Re: Hafele & Keating, Einstein, Dingle,Cocke, and Scott Murray

From: John Kennaugh (JKNG_at_kennaugh2435hex.freeserve.co.uk)
Date: 08/03/04 

Date: Tue, 3 Aug 2004 19:03:26 +0100

Been away for a week

Harry <harald.vanlintel@epfl.ch> writes
>This is the third "last time" and there is a saying - so this is it!
>Maybe words don't help...
>
>John Kennaugh <JKNG@kennaugh2435hex.freeserve.co.uk> wrote in message
>news:<6cXaSyVDKYABFw4y@kennaugh2435hex.freeserve.co.uk>...
>> Harry writes
>> >OK then - still one more remark. ;-)
>> >
>> >You wrote that in the inertial frame, g=0 AND a little further, that a=g
>> >whereby a = NOT 0....
>> >In the inertial frame, all definitions are relative to that frame: g is NOT
>> >relative to the earth but to the fictive inertial frame.
>> >Thus NOT g=0 but g= GM/R^2 and in free fall, a= v^2/R and a=g.
>> >
>> >Your weight is taken in the rotating frame of the earth. Just measure it in
>> >an inertial frame - that's possible using a low-flying airplane...
>>
>> I think you are confused.
>
>You snip without comment your self contradiction and then claim that I
>am confused! :-))
>
>> The reference frame is the one to which all
>> observations are referred. The reference frame from which I am making my
>> observations has within it something which is rotating which I am
>> observing. I cannot stop it rotating and weigh it and say that that is
>> its weight in my FoR. It isn't.
>
>You can and it is.

If I stop It rotating its weight becomes mMG/R^2. If it isn't rotating
and that is its weight it will move towards the centre of the earth and
accelerate at MG/R^2. You cannot say that in the rotating FoR it stays a
fixed distance from the centre of the earth while in the ECI it moves
towards the centre and hits the earth. You cannot change the net outcome
by changing the FoR.

  Either it stays a fixed distance from the centre of the earth or it
doesn't. In neither FoR does the mass get nearer to the centre of the
earth. Therefore in both FoR the weight is zero. Weight = mg, Mass
hasn't changed, therefore in both FoR g is zero.

When one talks of a FoR it is in effect the reference frame you consider
to be stationary. The whole concept of a "Rotating Frame of Reference"
is a nonsense. If an observer in a rotating FoR assumed he was
stationary i.e. did not acknowledge his rotation then he would develop
an entirely different set of laws of Physics as per the geocentric
universe. How would he explain a geostationary satellite? Perhaps that
the effect of gravity diminished with height and becomes zero at a
particular height.

We abandoned the geocentric model of the universe because the laws of
physics make more sense if we accept that the earth is rotating. That is
the same as referring everything to a non rotating inertial FoR. I
really do not understand at all what you mean when you talk about the
'rotating frame'. As far as I can see there is no such thing.

> But it's easy to describe the weight in the
>rotating frame,

I do not wish to describe anything in the 'rotating frame'. I refer
everything to the non rotating ECI FoR. In the ECI FoR the mass does not
get progressively nearer to the centre so its weight in that FoR must be
zero so g in that FoR must be zero.

> as expressed in values of the stationary frame: F=
>m*(g-a). That weight is 0 in free-fall, which only works for g=g0, and
>g= NOT 0 as you claim in free-fall.

Don't understand.

>
>> You have to observe it as it is with all
>> the forces acting upon it as it is. i.e. I can see that that mass is
>> weightless because of the way it is behaving so for that mass its
>> acceleration towards the centre of mass (down) is zero both from the
>> point of view of an observer travelling with it and from my point of
>> view in my non rotating ECI FoR.
>
>Weightless = no force, that's OK.

No weightless means that all the force is being utilised to make the
mass accelerate in a circle so that there is none left so g = 0 so
weight = m x 0 = 0 = weightless.

>Zero acceleration relative to the rotating frame = 10m/s^2 in ECI
>frame.

I do not know what you mean by 'rotating frame'

>
>> If it is free from restraint and its
>> acceleration is zero then g = 0. That is what g means.
>
>Hint: if in the inertial frame a=0 than the object is in inertial
>motion. Just answer this simple question: Is it?

The mass is not in inertial motion. Never was in any FoR. That is why
you refer everything to the ECI FoR. It is the hypothetical observer who
is inertial. The mass doesn't become inertial by observing it from an
inertial frame but the forces acting upon it make consistent sense only
when considered from the POV of the ECI FoR or even better the sun
centred inertial FoR. 

-- 
John Kennaugh
to email convert the number from hex to decimal

Relevant Pages