this experiment, Bob can analyze
them too, and he should predict the SAME result even if his analysis is a
bit different--for example, if Alice is standing on a platform and lets go
of a ball, the ball will hit the platform, from Alice's point of view this
is due to a gravitational force and from Bob's point of view this is due to
the platform accelerating up towards the ball, but either way the actual
prediction is the same. So, to say that Bob should observe the same results
of any local experiment (provided he is approximating everything to first
order) regardless of whether he's moving inertially in flat spacetime or
free-falling in gravity is physically equivalent to saying Alice should
observe the same results of any local experiment (again ignoring
second-order and higher effects) regardless of whether she's accelerating
through Bob's region in flat spacetime, or passing through his region
because he's in free-fall while she is not (say, she's standing on a
platform resting on a pole embedded in the Earth below, while Bob falls
past her).

Jesse

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