On 14 February 2014 10:38, meekerdb <[email protected]> wrote: > On 2/13/2014 11:02 AM, Jesse Mazer wrote: > > 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. > > > As an example SR is used to calculate the size of proton bunches in the > LHC even though they are at 9999% of the speed of light and subject to > enormous acceleration at the turning magnets. >
Wow, and there I was thinking we'd never achieve FTL travel! :-) (Sorry) -- 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 [email protected]. To post to this group, send email to [email protected]. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/groups/opt_out.
