Followup: pi on curved surfaces

[DaveC426913]

Is pi = 3.14159... only true in our flat universe?

We talk about whether our universe is open or closed - positive or negative curvature - and scientists have concluded that it is very nearly flat.

If the universe did indeed have a large positive curvature, would that result in a different value for C/d?

I'm asking of the "curvature of the universe" is the same thing as the "curvature of non-Euclidean surfaces".

 

[haruspex]

Wouldn't it be more a question of the relationship becoming nonlinear?

 

[TeethWhitener]

Keep in mind that a defining property of manifolds in general relativity is that they are locally Euclidean; i.e., each point in a manifold has a neighborhood that is homeomorphic to Euclidean space. Of course, the radius of that neighborhood can be quite tiny if the space is very curvy.

 

[DaveC426913]

Keep in mind that a defining property of manifolds in general relativity is that they are locally Euclidean; i.e., each point in a manifold has a neighborhood that is homeomorphic to Euclidean space. Of course, the radius of that neighborhood can be quite tiny if the space is very curvy.

Yes. This is what I was getting at. Pi has the value it has, provided it is plotted on a flat surface, which it will in a universe that is curved on very large scales. But if the universe were - instead of billions of light years in diameter - a mere one light year in diameter, we would measure pi to be different (smaller, but tiny fraction).

Yes?

 

[DaveC426913]

Wouldn't it be more a question of the relationship becoming nonlinear?

You mean different values at different points? Let's assume it has constant curvature.

 

[jbriggs444]

[

You mean different values at different points? Let's assume it has constant curvature.

Different values for different circle radii.

If you graph circumference versus diameter you may get the same function for all possible center points (e.g. if curvature is constant) but the one graph you get will not be a straight line plot with slope equal to pi.

 

[TeethWhitener]

Yes. This is what I was getting at. Pi has the value it has, provided it is plotted on a flat surface, which it will in a universe that is curved on very large scales. But if the universe were - instead of billions of light years in diameter - a mere one light year in diameter, we would measure pi to be different (smaller, but tiny fraction).

Yes?

I imagine if the universe were that small and closed, there would definitely be observable deviations from Euclidean geometry. However, if you're just talking about drawing a circle on an average-sized piece of paper, space would have to be much much curvier. After all, maps of local areas on Earth's surface tend to be flat, without appreciable curvature of latitude and longitude lines.

 

[lavinia]

Is pi = 3.14159... only true in our flat universe?

We talk about whether our universe is open or closed - positive or negative curvature - and scientists have concluded that it is very nearly flat.

If the universe did indeed have a large positive curvature, would that result in a different value for C/d?

I'm asking of the "curvature of the universe" is the same thing as the "curvature of non-Euclidean surfaces".

I am not sure what you are asking here but it seems that the previous thread answered your question. When a surface is curved, there is no fixed ratio between the circumference of a geodesic circle and its radius. As Mathwonk pointed out in the previous thread, for surfaces, the length of a geodesic circle is

$L = 2πr(1 - 1/6r^2K_0) + O(r^4)$ where $K_0$ is the Gauss curvature at $r = 0$.

This means that if $K_0$ is positive, then the length of the geodesic circle will be smaller that $2πr$ for sufficiently small $r$ and the length approaches $2πr$ in the limit as $r \rightarrow 0$. This is true even if the Gauss curvature is constant.

 

[DaveC426913]

I am not sure what you are asking here but it seems that the previous thread answered your question.

This followup is about whether 'the curvature of our universe' - not some hypothetical surface-of-a-basketball geometry - would affect pi.

I imagine if the universe were that small and closed, there would definitely be observable deviations from Euclidean geometry. However, if you're just talking about drawing a circle on an average-sized piece of paper, space would have to be much much curvier.

This is exactly what I'm trying to confirm, yes.

After all, maps of local areas on Earth's surface tend to be flat, without appreciable curvature of latitude and longitude lines.

Mapping is a kludge. There's a whole science of ways to put a conic peg into a trapezoidal hole.

 

[lavinia]

This followup is about whether 'the curvature of our universe' - not some hypothetical surface-of-a-basketball geometry - would affect pi.

There is no basketball here. The formula works for any point on any surface.

 

[DaveC426913]

There is no basketball here. The formula works for any point on any surface.

Yes.

To be clear, I'm merely trying to confirm that, when we say the "curvature" of the universe is nearly zero, we are talking about that same thing. I'm hearing the answer is yes.

 

[pbuk]

This followup is about whether 'the curvature of our universe' - not some hypothetical surface-of-a-basketball geometry - would affect pi.

No, π is defined with reference to Euclidean geometry so the curvature of our universe (which makes it non-Euclidean) cannot affect it.

A better question would be whether the curvature of the Universe affects the measurement of the ratio of the diameter of a circle to its circumference, and the answer to this is yes. However DaveC426913 you only seem to be considering the global curvature of the Universe, when you measure things you are concerned with local geometry and general relativity tells us (and it has been confirmed by measurements) that space is really curved. I am not aware of any experiments involving circular geodesics (if you got really near to a black hole this might be possible if you had infinite time), but gravitational lensing is a simple demonstration that our Universe is non-Euclidian.

 

[TeethWhitener]

This is exactly what I'm trying to confirm, yes.

Lavinia gave you the formula for figuring out how big the effect would be. The Gaussian curvature of a sphere of radius r is 1/r². So let's say you drew a circle with a radius of 1 meter and the universe was a closed sphere with radius = 1 light year (~10¹⁶ meters). Using lavinia's formula, the deviation of the circumference/diameter ratio from pi for the circle you drew would be ~π(1-10^-32), completely unmeasurable.

Edit: the circumference/diameter ratio itself, not the deviation from pi.

 

[jbriggs444]

To be clear, I'm merely trying to confirm that, when we say the "curvature" of the universe is nearly zero, we are talking about that same thing. I'm hearing the answer is yes.

Guessing what you are going after...

The "intrinsic curvature" in a geometry can be measured purely from within the context of the geometry. It does not depend on some kind of embedding where, for instance, the 2-dimensional surface of a basketball is embedded in a 3-dimensional Euclidean geometry. The ratio of the on-the-surface circumference of a particular circle to its on-the-surface diameter is such an internal measurement. The result reflects intrinsic curvature. We are talking about intrinsic curvature.

The "extrinsic curvature" in a geometry depends on an embedding. For instance, you can have a two-dimensional surface of a basketball curved into a three dimensional spherical shell. You have to step outside the geometry to measure this sort of curvature. A measure of the through-the-center diameter of a basketball would reflect extrinsic curvature. We are not talking about extrinsic curvature.

A better example is a torus -- the surface of a torus can have no intrinsic curvature. It can be flat from an intrinsic point of view but yet be curved when viewed exernally.

Intrinsic curvature is "real". Extrinsic curvature is only a feature of the model that is used. Our universe has intrinsic curvature. In our universe, the ratio of the circumference of a circle to its diameter is not a constant and is not equal to pi -- or would not be if we had the ability to perform a sufficiently precise direct measurement. But the curvature is small enough that we have no ability to perform such direct measurements. Instead, we can rely on indirect experiments to measure the curvature of our universe and calculate that the circumference to diameter ratio for any circles we can construct would not deviate measurably from pi.

 

[DaveC426913]

It does not depend on some kind of embedding where, for instance, the 2-dimensional surface of a basketball is embedded in a 3-dimensional Euclidean geometry.

Intrinsic curvature is "real". Extrinsic curvature is only a feature of the model that is used. Our universe has intrinsic curvature.

No, I totally get that.

Sometimes scientists talk about an open or closed universe, and whether ours will stop and recollapse in a Big Crunch. I may have read it wrong, but I thought they'd said a saddle-shaped universe was open and would not collapse. (Guess it's time to bone up on the newest cosmology) I was checking to see I hadn't conflated two things.

In our universe, the ratio of the circumference of a circle to its diameter is not a constant and is not equal to pi -- or would not be if we had the ability to perform a sufficiently precise direct measurement. But the curvature is small enough that we have no ability to perform such direct measurements. Instead, we can rely on indirect experiments to measure the curvature of our universe and calculate that the circumference to diameter ratio for any circles we can construct would not deviate measurably from pi.

OK, I'm comfortable now that the phrase 'curvature of our universe' is near zero' is to be taken literally, as in: parallel lines would converge over cosmic distances.

 

[jbriggs444]

No, I totally get that.

Apologies for that then.

Sometimes scientists talk about an open or closed universe, and whether ours will stop and recollapse in a Big Crunch. I may have read it wrong, but I thought they'd said a saddle-shaped universe was open and would not collapse. (Guess it's time to bone up on the newest cosmology) I was checking to see I hadn't conflated two things.OK, I'm comfortable now that the phrase 'curvature of our universe' is near zero' is to be taken literally, as in: parallel lines would converge over cosmic distances.

For an open (saddle shaped) universe, "parallel" lines never converge. And some lines which are not quite "parallel" also fail to converge.

 

[DaveC426913]

Here's an example I found that conflates the two concepts (closed meaning will collapse, and closed meaning parallel lines converge):

If the Universe was closed, the two lasers over time would converge.
If the Universe was open, the two lasers would diverge.

I do not know if these are correct.

 

[jbriggs444]

Here's an example I found that conflates the two concepts (closed meaning will collapse, and closed meaning parallel lines converge):
I do not know if these are correct.

In a closed universe, it does not matter which way you aim your lasers. All beams converge on the singularity corresponding to the collapse.

In an open universe, two lasers pointed in the same direction will have their beams diverge. There is some slop in the required angle, beams which are aimed to very slightly converge will end up diverging instead.

 

[DaveC426913]

Sorry, I just realized my earlier post was contextually ambiguous.

Open universe: one that will expand forever.
Closed universe: one that will eventually collapse.
Open universe: straight lines will diverge.
Closed universe: straight lines will converge.
Are these distinct meanings of open/closed, or are they the same thing?

In a closed universe, it does not matter which way you aim your lasers. All beams converge on the singularity corresponding to the collapse.

It sounds like you are saying they are the same.

 

[DaveC426913]

I lay awake last night rolling this over in my head. I get it now.

In any universe with any size or curvature, you can still derive pi easily enough.

As the size of your circles grow smaller, their C/d will converge on pi. This is what is meant by 'Locally, C/d will always equal pi'.

Pi is a universal constant.

 

[pbuk]

In any universe with any size or curvature, you can still derive pi easily enough.

π is defined in the Euclidean plane so is independent of the geometry of any physical universe.

As the size of your circles grow smaller, their C/d will converge on pi. This is what is meant by 'Locally, C/d will always equal pi'.

This isn't my field (please correct any mistakes someone) but I am not sure if that is true: in a smooth manifold I think it is true for the tangent space but not for any finite area of the space itself? [Edit: scale invariance nonsense deleted]

Pi is a universal constant.

This is not an expression we use in mathematics. Mathematics is not concerned with any real universe, something is either a constant or not, and π is a constant. Do you think the value of sin π depends on what shape the universe happens to be?

But I think I know what you mean...

 

[lavinia]

π is defined in the Euclidean plane so is independent of the geometry of any physical universe.

While π can be defined using Euclidean geometry it is can also be defined in any geometry on a surface using the limit quoted in post #8. Further it is an invariant of all compact Riemannian manifolds For instance for surfaces it is the ratio of the total Gauss curvature (its integral over the surface) to twice the Euler characteristic. This ratio is independent of any geometry. So in this sense π is a universal constant.

This isn't my field (please correct any mistakes someone) but I am not sure if that is true: in a smooth manifold I think it is true for the tangent space but not for any finite area of the space itself? [Edit: scale invariance nonsense deleted]

See post #8. In the limit as $r \rightarrow 0$ the length of a polar "circle" divided by its radial diameter is π.

 

[DaveC426913]

This is not an expression we use in mathematics. Mathematics is not concerned with any real universe, something is either a constant or not

In fact, the real universe is the context in which this originally came up. The question I'd been grappling with was what C/r would be in universes unlike our own.

 

[Physicalchemist]

If C/d was not pi in curved space, assuming we say it is pi', then pi' =/= pi and the infinite series of pi, like the slow convergent one
4(-1)^n/(2n+1)from n=0 -> infinity, would be incorrect in measuring the circle stuff. also, it would mean that e^(pi*i) is -1, but e^(pi'*i) is also -1 on the definition of e^ix = cos(x) + isin(x)