On Thu, Feb 13, 2014 at 12:22 PM, Edgar L. Owen <edgaro...@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 -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/groups/opt_out.
