Discussion:
Free fall
(too old to reply)
Luigi Fortunati
2024-01-21 09:04:19 UTC
Permalink
The elevator in free fall in the gravitational field is an inertial
reference system.

Is the elevator in free fall in the magnetic or electric field also an
inertial reference system?

Luigi Fortunati
Mikko
2024-01-21 13:42:49 UTC
Permalink
Post by Luigi Fortunati
The elevator in free fall in the gravitational field is an inertial
reference system.
Is the elevator in free fall in the magnetic or electric field also an
inertial reference system?
That depends on the magnetic and electric properties of the elevator.

Mikko
Luigi Fortunati
2024-01-24 03:46:45 UTC
Permalink
Post by Mikko
Post by Luigi Fortunati
The elevator in free fall in the gravitational field is an inertial
reference system.
Is the elevator in free fall in the magnetic or electric field also an
inertial reference system?
That depends on the magnetic and electric properties of the elevator.
The elevator is metallic, the body in the elevator is the same metal as
the elevator.

The field is generated by an electromagnet.

Luigi Fortunati

[[Mod. note --

The general principal is that "free fall" means no non-gravitational
forces are acting. So there are two basic ways to have "free fall":
either (1) there are no electric and magnetic fields present, and/or
(2) the free-falling bodies have no electric charges or magnetic
dipoles/quadrupoles/etc (either "native" or induced).

In particular, if there's a magnetic field present, then (2) implies
the absence of any diamagnetic or ferromagnetic materials. That means,
for example, no iron, steel, copper, or a bunch of other common metals.
-- jt]]
Luigi Fortunati
2024-01-25 08:32:21 UTC
Permalink
Post by Mikko
The elevator in free fall in the gravitational field is an inertial reference system.
Is the elevator in free fall in the magnetic or electric field also an inertial reference system?
That depends on the magnetic and electric properties of the elevator.
The elevator is metallic, the body in the elevator is the same metal as the elevator.
The field is generated by an electromagnet.
Luigi Fortunati
[[Mod. note --
The general principal is that "free fall" means no non-gravitational
forces are acting.
An even more general principal is that "free fall" means "fall without obstacles".

No one can deny that, in the most remote space far from all gravity, a metal elevator falling towards an electromagnet is in "free fall".

And a metal robot in a metal elevator resting on an electromagnet (without being able to look outside) cannot know if there is a motor that is accelerating the elevator or if there is an electromagnet that is attracting it towards the floor (principle of equivalence).

Luigi Fortunati
Jonathan Thornburg [remove -color to reply]
2024-01-26 08:18:39 UTC
Permalink
Post by Luigi Fortunati
An even more general principal is that "free fall" means "fall without obstacles".
No one can deny that, in the most remote space far from all gravity,
a metal elevator falling towards an electromagnet is in "free fall".
This is mistaken: In this context a magnetic field counts as an "obstacle",
and a magnetic material interacting with a magnetic field is *not* in
free fall.
Post by Luigi Fortunati
And a metal robot in a metal elevator resting on an electromagnet
(without being able to look outside) cannot know if there is a motor
that is accelerating the elevator or if there is an electromagnet that
is attracting it towards the floor (principle of equivalence).
A metal elevator will partially, but not fully, screen the magnetic field,
so there will still be a nonzero magnetic field inside the elevator.
This means that the robot can easily tell the difference: just see if,
inside the elevator, there's a differential acceleration between test
masses inside made of iron (ferromagnetic) vs oxygen (diamagnetic) vs
helium (non-magnetic).
--
-- "Jonathan Thornburg [remove -color to reply]" <***@gmail-pink.com>
currently on the west coast of Canada
"what I still don't understand to this day is why a suicide bomber is
cowardly and deceitful, and the bomber pilot who throws bombs at innocent
people from a height of five kilometers is courageous and brave."
-- Volker Pispers (German comedian)
Mikko
2024-01-26 10:03:16 UTC
Permalink
Post by Luigi Fortunati
Post by Mikko
The elevator in free fall in the gravitational field is an inertial reference system.
Is the elevator in free fall in the magnetic or electric field also an
inertial reference system?
That depends on the magnetic and electric properties of the elevator.
The elevator is metallic, the body in the elevator is the same metal as the elevator.
The field is generated by an electromagnet.
Luigi Fortunati
[[Mod. note --
The general principal is that "free fall" means no non-gravitational
forces are acting.
An even more general principal is that "free fall" means "fall without obstacles".
Not really. It is the same or less general depending on the interpretation
of the word "obstacle".

If there is a non-gravitational interaction that does not prevent the fall
the fall is not a free fall but one may say that there is no obstacle.

Mikko
Tom Roberts
2024-01-28 00:23:06 UTC
Permalink
Post by Luigi Fortunati
[[Mod. note -- The general principal is that "free fall" means no
non-gravitational forces are acting.
Yes.
Post by Luigi Fortunati
An even more general principal is that "free fall" means "fall
without obstacles".
No. Any "obstacle" would necessarily exert a force on the object, but
some forces are not obstacles. The moderator's statement is more
general, and is indeed the accepted meaning of "freefall" in physics.

Do not attempt to redefine the meanings of common words and phrases
-- that will prevent you from communicating with other people,
especially physicists.
Post by Luigi Fortunati
No one can deny that, in the most remote space far from all gravity,
a metal elevator falling towards an electromagnet is in "free fall".
Not true: every physicist would deny that, because the electromagnet
exerts an electromagnetic force on the elevator, making it NOT be
in freefall.

You REALLY need to get a book on basic physics and STUDY it. Your
repeated questions around here are not effective in your learning
physics. Better yet, enroll in a physics course at a local college or
university, so you will have an instructor to discuss these issues.

Tom Roberts

[[Mod. note -- I second these recommendations. A few of the (many)
excellent introductory physics books available are those described in:
https://en.wikipedia.org/wiki/Physical_Science_Study_Committee
https://en.wikipedia.org/wiki/Harvard_Project_Physics
https://en.wikipedia.org/wiki/Fundamentals_of_Physics
-- jt]]
Luigi Fortunati
2024-01-29 08:20:10 UTC
Permalink
Post by Tom Roberts
Post by Luigi Fortunati
No one can deny that, in the most remote space far from all gravity, a metal elevator falling towards an electromagnet is in "free fall".
Not true: every physicist would deny that, because the electromagnet
exerts an electromagnetic force on the elevator, making it NOT be
in freefall.
In the gravitational field the force of gravity acts on the elevator in the same way as, in the electromagnetic field, the electromagnetic force acts on the metal elevator.

Force that accelerates is one and force that accelerates is the other!

Luigi Fortunati
Luigi Fortunati
2024-01-30 08:27:37 UTC
Permalink
Post by Tom Roberts
Post by Luigi Fortunati
No one can deny that, in the most remote space far from all gravity, a metal elevator falling towards an electromagnet is in "free fall".
Not true: every physicist would deny that, because the electromagnet
exerts an electromagnetic force on the elevator, making it NOT be
in freefall.
Gravity also exerts a force on the elevator.

Force is electromagnetic and force is gravitational.

Luigi Fortunati
Tom Roberts
2024-01-31 07:59:59 UTC
Permalink
Post by Luigi Fortunati
Post by Tom Roberts
Post by Luigi Fortunati
No one can deny that, in the most remote space far from all
gravity, a metal elevator falling towards an electromagnet is in
"free fall".
Not true: every physicist would deny that, because the
electromagnet exerts an electromagnetic force on the elevator,
making it NOT be in freefall.
Gravity also exerts a force on the elevator.
Not in GR, which is now the generally-accepted theory of gravitation.

But to forestall your (baseless) objection, the standard definition of
freefall says no NON-GRAVITATIONAL forces are acting on the object.

I repeat: posting zillions of questions here is not helping you learn
very basic physics. Get a good textbook and STUDY, or better, enroll in
a physics course at a college or university.

Tom Roberts
Luigi Fortunati
2024-02-01 16:44:37 UTC
Permalink
Post by Tom Roberts
Post by Luigi Fortunati
Gravity also exerts a force on the elevator.
Not in GR, which is now the generally-accepted theory of gravitation.
The fact that it is "generally" accepted does not make it an absolute
truth (absolute truths do not exist).
Post by Tom Roberts
But to forestall your (baseless) objection, the standard definition of
freefall says no NON-GRAVITATIONAL forces are acting on the object.
This standard definition is as legitimate as the one that says: a
metallic body that "falls" towards an electromagnetic pole accelerates
in the same way as the body that "falls" towards a similar center of
gravity.

The two accelerations are indistinguishable from each other (one
acceleration is as good as the other).

Are you able to demonstrate that it is possible to distinguish one
acceleration from another?

Luigi Fortunati


[[Mod. note --
The problem with the "magnetic free-fall" definition is that the
acceleration differs from one metallic body to another. For example,
a copper body will experience different magnetic forces -- and hence
a different acceleration -- than an iron body.

In contrast, with the standard definition of free-fall, free-fall is
*universal*, i.e., *all* (non-spinning test) bodies have the same
(zero) non-gravitational forces, and (experimentally) they all have
the same free-fall acceleration.
-- jt]]
Luigi Fortunati
2024-02-02 11:39:55 UTC
Permalink
Post by Luigi Fortunati
[[Mod. note --
The problem with the "magnetic free-fall" definition is that the
acceleration differs from one metallic body to another. For example,
a copper body will experience different magnetic forces -- and hence
a different acceleration -- than an iron body.
In contrast, with the standard definition of free-fall, free-fall is
*universal*, i.e., *all* (non-spinning test) bodies have the same
(zero) non-gravitational forces, and (experimentally) they all have
the same free-fall acceleration.
-- jt]]
Ok, let's call "free fall" only the gravitational one which is universal (the same for everyone) and "non-free fall" the one with the addition of electromagnetic acceleration.

On the scaffolding there are the painter and the metal robot (with the same mass) and underneath there is an electromagnet with a force equal to the gravitational one.

Obviously, the electromagnet only acts on the robot and not on the painter.

When they are on the scaffolding, the weight force of the robot is double (despite having the same mass as the painter) because it is subject to two forces: gravitational and electromagnetic.

When they fall, the robot's acceleration is double that of the painter because it is generated by two forces instead of just one.

Therefore, let's call that of the painter "free fall" (without electromagnetic acceleration) and that of the robot "non-free fall " (because there is the addition of electromagnetic acceleration).

And now we see the effects of "free fall" and " non-free fall ".

In the "free fall" of the painter, Einstein says that the gravitational force disappears (because it is not a real force) and I ask: does the gravitational acceleration of the painter in free fall also disappear like the gravitational force or does the force disappear and not the acceleration ?

And during the robot's "non-free fall", which force (gravitational or electromagnetic) disappears and which does not? And which acceleration disappears and which does not?

Luigi Fortunati
Luigi Fortunati
2024-02-05 08:10:44 UTC
Permalink
There is a well that reaches exactly the center of the Earth and there
are two elevators: one stationary at the bottom of the well (elevator A)
and the other in free fall (elevator B).

Are the two elevators inertial reference systems of the same type or is
one reference system more inertial than the other?

Luigi Fortunati
Mikko
2024-02-06 08:26:51 UTC
Permalink
Post by Luigi Fortunati
There is a well that reaches exactly the center of the Earth and there
are two elevators: one stationary at the bottom of the well (elevator A)
and the other in free fall (elevator B).
Are the two elevators inertial reference systems of the same type or is
one reference system more inertial than the other?
Luigi Fortunati
The elevator B is moving so it is affected by air resistance and other
friction effect. It also meets the fast varying tidal effects from
density variations in the matter around the well.

If these effects can be kept small enough that they cannot be detected
in B then B is in free fall.

If the well is deep enough that the elevator A does not touch its
bottom the elevator A is in free fall.

If both elevators are in free fall in the above sense then there is
no detectable difference in the conditions inside of the elevator.

When the two elevators collide neither of the elevators is no longer
in free fall.
--
Mikko
Luigi Fortunati
2024-02-06 12:21:12 UTC
Permalink
Post by Mikko
Post by Luigi Fortunati
There is a well that reaches exactly the center of the Earth and there
are two elevators: one stationary at the bottom of the well (elevator A)
and the other in free fall (elevator B).
Are the two elevators inertial reference systems of the same type or is
one reference system more inertial than the other?
The elevator B is moving so it is affected by air resistance and other
friction effect. It also meets the fast varying tidal effects from
density variations in the matter around the well.
If these effects can be kept small enough that they cannot be detected
in B then B is in free fall.
If the well is deep enough that the elevator A does not touch its
bottom the elevator A is in free fall.
If both elevators are in free fall in the above sense then there is
no detectable difference in the conditions inside of the elevator.
When the two elevators collide neither of the elevators is no longer
in free fall.
You're twisting everything I wrote.

I wrote that the well ends at the center of the Earth (the bottom of the
well is at the center of the Earth).

That elevator A is STOPPED at the bottom of the well and, therefore, is
stationary at the center of the Earth.

That elevator A stays still not because the bottom of the shaft prevents
it from falling but because (at the center of the Earth) there is no
force of gravity and no space-time curvature that sets it in motion to
go somewhere: it stays still because nothing and no one pushes him
somewhere.

That elevator B is in free fall without friction and without resistance.

That elevator B is falling (in free fall) towards elevator A which it
will collide with in the future.

I would like to know *now* (before their clash) if they are both
inertial reference systems and if they are in the same way or if one is
more inertial than the other.

I hope I have been clear.

Luigi Fortunati
Mikko
2024-02-06 14:26:36 UTC
Permalink
Post by Luigi Fortunati
Post by Mikko
Post by Luigi Fortunati
There is a well that reaches exactly the center of the Earth and there
are two elevators: one stationary at the bottom of the well (elevator A)
and the other in free fall (elevator B).
Are the two elevators inertial reference systems of the same type or is
one reference system more inertial than the other?
The elevator B is moving so it is affected by air resistance and other
friction effect. It also meets the fast varying tidal effects from
density variations in the matter around the well.
If these effects can be kept small enough that they cannot be detected
in B then B is in free fall.
If the well is deep enough that the elevator A does not touch its
bottom the elevator A is in free fall.
If both elevators are in free fall in the above sense then there is
no detectable difference in the conditions inside of the elevator.
When the two elevators collide neither of the elevators is no longer
in free fall.
You're twisting everything I wrote.
I wrote that the well ends at the center of the Earth (the bottom of the
well is at the center of the Earth).
That elevator A is STOPPED at the bottom of the well and, therefore, is
stationary at the center of the Earth.
That elevator A stays still not because the bottom of the shaft prevents
it from falling but because (at the center of the Earth) there is no
force of gravity and no space-time curvature that sets it in motion to
go somewhere: it stays still because nothing and no one pushes him
somewhere.
That elevator B is in free fall without friction and without resistance.
That elevator B is falling (in free fall) towards elevator A which it
will collide with in the future.
I would like to know *now* (before their clash) if they are both
inertial reference systems and if they are in the same way or if one is
more inertial than the other.
I hope I have been clear.
Luigi Fortunati
The elevator at the bottom is not inertial. If the well were deeper the
elevator could but the bottom prevernts that.
--
Mikko
Luigi Fortunati
2024-02-11 08:54:07 UTC
Permalink
Post by Mikko
The elevator at the bottom is not inertial. If the well were deeper the
elevator could but the bottom prevernts that.
The bottom doesn't impede anything and, to prove it, in my new animation https://www.geogebra.org/m/mdymaxsb I totally eliminate it.

Now, the well goes from one part of the Earth to the other.

Initially, elevator A is constrained near the surface and elevator B is constrained to the center of the earth: neither of the 2 elevators can move.

If you remove the constraints with the appropriate button, elevator A accelerates in free fall and elevator B remains stopped in its place.

Question: Is the unconstrained elevator B that stays in its place an inertial or accelerated reference frame?

Luigi Fortunati
Mikko
2024-02-12 11:53:21 UTC
Permalink
Post by Luigi Fortunati
Post by Mikko
Post by Luigi Fortunati
Post by Mikko
If the well is deep enough that the elevator A does not touch its
bottom the elevator A is in free fall.
I wrote that the well ends at the center of the Earth (the bottom of the
well is at the center of the Earth).
That elevator A is STOPPED at the bottom of the well and, therefore, is
stationary at the center of the Earth.
The elevator at the bottom is not inertial. If the well were deeper the
elevator could but the bottom prevernts that.
The bottom doesn't impede anything and, to prove it, in my new
animation https://www.geogebra.org/m/mdymaxsb I totally eliminate it.
The new problem is essentially different from the one discussed above.
As can be seen, the elevator at the centre of Earth (the labels are
swapped in the new problem, so now it is B) so that its centre is at
the centre of Earth, which would be impossible in the original problem
as half of it would be under the bottom of the well.

The new problem is better if the intent was to make both elevators
as inertial as possible. The additional constraint that prevents
the collision at the centre of Earth causes a small deviation from
inertiality but its direction is sideways so it does not affect
the most important motions, which are in the vertical direction.
--
Mikko
Luigi Fortunati
2024-02-12 14:42:34 UTC
Permalink
Post by Mikko
The new problem is essentially different from the one discussed above.
As can be seen, the elevator at the centre of Earth (the labels are
swapped in the new problem, so now it is B) so that its centre is at
the centre of Earth, which would be impossible in the original problem
as half of it would be under the bottom of the well.
The new problem is better if the intent was to make both elevators
as inertial as possible. The additional constraint that prevents
the collision at the centre of Earth causes a small deviation from
inertiality but its direction is sideways so it does not affect
the most important motions, which are in the vertical direction.
To avoid any possible collision (so no one gets hurt), in my animation
https://www.geogebra.org/m/etm68buf I left only one elevator.

Well, is elevator A stopped in his place (before and after the cables
broke) an accelerated or inertial reference?

Luigi Fortunati
Luigi Fortunati
2024-03-15 08:11:39 UTC
Permalink
In free fall, can you go anywhere freely or are there constraints that
prevent this?

Of course you can't fall straight up and you can't fall sideways.

In free fall you can only go in one direction (the vertical one) and in
only one versus (downward).

The elevator (in free fall) and everything inside it are forced to fall
(always) vertically and (always) downwards.

So there is a constraint.

And, in free fall, can one move in a straight and uniform motion?

No, in free fall the motion is always accelerated.

The elevator (in free fall) and everything inside it are forced to
always accelerate.

So there is another constraint.

So why call it "free fall" and not "forced fall"?

Luigi Fortunati.

[[Mod. note -- The "free" in "free fall" means that no non-gravitational
forces are acting on the falling body. It's a statement about what forces
are (not) acting on the body, not about the uniqueness or non-uniqueness
of the resulting motion. -- jt]]
Mikko
2024-03-15 19:21:00 UTC
Permalink
Post by Luigi Fortunati
In free fall, can you go anywhere freely or are there constraints that
prevent this?
There are constraints. A free fall requires empty space.
A free fall to the end of the empty space ends there.
Post by Luigi Fortunati
Of course you can't fall straight up and you can't fall sideways.
You can if you have initial motion in that direction.
--
Mikko
Luigi Fortunati
2024-03-24 08:41:44 UTC
Permalink
Post by Mikko
Post by Luigi Fortunati
Of course you can't fall straight up and you can't fall sideways.
You can if you have initial motion in that direction.
The fall (obviously) is an acceleration: you do not fall at a constant
speed.

If you fire a shot upwards, the projectile has an initial acceleration
upwards (in the barrel of the gun) but this initial movement no longer
has any influence when the projectile exits the barrel and is in free
fall.

In this last phase, the projectile "falls" (i.e. accelerates) always
downwards and never in another direction.

Luigi Fortunati
Luigi Fortunati
2024-03-22 06:13:41 UTC
Permalink
Post by Luigi Fortunati
In free fall, can you go anywhere freely or are there constraints that
prevent this?
Of course you can't fall straight up and you can't fall sideways.
In free fall you can only go in one direction (the vertical one) and in
only one versus (downward).
The elevator (in free fall) and everything inside it are forced to fall
(always) vertically and (always) downwards.
So there is a constraint.
And, in free fall, can one move in a straight and uniform motion?
No, in free fall the motion is always accelerated.
The elevator (in free fall) and everything inside it are forced to
always accelerate.
So there is another constraint.
So why call it "free fall" and not "forced fall"?
Luigi Fortunati.
[[Mod. note -- The "free" in "free fall" means that no non-gravitational
forces are acting on the falling body. It's a statement about what forces
are (not) acting on the body, not about the uniqueness or non-uniqueness
of the resulting motion. -- jt]]
What makes gravitational forces different from non-gravitational
forces?

Luigi Fortunati

[[Mod. note -- That's a very good question!

From the perspective of Newtonian mechanics, we can operationally
define "mass" (more precisely, "inertial mass") via Newton's 2nd law
*without* involving gravitation at all. That is, we can apply the same
force to different objects [e.g., attach an ideal spring to the objects,
and apply enough force to stretch or compress the spring by some
standard amount], measure the objects' accelerations with respect to
an inertial reference frame, and define
m = F/a
for each body.

Now let's introduce an ambient gravitational field. For example, we
could consider vertical motion in a given place near the Earth or some
other massive body. If we ask what gravitational forces act on different
bodies, we find experimentally that these forces are all precisely
*proportional* to the bodies' inertial masses, i.e.,
F_grav = g m
where g is the *same* for all bodies in a given ambient gravitational
field (e.g., in the same place near the Earth). That is, the gravitational
force on a body with inertial mass 2 kg is (a) precisely twice that on
a body with inertial mass 1 kg, and (b) the *same* independent of the
composition of the body. As an example of (b), let's suppose we have
3 test bodies, each with an inertial mass of 1 kg, but the 1st test body
is made of iron, the 2nd test body is made of bismuth, and the 3rd test
body is made of helium. Experimentally, the gravitational forces acting
on these three test bodies (in the same ambient gravitational field,
e.g., in the same place near the Earth's surface) are all the *same*.

In contrast, for other types of forces we do *not* have proportionality
to inertial mass, nor do we have independence of composition. For example,
if we have an ambient magnetic field, the magnetic forces acting on our
three test bodies will be (very) different.

Corresponding to the above difference in *forces*, if we apply apply
Newton's 2nd law to *motion* under gravitation vs other forces, we find
quite different results:

For motion under the influence of gravitation alone (i.e., motion
where there are no non-gravitational forces, i.e., what I've described
as "free fall"), we find
a = F/m = gm/m = g
i.e., there is a *universal* free-fall gravitational acceleration,
independent of the free-falling body's mass or composition. For example,
our iron, bismuth, and helium test bodies will all have the *same*
free-fall gravitational acceleration.

In contrast, for motion under the influence of non-gravitational forces,
there is *not* a universal acceleration. For example, in the presence
of an ambient magnetic field, our iron, bismuth, and helium test bodies
will have (very) different accelerations.

It's the *universality* of free-fall acceleration (which, via Newton's
2nd law, is equivalent to the *proportionality* of force to inertial mass)
that distinguishes gravitational from non-gravitational forces, and that
motivates defining "free-fall" as the absence of non-gravitational
forces.
-- jt]]
Luigi Fortunati
2024-03-24 20:29:00 UTC
Permalink
Post by Luigi Fortunati
What makes gravitational forces different from non-gravitational
forces?
[[Mod. note -- That's a very good question!
That is, the gravitational force on a body with inertial mass 2 kg
is (a) precisely twice that on a body with inertial mass 1 kg,
and (b) the *same* independent of the composition of the body.
-- jt]]
It is certainly true the (b) which makes gravitational forces different
from non-gravitational ones.

But (if I'm not mistaken) (a) also applies to the electromagnetic force
which, on a 2-gram body of any material, is exactly double that on a
1-gram body of any material.

Is that so?

Luigi Fortunati

[[Mod. note -- Perhaps.

That is, let's call your 1-gram body "A", and your 2-gram body "B".
We can think of B as a pair of one-gram halves (call them "B1" and "B2")
glued together. The question is, does the presence of B1 change the
electromagnetic (EM) field at B2's location, or vice versa, by an
amount large enough that we need to care about it? If *not*,
then the EM force acting on B will be the sum of
(a) the EM force acting on B1 alone (i.e., if B2 were NOT there), and
(b) the EM force acting on B2 alone (i.e., if B1 were NOT there).
Assuming that B is small enough that the EM field doesn't vary
significantly across B's diameter, we should have (a) = (b), so in
this case the EM force acting on B should be twice the EM force
acting on A.

But, if the presence of B1 *does* change the EM field at B2's location
by a significant amount, then the EM force acting on B will *not*
equal the sum of (a) and (b) above, and the EM force acting on B will
*not* be twice the EM force acting on A.

The underlying idea here is that if we want proportionality to the body's
mass, we want the body to be an electromagnetic "test body",
https://en.wikipedia.org/wiki/Test_particle
That is, we want the perturbation in the electromagnetic field induced
by the test body's presence to be very small (small enough to neglect).

In some situations electromagnetic forces are "screened", i.e., the
interior of a solid body has a different electromagnetic field than the
surface, so the net force on the body is NOT proportional to the body's
mass, but instead closer to proportional to the body's surface area.
This is often the case for oscillating electromagnetic fields
(e.g., radio waves or light), where electrical conductors screen the
oscillating fields. In such a case (where the interior has a different
electromagnetic field configuration), it's *not* clear that a body with
twice the mass will experience twice the electromagnetic force.

Another example would be superconductors in a static magnetic field:
Provided the field isn't too strong, a superconductor will screen the
magnetic field, "expelling it" so that there's zero magnetic field
inside the superconductor. I don't know offhand whether the net
(electro)magnetic force acting on such a superconductor would or
would not be proportional to the superconductor's mass.

Annother interesting example is be light pressure on a mirror: If we
reflect a light beam (a suitable electromagnetic field oscillating at
between 4e14 and 7e14 cycles/second) off a mirror (say a solid cube of
some electrically-conductive material), the light exerts a (small) force
on the mirror. If we change from (say) a 1cm x 1cm x 1cm cube of mirror
material, to a 2cm x 2cm x 2cm cube of mirror material, we have 8 times
the mass, but only 4 times the surface area and hence only 4 times the
force exerted by the light. So in this case the force exerted by the
electromagnetic field is NOT proportional to the body's mass.

For gravity there's no "screening", so being a "test body" is easy.
-- jt]]
Tom Roberts
2024-03-25 08:14:06 UTC
Permalink
Post by Luigi Fortunati
[Moderator] That is, the gravitational force on a body with
inertial mass 2 kg is (a) precisely twice that on a body with
inertial mass 1 kg,
But (if I'm not mistaken) (a) also applies to the electromagnetic
force which, on a 2-gram body of any material, is exactly double
that on a 1-gram body of any material. Is that so?
Not at all! -- If one is charged and one is not, the electromagnetic
forces on them will be VERY different.

I repeat: your approach of making false statements here and expecting to
be corrected is not working, and you are CLEARLY not learning much
physics, if any. You NEED to get some good textbooks and STUDY. Better
yet, enroll in a college or university physics course, so you will have
an instructor with whom you can discuss your many misconceptions and
confusions.

Tom Roberts
Luigi Fortunati
2024-03-25 13:04:43 UTC
Permalink
Post by Luigi Fortunati
[[Mod. note -- Perhaps.
That is, let's call your 1-gram body "A", and your 2-gram body "B".
We can think of B as a pair of one-gram halves (call them "B1" and "B2")
glued together. The question is, does the presence of B1 change the
electromagnetic (EM) field at B2's location, or vice versa, by an
amount large enough that we need to care about it? If *not*,
then the EM force acting on B will be the sum of
(a) the EM force acting on B1 alone (i.e., if B2 were NOT there), and
(b) the EM force acting on B2 alone (i.e., if B1 were NOT there).
Assuming that B is small enough that the EM field doesn't vary
significantly across B's diameter, we should have (a) = (b), so in
this case the EM force acting on B should be twice the EM force
acting on A.
But, if the presence of B1 *does* change the EM field at B2's location
by a significant amount, then the EM force acting on B will *not*
equal the sum of (a) and (b) above, and the EM force acting on B will
*not* be twice the EM force acting on A.
This thing you say also applies to the gravitational field: if the presence of B1 changes the gravitational field at the position of B2 by a significant amount, then the gravitational force acting on B is not equal to the sum of (a) and (b ) and the gravitational force acting on B is not double the gravitational force acting on A.

The proportionality between force (whether gravitational or EM) and mass depends on each individual particle because it is precisely the particle that intercepts its share of the field force (whether gravitational or EM).

Obviously (and to avoid any misunderstanding) the comparison must be made between equal materials and, that is, if the material of the first body is XYZ, that of the second body is also XYZ (if the material of the first is iron, that of the second is also iron , if one is wood the other is also wood).

Luigi Fortunati
Luigi Fortunati
2024-03-25 08:14:06 UTC
Permalink
What makes gravitational forces different from non-gravitational forces?
Luigi Fortunati
[[Mod. note -- That's a very good question!
That is, the gravitationalforce on a body with inertial mass 2 kg is (a) precisely twice that on a body with inertial mass 1 kg, and (b) the *same* independent of the composition of the body.
-- jt]]
It is certainly true the (b) which makes gravitational forces different
from non-gravitational ones.

But (if I'm not mistaken) (a) also applies to the electromagnetic force
which, on a 2 gram body of any material X, is exactly double that on a 1
gram body of material X.

Is that so?

Luigi Fortunati
Tom Roberts
2024-03-17 20:49:22 UTC
Permalink
Post by Luigi Fortunati
In free fall, can you go anywhere freely or are there constraints
that prevent this? Of course you can't fall straight up and you
can't fall sideways.
As I keep saying, this depends on the meanings of the words you use.
Your wishy-washy words are a major part of your failure to
understand very basic physics.

If by the "direction of fall" you mean the 3-velocity relative to
ground-based coordinates, that can be pointed in any direction. If you
mean the 3-acceleration relative to ground-based coordinates, that can
only be pointed straight down. Note the former is the usual meaning for
"direction" of any motion, including falling.

Hint: throw a ball straight up. It is moving upward, it is going upward,
and its 3-velocity (relative to ground-based coordinates) is directed
upward. It is, of course, in free fall (neglecting air resistance). So
one COULD say "it is falling upward", but that is such poor terminology
that no physicist would way that.
Post by Luigi Fortunati
In free fall you can only go in one direction (the vertical one) and
in only one versus (downward).
This is just plain not true (here you use words with more definite
meanings). You can be GOING up or sideways -- that depends on the
initial conditions of your trajectory. Because "going" explicitly refers
to velocity, not acceleration. Your acceleration (relative to
ground-based coordinates) is always downward.
Post by Luigi Fortunati
The elevator (in free fall) and everything inside it are forced to
fall (always) vertically and (always) downwards.
Nope. See above.
Post by Luigi Fortunati
So there is a constraint.
There is a constraint on the 3-acceleration (relative to ground-based
coordinates): downward. There is no constraint on the direction of
3-velocity (relative to ground-based coordinates), because this depends
on the initial conditions; of course its direction will point
increasingly downward as time goes on.
Post by Luigi Fortunati
And, in free fall, can one move in a straight and uniform motion?
Yes, RELATIVE TO COORDINATES ACCELERATING DOWNWARD WITH YOU. No,
relative to ground-based coordinates.

[This opens the door to the "local vs. global"
distinction in GR. You have no hope of appreciating
the subtleties involved until you STUDY.]
Post by Luigi Fortunati
No, in free fall the motion is always accelerated.
Again, this depends on what you mean by those words. In Newtonian
mechanics this is true. But we live in a post-GR world, and presuming
Newtonian mechanics is not appropriate. In GR, of course, an object in
free fall follows a geodesic through spacetime, with ZERO proper
acceleration.
Post by Luigi Fortunati
So why call it "free fall" and not "forced fall"?
Because as the moderator said, it means that no NON-GRAVITATIONAL force
acts on the object. "It's a statement about what forces are (not) acting
on the body, not about the uniqueness or non-uniqueness of the resulting
motion."

As I keep telling you, your approach of making false statements in this
newsgroup is utterly failing to teach you basic physics. You MUST get
some good textbooks and STUDY. Better yet, take a college or university
course in physics so you'll have an instructor with whom to discuss your
confusions.

Tom Roberts
bertietaylor
2024-04-05 11:23:52 UTC
Permalink
Post by Luigi Fortunati
The elevator in free fall in the gravitational field is an inertial
reference system.
Is the elevator in free fall in the magnetic or electric field also an
inertial reference system?
Luigi Fortunati
That will depend upon the corresponding momentum of the gravitons
involved in the free fall. The fabric of spacetime will bend accordingly,
to account for the apparent distortions to the so-called magnetic and
electric fields.

bt

[[Mod. note -- We don't need the concept of "graviton" here -- a
classical picture is fine.
-- jt]]

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