Newton's 3rd law is wrong

I realize the enormity of the title of this article that refutes a law
that is more than three centuries old and that has never been contested
until now, but before presenting this work, I have carefully checked
all the concepts, numbers and formulas and I am ready to present this
article of mine in all other possible venues if it is rejected by
Physics Review.

Mine will be a logical, physical and mathematical demonstration.

Newton writes: If a horse pulls a stone tied to a rope, the horse is
also equally pulled towards the stone: in fact the rope stretched
between the two parts, by the same attempt to loosen, will push the
horse towards the stone and the stone towards the horse; and it will
impede the advance of the one by as much as it will promote the advance
of the other.

But this is not always true!

Indeed, it is true only in the one case in which the horse moves with a
rectilinear and uniform motion.

The explanation is simple.

The rope is subjected to two opposing forces, on one side there is the
action of the horse that pulls to the right and on the other there is
the reaction of the stone that pulls to the left.

If the two opposing forces are equal, the rope moves in a rectilinear
and uniform motion but if the horse accelerates, the rope also
accelerates and, therefore, during the acceleration the action of the
horse is necessarily greater (and not equal) to the reaction of the
stone and, therefore, what Newton says is true *only* when the horse
does not accelerate.

[[Mod. note -- You're missing some forces here.  If you account for
*all* the forces, Newton's 3rd law remains correct.

Let's look at this case (the horse, rope, and stone are all accelerating)
a bit more carefully.  For convenience of exposition, I'll take the
velocity and acceleration of all three bodies to be to the right.
And, I'll idealise the rope as having a constant length, i.e., as
not stretching.  (This means that the accelerations of the horse,
rope, and stone are all the same.)

Then we have the following free-body diagrams for the horizontal forces:

  <---stone------>   <------rope------->   <-------horse--------->

Each of the three objects has *two* distinct forces acting on it:
Stone:
  F_stone_left: friction of the ground on the stone
                (force pulling left to the left)
  F_stone_right: rope tension at the stone's end of the rope
                 (force pulling right on the stone)
Rope:
  F_rope_left = rope tension at the stone's end of the rope
                (force pulling left on the rope)
  F_rope_right = rope tension at the horse's end of the rope
                 (force pulling right on the rope)
Horse:
  F_horse_left = rope tension at the horse's end of the rope
                 (force pulling left on the horse)
  F_horse_right = force applied to horse by horse's legs acting on ground
                  (force pulling right on the horse)

By Newton's 2nd law,
  F_stone_right - F_stone_left = m_stone a (1)
  F_rope_right  - F_rope_left  = m_rope  a (2)
  F_horse_right - F_horse_left = m_horse a (3)
where /a/ is the (common) acceleration.

Since we've assumed /a > 0/ (acceleration to the right), we can infer
several things:

Since the stone is accelerating to the right, there must be a net force
to the right acting on it, i.e., we must have
  F_stone_right > F_stone_left. (4)

And, since the rope is accelerating to the right, there must be a net
force to the right acting on it, i.e., we must have
  F_rope_right > F_rope_left. (5)

And, since the horse is accelerating to the right, there must be a net
force to the right acting on it, i.e., we must have
  F_horse_right > F_horse_left. (6)

Notice that Newton's 3rd law only relates the forces of bodies that
are directly exerting forces on each other.  That is, Newton's 3rd law
says that
  F_stone_right = F_rope_left, (7)
and that
  F_rope_right = F_horse_left. (8)
But in general Newton's 3rd law does NOT itself say anything about the
relationship between F_stone_right and F_horse_left, becauase the stone
and the horse aren't directly exerting forces on each other.

Combining (4), (5), (6), (7), and (8), we have that
  F_horse_right > F_horse_left (this is just (6) again)
  = F_rope_right (by (8))
    > F_rope_left (by (5))
      = F_stone_right (by (7))
> F_stone_left (by (4)
i.e., we conclude that
  F_horse_right > F_stone_left (9)
In other words, the force-to-the-right applied to the horse by the
horse's legs acting on ground must be larger than the friction
force-to-the-left of the ground acting on the stone.  This is just
what we'd expect for acceleration to the right.

*If* we approximate the rope as having zero mass, then (2) says that
the rope tension is the same at both ends, i.e.,
  F_rope_right = F_rope_left. (9)
We can then combine (7), (8), and (9) to infer that
  F_stone_right = F_horse_left

Regardless of whether we approximate the rope as having zero mass,
or we treat the rope's mass as nonzero, in either case there's no
violation of Newton's laws.  In particular, in either case the forces
F_stone_right and F_horse_left *act on different objects* (the stone
and the horse, respectively).  Each object (still) has a net force
acting on it which points to the right.
-- jt]]

Newton also says: If some body, colliding with another body, will in
some way have changed with its force the motion of the other, in turn,
due to the opposing force, will undergo an equal change in its own
motion in the opposite direction.

[[Mod. note -- This statement is a bit ambiguous: I can't tell what
you mean by "motion".  Are you referring to velocity?  Acceleration?
Force?  Linear momentum?  Angular momentum?
-- jt]]

It is true and it is also obvious: the motions are modified in opposite
directions to the same extent because the sum of the forces that push
to the right is always exactly equal and opposite to the sum of the
forces that push to the left.

But if all this is true, it is not equally true that all the forces
that push to the right are actions of body A on body B and all those
that push to the left are reactions of body B on body A.

I will demonstrate this with two animations and the related
calculations.

In the first animation I will show what, in all likelihood, was the
main cause of the error that underlies the third law: believing that a
property valid for single particles can be extended with impunity also
to bodies.

In the animation https://www.geogebra.org/m/d667egkq we can (with the
appropriate sliders) collide and rotate the two particles A and B to
realize that the action and the reaction are *always* equal and
opposite because it is geometrically impossible to make the blue area
(action) greater or less than the red area (reaction): no particle A
can ever act on particle B more or less than particle B reacts on
particle A.

All true, as the third law prescribes.

But if we transfer this law from particles to bodies, it is no longer
correct and I demonstrate this with the second animation
https://www.geogebra.org/m/gd8uqyff where there is body A (mass m=2)
formed by the two particles A1 and A2 and body B (mass m=1) formed by
the particle B1 alone.

With the appropriate slider we can collide the two bodies to see how
(during the collision) the 4 forces F1, F2, F3 and F4 arise and grow
until the maximum compression.

The vector sum of all these forces is zero (and this is why the total
momentum is conserved) but not all these forces are actions and
reactions between bodies A and B.

So, what are the action and reaction forces between bodies A and B?

Certainly the forces F1=+1 and F2=-1 which cancel each other out are.

Certainly *not* is the force F4=-0.67 which is internal to body A and
does not concern body B.

The force F3=+0.67 is only at half so because the particle A2, pushing
towards the right, acts on everything it finds on the right and,
therefore, it is an internal force in body A for the part +0.33 which
pushes against the particle A1 and it is a force of action against body
B for the remaining part +0.33 which acts according to the 2nd law F=ma
against the particle B1 of body B which undergoes it and moves away
without reacting.

Therefore, mathematically, the forces of body A that act on body B are
2 for a total of +1.33 (the force F1 and half of the force F3), while
the reaction force of body B on body A is only one: the force F2=-1.

Physically it is even easier to explain.

During the collision, particle A1 can increase its thrust to the right
against particle B1 thanks to the help of the particle A2 behind it
which, with its thrust in the same direction, increases the force of A1
against body B, while particle B1 cannot do the same (it cannot
increase its reaction against body A) because it does not receive any
help from the particles behind it which are not there!

This is why the action of body A on body B (+1.33) is greater (and not
equal) to the reaction of body B on body A (-1).

Luigi Fortunati