On Monday, December 18, 2017 at 8:36:29 PM UTC, Brent wrote: > > > > On 12/17/2017 2:39 PM, [email protected] <javascript:> wrote: > > > > On Sunday, December 17, 2017 at 12:21:27 AM UTC, Brent wrote: >> >> >> >> On 12/16/2017 2:59 PM, [email protected] 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 [email protected]. To post to this group, send email to [email protected]. Visit this group at https://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
