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non-inetial: part IV




I want to start the non-inertial frame thread from a
clean slate. There has been so much quoting of quotes that
its too easy to lose track of what has been discussed. So I'll
repeat my basic claim with quasi-proof. There are equations
here and in the follow up post; I don't know how well they are
going come out since there exist undetermined carriage
returns and ascii is lousy for writing equations; please
bear with it.

I claim that one can apply the 2nd law in a non-inertial frame
as long as one includes terms that represent what I call
"kinematical" forces. Others call them "fictitious" , others
"pseudo" others "apparent". In order to be neutral between
Marlow and me, I'll call them "apparent" for the time being.

Proof: All the steps leading up to equation 4-120 in Goldstein's
2nd edition. (note: the centrifugal and coriolis terms may have
to be replaced with different "apparent" force terms if the case
warrants it.).

If anybody disagrees with this equation let them say why and
where it is wrong. I repeat it here, primed quantities are
quantities measured in the non-inertial frame of reference
( I have paraphrased it), I also assume people will
understand where vector quantities
are implied (ascii is lousy for equations):

F' = sum of forces, ...(1)
where F' is ma' and is the effective sum of forces in the
non-inertial frame. Calling F', ma' , is what Marion (3rd ed.)
does with the F', see page 342.

Now let's prove that a consequence of this is that we may apply
the work-energy theorem in the non-inertial frame. (We might do
this in order to solve for the final speed of the object in
question, as measured in the non-inertial frame, i.e. v'_f;
where _f denotes subscript)

"multiply" (technically dot it into both sides) both sides of (1)
by dl', where dl' is a directed line segment, which is part
of the trajectory of the object (for whom we are applying
the 2nd law)

Now intergrate both sides along the trajectory from the intial
to the final postion. Giving,

Int ( F'*dl') = Int (sum of forces * dl') . . . (2)

Note: the rhs of (2) is the total work done by all the forces and
includes work terms from the apparent forces. In fact, if there
are no "real"forces presents then this is work done solely
by apparent forces.

Now, following the same mathematical manipulations that one
does to prove the work-energy theorem in inertial frames we
replace
F'*dl' with m*v'*dv'. This follows from the fact that
F'=ma'=mdv'/dt.
Hence we get,
Int(m*v'*dv')
= work done by both real and apparent forces . . . (3)

The integration is trivial and we arrive at

0.5*m*v'_f*v'_f - 0.5*m*v'_i*v'_i
= work done by all forces. . .(4)

where v'_f = final velocity and v'_i = initial velocity.

in others words

Delta (kinetic energy)' = work done by real and apparent forces,

All quantities computed in the non-inertial frame.

hence we have work-energy theorem applied in the
non-inertial frame.

Note: in order for this to be correct and consistent we must
include in the work terms of the apparent forces. i.e. those
terms produce measurable effects (effects measured in the
prime frame only!!). Namely, they contribute to the quantity
that is the lhs of (4).

Conclusion: applying the 2nd law as the equations of motion
in the non-inertial frame correctly produces the trajectory
of the object as measured in the non-inertial frame as long
as apparent forces are included in the sum of forces term.

I'll give a specific example, worked out in detail, of the work
produces the correct change in "kinetic" energy in a future post.

Joel Rauber
rauberj@mg.sdstate.edu