(July 23, 2018 at 12:47 pm)Jehanne Wrote: Oh, I agree, absolutely, that the mathematics of tensor calculus on which General Relativity is based is well defined and extremely accurate. An orbit with an eccentricity of zero with a well-measured diameter times pi that is not equal to the circumferential distance of the orbit is much harder to grasp.
Hmmm...it follows quite readily from the metric. The relevant part has a space part
ds^2 = f® dr^2 + r^2 d(theta)^2
Where f®>1 for all r.
The measured radius is then the integral of sqrt(f®) dr from 0 to R. This will always be larger than R.
The circumference at r=R will be the integral of R d(theta) from 0 to 2*pi, so 2*pi*R. Since the measured radius is more than R, we get that
2*pi*R, the measured circumference, is smaller than 2*pi*(measured radius).
What precise difficulty are you having?
