No perfect circles in space...

[I_am_not_mafia]

Good find.

The point about there not being natural circles in space is useful when religionists talking about Maths as something that exists in its own right rather than a way of approximating reality invented by humans. I always argued that you don't get a perfect circle because nothing can be smaller than the Planck length and because of quantum fluctuations, but you don't even need to be that exact because the distortion of spacetime is enough to stop any circle that exists from being perfect.

The section of one physical supplanting another is also useful when talking to religionists about evolution. There is so much evidence for the theory of evolution through natural selection that any hypothesis that they come up with also needs to explain this evidence. Much like how Einstein's theory of relativity can be used to come up with the same results as Newton's calculations when used at a similar level of approximation.

[polymath257]

Please do; if you read my OP, the authors of 21st Century Astronomy (all PhDs) say that the distance would be shorter.

My bad. I had misremembered the Schwartzchild solution (the inverse is on the radial piece, not the time piece). Yes, the circumference would be shorter than 2*pi*r.

[Jehanne]

Please do; if you read my OP, the authors of 21st Century Astronomy (all PhDs) say that the distance would be shorter.

My bad. I had misremembered the Schwartzchild solution (the inverse is on the radial piece, not the time piece). Yes, the circumference would be shorter than 2*pi*r.

It is hard to imagine in one's head those type of circles with respect to a planetary orbit in 3 dimensions with zero eccentricity.

[Anomalocaris]

The key seems to be for a static planar circle not moving with respect to the observer, its radius must be 2 X pi X R. But when a real object is given the motion required to place it in what classical physics would describe to be a circular orbit around the sun, the motion of the object puts it in accelerated frame of reference with respect to the sun or any notional static points along its orbit. So an real orbit is not exactly analogous to a static planar circle.

[Jehanne]

Those who would imagine the Universe just "popping into existence" need to stop doing that and perhaps start focusing on this problem instead.

[polymath257]

My bad. I had misremembered the Schwartzchild solution (the inverse is on the radial piece, not the time piece). Yes, the circumference would be shorter than 2*pi*r.

It is hard to imagine in one's head those type of circles with respect to a planetary orbit in 3 dimensions with zero eccentricity.

The way I do it is by thinking of distance as a number. As we go out from the origin, that number runs slower than might be 'expected' for the radius. So, we get a radius that is lower than would be expected for the given circumference.

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Those who would imagine the Universe just "popping into existence" need to stop doing that and perhaps start focusing on this problem instead.

I'm not sure I see a problem here at all. The math is well-defined and accurately predicts observations. Curvature is defined *internal* to the spacetime manifold, not from any embedding.

The only issue seems to be the difficulty in 'imagining' that distance doens't work in the way Euclid expected. That is overcome through practice.

[Jehanne]

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.

[polymath257]

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?

[Jehanne]

The math is fine (although, you need a graphic); it's the visualization that is difficult.

[polymath257]

The math is fine (although, you need a graphic); it's the visualization that is difficult.

practice, practice, practice. Difficult is not the same as impossible.

I'd start by trying to imagine our 3D space as actually being the sphere in 4D with a very large curvature.

If that is the case, then no matter which direction you choose to go, you will eventually 'come around' to your starting point. Just like on a 2D sphere in 3D.

Can you imagine that?

If so, then think about what has to happen with spheres inside our space of increasingly large radii. When the radius gets to 'half way around', then the 'circle' is just a point and the 'circumference' is zero.

if not, then try harder.