Why classics assumed that the force constant in two different references frames?

AI Thread Summary
The discussion revolves around the assumption of constant force in different reference frames according to Newton's laws before the advent of modern physics. It highlights that while acceleration (a) is derived from position (x), it can still be constant in uniformly moving frames due to the nature of differentiation. The transformation equations show that the acceleration remains the same across frames because the term involving frame speed (v) becomes zero when differentiated twice. Participants clarify that the relationship between acceleration and position does not imply that acceleration must vary with position. The conversation concludes with an acknowledgment of the mathematical understanding, even if the logical interpretation remains challenging.
najat
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hi every one :)

i need a small help please ...
we have tow frames and tow observers , let suppose there is a force on an object in one of the frames , so from Newton law:
f=ma

"a" depend on "x" distance which is not constant in the other frame , so why they assumed that it is constant as the mass ?!
of course i am talking about the period before modern physics of Einstein and the electromagnetic theory.

thanks a lot ...
 
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it is true only for two uniformly moving observers.for them,
x'=x-vt
y'=y
z'=z
t'=t,where v is constant.double differentiation w.r.t. time shows the equality of forces.it is non relativistic version.
 
thanks andrien for the reply :)

one more quastion please:
"v" in that equations is the frame speed
what about the "a" for the body under the force ? this is what confuse me
a=dv/dt
v=x/t
so v depend on x ! ... how "a" can be constant?
 
a is the second derivative of x. This does not mean that a depends on x. (it may but it does not have to). Same way as the velocity does necessary depend on position even though it is v=dx/dt (and not v=x/t). There is such a thing as motion with constant velocity, isn't it?

Back to the original question, they don't assume it, it follows from the transformation equations (see andrien's post).
The acceleration in the moving frame is a^'=\frac{d^2x'}{dt^2}=\frac{d^2(x-vt)}{dt^2}=\frac{d^2x}{dt^2}=a
This is so because when you take the second derivative in respect to time of the term vt the result is zero (the first derivative is v which is a constant).
 
thank u nasu ...i v got it mathematically ...but not logically :)
i know i have to train my brain to imagine it ...
have a nice day ^_^
 
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