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From: Luigi Fortunati <fortunati.luigi@gmail.com>
Newsgroups: sci.physics.research
Subject: Re: Free fall
Date: Sun, 24 Mar 2024 13:29:00 PDT
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Approved: Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com (sci.physics.research)
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Luigi Fortunati il 21/03/2024 17:13:41 ha scritto:
> 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]]