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

On Feb 27, 4:16 am, "Ken S. Tucker" <dynam...@xxxxxxxxxxxx> wrote:
On Feb 26, 8:34 pm, Tom Roberts <tjroberts...@xxxxxxxxxxxxx> wrote:

Daryl McCullough wrote:
pmbsays...
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

The geodesic equation is not defined for m = 0 (i.e. not defined by
setting proper mass to zero). It’s defined by setting the 4-
acceleration of a particle to zero

specified, it's spec'd only for acceleration,
NOT FORCE.

Geodesic => 4-force/4-acceleration = 0
Geodesic => inertial force/coordinate acceleration != 0


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

No. This is wrong. The mass of the particle is not in this equation
and its should represent covariant components, not contravariant
components.


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.

When Pete stands on his bathroom scale (and gets depressed after
seeing he gained 10 lbs) the gravitational force is not zero.
Therefore dP^u/dtau is not zero.

Pete
.

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