Re: Experimental Evidence for Special Relativity
- From: John Kennaugh <JKNG@xxxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Sat, 25 Oct 2008 12:40:19 +0100
Dr. Henri Wilson wrote:
On Thu, 23 Oct 2008 21:28:40 +0100, John Kennaugh
<JKNG@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
Dr. Henri Wilson wrote:On Tue, 21 Oct 2008 16:32:19 +0200, "Paul B. Andersen"
<paul.b.andersen@xxxxxxxxxxxxxxxx> wrote:
John Kennaugh wrote:
[....]
Paul did not express himself very well but I think I understand his
point. I don't think either of you are going to see it from the PoV of
the other so let's try a different approach. Let us suppose you have a
circular path which is rotating (you'll have to view this with a fixed
pitch font)
s f
o o
o o
o o
o o
o o
Light starts at s and ends at f.
There is one IFoR where s and f are the same point that is an IFoR
travelling at v from left to right where v is the tangential speed. In
that FoR s is stationary when the light is emitted so both theories say
that light will travel at c.
x s o
x o x o
x o x o
x o x o
x + o
In this FoR the circle is moving from Right to left starting in position
shown by the 'o' circle and finishing with the 'x' circle. Point s is
both start and finish. (+ is a point on both o and x circles)
This is the same as using the rotating frame....except that rotation is
absolute and translation is not.
Neither s nor f move in your IFoR. So sorry John, I think you are reverting to
aether theory.
No aether need apply I assure you.
Imagine the following: Suppose you have a short, perfect cylinder mirrored on the inside. You could use it as the basis of a Sagnac experiment. The splitter would send the light at a grazing angle in both directions, the light path would spiral around and be combined when they reach the other side. When the table is rotated you get a fringe shift.
Question - would it make any difference to the outcome if the cylinder itself can rotate about its centre independent of the table and is made to rotate in say, the opposite direction to the table, or twice as fast as the table, or at any other speed?
I believe the answer is 'no' in which case the rotation of the path is irrelevant. If v<<c then in the IFoR in which the light is emitted (always the most relevant when analysing something in terms of Ballistic theory) the start and finish points are one and the same point. The circle itself is moving relative to that IFoR (and rotating but that is irrelevant) which lengthens the path in one directions and shortens it in the other.
--
Light going CW has to start going around the RHS of the loop when it is
near its furthest position to the right and ends up travelling around
the LHS of the loop as it nears its maximum excursion left. Light going
ACW starts going around the LHS of the loop when the loop is near its
furthest position to the right and ends up going around RHS of the loop
as it approaches its furthest point is left. The speed is c in both
directions according to both theories. The paths are different lengths
but far too complex for me to draw with ascii characters.
x # o
x o x o
x# o x #o
x o x o
x # x o
This shows 4 points # on the path of the CW light.
x # o
x o x o
x #o x# o
x o x o
x # x o
This shows 4 points on the ACW path.
The ACW path is at all times inside the CW path and is shorter. Both
theories say that in this FoR light travels at c. Both theories say
there is a fringe shift and agree as to what it is.
What if you regard the rotating ring from the point of view of a completely
separate inertial frame? ..but that is effectively what we do when we use the
nonR frame.
John Kennaugh
.
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