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

On Feb 27, 1:28 pm, "Ken S. Tucker" <dynam...@xxxxxxxxxxxx> wrote:
On Feb 27, 4:45 am,pmb<pm...@xxxxxxxxxxx> wrote:





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.

Pete, recall the Equivalence Principle (EP), and
find the gravitational acceleration is independant
of the nature of mass, (+/- or zero mass), use an
accelerating elevator to observe things falling.

Yes. The gravitational acceleration is independent of the mass. But
that’s not what you wrote. In essence what you wrote is Force =
acceleration. That is wrong.

To simplify I’ll explain in Newtonian terms; the gravitational force
on particle of mass m is given by

F = Mg

Where M = passive gravitational mass of particle and g is the
acceleration due to gravity. Newton’s law says

F = ma

Where m = inertial mass of particle and a = acceleration of the
particle. Thus we have

Mg = ma

Equivalence principle: passive gravitational mass = inertial mass --->
M = m

a = g

What I was saying above is that you attempted to write

F = a = g

And I suggested that you write F = mg = mg

Do you think there is a difference, between an
ordinary derivative dP^u/ds and the partial diff
&P^u/&s.

Yes.

Pete
.