On Monday, December 18, 2017 at 8:36:29 PM UTC, Brent wrote: > > > > On 12/17/2017 2:39 PM, agrays...@gmail.com <javascript:> wrote: > > > > On Sunday, December 17, 2017 at 12:21:27 AM UTC, Brent wrote: >> >> >> >> On 12/16/2017 2:59 PM, agrays...@gmail.com wrote: >> >> There's a problem applying SR in this situation because neither the >> ground or orbiting clock is an inertial frame.AG >> >> >> An orbiting clock is in an inertial frame. An inertial frame is just one >> in which no forces are acting (and gravity is not a force) so that it moves >> with constant momentum along a geodesic. Although it's convenient for >> engineering calculations, from a fundamental veiwpoint there is no separate >> special relativity and general relativity and no separate clock >> corrections. General relativity is just special relativity in curved >> spacetime. So clocks measure the 4-space interval along their path - >> whether that path is geodesic (i.e. inertial) or accelerated. >> > > *Interesting way to look at it. So free falling in a gravity field is an > extension of SR. But the thing I find puzzling is that in GR the curvature > of space-time is caused by the presence of mass, yet I can draw the path of > an accelerated body as necessarily a curve in a space-time diagram. I am > having trouble resolving these different sources of curvature. AG* > > > An accelerated body, i.e. one a force is acting on (a rocket, you standing > on the ground) is following a curved path that is more curved than the > "straightest" path. I put "straightest" in scare quotes because in the > curved spacetime the "straight" path is a geodesic which is still > curved...it's just the straightest possible path in the given spacetime. > > It is not true that "I can draw the path of an accelerated body as > necessarily a curve in a space-time diagram". >
*I was referring to flat space-time, for example as a two dimensional representation using x and t coordinates. If one models an accelerating particle, it necessarily moves on a curved path. So I think this is very suggestive; that one can substitute acceleration for space-time curvature induced by gravity or mass-energy, insofar as gravity produces an acceleration field. AG * > In general, if you drew a straight line in some coordinate representation > of a curved spacetime, it would correspond to an accelerated (non-geodesic) > path. Imagine drawing a straight line past the Earth. It would take > thrust to fly a rocket along that path. Of course you could construct a > coordinate system around the Earth such that straight lines on the diagram > corresponded to geodesics, but it would be a very messy and distorted > coordinate system. > > Brent > -- 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 https://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
