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.

So please take this into consideration and respond.

Edgar



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<javascript:>
> > 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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