In the formula F=ma force is related to acceleration.

If acceleration exists in all references, the force is real, otherwise,
if it exists in one reference and "disappears" in the other, then the
force is "apparent".

On the other hand, in the formula F=m1*m2/d^2 there is NO acceleration
and this is the difference between the second formula and the previous
one.

In F=m1*m2/d^2 the force depends exclusively on the presence of the two
masses m1 and m2 (which always exist) and on the square of their
distance (which always exists).

Therefore, if the masses m1 and m2 and the distance d^2 never disappear
(in any reference) the force F=m1*m2/d^2 is always real.

And what is the force F=m1*m2/d^2? It's gravity.

So gravity is not "apparent" because (the formula says so and not me)
its force does not disappear by changing the SDR.

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In article <s4qdb4$1tfg$1@gioia.aioe.org>, Luigi Fortunati <fortunati.luigi@gmail.com> writes:

In the formula F=ma force is related to acceleration.

If acceleration exists in all references, the force is real, otherwise,
if it exists in one reference and "disappears" in the other, then the
force is "apparent".

You can always transform an acceleration away by using an accelerating reference frame

On the other hand, in the formula F=m1*m2/d^2 there is NO acceleration
and this is the difference between the second formula and the previous
one.

It looks like you forgot the gravitational constant; that gives the
right-hand side the dimensions of ma.

In F=m1*m2/d^2 the force depends exclusively on the presence of the two masses m1 and m2 (which always exist) and on the square of their
distance (which always exists).

And on the gravitational constant.

Therefore, if the masses m1 and m2 and the distance d^2 never disappear
(in any reference) the force F=m1*m2/d^2 is always real.

Acceleration is always "real". (That is actually still a puzzle:
acceleration relative to what? As far as I know, no-one has ever shown quantitatively that Mach's principle can explain that.)

And what is the force F=m1*m2/d^2? It's gravity.

Right, which is why the expression usually includes the gravitational
constant.

So gravity is not "apparent" because (the formula says so and not me)
its force does not disappear by changing the SDR.

An acceleration is real; you can feel it. You can find an accelerating reference frame and transform it away (at least locally).

The essence of general relativity is that gravitation, in some sense, IS acceleration, but as far as I know no-one knows why the gravitational
constant has a non-zero value or has the value it has (apart from weak-anthropic explanations).

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On 10 Apr 2021 09:40:11 GMT, Luigi Fortunati wrote:

Let me try a simpler, and clearer, answer.

In the formula F=ma force is related to acceleration.

I always write this formula as a=F/m in order to stress the fact that if a force F acts on a body/particle of mass m, then the particle will get an acceleration a given by a=F/m. This is the Newton's second law.

F is the CAUSE (of the acceleration) and a is the EFFECT. I put the effect
on the left hand side, and the cause on the right hand side,

I DO NOT take the formula F=ma to be DEFINITIONS for F !!!

In a=F/M, F is ANY force, of any nature, coming from various sources, for example:

- a spring acting on the body or particle: k*x (x is the displacement from
the equilibrium position
- a rope - F = tension in the rope
- a surface with which the body is in contact
- an static electric force (Coulomb) Q1*Q2/d^2
- or gravitational force: m1*m2/d^2
- or friction, etc.

Correspondingly, if on the body/particle acts, for example, only a spring,
then the Newton's second law gives that the acceleration will be
a=F/m=k*x/m

Please note that the Newton's second law is only valid with respect to
certain frames of reference. But about this - maybe later.

IF you believe this was useful for you, please let me know and I will
answer your questions and then I will continue.

If acceleration exists in all references, the force is real, otherwise,
if it exists in one reference and "disappears" in the other, then the
force is "apparent".

On the other hand, in the formula F=m1*m2/d^2 there is NO acceleration
and this is the difference between the second formula and the previous
one.

In F=m1*m2/d^2 the force depends exclusively on the presence of the two masses m1 and m2 (which always exist) and on the square of their
distance (which always exists).

Therefore, if the masses m1 and m2 and the distance d^2 never disappear
(in any reference) the force F=m1*m2/d^2 is always real.

And what is the force F=m1*m2/d^2? It's gravity.

So gravity is not "apparent" because (the formula says so and not me)
its force does not disappear by changing the SDR.

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Alex <alexlREMOVE@removesolnet.ch> wrote:

In the formula F=ma force is related to acceleration.

I always write this formula as a=F/m in order to stress the fact that if a force F acts on a body/particle of mass m, then the particle will get an acceleration a given by a=F/m. This is the Newton's second law.

I prefer to write: dv/dt = F/m

F is the CAUSE (of the acceleration) and a is the EFFECT. I put the effect
on the left hand side, and the cause on the right hand side,

You might therefore like this paper:

"How to be causal: time, spacetime, and spectra",
Eur. J. Phys. 32, 1687?1700 (2011)
doi:10.1088/0143-0807/32/6/022

with a longer version at arxiv:1106.1792


Mind you, some physicsts find the proposal intensely annoying,
and they disagree with it quite strongly - despite its deliberately
limited scope (i.e. on the interpretation of temporal differential
equations). There are thus several extra appendices in the arxiv
version addressing various points raised.

#Paul

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