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).

On 16-Sept-24 3:01 pm, Luigi Fortunati wrote:
> 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.

The only way to refute a theory without the use of experimental evidence
is to show that it is internally inconsistent.

Since you have provided neither experimental evidence nor a
demonstration of an internal inconsistency, what you've presented is not
a refutation.

Sylvia.

Luigi Fortunati il 15/09/2024 19:01:35 ha scritto:
> [[Mod. note -- 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)

This alternation of greater and equal cannot be correct because where there is "greater" it means that there are net forces (and accelerations) and where there is "equal" there are not.

The three bodies move as a single body and, therefore, nowhere can there be areas (small or large) that accelerate together with areas that do not accelerate.

> [[Mod. note -- *If* we approximate the rope as having zero mass, then (2) (F_rope_right - F_rope_left = m_rope a)
> says that
> the rope tension is the same at both ends, i.e.,
> F_rope_right = F_rope_left. (9)

No! If the mass decreases, (2) says something else.

It says that F_rope_right - F_string_left = m_rope_verysmall*acceleration, because very small and zero are not the same thing at all!

You cannot make the acceleration disappear by decreasing the mass!

The rope, no matter how small its mass, always accelerates exactly as much as the horse and the stone.

> 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]]

This sentence is not mine, it is Newton's and you can find it in his book "Principles of natural philosophy" under "law III".

He means to say (I think) that: "the collision of body A determines a variation of the linear motion of body B equal to the variation of the linear motion of body A in the opposite direction".

Luigi Fortunati

> Luigi Fortunati il 15/09/2024 19:01:35 ha scritto:
>> [[Mod. note -- 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)
>
> This alternation of greater and equal cannot be correct because where
> there is "greater" it means that there are net forces (and
> accelerations) and where there is "equal" there are not.

Those formulas are for situation where there is acceleration and
therefore net forces. The equalities only apply to forces from the
ends of the same interaction. The force at on the horse side end of
the rope must be equal to the froce on the rope side end of the horse
because there is no mass between the horse and the rope. Likewise
there is no mass between the rope and the stone.

> The three bodies move as a single body and, therefore, nowhere can
> there be areas (small or large) that accelerate together with areas
> that do not accelerate.

They do not move like a single rigid body. In particular, there are areas
of the horse that do not accelerate (hoofs when they touch the ground) and
areas that do accelerate (hoofs when they don't touch the ground). When
the force in the rope vaires the length of the rope varies so the
accleretions at the two ends of the rope differ.

[[Mod. note -- In my analysis I idealized the rope as non-stretching,
so that the stone, rope, and non-hoof parts of the horse all share
a common acceleration. -- jt]]

>> [[Mod. note -- *If* we approximate the rope as having zero mass, then
>> (2) (F_rope_right - F_rope_left = m_rope a)
>> says that
>> the rope tension is the same at both ends, i.e.,
>> F_rope_right = F_rope_left. (9)
>
> No! If the mass decreases, (2) says something else.

The accleration of the rope is roughly constant and fully determined
by the acclereations of the horse and stone. Therefore, wen the mass
decreaces so does the difference of F_rope_right and F_rope_left. The
difference is zero if the mass of the rope is zero, regardless of
acceleration.

> ...

>> 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]]
>
> This sentence is not mine, it is Newton's and you can find it in his
> book "Principles of natural philosophy" under "law III".

It isn't Newton's as Newton wrote his "Principia" in Latin. It isn't
Motte's, either, but uses the word "motion" in the same sense. The
full term is "quantity of motion" but Motte uses plain "motion" as a
shorter term for the product of mass and speed. In today's English the
usual word is "momentum".

> He means to say (I think) that: "the collision of body A determines a
> variation of the linear motion of body B equal to the variation of the
> linear motion of body A in the opposite direction".

Instead of "variation" Motte uses "change". Newton's focus is on initial
and final states although conservation of momFentum also apply during
the collision.

--
Mikko

Mikko il 21/09/2024 19:21:13 ha scritto:
>> Luigi Fortunati il 15/09/2024 19:01:35 ha scritto:
>>> [[Mod. note -- 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)
>>
>> This alternation of greater and equal cannot be correct because where there is "greater" it means that there are net forces (and accelerations) and where there is "equal" there are not.
>
> Those formulas are for situation where there is acceleration and
> therefore net forces. The equalities only apply to forces from the
> ends of the same interaction.

Who tells you that equalities (where there are no net forces and not even acceleration) only apply to the ends of the rope?

I'll tell you: the 3rd law tells you and you repeat it trusting that the ends of the rope have particular characteristics that allow them not to accelerate while all the other points of the rope accelerate.

But this is not true at all because the ends of the rope accelerate exactly like everything else, proving that on the entire rope (ends included) a force acts to the right (that of the horse) greater than that which acts to the left (that of the stone).

> The force at on the horse side end of
> the rope must be equal to the froce on the rope side end of the horse
> because there is no mass between the horse and the rope. Likewise
> there is no mass between the rope and the stone.

This is really bizarre!

What does it mean that there is no mass at the end of the rope?

What is there that acts as a bond between the rope and the horse and between the rope and the stone: a hole? A void?

It is obvious that the bond is between mass and mass, without jumps and without interruptions.

>> The three bodies move as a single body and, therefore, nowhere can there be areas (small or large) that accelerate together with areas that do not accelerate.
>
> They do not move like a single rigid body. In particular, there are areas
> of the horse that do not accelerate (hoofs when they touch the ground) and
> areas that do accelerate (hoofs when they don't touch the ground). When
> the force in the rope vaires the length of the rope varies so the
> accleretions at the two ends of the rope differ.

I did not say that they move as a *rigid* body but as a *single* body where there are no interruptions, no voids and no detached parts.

> [[Mod. note -- In my analysis I idealized the rope as non-stretching,
> so that the stone, rope, and non-hoof parts of the horse all share
> a common acceleration. -- jt]]
>
>>> [[Mod. note -- *If* we approximate the rope as having zero mass, then (2) (F_rope_right - F_rope_left = m_rope a)
>>> says that
>>> the rope tension is the same at both ends, i.e.,
>>> F_rope_right = F_rope_left. (9)
>>
>> No! If the mass decreases, (2) says something else.
>
> The accleration of the rope is roughly constant and fully determined
> by the acclereations of the horse and stone. Therefore, wen the mass
> decreaces so does the difference of F_rope_right and F_rope_left. The
> difference is zero if the mass of the rope is zero, regardless of
> acceleration.

All this that you are saying here has nothing to do with the acceleration of the rope that depends *exclusively* on the horse and the stone.

The rope accelerates to the right *exclusively* because the horse's force (action) pulls it to the right *more* than the reaction of the stone pulls it to the left.

Otherwise, the rope would go forward at a constant speed, without accelerating.

Obviously.

Luigi Fortunati

Newton says: If a horse pulls a stone tied to a rope, the horse is also
pulled equally towards the stone.

It's like tug-of-war: someone pulls the rope to the right, someone else
pulls it to the left.

The poor rope is at the mercy of whoever pulls more: how do we know who
pulls more? Let's look at the rope!

If the rope moves at a constant speed, then they both pull with the
same force.

If the rope accelerates to the right, then the force that pulls to the
right pulls more and if it accelerates to the left it is the opposite.

No one would ever dream of saying that the opposing forces that pull
the rope to the right and to the left are always equal and opposite!

Only Newton's law III says so.

By the 3rd law, no one would ever win at tug-of-war!!

Luigi Fortunati

[[Mod. note -- As I explained in a moderator's note on 2024-Sep-16,
you're misunderstanding what Newton's 3rd law says. Newton's 3rd law
says that
(1) The force the rope applies to the stone is equal in magnitude and
opposite in direction to the force the stone applies to the rope,
AND
(2) the force the horse applies to the rope is equal in magnitude and
opposite in direction to the force the rope applies to the horse.

But Newton's 3rd law does NOT say that (1) = (2). In fact, if the horse,
rope, and stone are all accelerating to the right, then (2) > (1), with
the difference (which is typically fairly small) being given by
m_rope * a = (2) - (1)
-- jt]]

Moderator 28/09/2024 13:19:13 ha scritto:

> [[Mod. note -- As I explained in a moderator's note on 2024-Sep-16, you're misunderstanding what Newton's 3rd law says. Newton's 3rd law says that (1) The force the rope applies to the stone is equal in magnitude and opposite in direction to the force the stone applies to the rope, AND (2) the force the horse applies to the rope is equal in magnitude and opposite in direction to the force the rope applies to the horse.

This is why the third law error has never been discovered.

Because we trusted action separated from reaction, as if action could act here and reaction (independently) there.

But no, action always fights against reaction and tries to make its own reasons prevail over those of the other.

The action of one body does not go against another body but against the reaction of the other body and vice versa.

The difference is fundamental because in an indirect clash there is never a winner, while in a direct clash there is (usually) a winner.

In a direct clash with reaction, action wins if it is stronger and loses if it is weaker.

But as usual, I can express myself better with an example taken from real life.

We have to build a road but there is a large stone that is in the way and, therefore, we have to find a way to move it.

Let's take a thick rope, tie one end to the horse and the other to point P of the stone.

Our idea is that, if we can move point P of the stone, we will also be able to move the entire stone.

We encourage the horse to move, it pulls the rope and the rope pulls point P which does not move.

What happened? What happened is that (at point P) the force coming from the horse (which is equal to +8) is not enough because the friction of the stone can only be overcome with a force greater than +10.

The horse does not make it and its action +8 on point P is cancelled out by the reaction -8 of the rest of the stone (always on point P) and point P does not move, just as neither the horse nor the stone move.

There are no net forces, there are no accelerations (the velocity remains equal to v=0) and the action of the horse meets a perfectly equal and opposite reaction everywhere.

Just as the third law states.

Then we replace the weak horse with a stronger pack horse that can exert a force equal to +12 and point P starts moving, that is, it accelerates from speed v=0 to speed v>0.

What happened? What happened was that, at point P, the action +12 of the horse prevailed over the reaction -10 of the rest of the stone and point P moved forward together with the whole stone!

This is how action and reaction work, with a direct clash where in the first case (with the weak horse) they tied and in the second case (with the stronger horse) the action won.

If it were as the third law states, how could we have moved the stone to build the road if at point P the rightward action of the horse was (always and in any case) equal and opposite to the leftward reaction of the rest of the stone?

It is at point P that the action of the horse and the reaction of the rest of the stone confront each other.

Luigi Fortunati

In the animation https://www.geogebra.org/m/h7zhvtkj there is a thin
sliding wall stopped at the point x=0 and there are two bodies A and B
whose masses we can make equal with the appropriate "Same mass" button
or variable with the appropriate slider.

When the two masses are equal, the sliding wall remains still both
before and after the collision, because the two opposing forces it
receives are equal.

When one mass is greater than the other, after the collision the
sliding wall starts moving (accelerates) in the direction of the
greater mass.

Click on the "Collision" button to see what happens at the moment of
the collision, where for just an instant two opposing blue and red
forces are activated.

Is it correct to say that the blue force acts on the wall and *also* on
body B, and that the red force acts on the wall and *also* on body A?

Luigi Fortunati

Arm wrestling: initially, hand A pushing to the right and hand B pushing to the left are still because the action and reaction are equal and opposite.

If the two opposing forces remained *always* equal (as Newton's third law states), they would never start moving to the right or even to the left: they would remain eternally still.

Ibex A and B fight horns against horns and are still because the action of ibex A to the right is perfectly equal and opposite to the reaction of ibex B to the left: how could they both start moving (accelerate) to the right or to the left if neither of the two forces prevailed over the other? They couldn't and they would remain eternally still!

Team A pulls the rope to the right with the same force with which team B pulls it to the left and they remain still until when? Until the balance of the two opposing forces is broken and one of the two forces prevails.

The rope between the horse and the stone moves with uniform speed until the force of the horse pulling to the right is equal and opposite to the friction of the stone on the ground pulling to the left but when its speed increases (acceleration) the force of the horse cannot be equal to the resistance of friction!

In short, the perfect balance between action and reaction exists only when the two bodies (together) are still or move with uniform and rectilinear speed but not when they accelerate!

The young student who comes across this discussion will ask himself: are these simple considerations right or wrong?

Since they are contrary to the teachings he receives, he should immediately think that Newton is right and, therefore, that these considerations must be easily contestable by those who know more.

But then he would also ask himself: how come the discussion ran aground without anyone being able to refute it?

And the doubt would remain.

Luigi Fortunati

I am not the first to say that Newton's third law is wrong.

Einstein said it before me (implicitly), with his General Relativity.

With my animation https://www.geogebra.org/m/v33hu4en I show what
Newton said.

Its gravity corresponds to the sum of the ratios between the mass <M>
(4) of the Earth and the mass <m> (2) of the Moon (numbers chosen for
ease of exposition): every single particle of the Earth interacts with
every single particle of the Moon, so that the total number (4) of the
red forces exerted by the Earth on the Moon is exactly equal to that of
the blue forces exerted by the Moon on the Earth (8 forces in total
corresponding to the product of the masses Mxm 4x2).

For this reason, according to Newton, the action of the Earth on the
Moon is perfectly equal and opposite to the reaction of the Moon on the
Earth, as prescribed by the third law.

Instead, Einstein argues that there is no force between the particles
of the Earth and those of the Moon, and that the action between the two
bodies is due to the space-time curvature of one in contrast to the
different space-time curvature of the other.

But the two curvatures are not equal!

And therefore, even for Einstein, the gravitational equality between
the two opposing bodies no longer exists, except in the one case in
which the two masses and the two curvatures are equal.

And never when they are different.

Luigi Fortunati

On 12/2/24 2:10 AM, Luigi Fortunati wrote:
> I am not the first to say that Newton's third law is wrong.

Hmmm. Your arguments are all wrong, because you keep omitting important
effects. (You have repetitively posted so many invalid arguments that
many/most participants around here simply ignore your posts.)

> Einstein said it before me (implicitly), with his General Relativity.

No, he did not. Your notions about GR are seriously wrong.

> With my animation https://www.geogebra.org/m/v33hu4en I show what
> Newton said.

I never click on such links.

> [...]
> Instead, Einstein argues that there is no force between the particles
> of the Earth and those of the Moon, and that the action between the two
> bodies is due to the space-time curvature of one in contrast to the
> different space-time curvature of the other.

That is not correct. Using the spacetime curvature interpretation of GR,
it is the curvature due to ALL components of the solar system that
determines all of their orbits. It is possible to only APPROXIMATELY
separate it as you assume, and that approximation destroys your argument.

> But the two curvatures are not equal!

Of course not! Granting your separation, their effects are not equal,
either. The effect of earth on moon is huge and causes the moon to orbit
around the earth. The effect of moon on earth is MUCH smaller, and
merely makes it wiggle a little bit as it orbits the sun.

Consider a solar system with just sun, earth, and moon (i.e. ignore
everything else). The earth does not orbit around an ellipse, not even
approximately -- it is the earth-moon barycenter that orbits around the
(approximate) ellipse you are thinking of. The earth wiggles around that
(approximate) ellipse. The wiggles have period ~ 29.5 days,
corresponding to the moon's orbit. (This is exact in Newtonian
mechanics, but only approximate in GR -- NM is linear while GR is not.)

> And therefore, even for Einstein, the gravitational equality between
> the two opposing bodies no longer exists, ]...]

Hmmm. Newton's third law discusses FORCES, not "gravitational equality".
In Newtonian mechanics, because earth and moon have such different
masses, the effects of equal forces on them are most definitely NOT
equal. In the spacetime curvature interpretation of GR there are no
gravitational forces, and one simply cannot apply any of Newton's laws.
(But one can apply the Newtonian approximation to GR, and all three of
Newton's laws apply within that approximation.)

Hint: it is outrageously arrogant to think you alone can see an error in
a theory that has stood the test of time for hundreds of years and
inspection by tens of thousands of physicists. There is a reason that no
journal articles have been published on this....

Tom Roberts

Tom Roberts il 12/12/2024 09:20:30 ha scritto:
>> With my animation https://www.geogebra.org/m/v33hu4en I show what
>> Newton said.
>
> I never click on such links.

My Geogebra works are the heart of my proofs and thought experiments.

And you make your judgments without even looking at them.

>> [...]
>> Instead, Einstein argues that there is no force between the particles
>> of the Earth and those of the Moon, and that the action between the two
>> bodies is due to the space-time curvature of one in contrast to the
>> different space-time curvature of the other.
>
> That is not correct. Using the spacetime curvature interpretation of GR,
> it is the curvature due to ALL components of the solar system that
> determines all of their orbits.

What do the curvatures due to all the other components of the solar
system have to do with it?

And while you're at it, why don't you add all the other galaxies in the
universe?

>> But the two curvatures are not equal!
>
> Of course not! Granting your separation, their effects are not equal,
> either. The effect of earth on moon is huge and causes the moon to orbit
> around the earth. The effect of moon on earth is MUCH smaller, and
> merely makes it wiggle a little bit as it orbits the sun.

The Earth's effect on the Moon is *much* larger than the Moon's effect
on the Earth because Earth's spacetime (which is more curved than the
Moon's) acts on the Moon *more* than the Moon's reacts on the Earth.

> Consider a solar system with just sun, earth, and moon (i.e. ignore
> everything else). The earth does not orbit around an ellipse, not even
> approximately -- it is the earth-moon barycenter that orbits around the
> (approximate) ellipse you are thinking of. The earth wiggles around that
> (approximate) ellipse. The wiggles have period ~ 29.5 days,
> corresponding to the moon's orbit. (This is exact in Newtonian
> mechanics, but only approximate in GR -- NM is linear while GR is not.)

This is beside the point: the motion of the Earth-Moon barycenter
depends on forces *external* to the Earth and the Moon, while the third
law talks about forces *internal* to the Earth-Moon system.

>> And therefore, even for Einstein, the gravitational equality between
>> the two opposing bodies no longer exists, ]...]
>
> Hmmm. Newton's third law discusses FORCES, not "gravitational equality".

No, Newton's third law talks about "action and reaction" and not just
forces.

> In Newtonian mechanics, because earth and moon have such different
> masses, the effects of equal forces on them are most definitely NOT
> equal. In the spacetime curvature interpretation of GR there are no
> gravitational forces, and one simply cannot apply any of Newton's laws.

The third law can (and should) be applied in General Relativity because
it does not speak of forces but of action and reaction.

The spacetime of General Relativity (with its curvature) acts between
the Earth and the Moon as the tension in Newton's string acts between
the horse and the stone.

> (But one can apply the Newtonian approximation to GR, and all three of
> Newton's laws apply within that approximation.)
>
> Hint: it is outrageously arrogant to think you alone can see an error in
> a theory that has stood the test of time for hundreds of years and
> inspection by tens of thousands of physicists. There is a reason that no
> journal articles have been published on this....

I don't think I can see a mistake, I *prove* that the mistake is there
(for those who watch my animations).

An error that neither you nor anyone else have been able to detect.

Luigi Fortunati.

On 12/16/24 4:24 AM, Luigi Fortunati wrote:
> Tom Roberts il 12/12/2024 09:20:30 ha scritto:
>>> With my animation https://www.geogebra.org/m/v33hu4en I show what
>>> Newton said.
>> I never click on such links.
>
> My Geogebra works are the heart of my proofs and thought experiments.
> And you make your judgments without even looking at them.

But your statements here show a major lack of understanding, and invalid
attempts to intermix Newtonian mechanics and General Relativity.

>>> [...]
>>> Instead, Einstein argues that there is no force between the particles
>>> of the Earth and those of the Moon, and that the action between the two
>>> bodies is due to the space-time curvature of one in contrast to the
>>> different space-time curvature of the other.
>>
>> That is not correct. Using the spacetime curvature interpretation of GR,
>> it is the curvature due to ALL components of the solar system that
>> determines all of their orbits.
>
> What do the curvatures due to all the other components of the solar
> system have to do with it?
> And while you're at it, why don't you add all the other galaxies in the
> universe?

Yes, in principle one should do that. In practice their effects are
completely negligible.

> The Earth's effect on the Moon is *much* larger than the Moon's effect
> on the Earth because Earth's spacetime (which is more curved than the
> Moon's) acts on the Moon *more* than the Moon's reacts on the Earth.

Hmmm. To claim that you must assume separation. But as I have said
several times before, GR is not separable like that. Removing the earth
from consideration affects the moon's path enormously; removing the moon
affects the earth's path by only a small amount.

>> [...]
>> Hmmm. Newton's third law discusses FORCES, not "gravitational equality".
>
> No, Newton's third law talks about "action and reaction" and not just
> forces.
But TO NEWTON, "action and reaction" here refer to forces, not some
nebulous and undefined notion of yours. Your arguments here are based on
PUNs. (Nomenclature has changed over 300 years, and translation from
Latin adds additional ambiguity and potential for confusion.)

> [...]
> The third law can (and should) be applied in General Relativity because
> it does not speak of forces but of action and reaction.

Strictly speaking, GR does not use forces to model gravitational
interactions -- it uses geometry [#]. But see above for the PUNs you use
here, which destroy your claims.

[#] in the geometrical interpretation of GR, which we are using.

> The spacetime of General Relativity (with its curvature) acts between
> the Earth and the Moon as the tension in Newton's string acts between
> the horse and the stone.

That is not how GR actually models this. Geometry (spacetime) does not
"act", it just is. Moreover, GR is not separable, as you implicitly
assume here.

GR uses very different concepts from Newtonian mechanics. But you keep
attempting to use Newtonian concepts to "describe" GR -- that's invalid.

> [...]
> I don't think I can see a mistake, I *prove* that the mistake is there
> (for those who watch my animations).
> An error that neither you nor anyone else have been able to detect.

Because there is no error, except by you.

I repeat: it is outrageously arrogant to think that you alone can see an
error in a theory that has stood the test of time for hundreds of years
and inspection by tens of thousands of physicists. As I have said
before, you REALLY need to take a course in physics at a college or
university....

Tom Roberts

In article <vk0k8g$2p4uk$1@dont-email.me>, Luigi Fortunati wrote:
> The cables break and the elevator goes into free fall.
[[...]]
> for the first law, Einstein says that a body in the elevator
> in free fall is at rest with respect to the elevator itself.

Not quite: a body *sufficiently close to the elevator's center of mass*
is *unaccelerated* with respect to the elevator itself. "Unaccelerated"
means constant velocity, but that velocity need not be zero.

> So, why does a body placed below the center of gravity of a
> free-falling elevator accelerate downwards, and if it is above the
> center of gravity, it accelerates upwards?

Assuming this elevator is near the Earth's surface, the answer to
your question is "tidal forces". In more detail...

[For the following, I'll use the sign convention that the Newtonian
"little g" vector point down, and has *positive* magnitude.]

We observe that the Newtonian "little g vector" (as measured by observers
stationary with respect to (wrt) the Earth's surface; let's call this
g_wrt_Earth) varies with position. In this case (measuring near to,
and above, the Earth's surface), g_wrt_Earth always points roughly
down, but decreases in magnitude as we go up in altitude away from the
Earth's surface.

If we now go to a free-falling local inertial reference frame (FFLIRF),
say one accelerating down with acceleration g_elevator wrt the Earth's
surface -- we'll measure a little-g vector (as measured wrt the FFLIRF
-- let's call this g_wrt_FFLIRF) of
g_wrt_FFLIRF(position) = g_wrt_Earth(position) - g_elevator (1)
where I've explicitly shown which terms are position-dependent.

Since g_elevator is (by the definition of FFLIRF) precisely g_wrt_Earth
at the elevator's center-of-mass position, by equation (1), g_wrt_FFLIRF
must be zero at this position. But since g_wrt_Earth varies with position,
g_wrt_FFLIRF must also vary with position. That is, g_wrt_FFLIRF will in
general be nonzero for positions away from the elevator center-of-mass.

In this case (elevator near to, and above, the Earth's surface),
g_wrt_Earth points down everywhere in the elevator, but at the TOP of the
elevator
|g_wrt_Earth(top of elevator)|
< |g_wrt_Earth(elevator center of mass)| (2)
where I'm using | | to denote the magnitude of a vector.
By equation (1), this means that g_wrt_FFLIRF(top of elevator) points UP.
Similarly, at the BOTTOM of the elevator,
|g_wrt_Earth(bottom of elevator)|
> |g_wrt_Earth(elevator center of mass)| (3)
so equation (1) tells us that g_wrt_FFLIRF(bottom of elevator) points
DOWN.

This means that test masses placed in different parts of the elevator
have a "tidal" acceleration wrt each other. Alternatively, if we hold
on to test masses in different parts of the elevator (i.e., we hold them
at fixed positions wrt the elevator, so they are NOT in free-fall), then
we'll feel "tidal" forces acting on the masses. Tidal forces/accelerations
represent a deviation of our supposed FFLIRF from actually being inertial.

In article <vk8fh7$h93j$1@dont-email.me>, Luigi Fortunati wrote:
> Can we define the interior space of the elevator as "local" or is it
> too big?
>
> If it is too big, how big must it be to be considered "local"?
>
> If it is shown that there are real forces inside the free-falling
> elevator, can we still consider this reference system inertial?

Tha answer to all of these questions is the same, namely, "It depends
on your accuracy tolerance for measurements".

That is, it's easy to see that |g_wrt_FFLIRF(top of elevator)| and
|g_wrt_FFLIRF(bottom_of_elevator)| both vary with the size of the
elevator. In particular, if you make the elevator 10 times smaller,
both of these numbers will also be about 10 times smaller. So, if you
make the elevator small enough, then both of these numbers will be tiny
enough that we can approximate them as zero. That is, if the elevator
is small enough, then tidal accelerations/forces are negligible, i.e.,
the elevator is (to within our accuracy tolerance) an inertial reference
frame.

Similarly, the effect of any given nonzero g_wrt_FFLIRF compounds over
time, i.e., if we say that our experiments are only going to last for
some finite duration, then the shorter that time, the less the effect of
any given nonzero g_wrt_FFLIRF. Thus if our duration is short enough,
then we can neglect the effects of g_wrt_FFLIRF being zero, and the
elevator is (to within our accuracy tolerance) an inertial reference
frame.

More precisely, for any fixed accuracy tolerance, if we make the elevator
small enough and/or make our measurements for a short enough duration,
then the tidal accelerations/forces across the elevator will be small
enough to be negligible.

As Tom Roberts explained earlier in this thread, just how small the
elevator must be and/or just how short our measurement duration must be,
depends on how fast the local little-g field varies with position. For
example, if instead of our elevator being near to (and having a little-g
vector dominated by) the Earth, our elevator is instead near to the
event horizon of a billion-solar-mass black hole, then little-g varies
much less strongly with position than it does near the Earth. So, in
this case (near a huge black hole) the elevator can be larger, and/or
we can make measurements for a longer duration, and still stay within
the same fixed accuracy tolerance.

> Are tidal forces real?

They can do work, so they must be real.

To see this, fix a stick running from floor to ceiling in the elevator,
and place two beads on it, which can slide up and down with a (small)
bit of friction. If the friction is small enough (but still nonzero),
then the beads will move under the influence of the tidal forces, and
the friction will cause the beads and the stick to heat up. This
heating shows that the tidal forces are doing work, and hence that
tidal forces must be real.

[Digression: the same question arose in general relativity,
when well into the 1950s many researchers were unsure whether
or not gravitational waves were real. The two-beads-on-a-stick
gedanken experiment for gravitational waves was proposed by
Feynman and analyzed in detail by Bondi in 1957, proving that
gravitational waves could do work and must thus be real.]

See
https://en.wikipedia.org/wiki/Tidal_power
for some real-world examples of work done by tidal forces, with (in some
cases) power output measured in the hundreds of megawatts. These systems
all ultimately exploit the fact that g_wrt_Earth is NOT constant from one
edge of the tidal power basin to the other edge, i.e., that the (entire)
tidal power basin as NOT a local inertial reference frame.

--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
currently on the west coast of Canada
"[I'm] Sick of people calling everything in crypto a Ponzi scheme.
Some crypto projects are pump and dump schemes, while others are pyramid
schemes. Others are just standard issue fraud. Others are just middlemen
skimming off the top. Stop glossing over the diversity in the industry."
-- Pat Dennis, 2022-04-25

In article <vk0k8g$2p4uk$1@dont-email.me>, Luigi Fortunati asked
> Are tidal forces real?

In article <lt1tqbFapkrU1@mid.dfncis.de>, I replied
> They can do work, so they must be real.
>
> [[...]]
>
> See
> https://en.wikipedia.org/wiki/Tidal_power
> for some real-world examples of work done by tidal forces, with (in some
> cases) power output measured in the hundreds of megawatts. These systems
> all ultimately exploit the fact that g_wrt_Earth is NOT constant from one
> edge of the tidal power basin to the other edge, i.e., that the (entire)
> tidal power basin as NOT a local inertial reference frame.

Oops, that last sentence of mine was quite wrong. :(

The variation in g_wrt_Earth over the tidal power basin is generally
negligible for a tidal power system. I should have written this instead:

These systems
all ultimately exploit the fact that g_wrt_Earth varies with time
AND varies with position on the Earth's surface. The combination
of these variations causes ocean tides (which certainly do mechanical
work).

I should also point out this GREAT animation of worldwide tides at one
particularly-powerful frequency:

https://en.wikipedia.org/wiki/File:Global_surface_elevation_of_M2_ocean_tide.webm

--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
currently on the west coast of Canada
"[I'm] Sick of people calling everything in crypto a Ponzi scheme.
Some crypto projects are pump and dump schemes, while others are pyramid
schemes. Others are just standard issue fraud. Others are just middlemen
skimming off the top. Stop glossing over the diversity in the industry."
-- Pat Dennis, 2022-04-25