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

From: John Kennaugh (JKNG_at_kennaugh2435hex.freeserve.co.uk)
Date: 07/16/04 

Date: Fri, 16 Jul 2004 21:36:42 +0100

Harry writes
>John Kennaugh <JKNG@kennaugh2435hex.freeserve.co.uk> wrote in message
>> Harry writes
>> >"John Kennaugh" <JKNG@kennaugh2435hex.freeserve.co.uk> wrote in message

[.........]

>> Can we start by clarifying what we mean by 'potential' in case I am
>> understanding that wrongly.
>
>Maybe I am wrong, or there are different definitons. I mean it the
>same way as potential in Voltage.
>
>> Suppose you have two points
>>
>> A _
>> ^
>> h
>> B _|
>>
>> Distance h is small compared to the radius of the earth R so the force
>> of gravity at both A and B would be near enough the same. If you drop a
>> mass from A to B then it will accelerate at g and acquire an amount of
>> energy mgh. The potential energy of a mass m at A relative to B is the
>> amount of energy you would get if mass m fell from A to B - also mgh.
>
>In which frame are you, and to which frame apply some of the other
>discussions?

I'm not sure that frames come into it at all but I believe I am an
inertial observer stationary w.r.t the axis of the earth. I think the
above is a completely general statement provided h is the vertical
displacement and g is the measured acceleration at that point in the
universe and provided g is near enough the same for A and B.

>>From the below I know, but in earlier remarks you cited equations that
>apply to the rotating frame. I think that's a main cause of confusion.

I don't really understand what you mean by the rotating frame. Speed is
relative, there is no such thing as absolute linear motion but you can
tell whether or not you are rotating. I don't see that it matters
whether I am a non rotating observer viewing the earth or whether I am
an observer on the earth well aware that the earth is rotating. The
equations are the same.

>In the rotating frame, there is also a centrifugal force and a
>Coriolis acceleration,

It is rather a long time since I was at school but having thought about
it there is not and never was, in physics, anything called centrifugal
force. i.e. a force away from the centre of rotation.
In popular culture there is. One sees the mass on a piece of string
apparently pulling away from the centre but if you cut the string it
does not fly in a direction away from the centre it goes in a straight
line at a tangent to the circle. The direction it was travelling in. The
force is centripetal - towards the centre - forcing the mass to go in a
circle when its inertia wants it to go in a straight line.

>whereby the centrifugal force reduces g and the
>potential, while in the ECI frame the potential is purely from
>gravitation, and the force is reduced by the centripetal acceleration.
>Most textbooks relate to the rotating frame when they discuss "g".

This makes no sense to me.

A given 'object' has mass and that is a constant.

That mass may be anything from weightless (in orbit) of have a
measurable weight. Weight = mg where g is not a constant it is
acceleration due to gravity at the point you are measuring the weight.
You can compare g at different points by measuring the weight of the
same mass at those points.

  The force of gravity = G m(1)m(2)/d^2

G, the universal gravitational constant is a constant (obviously). If
we make m(2) = M the mass of the earth then the force on a mass m is

        Force of gravity = GmM/r^2

Where r is the radial distance from the centre of the earth. The force
of gravity is not affected by rotation or your FoR. v does not appear in
the formula.

        Force = mass x acceleration

        acceleration of mass m = Force / mass = GM/r^2

  Again totally independent of rotation or FoR.
  Acceleration is always in the direction of the force producing it.

Assuming the c of g of the earth is at its centre (it isn't quite but
keep it simple)

At the pole at mean sea level
  the total acceleration due to the force of gravity
                        = GM/(R(p)^2)
     As there is no centripetal acceleration this = g(p)

where g(p) is the local value of g at the pole.
independent of FoR.

At the equator (mean sea level)
total acceleration due to the force of gravity
       = GM/(R(e)^2)

Here there is centripetal acceleration so
        GM/(R(e)^2) = g(e) + 1/2 v^2/R(e)

The total acceleration due to the force of gravity has to be divided
between the centripetal acceleration and the local value of g = g(e).

  The weight of mass m = m g(e). If you let it fall it will accelerate at
g(e) towards the centre of the earth.

  The potential energy between top points at that location = m g(e) h
where h is the vertical displacement between the two points.

The gravity potential difference between those points is g(e) h.

Gravity is the only force involved. All acceleration is due to gravity.
All acceleration due to gravity is in the direction we call 'down' i.e.
towards the centre of mass of the earth.

Again I don't see how changing your FoR changes anything. If instead of
being an impartial (inertial) observer looking at the earth from space I
was located on the surface of the earth at the equator I would measure
the value of g as g(e). If I knew the dimensions/mass of the earth the
equation:

              GM/(R(e)^2) = g(e) + 1/2 v^2/R(e)

Would still apply because I am smart enough to know I am rotating.

>
>> The gravitational potential at A relative to B is ether what you would
>> have to multiply mass by to get potential energy or the energy which
>> would be produced by a unit mass accelerating from A to B.
>> What is confusing is that one tends to think of rotation causing
>> 'centrifugal force' a mechanical force acting outwards opposing gravity.
>> That is completely wrong.
>
>Ok. It is wrong for the ECI frame. Now I'm "with you"!

I understand ECI stands for Earth Centred Inertial Frame which is where
I am throughout. You make a statement later which worries me you say:

>That is only true in the rotating frame with centrifugal force, which
>you just denied.
> As I told you before, in that frame both masses have
>zero speed, so the equivalence perspective seems to work fine like
>that as well.

You seem to be saying that to an observer in the ECI frame all masses on
the surface of the earth appear stationary. This would imply the
observer is rotating with the earth. Surely 'I' stands for 'inertial' so
an observer in the ECI frame is not rotating with the earth the earth is
rotating about him. There is no *inertial* frame where both masses have
zero speed. One of us is confused and we need to sort out who it is :o)

>
>> The only force is gravity inwards causing both
>> centripetal acceleration of the mass towards the earth and the falling
>> acceleration of the mass towards the earth. Acceleration is always in
>> the direction of the force causing it and there is in this case only one
>> force - gravity.
>
>Fine.
>
>> A force acting on a mass produces a fixed amount of acceleration. If
>> some of that acceleration produced is in the form of centripetal
>> acceleration towards the centre of the earth then it leaves what remains
>> of that fixed amount to cause the mass to fall to the earth. Note - once
>> something is rotating it does not require any force to keep it rotating,
>> nor can it create a force. There are no other forces, (mechanical or
>> otherwise) other than gravity involved just different allocation of the
>> total acceleration which can be produced by gravity.
>>
>> If A and B are on the equator and are rotating about the axis of the
>> earth then the force of gravity is already causing a unit mass at A to
>> accelerate towards B at v^2/R (centripetal acceleration). When we let it
>> fall to B the net acceleration ge = go - (v^2/r) where go is the total
>> acceleration which gravity can produce MeG/(R^2) or what the
>> acceleration due to gravity would be if it was not rotating.
>
>Watch out, you just switched frames!

I don't understand what you are talking about.

>The gravity constant is only
>affected when you switch frames,

I have assumed the gravity constant G is constant. Nothing can affect
it.

> and the net g is then reduced by
>centrifugal force.

There is no such thing as centrifugal force in physics. The net g is
reduced because a fixed force (gravity) can only give a mass a fixed
total acceleration and if it is already accelerating it to make it
travel in a circle then g is the acceleration left over. Both g and
centrifugal acceleration are acceleration due to gravity in the
direction 'down'.

>You can't have it both ways and mix the descriptions!
>
>> The principle of equivalence says that gravitational acceleration is
>> indistinguishable from acceleration caused by mechanical forces but in
>> this case there are no mechanical forces anyway. Only gravity.

>
>OK, we're back in the ECI frame, as I proposed in my last mail to you.

As far as I am aware I never left it. I am throughout an inertial
observer stationary w.r.t the centre of the earth.

>
>> The gravitational potential between the equator and the pole is the
>> energy we would get if we took a unit mass from the equator to the pole.
>> The acceleration of a unit mass (of water) free to 'drop' from the
>> equator to the pole is zero which means that the gravitational potential
>> of both is the same.

>That is only true in the rotating frame with centrifugal force, which
>you just denied.
> As I told you before, in that frame both masses have
>zero speed, so the equivalence perspective seems to work fine like
>that as well.

There is no *inertial* frame in which both masses have zero speed.

>But I'm not used to working with rotating frames, and in
>the ECI frame the water at the equator is not in static equilibrium
>but accelerating towards the earth. I'm more at ease in inertial
>frames and therefore I discussed that calculation approach.

I don't know what you are talking about sorry.

>
>> As far as I can see Murray is right unless I have totally misunderstood
>> the term 'gravity potential'.
>
>Neither: we were both unclear about which frame of reference, and its
>implications.
>
>> Thank you for continuing this discussion. It is giving my brain a good
>> work out. I think it is coming clearer.
>
>For me it also was refreshing! - and off the record I made some
>progress with understanding the connections between energies and clock
>frequency. Thanks for that!

When you have it clear let me into the secret. 

-- 
John Kennaugh
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