# Followup: pi on curved surfaces

1. Dec 20, 2015

### 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".

2. Dec 21, 2015

### haruspex

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

3. Dec 21, 2015

### 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.

4. Dec 21, 2015

### DaveC426913

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?

5. Dec 21, 2015

### DaveC426913

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

6. Dec 21, 2015

### jbriggs444

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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.

7. Dec 21, 2015

### TeethWhitener

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.

8. Dec 21, 2015

### lavinia

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.

9. Dec 21, 2015

### DaveC426913

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

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

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

Last edited: Dec 21, 2015
10. Dec 21, 2015

### lavinia

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

11. Dec 21, 2015

### DaveC426913

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.

12. Dec 21, 2015

### MrAnchovy

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.

13. Dec 21, 2015

### TeethWhitener

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.

14. Dec 21, 2015

### jbriggs444

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.

15. Dec 21, 2015

### DaveC426913

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.

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.

16. Dec 21, 2015

### jbriggs444

Apologies for that then.

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

17. Dec 21, 2015

### DaveC426913

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.

18. Dec 21, 2015

### jbriggs444

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.

19. Dec 21, 2015

### 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?

It sounds like you are saying they are the same.

20. Dec 22, 2015

### 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.