SinghRP said:
To PeterDonis: Answer to "What makes you think that?
I get lost in mathematical jungle. I am a physicist, and I like to think in terms of physical models. I recall Feynman also: The glory of mathematics is that you don't have to say what you are talking about.
The genesis of POE is in:
(Gravitational mass) (Gravitational field intensity) = (Inertial mass) (Acceleration).
(I agree with your "corrected" stateent.) But my statement is just an interpretation the above.
Another expression:
(Electical charge) (Electrical field intensity) = (Inertial mass) (Acceleration).
I may interpret it as: an electrical charge at rest in a frame is equivalent to being in an "electrical field" in an acclerating frame.
Not quite. Mathematically,
M_{inertial}\ \vec{A} = M_{grav}\ \vec{g}_{grav}
where \vec{g}_{grav} is the gravitational field.
Since M_{grav} = M_{inertial}, you can divide through by M_{grav} to get:
\vec{A} = \vec{g}_{grav}
Which means that acceleration due to gravity is
universal, the same for all objects, regardless of their mass, or what they are made out of, or whatever. That's the principle character of "fictitious" or "inertial" forces such as the "g" forces that arise in an accelerating rocket. This means that gravity can be understand as locally equivalent to a fictitious or inertial force.
In contrast, if you start with the force due to electric fields, you have:
M_{inertial}\ \vec{A} = Q \vec{E}
where Q is the electric charge, and \vec{E} is the electric field.
If you do the same trick of dividing through by M_{inertial}, you find:
\vec{A} = \dfrac{Q}{M_{inertial}} \vec{E}
So the acceleration due to electrical forces is
not universal; the acceleration depends on the charge-to-mass ratio. So electrical forces
can't be interpreted as inertial, or fictitious forces, since they accelerate different objects in different ways.