Re: Newbie question about gravity

On May 22, 8:35 pm, "Bill Hobba" <rubb...@xxxxxxxx> wrote:
"Phil" <cms...@xxxxxxxxxxxxx> wrote in message
Bill,

Consider an object at perigee of a highly eccentric orbit, it will
rise above an orbit of zero eccentricity (which intersect with
perigee) with a relative acceleration proportional to [(V_he)^2 - (2GM/
r)]/r , where Vhe is the velocity of the object in the highly
eccentric orbit at perigee, 2GM/r is the square of the velocity of an
object which orbit is of zero eccentricity and intersects perigee of
the he orbit. So lets consider the physical situation where the
coordinate origins are congruent with perigee of the highly elliptical
orbit. The relative acceleration of the comoving coordinate systems
can't be explained by tidal effects.

For one thing, at perigee they are at the same potential, even so, the
relative acceleration of the coordinate systems are at a maximum at
perigee when the tidal effects are precisely zero.

Tidal effects can never be made zero by any method whatever -
that is precisely what is meant by space-time curvature - even at a point the
curvature is not zero if gravity is present.

I would agree that curvature is local but would you consider the
curvature at a point to be an expression of tidal acceleration? On
the other hand, it does occur to me that gravity need not result in
curvature in the case of a uniform field and that makes sense to me.
Is this what you mean, particularly, curvature/tidal acceleration is a
description of the nonuniformity of the field?

The precise statement is that
locally, by making the region smaller and smaller, we can find a coordinate
system that to a better and better approximation is inertial. It is like a
plane tangent to a sphere at a point. The curvature of the sphere is not
altered in any way by the tangent plane but the smaller the region about
that tangent point considered the more and more it can be considered as
being the same as the plane.

Then wouldn't it also be true that "making the region smaller"
accomplishes the same thing when accelerating in flat space-time? I
mean, aside from my accelerometer, how can I distinguish a curved path
in flat space-time from a straight path in curved space-time (wrt to
observations of dv/dt)? So provided accelerated frames reduce the
domain small enough, are they not also equally valid for the practice
of the STR?

Clearly, in a gravity field, all local freely falling frames
accelerate _equally_ wrt to a gravity centered non-rotating frame of
reference. Because this is so, all these freely falling coordinate
systems accelerate with respect to each other no matter how small the
domain because even differentials can relate proportionately to one
another even while they seem zero to observation. For example, twice
the curvature, twice the _finite_ number for dv/dt, right? And twice
the infinetismal number for 1/2 (dv/dt)dt^2, right?

One thing I could really use some help with is how one applies the
concept of curvature without knowledge of the field itself _and
especially_ its context in the field's own gravity centered non-
rotating coordinate system. For example, if I need to know the Mass
of the object causing the local field or my distance from it, or my
motion wrt to this system, or even which way is "down" or which way is
"up", then the relationship would seem tied to some special coordinate
system. Even so, it is quite unclear to me how the relationship can
adequately be described with out knowledge and use of the field's own
non-rotating system of coordinates. How does one adequately make
predictions without knowledge of the field's own coordinate system and
how one's own system of coordinates relate to it?

.

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