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Re: Ballistic Theory and the Sagnac Experiment

Subject: Re: Ballistic Theory and the Sagnac Experiment
From: Henri Wilson
Date: Wed, 22 Mar 2006 22:24:28 GMT
Newsgroups: sci.astro, sci.physics.relativity, sci.physics
On Wed, 22 Mar 2006 16:42:29 +0100, "Paul B. Andersen"
<paul.b.andersen@xxxxxxxxxxxxxxxx> wrote:

>Henri Wilson wrote:
>> On Mon, 20 Mar 2006 16:37:19 +0100, "Paul B. Andersen"
>> <paul.b.andersen@xxxxxxxxxxxxxxxx> wrote:
>> 

>> 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|

Paul, that is centripetal force. (you omitted the minus sign somewhere)

In the rotating frame, w = 0.

You obviously don't understand the basics.

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

The imaginary Centrifugal force has the same value as the centripetal force in
the non-rotating frame.  It's an 'instantaneous' thing.

>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.

But in the case of a rotating sphere, such as Earth, it is NOT always in that
plane.
You calculation applies to the force due to change in angular momentum in the
NonR frame. 
Coriolis is the imaginary equivalent in the R frame (where w is again zero).

>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.

Paul, in the R frame, w = 0. Do you disagree with that.

>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.

I am not confused. 

If u = -v, the sum of your two forces should be zero. The object in question is
standing perfectly still.

That is not what your analysis produces. You have made a mistake.

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

That is correct..and in this frame, w is always zero.

>> 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.

Centripetal force doesn't exist in the R frame because w = zero. It's imaginary
equivalent is used instead.

>The centripetal force that is a real measureable force
>_which is independent of frames of references_.
>
>But this is trivial, Henri.

It IS trivial to calculate but not many people can grasp the concept.
Centripetal force does NOT exist in the R frame.

>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.

But when v (NR frame) = - u (R Frame), you are standing still (NR frame) and
experience no radial forces.

Your maths says otherwise.

>
>> 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.

For a wheel, yes.

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

What you are calling 'Coriolis force' IS then radial. This is the 2D example of
Coriolis force....but you are ignoring the fact that it is an imaginary force
because w = 0 in the R frame.



>> 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?

Actually I thought the same as you did until it was explained to me on one of
the physics NGs some years ago. I think it might have been Tom Roberts who did
that.

>Read my analysis again, carefully.
>Remember that there is but one rotating frame, namly
>the one fixed to the carousel.

I know that Paul. ..and w = 0.

>And before you do, be sure to learn the correct definitions
>for the centrifugal and Coriolis forces.

Like I said, there appears to be general confusion.

>>>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.

Centripetal force doesn't exist in the R frame. You are refering to the
imaginary 'centrifugal force'.

>
>>>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.

I haev ben quite aware all along that the contribution due to 'v' is much
larger than that due to 'u'. I did that in my head.
However, in the iFoG, the 'c' part is equal in both directions and cancels.
...like a wheatstone bridge..

>> 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.

You have proved nothing at all...particulary, as I pointed out to Jerry, when
the deflection from the centre probably follows a 1/x^3 or 1/x^4 law.

>>>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". :-)

Your own explanation of Sagnac is precisely that relating to sound in air. Can
you provide a physical reason why light should behave like sound in air?

>>>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, I have told you why your claim is wrong. Do you want me to go through it
all again?

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