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Re: [Phys-l] PER folks and units of g



Ken Fox wrote:|
|I agree with Jeffrey's as the best solution. In my mind there is a
huge
| conceptual difference between a(g) and g. It is a wonderful point of
nature
| that they are numerically equal and even the units are equivalent
(pun
| intended). I tweak the brain by indicating that one day they will
appreciate
| this coincidence.

Perhaps one should here explicitly summarize the situation:

1) According to N2, the acceleration of a particle subject to an
external force field is given by:
a_F = F/m_i, where F is the force exerted on the particle by the
field, and m_i is the particle's inertial mass.

2) For an electric force field, this becomes:
a_e = e*q/m_i , where e is the electric field intensity (electric
force per unit charge), and q is the charge of the particle.

3) For a gravitational force field :
a_g = g*m_g/m_i , where g is the gravitational field intensity
(gravitational force per unit gravitational mass), and m_g is the
particle's gravitational mass (gravitational "charge").

4) The result in (3) reduces to a_g = g , ie the particle's
acceleration in a gravitational field is numerically equal to the
gravitational field intensity- the same value for all particles
subjected to this gravitational field, independent of the particle's
mass value(s). We can take this final step only because the ratio
m_g/m_i is the same for all particles, so that we can freely choose to
define this ratio to be 1. We cannot reduce the result in (2) in the
same way because the electric charge to inertial mass ratio (q/m_i )
varies in value from particle to particle.

5) This "coincidence" (m_g/m_i => 1, for all particles) was enshrined
by Einstein as the Principle of Equivalence and was exploited in his
1916 General Relativity model. . . .

Bob Sciamanda
Physics, Edinboro Univ of PA (Emeritus)
www.winbeam.com/~trebor
trebor@winbeam.com