Re: [Phys-L] gravity, weightlessness, etc.

[John Denker <jsd@av8n.com>]

On 01/19/2018 10:55 PM, Richard Tarara wrote:

> But your _sensation_ of your weight IS the force UP.

I agree with all of that except the "But".  We have to
be careful, because this is one of those glass-half-full
versus glass-half-empty scenarios.

 a) Sure, it is 100% correct to say that your nerve-cells
  respond to an upward force exerted *on* your feet.
 b) It is equally correct to say that they respond to a
  downward force exerted *by* your feet.

The third law guarantees that you can't have one without
the other, so it would be a Bad Idea to use (a) to argue
against (b) or vice versa.

Weight is a downward vector by convention, but the nerve
cells know nothing of this convention.  Conversely, by
convention laws such as F=ma are formulated in terms of
the force exerted /on/ (not by) the object.  So it's a
mess.  This is another part of the reason why I wouldn't
want to "define" weight in terms of force on the feet;
it's too easy to mix up the minus signs.

Taking another step down that road, in complicated situations
(e.g. fluid dynamics) you could in principle keep track of
the forces acting on each surface of each parcel of fluid.
I actually worked this out once.  It was a nightmare.  It is
verrrry much easier to keep track of the momentum flowing
in and out, which is what sane people do.

> At least you can maintain a Newtonian view if you always analyze 
> rotating systems from the stationary observer and then try to explain
> the sensations of someone in that rotating system.

That's fine as a starting point.  You can't teach everything
at once, but you have to start somewhere, and nonrotating
frames are the obvious place to start.

The rule is, you are free to choose whatever frame you like.
Other folks are free to choose differently.

In particular, pilots are going to use the frame attached
to the airplane.  It's really much simpler than asking some
Newtonian bystander to analyze the situation.  Students can
handle this just fine, even students who have never taken
high school physics:  2 gees in a steeply-banked turn, 4
gees at the bottom of a loop, 0 gees at the top of a loop,
et cetera.  You don't even need an airplane for this;  a
playground swingset suffices to give the general idea.

> Otherwise, as far as I can tell, you lose the the third law since 
> there is no object acting on you outwards to be the agent of the 
> centrifugal force.

That raises a super-important issue.

I consider the third law to be a statement that momentum is conserved.
And the fact is, momentum *is* conserved.  Period.  You cannot "lose"
the third law, no matter what reference frame (if any) you choose.
It would be better to say that in an exotic frame, the equation that
/expresses/ conservation of momentum gets ugly.

By "exotic" I mean to include rotating frames and other accelerated
frames, and other much-less-fancy things as well.  A good pedagogical
place to start is this:

In two dimensions, using rectangular Cartesian coordinates, conservation
of momentum is:
	m dx/dt = constant
	m dy/dt = some other constant

Now let's switch to polar coordinates.  No rotation, just the plain
old plane, with different coordinates.  Except maybe in exceptional
trivial cases, conservation of momentum is not
	☠ m dθ/dt = constant ☠
nor
	☠ m dr/dt = constant ☠

Obviously momentum is still conserved;  you just have to work harder
to formalize it.  There's no big mystery;  it's just work.

This has nontrivial real-world implications.  If you take off from
Juneau with a perfectly-aligned gyrocompass it will be off by a
huge amount by the time you get to Anchorage, for reasons having
nothing to do with the rotation of the earth, just because of the
polar coordinates that define the customary notion of North.

If you just sit there and watch the thing it "looks like" momentum
is not conserved (in this case the angular momentum of the gyro)
but really that's not what's happening at all.  The real momentum
really is conserved.

=========

> I think this may be why some people (I did when teaching) go for the
> 'apparent weight' approach.

That seems like a non-sequitur and a non-issue to my way
of looking at things.

There would be huge issues if you put g instead of δg on the
LHS of the equation of universal gravitation:

                      G M
	δg = - rhat -------			[1]
                      r^2

but there are multiple reasons for preferring the δg formulation,
starting with the fact that you don't need to bother with
"apparent" weight or "apparent" weightlessness.  You can set
g=0 relative to the frame comoving with the space station and
everybody is happy.  Equation [1] still holds, the third law
still holds, et cetera.  Also you don't run afoul of the
equivalence principle.  Also you can easily explain why the
terrestrial laboratory g is measurably different from what
the bogus (g=...) version of equation [1] would predict.
And so on.