Re: operational F, m, and a

["John S. Denker" <jsd@MONMOUTH.COM>]

At 04:21 PM 10/17/01 -0400, Michael Edmiston wrote:
> John Denker said:  If you define F in terms of ma, then F=ma becomes a
> tautology.  That means it becomes impossible in principle to even imagine an
> experiment to determine how accurately F equals ma.  It becomes impossible
> in principle to design an experiment to check for small violations of the
> F=ma law.
>
> I am still trying to decide whether I think that is really a problem or not.

I'm not sure it's a problem.  My intuition tells me it's a problem, but I
can't at the moment prove it, so I'm sticking (for the moment) to a lesser
statement:  The possibility of choosing to define F:=ma does not exclude
other choices.

> If we defined F=Gmm/r^2 then wouldn't that become the tautology.

Yes, it would.  This supports my point that defining F:=ma is a non-unique
choice.

> Don't we have exactly the same problem with the
> spring scale (fish scale) with the tautology F=kx?

Excellent question.  Actually, that's two questions, with two different
answers:

  1) Defining F:=kx would be just as problematic.  Actually more so, since
kx doesn't do a very good job of describing real springs.

  2) But if you read the fine print, the fish-scale approach does not
depend on F=kx.  Start like this:
  -- Have a fish scale with only one graduation-mark.  Extend the spring
so that the pointer lines up with this mark.  This arbitrarily defines the
unit force;  this is the primary standard.
  -- Use the foregoing unit force to pull on another fish scale.  Invoke
the third law of motion to argue that the two forces are equal and
opposite.  Put a mark on the second scale.  It is now a secondary standard
for the unit force.
 -- Use the two of them to pull on a third.  Use the vector sum laws to
calibrate the new scale, for all values from zero to two units.
 -- And of course we need to keep all our scales in the elastic regime,
far below the elastic limit.  Operationally, check for reproducibility and
lack of hysteresis.

You get the idea.  I'm using Newton's laws and the vector axioms to avoid
needing any new axioms such as F:=ma or F:=GmM/R^2.

My intuition is "the fewer axioms the better".  If you add a new axiom such
as F:ma, you run the risk of being inconsistent with the old axioms.  If
your axioms are inconsistent, you have created a theory that describes
nothing, a fancy book about the properties of the null set.  Disclaimer:  I
have not proved that F:ma is inconsistent with the other axioms of
physics;  intuition yes, proof no.


> Same subject... different line of thinking.  This question is operational in
> nature.  One criticism about F=ma is it isn't always correct, i.e. when
> relativity comes into play.  But F=kx suffers a similar problem... not from
> relativity, but from non-linearity in the spring, especially as we approach
> the elastic limit.  If F=kx is not totally true, how do you rule your scale,
> or is the scale only valid at just one position?  If we do rule the scale,
> say every 0.1 N from 0 to 10 N, and we do this in a linear fashion, how
> would we ever know if it is correct?  Note, if people try to answer this
> question I believe we should disallow hanging masses from the scale because
> then we would really be using the weight approach to defining force rather
> than the spring-scale approach.

Those are the right questions to ask.  Very incisive.

Yes, the F:=kx approach would have lots of problems.  That's why we need
the elaborate calibration procedure described above.  The primary standard
is valid at only one position.