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Re: [Phys-l] g...



At 08:32 -0500 11/19/06, Herb Gottlieb wrote:

1) Why "g " is not an acceleration .
Doesn't "g" refer to something that changes velocity as it travels?

No. g refers to the strength of the local gravitational field (and in the case of the earth's surface, it has a small correction normally incorporated for latitude and the earth's rotation, but this is small enough that it can be neglected for most situations).

2) What is the "irrelevant accident" of physics that you cite here?

It is the fact that inertial mass and gravitational mass appear to be equal. I know of no principle of physics that requires this to be true. Because of this fact, the proper field units for g, N/kg, reduce to m/s^2. But g is used in many cases where there is no acceleration occurring, such as in determining the force of gravity on an object (sometimes called its weight), mg. Calling g an acceleration is one of the things that I think makes the whole concept of g very confusing to introductory students.

3) Why isn't the same true of electricity?
Won't an isolated electron in outer space be accelerated similarly when
it happens to be in an electric field?

Yes, but it accelerates in proportion to both its charge and mass. The units of the electric field, in analogy with those of the gravitational field are N/C, which works out to m kg/C s^2. so an elctron and an anti-proton, both having the same charge, but different inertial masses will accelerate differently in an isolated electric field. The same particles, however, will accelerate identically in an isolated gravitational field, because the field quantity that they carry in this case, is the same as their inertial mass. If we think of the gravitational field unit as m kg-inertial/kg-grav s^2, then the symmetry with the electric field units becomes obvious. Because of the equivalence of inertial and gravitational mass all objects accelerate the same in a uniform, isolated gravitational field, but because of the non-equivalence of "electrical mass" and inertial mass, the same is not true of objects in an electric field.

Hugh
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Hugh Haskell
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<mailto:hhaskell@mindspring.com>

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