Re: [Phys-l] g...

[Hugh Haskell <hhaskell@mindspring.com>]

At 11:09  -0500 11/19/06, John Denker wrote:
> I need the next level of detail here.  Taken separately, several
> steps in that argument make sense.  Taken collectively, they don't
> add up;  they don't suffice to support the conclusion.
OK. I'm leaving out too many steps here. The reason I assert that g 
should not be considered an acceleration is because we can consider 
the general case of a field acting on an object through some 
mechanism, call is the "field-susceptible property." In the case of 
gravity, that is gravitational mass. In the case of electricity that 
is charge (for the moment, call that "electrical mass" merely to 
illustrate the analogy). In the general case the force on an object 
subject to that particular field is

Force = (Field susceptible property) x (Field strength)

Based on this, we can define the field strength in terms of its 
action on a susceptible object as

Field Strength = Force/(Field Susceptible Property)

with units corresponding to the units on the right hand side above as 
Newtons/(amount of Field Susceptible Property present).

According to N-2, the acceleration of said object if the field under 
discussion is the only one present is

acceleration = Force/(Inertial Mass)

and the units for this acceleration are the traditional m/s^2.

We can calculate the acceleration for objects in an generic field as

acceleration = (Field Susceptible Property) x (Field strength)/(Inertial Mass).

In a gravitational field this works out to

(gravitational acceleration) = (gravitational mass) x (grav. field 
strength)/(inertial mass)

and in an electric field it is

(electrical acceleration) = (electrical mass [AKA charge]) x (elect. 
field strength)/(inertial mass).

Now we can go to the laboratory and very carefully measure the 
acceleration of many different objects in a gravitational field (I 
will leave the details of constructing experiments that measure only 
the effects of the gravitational field on the objects to the reader). 
And we find that in every case the acceleration for the same field 
strength is the same, depending only on the strength of the field.

A reasonable conclusion here is that there is no distinction, at 
least in this set of experiments, between the gravitational mass and 
the inertial mass. So we drop the distinction, cancel the masses from 
the acceleration equation, and find that the acceleration is 
numerically equal to the strength of the gravitational field, as we 
have defined that quantity.

Of course a similar set of experiments in electric fields will give 
dramatically different results, showing that what we have been 
calling electrical mass is not equivalent to inertial mass, so we 
cannot cancel those two objects out of our acceleration equation. So 
we rename the electrical mass something else, say, "charge," give it 
some appropriate units, say, Coulombs, and then not that the 
acceleration in an electric field is the product of the field 
strength and the charge to mass ratio.

I use the electric field to illustrate that the formalism of finding 
accelerations in both is identical, but there is an important 
difference. I could have called gravitational mass gravitational 
charge to illustrate the similarity of the formalism and then I would 
have found that gravitational charge is indistinguishable from 
inertial mass, and proceeded as above.

I guess I need to concede that I didn't prove that the gravitational 
field strength is not an acceleration--I wasn't trying to do that. 
The point I was trying to make was that calling it an acceleration is 
not a useful way to introduce the subject, even though it is 
possible, because of the principle of equivalence, to reduce the 
generic expression for gravitational acceleration to

(gravitational acceleration) = (gravitational field strength).

The reason I don't think this is a good idea is that the 
gravitational field strength is very often used in connection with 
processes that have nothing to do with acceleration, such as what is 
commonly called weight (even though I also think weight is not a 
useful term, but that is a different argument for another day). It is 
this fact that causes students to become confused to the point that 
every year I have one or two who become convinced that all 
accelerations are equal to g. So I agree with those who insist on 
giving the units of g as N/kg, and would go even further to argue 
that subscripts for acceleration need not be used. Acceleration is 
acceleration, whether it comes from gravity or electricity or 
whatever. Students need to learn than gravity has this convenient 
property that enables many calculations to be made more simply that 
parallel ones in electricity because, in the case of gravity the 
"charge to mass" ratio is always simply 1. But aside from that 
simplification the process is the same for electricity as it is for 
gravity.

Now I have no idea why gravity has this unique property. I don't 
think it is required by any principle of physics that I know of, and 
it seems to me that making the case that the formalism for 
calculating forces due to generic fields is the same for all fields 
(neglecting things like vector products and tensor forces and the 
like that come up later), and one shouldn't become complacent because 
gravity *seems* different from the others. It isn't. It just has a 
principle of equivalence that the others don't.

It would not surprise me if it turned out that the principle of 
equivalence is somehow related to the fact that we have not yet 
figured out how to incorporate gravity into quantum mechanics.

Hugh
--

************************************************************
Hugh Haskell
<mailto:haskell@ncssm.edu>
<mailto:hhaskell@mindspring.com>

(919) 467-7610

When you are arguing with a stupid person, it is a good idea to make sure that
person isn't doing the same thing.
					Anonymous