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" <george@xxxxxxxxxxxxxxxxx> wrote: 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." 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. 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 See? 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? 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. 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. Paul |

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