🚀 go-pugleaf

On Tue, 02 Jan 2007 20:01:13 GMT, Tom Roberts <tjroberts137@sbcglobal.net>
wrote:

>Wolfgang G. Gasser wrote:
>> Two distant observers with each a (transversal) LIGHT CLOCK orbit at a
>> constant distance around their common center (assumed at rest).
>> 
>> Special relativity postulates that the light beams in both light clocks
>> propagate at c relative to both (almost inertially moving) observers.
>
>That is the key point -- these observers are NOT moving inertially. The 
>approximation is important (see below), and negates your claims.
>
>
>> This requires that for each observer the clock rate of the opposite clock
>> CAN be lower than the clock rate of his own clock.
>> Because of symmetry and unchanging distance, clock rate CANNOT differ.
>
>The error is in your gedanken, not in SR itself.
>
>For approximately inertial motion, the basic equations of SR are only 
>APPROXIMATELY valid. You ignored the approximation you, yourself, made.
>
>At a deeper level, however, such orbiting observers cannot be treated as 
>inertial observers, because a circle is different topologically from a 
>line. In SR's Minkowski spacetime, inertial observers move in straight 
>lines, not in circles, and the basic equations of SR are valid for 
>inertial observers and coordinates. Yes, for a very large radius a 
>circle is approximately a straight line for a small portion of its 
>circumference, but if one uses the fact it is a circle in any way, or 
>uses a large fraction of the circumference, then these are not inertial 
>observers, not even approximately.

So the GPS clocks that you claim prove SR correct move in straight lines do
they Snipper?

....You become more of a comedian with every post...

>
>In your gedanken, you are using the circular motion when you say 
>"because of symmetry and unchanging distance".
>
>
>The right way to consider this is to ask: for two observers on the 
>circumference of a rotating disk, how do they observe the tick rate of 
>each others' clocks? Assuming that each clock emits an EM signal at each 
>tick, and the other observer receives this signal and compares the rate 
>of the received signal to their own clock's tick rate, the answer is 
>that both received signals have the same rate as the local clock. This 
>answer is independent of radius, rotation rate, and separation of the 
>two observers around the circle.
>
>
>Francois Guillet wrote:
>> Special relativity doesn't apply to non inertial reference frames.
>
>This is not true. The basic equations of SR are written in terms of 
>inertial coordinates, and these equations are valid only in inertial 
>frames. But one can use calculus to determine transforms to non-inertial 
>coordinates, and SR remains valid using such coordinates and the 
>transformed equations.
>
>SR applies only to Minkowski spacetime, however. That is, neither 
>curvature of the manifold nor topology other than R^4 can be handled in SR.
>
>
>Tom Roberts