: That the Reimann tensor (not curvature tensor) does not vanish is irrelevant. : - I disagree. Spacetime curvature

Reimann tensor not curvature you're trying to force a paradigm I don't share.

:and tidal gravity are the same thing.

Nonsense, you're thinking globally. In order to understand GR you must think locally.

:Mathematically, : the curvature tensor (i.e. the Reimann tensor)

The Reimann tensor(NOT curvature tensor) you're trying to force a paradigm I don't share.

:can be calculated directly from the tidal force tensor.

There is no such thing as a tidal force "tensor". I suppose you can define one as gmn;sl, but you can't get the Reimann tensor from that. You can get it from gmn,s and gmn, but gmn,s is not a tensor.

:Thus local (i.e. infinitely small) measurements of tidal force can be : used to determine the curvature tensor.

You misunderstand the term local. Local does not mean small. It means near the origin of a particular coordinate frame where you put an observer and for a short enough time interval to consider the gravitational field of the frame static. You are doing remote(ultimately global) measurements on tidal force, then deducing the Reimann tensor. This you may do, but you must know the tidal force globally, and it must be the global force in that particular frame because tidal force is not a tensor. If you do a remote measure on a tidal force you do not always get the same value as a local inertial frame does(which is always zero.

:If the tidal force tensor

There is no such tensor.

:vanishes then the : curvature tensor vanishes.

No, If the tidal force vanishes globally then the Reimann tensor vanishes.

:However the existence of curvature

A nonzero Reimann tensor, you're trying to force a paradigm I don't share.

:is an invariant property of : spacetime so the tidal force tensor cannot vanish.

There is no such "tensor" and a globally zero Reimann tensor does not imply no tidal force. Tidal force is NOT a tensor. Take the gravitational field for a uniform rocket acceleration:

ds2 = (1 - az/c2)2dct2 - dz2 - dr2 - r2df2 The Reimann tensor here is zero, but In this frame(not all others of course) there are remote tidal forces. To realize this one simply needs to realize that this frame has an event horizon and so to get an object with mass to hover just above it according to this frame requires a force that diverges as we consider locations approaching it.

:Any body of finite size will be : subject to tidal distortions even in locally inertial frames.

Not according TO the local inertial frame. Only according to remote ones.

:Locally inertial means that : the metric, guv, is the Minkowski metric nuv. So in a locally : inertial frame guv = nuv. But this does not imply that that : curvature tensor vanishes.

Reimann tensor not curvature tensor. Your trying to force a paradigm I don't share.

:This also implies that the tidal force tensor does not vanish.

There is no such tensor so it implies no such thing.

: In fact an observer in orbit around a black hole will feel the tidal forces.

Not under a short enough proper time to be able to consider the frame local, the free falling observer then won't experience any tidal force.

:Even if the : observer is very small e.g. 10 cm.

: Since the question regarded the distortion of objects it is assumed that the object is : not a point particle.

SO what?

• Re: Does velocity play a factor? - tOM - November 16, 1999 - 03:51 UTC
  • Re: Does velocity play a factor? - And also From Me - November 16, 1999 - 03:54 UTC
    • Re: Does velocity play a factor? - Zephram_Cochrane/">Zephram Cochrane - November 16, 1999 - 08:15 UTC
    • Re: Does velocity play a factor? - Zephram_Cochrane/">Zephram Cochrane - November 16, 1999 - 08:24 UTC