On Tuesday, December 19, 2017 at 3:32:22 AM UTC, Brent wrote: > > > > On 12/18/2017 6:36 PM, [email protected] <javascript:> wrote: > > > > On Monday, December 18, 2017 at 8:48:08 PM UTC, Brent wrote: >> >> >> >> On 12/18/2017 12:19 AM, [email protected] wrote: >> >> >> >> On Sunday, December 17, 2017 at 10:39:18 PM UTC, [email protected] >> 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 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* >>> >> >> *Einstein must have figured that since gravity produces an acceleration >> field, and accelerating test particles move along curved paths in >> space-time, he could replace acceleration by inertial paths in a space-time >> curved by the presence of mass-energy. But now, when comparing test >> particles moving along different paths in space-time, he couldn't use the >> Lorentz transformation because the relative velocities of the frames are >> not necessarily constant. So how did he propose to find the correct >> transformation equations, and what are they? And what were the laws of >> physics, in this case gravity, that had to be invariant? AG* >> >> >> What's invariant is the measure along a path in spacetime - it's what an >> ideal clock measures. The relation between the measure along two different >> paths obviously depends on the lumpiness of the spacetime through which >> they travel. It's as if I headed north thru the Sierras while you sailed >> up the coast. There's no simple relation between our path lengths even if >> we travel between the same two points. >> > > *So what's invariant along along two paths with the same endpoints? * > > > It's not about two paths. The length of each path as measured using > Einstein's theory of the metric (i.e. as warped by mass-energy) is an > invariant. Just as the distance your car's odometer would measure driving > from NY to LA, it's some number and it depends on (a) the path you took and > (b) the topography along that path. The interesting point is that two such > paths between a pair of events are different durations as measured by > clocks carried along the trips. That's contrary to Newton, for whom time > was an invariant. > > *Not clear from what you write. But whatever it is, why is that deemed to > be invariant? * > > > Because it doesn't depend on what reference system you use in spacetime. > It's measuring a distance which is a real thing, not something > relative/subjective. > > *Shouldn't it be the laws of physics, in this case gravity, and hence the > field equations? AG * > > > It's the basis for them. They can be written in terms of an extremal > principle for the invariant path lengths. >
*Is this the method Einstein used to derive the field equations? This is one of my key interests in this subject; to understand the method he used to derive the field equations. If so, why is invariant path lengths such a crucial condition? I agree that physics seeks invariants, but why this particular one? AG* The Lorentz transformation is just the simple limiting case of flat, smooth > spacetime. It's useful because in a sufficiently small local region > spacetime is going to be flat and smooth. > > 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] <javascript:>. To post to this group, send email to [email protected] <javascript:>. Visit this group at https://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout. -- 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.
