On Tuesday, December 19, 2017 at 3:32:22 AM UTC, Brent wrote: > > > > On 12/18/2017 6:36 PM, agrays...@gmail.com <javascript:> wrote: > > > > On Monday, December 18, 2017 at 8:48:08 PM UTC, Brent wrote: >> >> >> >> On 12/18/2017 12:19 AM, agrays...@gmail.com wrote: >> >> >> >> On Sunday, December 17, 2017 at 10:39:18 PM UTC, agrays...@gmail.com >> 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 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 everything-li...@googlegroups.com <javascript:>. To post to this group, send email to everyth...@googlegroups.com <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 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.
