On Thu, Feb 13, 2014 at 7:41 PM, Edgar L. Owen <edgaro...@att.net> wrote:

> Jesse, Brent, Liz, et al,
>
> Free fall in a gravitational field is NOT acceleration. Standing on the
> surface of the earth IS acceleration because only then is the acceleration
> of gravity felt as such.
>

Yes, that's why I equated inertial motion in flat spacetime with freefall
in a gravitational field--"Bob" could be either one in my example. I also
equated accelerated motion in flat spacetime with non-freefall in a
gravitational field--"Alice" could be either one in my example, note where
I said she'd observe the same thing "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)."



> Now imagine that elevator is enormous, the size of a planet (but assume
> also in this thought experiment that it has no mass and thus has no
> gravitational effect).
>

The equivalence principle simply doesn't apply to large regions of space
where tidal forces can be observed, mathematically it only applies in the
infinitesimal neighborhood of a point in spacetime, though in practice if
your measuring instruments aren't too precise a reasonably small space like
an elevator should be OK (at least in the Earth's gravitational field--in
the gravitational field of a black hole even an elevator would be too large
because there'd be a significant tidal force between the top and bottom).
See the discussion about how tidal forces spoil any attempt to make the
equivalence principle work in a non-infinitesimal region at
http://www.einstein-online.info/spotlights/equivalence_principle

Jesse


> On Thursday, February 13, 2014 2:02:15 PM UTC-5, jessem wrote:
>>
>>
>>
>> On Thu, Feb 13, 2014 at 12:22 PM, Edgar L. Owen <edga...@att.net> wrote:
>>
>>> All,
>>>
>>> By the Principle of Equivalence acceleration is equivalent to
>>> gravitation.
>>>
>>
>> Too vague. A more precise statement is that in an observer in free-fall
>> in a gravitational field can define a "local inertial frame" in an
>> infinitesimally small neighborhood of spacetime around them, and that if
>> the laws of physics are expressed in the coordinates of this frame, they
>> will look just like the corresponding equations in flat SR spacetime,
>> though only in the first-order approximation to the equations (i.e.
>> eliminating all derivatives beyond the first derivatives). See for example:
>> http://books.google.com/books?id=ZfMWbQB2dLIC&lpg=PP1&pg=PA52 and
>> http://books.google.com/books?id=95Frgz-grhgC&lpg=PP1&pg=PA481 and
>> http://books.google.com/books?id=jjBMw0KFtZgC&lpg=PP1&pg=PA5
>>
>> Even though the curvature disappears in the first order terms, it remains
>> in the higher order terms, whereas curvature is really zero in all terms
>> for an accelerating observer in flat spacetime. So, the answer to your
>> question is that acceleration does not in itself cause spacetime curvature,
>> SR can handle acceleration just fine as discussed at
>> http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html ,
>> but this isn't a violation of the equivalence principle since the
>> mathematical formulation of the principle deals only with first-order terms.
>>
>> Jesse
>>
>>
>>  --
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