Re: Why Einstein geometrized away Newton's "force of gravity"

On Feb 26, 8:34 pm, Tom Roberts <tjroberts...@xxxxxxxxxxxxx> wrote:
Daryl McCullough wrote:
pmb says...
I'll put this [something about "inertial forces"] into
tensor form in the future.

I just don't understand what *benefit* there is in talking
about inertial forces. Mathematically, the difference is
a triviality: In the equation
m (d/dtau U^u + Gamma^u_vw U^v U^w) = F^u
you can, if you like, move the Gamma term to the other side
of the equation to get:
m d/tau U^u = F^u - m Gamma^u_vw U^v U^w
Then you can call the right side F_total^u.
but what is the advantage of doing that?

IMO.......
In the geodesic equation, m=0 is carefully
specified, it's spec'd only for acceleration,
NOT FORCE. In the approximate, planets wrt the
Sun are regarded as zero mass points, otherwise
one needs Guv=Tuv consideration.

I don't see any advantage to this algebraic manipulation. But there is a
major loss: the equation no longer reflects the underlying symmetry of
the theory. The original equation was written in terms of tensors
(before projecting it onto some coordinate system's basis), was
manifestly invariant under coordinate transforms, and directly and
manifestly displayed that symmetry of the theory. This last equation
loses that completely -- no term in it represents a tensor (i.e. obeys
the underlying symmetry of the theory), except the original F^u.

There is NO "F^u" (absolute aka proper acceleration
in GR), that's why it's called GR.
So use,

(d/dtau U^u + Gamma^u_vw U^v U^w) = F^u =0

and Pete standing on his bathroom scale,

d/dtau U^u =0 , therefore,

Gamma^u_vw U^v U^w = 0 , when Pete stands on his

bathroom scale, relative to Earth coordinates.

Why is that so difficult?
Regards
Ken S. Tucker
.

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