🚀 go-pugleaf

"xxein" <xxein@comcast.net> wrote in message 
news:cb9cc14c-124f-4c0b-a921-5e0f6be794a7@e4g2000hsg.googlegroups.com...
On Jan 30, 9:27 pm, "Jeckyl" <no...@nowhere.com> wrote:
> > On Jan 30, 5:30 am, Pentcho Valev <pva...@yahoo.com> wrote:
> >>http://hem.bredband.net/b153434/Works/Einstein.htm
> >> Divine Albert: "In a space which is free of gravitational fields we
> >> introduce a Galilean system of reference K (x, y, z, t), and also a
> >> system of co-ordinates K' (x', y', z', t') in uniform rotation
> >> relatively to K. Let the origins of both systems, as well as their
> >> axes of Z, permanently coincide. We shall show that for a space-time
> >> measurement in the system K' the above definition of the physical
> >> meaning of lengths and times cannot be maintained. For reasons of
> >> symmetry it is clear that a circle around the origin in the X, Y plane
> >> of K may at the same time be regarded as a circle in the X', Y' plane
> >> of K'. We suppose that the circumference and diameter of this circle
> >> have been measured with a unit measure infinitely small compared with
> >> the radius, and that we have the quotient of the two results. If this
> >> experiment were performed with a measuring-rod at rest relatively to
> >> the Galilean system K, the quotient would be [pi]. With a measuring-
> >> rod at rest relatively to K', the quotient would be greater than [pi].
> >> This is readily understood if we envisage the whole process of
> >> measuring from the "stationary" system K, and take into consideration
> >> that the measuring-rod applied to the periphery undergoes a Lorentzian
> >> contraction, while the one applied along the radius does not. Hence
> >> Euclidean geometry does not apply to K'."
>
> >> Why does Divine Albert believe that, "with a measuring-rod at rest
> >> relatively to K', the quotient would be greater than [pi]"? Obviously
> >> because "the measuring-rod applied to the periphery undergoes a
> >> Lorentzian contraction". Then why does the segment of the periphery,
> >> to which the measuring-rod is applied and which moves together with
> >> the measuring-rod, fail to undergo a Lorentzian contraction?
>
> The rod when measuring the radius has its long edge aligned in the 
> direction
> of a radial line, and so the long edge is not in the diretio of motion (it
> is a right angles to it) so you get no contraction in length of ruler (but
> you do get one in the width).
>
> When the rod is turned around to measure along the periphery, the long 
> edge
> of the rod aligned in the direction of motion and so it is contracted, but
> this time the width is not.- Hide quoted text -
>
> - Show quoted text -

xxein:  So?  Move it continuously around the perimeter/circumference
in d/steps.  Add them up.  Now measure each radian in the same d/steps
and add THEM up.  The ratio of the summations = pi.

This does not work in gravity, however.

Yes it does, geez you stupid shit.