Re: Experimental Evidence for Special Relativity
- From: hw@..(Dr. Henri Wilson)
- Date: Sat, 25 Oct 2008 21:25:37 GMT
On Sat, 25 Oct 2008 12:40:19 +0100, John Kennaugh
<JKNG@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
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?
No, rotation is absolute and can be measured absolutely. An ordinary gyroscope
can be used to demonstrate that.Translation on the other hand not absolute but
is relative. I'm sure we all agree oon that.
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.
No John, if you are sitting on the source, the circle doesn't appear to move or
rotate. You will be unaware that one path is longer than the other. This is one
of Paul's mistakes when he uses the rotating frame.
He seems to think that if one sits on a carousel, it never rotates no matter
how fast it spins in the nonR frame. This approach annihilates his SR Sagnac
explanation too....but he is too indoctrinated to realise and accept the truth.
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.
Henri Wilson. ASTC,BSc,DSc(T)
www.users.bigpond.com/hewn/index.htm.
.
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