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Re: Ballistic Theory and the Sagnac Experiment

Subject: Re: Ballistic Theory and the Sagnac Experiment
From: "Paul B. Andersen"
Date: Wed, 22 Mar 2006 16:42:29 +0100
Newsgroups: sci.astro, sci.physics.relativity, sci.physics
Henri Wilson wrote:
On Mon, 20 Mar 2006 16:37:19 +0100, "Paul B. Andersen"
<paul.b.andersen@xxxxxxxxxxxxxxxx> wrote:


Henri Wilson wrote:

On Fri, 17 Mar 2006 14:50:00 -0000, "George Dishman" <george@xxxxxxxxxxxxxxxxx>
wrote:


I have pointed out your mistake.
Walking around a carousel in opposite directions will soon tell you that I'm
right.

Right about what?
Let's have a closer look at your carousel.

You are obviously very confused about the role of
the centrifugal 'force', so let us be concrete and
calculate the forces.

I would advice you not to make a fool of yourself
and claim that my analysis is wrong.
Think before writing, and don't repeat stupidities
like:
"It is an imaginary force, 'centrifugal', in the rotating frame.
Its magnitude is the same as the centripetal force in
the non-rotating frame."


It is not surprising that the definition of centrifugal force is rather vague.
Each author seems to have his own opinion.

I am a little reluctant to answer this posting.
I am tempted to repeat that it is too stupid to respond to
because you are quibbling about elementary mechanics.

But I always end up with doing it anyway, so OK.

What is YOUR definition of centrifugal force, Paul? I think your 'coriolis
force' is my centrifugal force, and my 'coriolis force' is something you appear
to know nothing about. Maybe you live to close to the N pole.

There is but one definition of the centrifugal and Coriolis
pseudo forces in _mechanics_. You can look it up in any text book.
Or you can read my posting.
But OK, let me repeat it.

In a frame rotating with the angular velocity w (a vector)
we have the following pseudo forces acting on a body with mass m
which is moving with the velocity u (a vector) in the rotating frame:

The centrifugal force:
  F = m* (w X (w X r))  where r is the radius vector.
  The magnitude of this force will be m*w^2*|r|, and
  its direction will always be radially outwards.
  Expressed with the speed v = |w|*|r| : F = m*v^2/|r|

  Note that the centrifugal force does NOT depend on
  the velocity u in the rotating frame.

The Coriolis force:
  F = 2*m*(w X u)

  Note that the direction of this force always is
  in the plane of rotation, and is perpendicular to
  the velocity u. When u is tangential as in our case,
  the direction will be radial.

This NG is a sci group.
When you discuss in this group, you better use the normal
accepted definitions of centrifugal and Coriolis,
don't invent your own.


Much of your confusion is probably due to the fact that
you are not aware of which frame of reference you are using.
You seem to operate with three different rotating frames:
#1; the frame fixed to the carousel
#2: the rotating frame where you are stationary when you
    are running in one direction
#3: the rotating frame where you are stationary when you
    are running in the opposite direction.

In this case it should be obvious that the "rotating frame"
is the frame fixed to the carousel, and nothing else.

Twirl an object around by hand, on a string.
In the non-rotating frame, both the object and your hand rotate around the
barycentre. A centripetal force is required to accelerate both your hand and
the object towards the barycentre. These two CENTRIPETAL forces are opposite in
direction and  show up as a tension in the string = mv^2/r or MV^2/R.
Centrifugal force does not stricty exist in this frame, although in most cases
R is so small that even the best scientists and engineers (including myself)
tend to call the object's 'pull' on the string a 'centrifugal force'. This is
also convetient in the case of a balanced wheel where the barycentre is also
the centre of rotation.

The force the 'pull' is exerting on the object is acting towards the centre.
It is a centripetal force.
No other forces are acting on the object.

In the rotating frame, The object does not move but the same tension remains in
the string. How can this happen? It is due to the imaginary 'centrifugal
forces', of course.

But the centripetal force is the same.

The centripetal force that is a real measureable force
_which is independent of frames of references_.

But this is trivial, Henri.

The point you seem to miss completely is that when you are running
around in your carousel, you are not stationary in the "rotating frame"
which is fixed to the carousel. Your velocity in the rotating frame
is tangential c.

Coriolis force goes like this.

In the above example, if the string is shortened, the object's rotation rate
will increase due to conservation of momentum. In the rotating frame however, shortening the string is no big deal and should
do nothing except bring the object closer to the centre. However in practice,
an imaginary force pushes the object sideways during this process. That is the
imaginary CORIOLIS force. It explains why heavy air moving towards the earth's
poles forms a couple with light air moving towards the equator and we get the
familiar rotations around low and high pressure systems.

Right.
The Coriolis force is always perpendicular to the velocity.

So if the velocity of the object is radial, then
the Coriolis force is tangential.

But when you run around the carousel your velocity is tangential,
so the Coriolis force is radial!

I hope the students of Norway will benefit from your 'lesson'.

Students in Norway learn this the first year.
You obviously didn't.
Or have you forgotten what you once knew?

Read my analysis again, carefully.
Remember that there is but one rotating frame, namly
the one fixed to the carousel.
And before you do, be sure to learn the correct definitions
for the centrifugal and Coriolis forces.

or:
" 'centrifugal' often confused with 'centripetal'....
  and they have the same values anyway."

(In the example below the centripetal force is
ten million times greater than the centrifugal force.)

To be a valid analogy, you will have to run at a vastly
higher speed than the peripheral velocity of the carousel.
A 1 m radius I-FOG can detect the rotation of the Earth,
that's a peripheral speed of 2.3*10^-13 c.
But let us be generous, let the peripheral speed of
the carousel be as high as a 10^-4 part of your speed.
Let the radius of the carousel be r = 10m.
Let the speed with which you run be c = 10 m/s.
Let the peripheral velocity of the carousel be v = 0.001 m/s.


Let your mass be m = 100kg.

The centripetal forces the carousel exerts on
your feet are now:
Running with the rotation:
 The centripetal acceleration is (c+v)^2/r
 The force is Ff = m*(c+v)^2/r = 1000.20001 N
Running in opposite direction:
 The centripetal acceleration is (c-v)^2/r
 The force is Fb = m*(c-v)^2/r =  999.80001 N

Note that these are the actual centripetal forces
acting on your feet. These forces could be measured
and are obviously independent of which frame
you use to calculate them in.

Note that this suffice to address your problem.
The centripetal forces are:
Ff = m*(c+v)^2/r = 1000.20001 N
Fb = m*(c-v)^2/r =  999.80001 N

These forces do NOT depend on which frame of
reference you use to calculate them in.

The rest is only to demonstrate this fact:
calculated in the rotating frame fixed to
the carousel, the centripetal forces MUST
remain the same.

So what about the 'centrifugal' force?
Let us calculate the forces in the rotating frame.
The centripetal acceleration is c^2/r and
the component of the centripetal force causing
it is thus Fca = m*c^2/r = 1000 N
The centrifugal force is m*v^2/r = 0.00001N,
same in both directions.


That's not how you calculate centrifugal force.. It has the same magnitude as
the centripetal foreces.


Since the centrifugal
force is acting outwards, there must be a component
of the centripetal force counteracting this.
Fcf = = 0.00001N
So far, the two components of the centripetal
force are equal for both directions and amounts
to 1000.00001N.
So what's wrong? Why are the forces equal?
Because there is a third component of
the centripetal force, namely the one counteracting
the Coriolis pseudo force.
This force is 2m*(w X c) where w is the angular
velocity vector (spin vector) and c is the velocity
(vector) of the object. In our case the absolute
value of this force is m*v*c/r, and its direction
is radially outwards when you are running with
the rotation and radially inwards when you are
running in the opposite direction.
So the third component of the centripetal force
counteracting the Coriolis force is:
With the rotation:  Fcof = m*v*c/r = 0.2 N
Opposite direction: Fcob = - m*v*c/r = - 0.2 N

So the centripetal forces will be:
With the rotation:  Ff = Fca + Fcf + Fcof = 1000.20001
Opposite direction: Fb = Fca + Fcf + Fcob =  999.80001


If c=v then  Fb = zero.


The centripetal forces are the forces exerted on your
feet by the carousel. They are obviously the same
whether you calculate them in the stationary or in
the rotating frame.

Note also that:
m*(c+v)^2/r = m*c^2/r + 2*m*c*v/r + m*v^2/r

Of course the centripetal forces are different.
But not much.
Let us assume that there is a friction slowing
you down, and that the slowing during one revolution
is proportional to the centripetal force, we can write:
Slowing when going with the rotation:
delta_vf = k*1000.20001 N
Slowing when going in the opposite direction:
delta_vb = k*999.80001 N
Now we know that to explain the Sagnac in an
I-FOG, the difference between the two speeds
must be in the order of 2v.
thus, in our analogy:
delta_vf - delta_vb = k*0.4 N = 2v = 0.002 m/s
k = 0.005 m/Ns
delta_vf = 5.001 m/s
delta_vb = 4.999 m/s

See?
For the difference to be big enough, you would
have to slow down to half the speed.


I don't like your method or your maths.

Since it proves you wrong, it is reasonable
that you don't like it. :-)

But I note with interest that you do not even try
to refute neither my method nor my math.

You should burn all the books in Norway, they are obviously wrong.


In an I-FOG with a much smaller v/c ratio
it would be even worse.

The very idea is idiotic beyond belief.


The fact is, I have proved my point. There are two separate effects. 1. The
beam's 'energy centre' is thrown slightly off the 'fibre centre', thus
increasing path length by different amounts for the two beams. 2) Each beam will experience a different amount of 'wall drag' due to slightly
different amounts of centrifugal force pushing them towards the outside of the
fibre.

The fact is, I have proven why this point of yours
is idiotic beyond belief.



I have shown that to be wrong.

You have shown nothing whatsoever. Post the results
of your experiment. If you do manage to prove it wrong
then you have shown ballistic theory to be wrong.

The light beams DO NOT experience the same 'centrifugal' forces in both
directions. Right or wrong, George?

Wrong.
The centrifugal forces do not depend on velocity
(neither speed nor direction), and it is tiny.
It is however correct that the centripetal forces
are very slightly different.


That proves my point and I win the argument.
I never claimed that htis was the major cause of the sagnac effect.
The main reason it occurs is basically due to the fact that photons have AXES
that don't like turning corners. The interfere strangely with photons whose
axes point in different directions.

When you realize that one of your "explanations"
doesn't work, you always come up with an even more
idiotic "explanation". :-)

To explain the Sagnac, the light would have to slow
down to a fraction of it original speed in an I-FOG.
It doesn't.

Face it, Henri.
The Sagnac falsifies the ballistic theory.
No way out.


I suppose it proves LET too, eh Paul?...because that's what YOU use to explain
the effect.

Sagnac proves no theory, of course.
But it is in accordance with ether theories as well as SR.
The only theory it falsifies, is the ballsitic theory.

Paul

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