In the case of my animation
https://www.geogebra.org/m/mjnqb8vk
before the collision, the inertia of the mass m1 is not a vector in the reference of m1 but it is certainly a vector in the reference of the mass m2 because it moves horizontally to the right.

And the inertia of mass m2 is certainly a vector in the reference of m1 because it moves horizontally to the left.

It's correct?

Luigi Fortunati

On Monday, October 23, 2023 at 6:31:29=E2=80=AFAM UTC-5, Luigi Fortunati wrote:
> In the case of my animation
> https://www.geogebra.org/m/mjnqb8vk
> before the collision, the inertia of the mass m1 is not a vector in
> the reference of m1 but it is certainly a vector in the reference of
> the mass m2 because it moves horizontally to the right.
>
> And the inertia of mass m2 is certainly a vector in the reference of
> m1 because it moves horizontally to the left.
>
> It's correct?
>
> Luigi Fortunati

Inertia is a property of an object, not a dynamic variable. I think
what you mean is momentum. Momentum is part of a Lorentz invariant
4-vector:

p = {energy/c, x-momentum, y-momentum, z-momentum} (in momentum units)

The momentum parts comprise a vector in 3-space. This energy-momentum
4-vector is conserved in any collision:

p(m1 before) + p(m2 before) = p(m1 after) + p(m2 after)

This is true even for inelastic collisions as long as the energy lost to
friction/deformation is included in the the after energy components. This
is true relativistically provided the relativistically correct energy and
momenta are calculated.

Rich L.

On Wednesday, October 25, 2023 at 6:01:34=E2=80=AFAM UTC-5, Luigi Fortunati wrote:
> Richard Livingston il 24/10/2023 18:21:13 ha scritto:

> Yes, that's right, inertia is that property of bodies that makes them
> go straight at uniform speed.

No, inertia is the ability of a body to resist being accelerated. Its
quantity is what we think of as inertial mass. It is a scalar.

George Hrabovsky il 26/10/2023 11:12:16 ha scritto:
>> Yes, that's right, inertia is that property of bodies that makes them
>> go straight at uniform speed.
>
> No, inertia is the ability of a body to resist being accelerated. Its
> quantity is what we think of as inertial mass. It is a scalar.

If inertia is the ability of bodies to resist acceleration (which is a
vector), then it cannot be a scalar!

George Hrabovsky il 26/10/2023 11:12:16 ha scritto:
> No, inertia is the ability of a body to resist being accelerated.=20

Inertia is (first of all) a property of bodies and, therefore, every=20
material body always has this property, at any moment, both before,=20
during and after the collision (there are no bodies that have inertia=20
at certain moments and in others don't have it).

Instead, resistance to acceleration is not always there, it is only=20
there when acceleration occurs, because the two things are connected.

There is resistance to acceleration if there is acceleration and there=20
is no resistance to acceleration if there is no acceleration.

Inertia in the absence of any acceleration (which must not resist=20
acceleration) is a completely different thing from inertia in the=20
presence of acceleration (which must resist acceleration).

You can understand everything by looking at my animation
https://www.geogebra.org/m/mjnqb8vk

If you press the =E2=80=9CStart=E2=80=9D key and then the =E2=80=9CStop=E2=
=80=9D key before the=20
collision between the bodies m1 and m2, you will see that the two=20
bodies have their inertias ready to react to any future acceleration=20
but without doing anything as long as there are no accelerations at=20
which to oppose.

They are "potential" inertias that do not do what you say, because they=20
do not oppose anything, as there is nothing to oppose.

Then restart the animation and stop it when the bodies m1 and m2=20
collide.

In this case, there is acceleration: the body m1 accelerates m2 towards=20
the right and the inertia of the body m2 opposes and pushes m1 towards=20
the left.

For the body m1, this push of m2 is an acceleration towards the left=20
and the inertia of m1 rebels and pushes m2 towards the right.

In short, both bodies m1 and m2 undergo an acceleration and both oppose=20
each other with their inertia, i.e. with their ability to resist the=20
acceleration.

The inertias that are inactive before the collision (which do not=20
oppose anything) are completely different from the inertias that are=20
active during the collision (which oppose the acceleration).

And after the collision, the inertias return to being exactly as they=20
were before the collision, that is, inertias that do not have to oppose=20
anything because there is no acceleration to resist!

Luigi Fortunati

On 10/28/23 1:02 PM, Luigi Fortunati wrote:
> George Hrabovsky il 26/10/2023 11:12:16 ha scritto:
>>> Yes, that's right, inertia is that property of bodies that makes
>>> them go straight at uniform speed.
>> No, inertia is the ability of a body to resist being accelerated.
>> Its quantity is what we think of as inertial mass. It is a scalar.

Actually, inertia does both -- it resists acceleration when a force is
applied, and it makes an object move in a uniform straight line when no
force is impressed on the object.

> If inertia is the ability of bodies to resist acceleration (which is
> a vector), then it cannot be a scalar!

No. When a force is impressed upon a massive object, the object's
inertia resists in the opposite direction. So inertial CANNOT be a
vector if it is to act the same for all forces in all directions. Indeed
"inertia" is really another name for "mass", which is clearly a scalar.

Tom Roberts

On Saturday, October 28, 2023 at 1:02:17 PM UTC-5, Luigi Fortunati wrote:
> George Hrabovsky il 26/10/2023 11:12:16 ha scritto:
>>> Yes, that's right, inertia is that property of bodies that makes them
>>> go straight at uniform speed.
>>
>> No, inertia is the ability of a body to resist being accelerated. Its
>> quantity is what we think of as inertial mass. It is a scalar.
> If inertia is the ability of bodies to resist acceleration (which is a
> vector), then it cannot be a scalar!
This is wrong. If you multiply a vector by the scalar 1/2, then each component is half of what it was; you have resisted the vector by using a scalar.

On Saturday, October 28, 2023 at 3:28:43 PM UTC-5, Luigi Fortunati wrote:
> George Hrabovsky il 26/10/2023 11:12:16 ha scritto:
>> No, inertia is the ability of a body to resist being accelerated.
>
> Inertia is (first of all) a property of bodies and, therefore, every
> material body always has this property, at any moment, both before,
> during and after the collision (there are no bodies that have inertia
> at certain moments and in others don't have it).

But that does not make it a vector.
>
> Instead, resistance to acceleration is not always there, it is only
> there when acceleration occurs, because the two things are connected.

This is in error. Forces acting on a body accelerate the mass of the body. The mass is always there, and it is a measure of inertia. This is encapsulated in Newton's second law, F = m a, F is the applied force, m is the mass, and a is the acceleration. F and a are vectors, m is a scalar.
>
> There is resistance to acceleration if there is acceleration and there
> is no resistance to acceleration if there is no acceleration.

The mass is always there, otherwise how would the body know when to have mass or not?
>
> Inertia in the absence of any acceleration (which must not resist
> acceleration) is a completely different thing from inertia in the
> presence of acceleration (which must resist acceleration).

Why?
>
> You can understand everything by looking at my animation
> https://www.geogebra.org/m/mjnqb8vk

Your animation demonstrates the conservation of momentum and the effects of friction (a force caused by electromagnetism), so it has nothing to do with inertia intrinsically.

George

On Sunday, October 29, 2023 at 2:40:57=E2=80=AFAM UTC-5, Tom Roberts wrote:
> On 10/28/23 1:02 PM, Luigi Fortunati wrote:
> > George Hrabovsky il 26/10/2023 11:12:16 ha scritto:
> >>> Yes, that's right, inertia is that property of bodies that makes
> >>> them go straight at uniform speed.
> >> No, inertia is the ability of a body to resist being accelerated.
> >> Its quantity is what we think of as inertial mass. It is a scalar.
> Actually, inertia does both -- it resists acceleration when a force is
> applied, and it makes an object move in a uniform straight line when no
> force is impressed on the object.

My only objection to this, is that mass is component of the momentum, and
momentum is what makes an object travel in a straight line until; acted
upon by an external force (which is the time derivative of the momentum).

George

Inertia is a consequence of conservation of energy. Consider the
following argument:
-A particle of mass m_0 at rest has zero momentum and energy
E=m_0 c^2
-Quantum mechanically the wave function has frequency
\omega_0 = \frac{m_0 c^2}{\hbar}, and k=0.
-If you Lorentz transform to a moving frame with velocity beta,
the frequency becomes
\omega_1 = \omega_0 \gamma
and the wavenumber becomes
k = \omega_0 \gamma \beta
In other words, to promote an object at rest in one frame to
being at rest in a moving frame, you have to add energy. This
is the experienced as "inertia", the tendency for objects to
remain in constant motion unless disturbed. (By disturbed,
I mean energy is added or subtracted from the object.)

Rich L.

Tom Roberts il 28/10/2023 18:40:52 ha scritto:
>> If inertia is the ability of bodies to resist acceleration (which is
>> a vector), then it cannot be a scalar!
>
> No.

Maybe you meant to say: yes, because then you write...

> When a force is impressed upon a massive object, the object's
> inertia resists in the opposite direction.

If you say it resists in the opposite direction, then it is a vector,
because only vectors have a direction.

> So inertial CANNOT be a vector if it is to act the same for
> all forces in all directions.

Inertia doesn't have to act in all directions, it just has to be ready
to act in all directions, *before* external forces arrive!

But when an external force arrives, inertia reacts in only one
direction: the opposite one!

Just like friction does, it is ready to brake in all directions but
when activated it only brakes in the direction opposite to the motion.

Check out my animation
https://www.geogebra.org/m/mjnqb8vk

Before the collision, the body m2 is ready to act in all directions
because it does not know from which side the force will come.

He is ready to act but does not act.

It is only when the body m1 comes at it from the left that the inertia
of the body m2 reacts and it does so in only one direction, the one
opposite to the force it receives, i.e. towards the left!

> Indeed "inertia" is really another name for "mass",
> which is clearly a scalar.

Inertia is not a mass, it is a *property* of the mass which concerns
its behavior in the event of an external vectorial intervention.

Luigi Fortunati.

Richard Livingston il 29/10/2023 04:42:09 ha scritto:
> Inertia is a consequence of conservation of energy. Consider the
> following argument:
> -A particle of mass m_0 at rest has zero momentum and energy
> E=m_0 c^2
> -Quantum mechanically the wave function has frequency
> \omega_0 = \frac{m_0 c^2}{\hbar}, and k=0.
> -If you Lorentz transform to a moving frame with velocity beta,
> the frequency becomes
> \omega_1 = \omega_0 \gamma
> and the wavenumber becomes
> k = \omega_0 \gamma \beta
> In other words, to promote an object at rest in one frame to
> being at rest in a moving frame, you have to add energy. This
> is the experienced as "inertia", the tendency for objects to
> remain in constant motion unless disturbed. (By disturbed,
> I mean energy is added or subtracted from the object.)

All this concerns the inertia of a single mass that moves without
having to deal with the inertia of another mass with which it collides.

When in my animation
https://www.geogebra.org/m/mjnqb8vk
body m2 is suddenly hit by m1, does its inertia passively accept the
intrusion or does it rebel and opposes in the opposite direction?

Luigi Fortunati

[[Mod. note -- The problem with phrases like "passively accept" or
"rebel" or "oppose" is that it's hard to pin down their meanings.
For example, how should we operationally define "passive acceptance"?
Without a clear operational definition, it's hard to do a careful
analysis.

Newton's 2nd law is unambiguous: apply a (vector) net force F to
an object, and the object accelerates with a (vector) acceleration.
The acceleration vector is observed to be proportional to the net-force
vector, with a fixed proportionality constant (which we call the
"inertial mass", or just "mass" for short) for any given object.
Each of these phrases has a clear operational definition.
-- jt]]

Luigi Fortunati il 30/10/2023 06:07:04 ha scritto:
> All this concerns the inertia of a single mass that moves without having to deal with the inertia of another mass with which it collides.
>
> When in my animation
> https://www.geogebra.org/m/mjnqb8vk
> body m2 is suddenly hit by m1, does its inertia passively accept the intrusion or does it rebel and opposes in the opposite direction?
>
> [[Mod. note -- The problem with phrases like "passively accept" or
> "rebel" or "oppose" is that it's hard to pin down their meanings.
> For example, how should we operationally define "passive acceptance"?
> Without a clear operational definition, it's hard to do a careful
> analysis.
>
> Newton's 2nd law is unambiguous: apply a (vector) net force F to
> an object, and the object accelerates with a (vector) acceleration.
> The acceleration vector is observed to be proportional to the net-force
> vector, with a fixed proportionality constant (which we call the
> "inertial mass", or just "mass" for short) for any given object.
> Each of these phrases has a clear operational definition.
> -- jt]]

The operational definition according to Newton is exactly the same as what I wrote.

Newton says: <L The vis insita, or innate force of matter, is a power of
resisting, by which every body, as much as it lies, endeavors to
persevere in its present state, whether it be of rest, or of moving
uniformly forward in a right line. This force is proportional to the
body whose force it is; and differs nothing from the inactivity of the
mass, but in our manner of conceiving it. A body, from the inactivity of
matter, is not without difficulty put out of its state of rest or
motion. Upon which account, this vis insita, may, by a most significant
name, be called vis inertiae, or force of inactivity. But a body exerts
this force only, when another force, impressed upon it, endeavors to
change its condition>.

And I also wrote that this inertial force (in my animation) is exerted
*only* when the force of the body m1 tries to accelerate m2 or (which is
the same) when the force of the body m2 tries to slow down m1.

And then Newton continues thus: < And the exercise of this force may be
considered both as resistance and impulse; it is resistance, in so far
as the body, for maintaining its present state, withstands the force
impressed; it is impulse, in so far as the body, by not easily giving
way to the impressed force of another, endeavors, to change the state of
that another. Resistance is usually ascribed to bodies at rest, and
impulse to those in motion; but motion and rest, as commonly conceived,
are only relatively distinguished; nor are these bodies always truly at
rest, which commonly are taken to be so>.

And here too he says what I wrote and that is that (in my animation) the
inertial force of the body m1 is impulse and the inertial force of m2 is
resistance but he warns that the two things are interchangeable.

Luigi Fortunati.

On 10/30/23 4:34 PM, Luigi Fortunati wrote:
> Newton says: <L The vis insita, or innate force of matter, [...]

The translator used a PUN on the word "force".

In Newtonian mechanics, force is a vector while the "vis insita" is a
scalar, which today is called mass.

Tom Roberts

Tom Roberts il 01/11/2023 08:59:12 ha scritto:
>> Newton says: <L The vis insita, or innate force of matter, [...]
>
> The translator used a PUN on the word "force".
>
> In Newtonian mechanics, force is a vector while the "vis insita" is a scalar, which today is called mass.

The "vis insita" and the mass cannot be the same thing.

Newton says: "a body exerts this force (the vis intima) *only* when another force, impressed upon it, endeavors to change its condition".

The vis intima changes over time: it is only there when there is an external force (which it resists and opposes in the opposite direction) and it is not there at all when the external force is not there (because it does not have to resist anything )

Instead, mass is a quantity of matter that always exists and is always the same.

This is why, the "vis insita" and the mass are two different properties of bodies: the first is a non-force (when it does not act) and is a force (when it acts in opposition to an external force), the second is not, because it is never a force.

Luigi Fortunati.

George Hrabovsky il 29/10/2023 11:51:56 ha scritto:
>> Instead, resistance to acceleration is not always there, it is only there when acceleration occurs, because the two things are connected.
>
> This is in error. Forces acting on a body accelerate the mass of the body.

Certainly, but the forces that act on a body also have another consequence: they activate the body's resistance force!

And vice versa, the body's (inertial) resistance force is activated *only* when the external force intervenes.

Newton also says it: "A body exerts this force only, when another force, impressed upon it, endeavors to change its condition; and the exercise of this force may be considered both as resistance and impulse; it is resistance, in so far as the body, for maintaining its present state, withstands the force impressed; it is impulse, in so far as the body, by not easily giving way to the impressed force of another, endeavors, to change the state of that another. Resistance is usually ascribed to bodies at rest, and impulse to those in motion; but motion and rest, as commonly conceived, are only relatively distinguished; nor are these bodies always truly at rest, which commonly are taken to be so."

L’inerzia è sia impulso che resistenza.

> The mass is always there, and it is a measure of inertia. This is encapsulated in Newton's second law, F = m a, F is the applied force, m is the mass, and a is the acceleration. F and a are vectors, m is a scalar.

Of course, mass (which is the amount of matter) is always there because it is a scalar and has no direction.

Instead, inertia is not a quantity of matter but is the ability of bodies to resist, to oppose, to react: it is an action.

So, mass and inertia are not the same thing.

>> There is resistance to acceleration if there is acceleration and there is no resistance to acceleration if there is no acceleration.
>
> The mass is always there, otherwise how would the body know when to have mass or not?

The body doesn't need to know when it has mass and when it doesn't, because mass is *always* there.

Instead, he needs to know when it's time to resist and when not.

>> Inertia in the absence of any acceleration (which must not resist acceleration) is a completely different thing from inertia in the presence of acceleration (which must resist acceleration).
>
> Why?

You can understand this by looking at my animation where the bodies m1 and m2 (before the collision) do not have to resist anything because no one is accelerating them and, instead, during the collision they both have to resist because they are accelerating each other.

>> You can understand everything by looking at my animation https://www.geogebra.org/m/mjnqb8vk
>
> Your animation demonstrates the conservation of momentum and the effects of friction (a force caused by electromagnetism), so it has nothing to do with inertia intrinsically.

My animation demonstrates that there is a simultaneity between inertia and electromagnetic forces.

In fact, the electromagnetic forces are activated only during the collision, that is, precisely at the moment in which both bodies m1 and m2 need a force to be able to resist the mutual external acceleration.

This means that it is precisely the bodies m1 and m2 that use *their* electromagnetic forces (what else if not?) to be able to oppose each other.

In fact, in the lower part of my animation (despite there being contact between m1 and m2) no electromagnetic force is activated because the inertias do not activate them (not needing them).

Luigi Fortunati

On Thursday, November 2, 2023 at 3:17:24 AM UTC-5, Luigi Fortunati wrote:
This will be my last response on this topic, as you refuse to attempt to understand reality. As such, this is becoming a waste of my time.

> Certainly, but the forces that act on a body also have another consequence: they activate the body's resistance force!

Such an activation requires a mechanism. In a spring we have the Hooke's law force that activates only when you displace a particle on a spring from equilibrium. You suggest no such mechanism, and there is no physical evidence for any such mechanism in matter other than mass. Mass is completely well understood in classical mechanics as the quantity of inertia, thus it goes by the name of inertial mass. Your refusal to accept this tells me that you are no longer dealing with reality.
>
> And vice versa, the body's (inertial) resistance force is activated *only* when the external force intervenes.

Nonsense.

> Newton also says it: "A body exerts this force only, when another force, impressed upon it, endeavors to change its condition; and the exercise of this force may be considered both as resistance and impulse; it is resistance, in so far as the body, for maintaining its present state, withstands the force impressed; it is impulse, in so far as the body, by not easily giving way to the impressed force of another, endeavors, to change the state of that another. Resistance is usually ascribed to bodies at rest, and impulse to those in motion; but motion and rest, as commonly conceived, are only relatively distinguished; nor are these bodies always truly at rest, which commonly are taken to be so."

You suggest that inertia is a force. This makes no sense as solving the equations of motion we would be multiplying by a new force and the end result would be in units of Newton's squared, and they are not.
>

> You can understand this by looking at my animation where the bodies m1 and m2 (before the collision) do not have to resist anything because no one is accelerating them and, instead, during the collision they both have to resist because they are accelerating each other.

Your animations are not solutions of the equations, they are an artist rendering. Thus, they are useless as a model.

> My animation demonstrates that there is a simultaneity between inertia and electromagnetic forces.
>
> In fact, the electromagnetic forces are activated only during the collision, that is, precisely at the moment in which both bodies m1 and m2 need a force to be able to resist the mutual external acceleration.

You clearly have no understanding of the electromagnetic basis of friction. The reason why you have friction is because of the interaction of atoms and ions with each other electromagnetically. Atoms want to stick together, so as an object moves it experiences a force that tries to get the object to stay where it is. You diagram does not even attempt to show how that works.
>
> This means that it is precisely the bodies m1 and m2 that use *their* electromagnetic forces (what else if not?) to be able to oppose each other.

No, it is their differences in their initial states of momentum.
>
> In fact, in the lower part of my animation (despite there being contact between m1 and m2) no electromagnetic force is activated because the inertias do not activate them (not needing them).

As I said above, I am done wasting my time on this. You will never accept reality. Good luck.

George

George Hrabovsky il 02/11/2023 14:08:07 ha scritto:
>> Newton also says it: "A body exerts this force only, when another force,
>> impressed upon it, endeavors to change its condition; and the exercise of
>> this force may be considered both as resistance and impulse; it is
>> resistance, in so far as the body, for maintaining its present state,
>> withstands the force impressed; it is impulse, in so far as the body, by not
>> easily giving way to the impressed force of another, endeavors, to change
>> the state of that another. Resistance is usually ascribed to bodies at rest,
>> and impulse to those in motion; but motion and rest, as commonly conceived,
>> are only relatively distinguished; nor are these bodies always truly at
>> rest, which commonly are taken to be so."
>
> You suggest that inertia is a force.

It is Newton who expressly says so!

In fact, he says: "A body exerts this force (inertia, vis intima)
*ONLY* when another force, impressed upon it, endeavors to change its
condition".

Therefore, inertia becomes a force only when external conditions
require it (not always).

This is what happens in my animation *ONLY* during the collision: what
does the body m2 do on that occasion? It opposes (resists) the other
force (that of m1) which strives to change its condition of rest.

And m1 does the same when another force (that of m2) tries to change
its condition of rectilinear and uniform motion.

The inertia (the vis intima) of m1 and m2 is not always a force, it
becomes *ONLY* when each of the two becomes an *external* force for the
other.

You're disputing Newton too, not just me.

Luigi Fortunati.

I want to highlight an error of judgment that we all make and that my animation
https://www.geogebra.org/m/mjnqb8vk
seems to confirm it.

In fact, during the collision, the two arrows seem to indicate the presence of a blue force that acts *only* on the body m2 and of a corresponding red force that acts *only* on the body m1 (moreover only at the two points A and P) .

Of course, this is not the case.

The two bodies m1 and m2 are not two single and compact blocks but are a set of billions and billions of particles connected to each other by a network of forces that I have schematized in my new animation
https://www.geogebra.org/m/sqmsc4ey
where it is clearly seen that (in the collision) the blue and red forces arise and develop within the two bodies.

The blue forces start from the zero level of the body m1, where the particles W, T and V (which move forward inertially) push the corresponding particles of level 1 who are slowing down.

This push (the blue force) increases in the transition from the first to the second level because those of level 1 are added to the pushes of the level zero particles and so on, increasing their strength more and more until level 4, where they reach their maximum .

Therefore, the blue forces, even before arriving at the body m2, exert their action on the particles of the body m1 (particles of the body m1 pushing other particles of the body m1).

And the same goes for the red forces which are forces that the particles in the body m2 exert on other particles in the body m2, before arriving at the body m1.

All this means that it is not at all true that (during the impact) the blue force directed to the right acts *only* on the body m2 and that the red force directed to the left acts *only* on the body m1.

A red *external* force directed to the left and a blue *internal* force directed to the right act on body m1 and a blue *external* force directed to the right and a red *internal* force directed to the left act on body m2.

Forces always act in pairs, as the third principle teaches: there are no forces (actions) without the corresponding opposing forces (reactions).
Luigi Fortunati.

On Monday, October 23, 2023 at 7:31:29 PM UTC+8, Luigi Fortunati wrote:
> In the case of my animation
> https://www.geogebra.org/m/mjnqb8vk
> before the collision, the inertia of the mass m1 is not a vector in the reference of m1 but it is certainly a vector in the reference of the mass m2 because it moves horizontally to the right.
>
> And the inertia of mass m2 is certainly a vector in the reference of m1 because it moves horizontally to the left.
>
> It's correct?
>
> Luigi Fortunati
this is kind of speculative:

given the photo of the Eagle nebula. What if a nebula was a kind of
space where Dark Matter/Energy becomes c speed matter? Black holes are a
space where c speed energy never returns. Does this inversion actually
exist.

What is the form of the "ball like" emitted mass from this nebula? Is
it a hydrogen ball yet to be a sun. Is it a Granite ball yet to be a
planet? Is Jupiter yet to be a solar system?

I would hope to dream of a larger cosmos. Is the emission a class of
pure electron mass constuction?

On 10/23/23 6:31 AM, Luigi Fortunati wrote:
> [...]

[The context of this question is clearly Newtonian mechanics.
But my answer holds for relativistic mechanics as well.]

To definitively answer the question "is inertia a vector", one must find
"inertia" in some equation(s). Unfortunately, "inertia" does not appear
in any equation of mechanics. So the question is meaningless, or at
least unanswerable.

[This includes Newton's original "vis insita".]

Note: do not be confused by "moment of inertia" -- look at its
definition and you'll see it is misnamed, and is really the second
moment of mass.

In modern physics,the closest quantity to "inertia" is mass, which is
clearly a scalar (i.e. not a vector).

Tom Roberts

On Monday, October 23, 2023 at 6:31:29=E2=80=AFAM UTC-5, Luigi Fortunati wrote:
> In the case of my animation
> https://www.geogebra.org/m/mjnqb8vk
> [etc.]

There are two answers to consider.
(This may be the last note I send out, through the Google portal, if it
gets through.)

[[Mod. note -- Yes, apparently google is shutting down its "google groups"
usenet portal. You (and anyone else using google groups) will probably
need to install a usenet client program, commonly called a "newsreader".
See
https://en.wikipedia.org/wiki/Comparison_of_Usenet_newsreaders
for a list of some common newsreader programs. As another alternative,
recall that articles for sci.physics.research can emailed directly to the
submission system at: sci-physics-research@astro.multivax.de
-- jt]]

First...

in the mechanics of a system whose dynamics is described by a Lagrangian
L(q,v,t) that is a function of the configuration space coordinates q =
(q^a: 0 < a <= N) and its corresponding velocities v = (v^a = dq^a/dt: 0 <
a <= N), these play the role of the "kinematic variables", with v^a =
dq^a/dt being the "kinematic equation", while the Lagrangian is
essentially a device used to generate the corresponding "dynamic
variables" - the momentum p_a = @L/@v^a and force f_a = @L/@q^a, as the
partial derivatives (using "@" to denote the partial derivative
operator) ... along with the "dynamic equation" dp_a/dt = f_a, as the
Euler-Lagrange equation.

The two sets of equations
Kinematic: dq^a/dt = v^a
Dynamic: dp_a/dt = f_a
and their corresponding set of variables provide a general framework for
the system's dynamics, while the
Constitutive Relations: f_a = @L/@q^a, p_a = @L/@v^a
fill in the gap on the detailed structure of the system being described.

If you rewrite this as a second order equation, with acceleration
components a = (a^a: 0 < a <= N), it would take the form:

dp_a/dt = f_a => f_a = m_{ab} a^b + s_{ab} v^b + @/@v^a (@L/@t)

(summation convention used here and below)

where

s_{ab} = @p_a/@q^b = @f_b/@v^a

and

m_{ab} = @p_a/@v^b = @p_b/@v^a = m_{ba}

being the *Coefficients Of Inertia*.

The remaining second order derivatives

k_{ab} = -@f_a/@q^b = -@f_b/@q^a = -k_{ba}

play the analogue of spring coefficients, particularly if the system is
near an equilibrium point.

So, in this sense, the inertia is a rank (0,2) configuration space tensor.

For a system composed of a single body, configuration space is confused
with geometric space, with N = 3 and the configuration coordinates (q^1,
q^2, q^3) being the body's spatial coordinates. In that context, the
coefficients of inertia form a 3 x 3 rank (0,2) tensor with respect to
spatial coordinates.

In a non-relativistic setting, in Cartesian coordinates, it is diagonal
and constant. For instance, a conservative system has constitutive laws of
the following form:

p_a = m delta_{ab} v^b, f_a = -@U/@q^a, U = U(q)

delta_{ab} = Kronecker delta (1 if a = b, 0 if a != b).

For non-Cartesian coordinates, it is not constant and need not be
diagonal. For cylindrical coordinates, the coefficients of inertia would
be a matrix of the form m = diag(m, m, m r^2), where r is the radial
coordinate.

In a relativistic setting, the v-dependent part of the Lagrangian has the
form

L = m v^2/(1 + sqrt(1 - A v^2)) + ..., A = 1/c^2
"sqrt" denotes the square root operation

so that

p_a = @L/@v^a = m delta_{ab} v^b/sqrt(1 - A v^2).

The coefficients of inertia then have the form

m_{ab} = M delta_{ab} + M A v_a v_b/(1 - A v^2)
v_a = delta_{ab} v^b
M = m/sqrt(1 - A v^2)

This decomposes into the "longitudinal" and "transverse" mass.

For a system with two or more bodies, the configuration space has N = 3n,
where n is the number of bodies, and the coefficients of inertia - as a
matrix - reduce to a block diagonal form consisting of n 3 x 3 blocks of
similar form as those just described.

Second ...

In Relativity, under an infinitesimal boost (by an infinitesimal boost
velocity upsilon), the total energy E and momentum p = (p_x, p_y, p_z)
transform as:

delta(E) = -upsilon.p,
delta(p) = -upsilon M
M = A E, A = 1/c^2,
().() = inner product

The transform has the following invariant:

E^2 - p^2/A

which characterizes the "rest mass" m, if it is positive, with:

E^2 - p^2/A = m^2/A^2.

The non-relativistic correspondence to this is arrived at by decomposing
the total energy E into the kinetic energy H and mass-energy mc^2 as:

E = H + m/A, i.e. M = m + A H.

The corresponding transforms, assuming m is invariant are

delta(m) = 0
delta(H) = -upsilon.p,
delta(p) = -upsilon M,
delta(M) = -A upsilon.p.

The transform has the following two invariants:

p^2 - 2 M H + A H^2 = p^2 - 2 m H - A H^2
m = M - A H

In the non-relativistic limit, this becomes:

p^2 - 2 M H = p^2 - 2 m H
m = M

and the transforms become

delta(H) = -upsilon.p,
delta(p) = -upsilon M,
delta(M) = 0.

The inertia "M" - in both the relativistic and non-relativistic settings -
is a component of the 5-vector (H, p_x, p_y, p_z, M). For relativity, the
5-vector splits into a 4-vector (E = H + M/a, p_x, p_y, p_z) and a 1-
vector (m = M - A H). The condition that the 1-vector be the norm of the
4-vector (with respect to the Minkowski metric) is:

p^2 - 2 M H + A H^2 = 0.

This can be actually be generalized to:

p^2 - 2 M H + A H^2 = -2 m U + A U^2

by allowing H to have an "internal" part, U:

H = m v^2/(1 + sqrt(1 - A v^2)) + U.

In that case, the invariant M - A H no longer coincides with the rest
mass, but has the form

M - A H = mu = m - A U,

the rest mass, itself, now decomposing into an invariant mass "mu" plus a
contribution from the internal energy:

m = mu + A U.

The non-relativistic counterpart, corresponding to setting A = 0, would
then be:

H = 1/2 m v^2 + U,
M = mu = m.

This is, strictly, *more* general than Special Relativity - the symmetry
group for the two invariants above is not the Poincare' group, but the
one-dimensional extension of it (which is where the linear invariant,
"mu", comes in). The latter is what has, as its non-relativistic limit,
the Bargmann group. The Bargmann group is the central extension of the
Galilei group and is more properly considered to be the symmetry group for
non-relativistic theory, than the Galilei group is, because Galilei
corresponds only to the case m = 0.

In Relativity, the coordinates have the following transform under
infinitesimal boosts:

delta(t) = -A upsilon.r,
delta(r) = -upsilon t,
r = (x, y, z).

Correspondingly, one also has the following transforms for the coordinate
differentials:

delta(dt) = -A upsilon.dr,
delta(dr) = -upsilon dt,
dr = (dx, dy, dz).

This leads to the invariants:

dt^2 - A dr^2 = dt^2 - A (dx^2 + dy^2 + dz^2)
E dt - p.dr

which pairs off, naturally, with the 4-vector (E, p) = (E, p_x, p_y, p_z).

There is no non-relativistic version of this, and if you attempt to create
one by replacing the total energy E with the kinetic+internal energy H as:

H dt - p.dr

then you would find that it has the following as its transform

delta(H dt - p.dr) = (M - A H) upsilon.dr = mu upsilon.dr.

It is not invariant, but strongly suggests the inclusion of another
coordinate "u" with the transform:

delta(u) = -upsilon.r,
delta(du) = -upsilon.dr,

that would make it invariant with:

delta(H dt - p.dr + mu du) = 0

This applies both relativistically and non-relativistically. In the non-
relativistic case, the extra coordinate yields a 4+1 dimensional geometry:
Bargmann geometry. The relativistic version of it has no standard name. It
is characterized by the invariants:

dx^2 + dy^2 + dz^2 + 2 dt du + A du^2,
ds = dt + A du

where "s" plays the role of an absolute time - a lingering vestige of the
absolute time of non-relativistic theory.

The inertia "m" (more accurately, "mu") then plays the role of being the
conjugate to the "u" coordinate. Requiring the dynamics of a system
described by a Lagrangian to be independent of the u-coordinate then leads
- by way of the Noether Theorem - the constancy of m.

An example of a Lagrangian for a free body would be given by the action

S = integral L dt = integral m (dr^2 + 2 dt du + A du^2)/ds

i.e.

L = m/2 (|dr/dt|^2 + 2 du/dt + A (du/dt)^2)/(1 + A du/dt)

where "m" is written in as a Lagrange multiplier.

The Euler-Lagrange equations can be reduced to:

d/dt (M dr/dt) = 0,
dm/dt = 0,
|dr/dt|^2 + 2 du/dt + A (du/dt)^2 = 0,
M = m/sqrt(1 - A |dr/dt|^2) = +/- m/(1 + A du/dt)

An alternative that includes an "internal energy" U would be

S = integral m/2 (dr^2 + 2 dt du + A du^2)/ds + U (ds - dt - A du)

written with "s" as the parameter:

S = integral
m/2 (|dr/ds|^2 + 2 dt/ds du/ds + A (du/ds)^2) ds
+ U (1 - dt/ds - A du/ds) ds

where "U" is written in as a second Lagrange multiplier.