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Re: Ballistic Theory and the Sagnac Experiment

Subject: Re: Ballistic Theory and the Sagnac Experiment
From: "Paul B. Andersen"
Date: Mon, 20 Mar 2006 16:37:19 +0100
Newsgroups: sci.astro, sci.physics.relativity, sci.physics
Henri Wilson wrote:
On Fri, 17 Mar 2006 14:50:00 -0000, "George Dishman" <[email protected]>

"Henri Wilson" <HW@..> wrote in message news:[email protected]

On 15 Mar 2006 05:12:24 -0800, "George Dishman" <[email protected]>

Henri Wilson wrote:

On 14 Mar 2006 02:53:58 -0800, "George Dishman" <[email protected]>
No, my claim is and always has been that Ritz's
ballistic theory predicts a null result for Sagnac
and that remains true. You haven't developed an

One of your main arguments has always been that in the rotating frame, both
beams always move identically.

I have stated a simple fact - Ritz's theory postulates
they move at c relative to the source.

and you claim they move identically around the circle in exactly the same but
opposite ways.

Not me, ballistic theory says that.

Yes George, stop being funny. You claimed the BaTh says....

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

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
"It is an imaginary force, 'centrifugal', in the rotating frame.
 Its magnitude is the same as the centripetal force in
 the non-rotating frame."
" '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.

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. 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

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

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

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

The very idea 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?

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.

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.


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