Re: Experiments on the validity of Relativity

Il 19/05/2024 00:40, Luigi Fortunati ha scritto:

[[Mod. note --
In your animation, while A and B are initially stationary (before the
animation starts):
* in the coordinate system of your animation, A's center of mass
   has zero coordinate acceleration
* in the coordinate system of your animation, B's center of mass
   has zero coordinate acceleration
* A's center of mass has nonzero proper acceleration
* B's center of mass has nonzero proper acceleration

These two latest statements are equivocal.

You say that the centers of mass A and B (which are stationary) have
their nonzero proper acceleration and this is in contrast with what I
have studied, that is, that acceleration is the variation of velocity.

So can you explain to me how two points whose velocity is always the
same (i.e. zero) can have an acceleration (whether proper or not)?

Can there be non-zero acceleration if velocity is zero and remains zero?

Obviously not.

So, your "proper acceleration" is not a real acceleration.

Luigi Fortunati

[[Mod. note --
*Coordinate* acceleration is indeed "the variation of velocity".

But, *proper* acceleration is something different:  An observer's
*proper* acceleration is defined as the acceleration measured by an
(ideal) accelerometer she carries with her.  Another way to say this
is that an observer's proper acceleration is her acceleration
*relative* to a freely-falling inertial reference frame (IRF) at her
current location.

Another way to think of this is to imagine that the observer carries
a small rocket with her.  The observer should let go of the rocket, so
that there are no forces acting on the rocket except for the rocket's
own thrust and any gravitational fields, and then adjust the rocket
thrust until the rocket matches the observer's motion (i.e., the rocket
remains stationary *relative to the observer*).  If the rocket thrust
required to keep the rocket stationary *relative to the observer* is
$F$, then the observer's proper acceleration is $F/m$, where $m$ is the
(current) mass of the rocket.


Before considering your two planets, it might be useful to first
consider a few simpler situations.

#1. If you're in free-fall, what's your proper acceleration?
    In this case you and the freely-falling IRF are both freely-falling
    at the same location, so your acceleration with respect to that IRF
    is zero, and your rocket doesn't need any thrust to match your motion
    and stay stationary next to you.  This means that your proper
    acceleration is zero.  This is important: the proper acceleration
    of a freely-falling observer is *zero*.

#2. If you are on the Earth's surface, stationary with respect to the
    Earth's surface), what is your proper acceleration?  To answer this,
    we need to think about what a freely-falling IRF's motion would be
    at your location.  That IRF would be accelerating downwards at (about)
    9.8 m/s^2 relative to the Earth's surface.  Therefore,
    *relative to that freely-falling IRF*, you (stationary with respect
    to the Earth's surface) are accelerating *upwards* at 9.8 m/s^2.
    Equivalently, to hang stationary in mid-air next to you (not
    accelerating with respect to you), your rocket needs to be thrusting
    *upwards* so as to deliver an acceleration of 9.8 m/s^2 *upwards*
    relative to free-fall.  This last sentence is important, so let me
    repeat it: in order to hang stationary in mid-air next to you, your
    rocket needs to be thrusting *upwards* so as to deliver an acceleration
    of 9.8 m/s^2 *upwards relative to free-fall* (at your location).
    Therefore, your proper acceleration is 9.8 m/s^2 *upwards*.


Now let's consider your two planets A and B, in the initial condition
where they are held apart (stationary) by some external forces.  (E.g.,
maybe there's a strut between them; the strut is in compression due to
mutual gravitational attraction of A and B.)

This situation is precisely analogous to my scenario #2 above.  We've
said that A and B are being held apart by external forces, so they're
stationary with respect to each other.  This means that A's center of
mass is *not* in free-fall.

If you are located at ("holding on to") A's center of mass, what does
your accompanying rocket need to do to stay stationary next to you?
(Remember, we've said that the only forces acting on the rocket are
it's own thrust and any gravitational fields.)  In order to stay
stationary next to you, your accompanying rocket needs to thrust
directly *away from B* in order to counteract B's gravitational field.

Therefore, your proper acceleration (which is exactly the same as the
proper acceleration of A's center of mass) is nonzero.  In fact, your
(and A's-center-of-mass's) proper acceleration points *away from B*,
and (assuming A and B are both spherical) has a magnitude $GM/R$ where
$G$ is the Newtonian gravitational constant, $M$ is the mass of planet B,
and $R$ is your distance from B's center of mass.
-- jt]]