🚀 go-pugleaf

Michael Weiss (mkw21@cus.cam.ac.uk) wrote:
> On 20 Mar 1995 son3776@utdallas.edu wrote:

> > Considering that we can't detect any such curvature in spacetime because
> > we ourselves are affected by the same curvature to such an extent that if
> > we travel with a relative velocity of zero in relation to the drawn
> > circle there won't be any noticeable distortions, we will measure pi (to
> > a degree of accuracy).

> You keep saying this, and I keep telling you it's wrong, and I haven't
> heard you respond yet.  The claim you make is:

>    "We don't notice the spacetime curvature because we 'curve' with it"

> and that is *false*.  Flat-out false.  If you make a large circle
> (defined as "a locus of points all of which are measured to be
> equidistant from a common point") deep in outer space and then

>    (a)  Measure the distance around the circle
>    (b)  Measure the distance across the circle
>    (c)  Divide (a) by (b)

> You.  Will.  Not.  Get.  Pi.

And I'm sure you've done this before . . .

But seriously, the circle that's drawn is part of spacetime.  We are part
of space time.  The instruments we measure are part of space time.  To
use a planar analogy, draw a circle, us, a ruler measuring the
diameter and a tape measure measuring the circumference on a stretchable
sheet.  Stretch the sheet into a curve and you'll see that we, the
circle, the ruler and the tape measure have stretched.

The ruler still measures the same length for the diameter, and the tape
measure still measures the same length for the circumference.  If the
ratio was pi to being with, it'll still be pi.

--
"Somewhere else the tea's getting cold." -- The Doctor