# Could PI have a different value in a different universe?

PI is determined by the circumference of a circle.

What bothers me is why is it any particular constant number? What I mean is, would a circle as defined as a perfect circle in any universe lead to a different value of PI?

Would all universes where a circle could be constructed by "people" there also lead to the value of PI?

If its true then it leads to the conclusion that PI is some sort of constant value constant to all universe. What is the meaning of that?

Science fiction references.

In science fiction Pi sometimes has a different value in different universes, for example Greg Bear's "The Way", it says "Gates are capped with cupolas formed from Space-time itself. As distortions in space-time geometry, their nature can be calculated by 21st century instruments laid on their 'surfaces'. The constant Pi, in particular, is most strongly affected.".

A message is found encoded within Pi, in the novel by Carl Sagan, "Contact" "Ellie, acting upon a suggestion by the senders of the message, works on a program which computes the digits of pi to record lengths in different bases. Very far from the decimal point (1020) and in base 11, it finds that a special pattern does exist when the numbers stop varying randomly and start producing 1s and 0s in a very long string.".

• [Note: said question is now closed] This question seems like it answers yours (do people want to close as duplicate?). In particular, the main idea in the answers of that question is: $\pi$ is defined in terms of mathematics, not physics. It doesn't "change" depending on "what universe" we are in - math does not depend on reality. – Zev Chonoles Jul 22 '11 at 4:22
• 7 is also a constant value, constant to all universes. So? – Gerry Myerson Jul 22 '11 at 4:23
• I still think there is a "so" here but can't think how to articulate it. – Phil Jul 22 '11 at 4:29
• Imagine a universe where small-scale structure is based on the $p$-adic numbers instead of the real numbers. Perhaps then some number other than pi would be the one attracting the kooks. – GEdgar Jul 22 '11 at 14:51
• @dan: topology won't help you there. Length is not intrinsic to topological spaces. To measure length you need a metric space. To be able to have rigid motion (rotate a curve 360 degrees) you need some sort of symmetry: so you are down to homogeneous spaces essentially. Then if you want the ratio to be independent of the initial length of the segment $AB$ (this is not true in spherical or hyperbolic geometry), then you need scaling invariance. Operations one takes for granted in Euclidean geometry may not be well-defined in other geometries or topologies. – Willie Wong Jun 21 '12 at 7:44

## 7 Answers

Physically, the ratio of a circle's circumference to its diameter $C/d$ is not really $\pi$. General relativity describes gravity in terms of the curvature of spacetime, and roughly speaking, if you take $(C/d-\pi)/A$, where $A$ is the circle's area, what you get is a measure of curvature called the Ricci scalar.

But even if you're doing general relativity, you don't just go around redefining $\pi$. The thing is, $\pi$ occurs in all kinds of contexts, not just as $C/d$. For instance, you could define $\pi$ as $4-4/3+4/5-4/7+\ldots$, which has nothing to do with the curvature of space.

So if you define $\pi$ as $C/d$, you don't even get a consistent value within our own universe, whereas if you define it as $4-4/3+4/5-4/7+\ldots$, you get an answer that is guaranteed to be the same in any other universe.

Another way of looking at it is that $\pi$ is not the $C/d$ ratio of a physical circle, it's the $C/d$ ratio of a mathematically idealized circle that exists in certain axiomatic systems, such as Euclidean geometry. Viewed this way, it doesn't matter that our universe isn't actually Euclidean.

• +1 for mentioning idealised circles. This is probably a more important point to make than saying that $\pi$ can be defined by infinite series or the least positive root of the transcendental function $\sin$ or the square of the integral $\int_{-\infty}^{\infty} \exp(-x^2) \, dx$, etc. – Zhen Lin Jul 22 '11 at 7:37
• +1 for mentioning the mathematically idealized circle vs physical circle. That actually cleared some of my misconceptions about $\pi$ being defined physically. – Airdish Mar 15 '16 at 14:55

This is a complement of other answers. One can define a value $\pi_p$ as $\pi$ in $\ell_p$. $\ell_p$ is two dimensional space with a metric as follows: $$d\left((x_1,y_1),(x_2,y_2)\right)=\left(|x_1-x_2|^p+|y_1-y_2|^p\right)^{1/p}$$ for $1\leq p \leq \infty$. Then the circle $C_p$ is defined as all points $(x,y)$ such that: $$\left(|x|^p+|y|^p\right)^{1/p}=1$$ The diameter of this circle is $2$. And therefore we can define $\pi_p$ as the half of circumference of the circle. It can be seen that $\pi_1=4$, $\pi_2=\pi$ and $\pi_\infty=4$. In following image, you can see the value $\pi_p$ versus p: Reference: Look at the following article $\pi _{p}$ the Value of π in $\ell _p$.

• How to be able to read the article mentioned in the reference? Could you explicitly write the formula for pi(p), please. – user534397 Nov 9 '18 at 6:52
• @user534397 Here is another link to the same paper. The formula is the following definite integral: $$\pi_p=4\int_0^{2^{-1/p}}(1+|x^{-p}-1|^{1-p})^{1/p}\,dx.$$ – r.e.s. Apr 16 at 13:34

That there will be a number $\pi$ is a mathematical fact. But whether the significant number would be the same is a more interesting question. Some people in our own universe would prefer that the constant had been chosen to be $2\pi$ i.e. $6.28 ...$ instead of $3.14 ...$ as it would reduce the number of factors of 2 in some formulae.

It would also be possible to imagine, in a higher dimensional universe, that the basic round object might be, say, a 3-sphere, with the significant constant would be defined in relation to its geometry rather than the geometry of a circle.

Living in a world which was non-euclidian (e.g. on the surface of a sphere) would make other numbers geometrically significant, but there would still be $\pi$ = $3.14 ...$ sitting in the background.

• I do live on the surface of a sphere - don't you? – Gerry Myerson Jul 22 '11 at 6:45
• Are there no mountains, valleys, and oceanic trenches on your planet, M. Myerson? ☺ – JdeBP Jul 22 '11 at 14:43
• @Gerry: Indeed, but for practical purposes in ancient times the curvature was not sufficiently evident, and the idealised form of local geometry was/is Euclidean. I'm sure you realise I was thinking about the situation where there was a measurable effect from the curvature of space (so objects within the space would not necessarily be flat). I live on an approximate 2-sphere which is an object in space. But the question would be if 3-dimensional space had a sufficient locally measurable curvature (e.g. a 3-sphere) what would the significant number be? – Mark Bennet Jul 22 '11 at 16:33

When I think of different 'universes,' I imagine places that are fundamentally different than our own. Because pi is just the ratio of the circumference to the diameter, that won't change so long as the behavior of the 'metric of the universe' doesn't change.

But suppose that we considered the 'taxi-cab universe,' where the pertinent metric is the taxicab metric (which I have also called the Manhattan Metric, which is nicely alliterative). In such a universe, a circle looks to us to be a square. But within the metric, a circle with radius 4 would have circumference 32. So taxicab-pi would be 4. How nice and even.

I used that as an example, but really it's still just a mathematical creation. One could more or less analyze many different geometries, topologies, manifolds, etc. And to each might be associated some different way of relating a 'circle' (whatever that may mean) to the metric.

About Contact: if $\pi$ is normal in base 11 (never proven but almost surely true), the "special pattern [...] of 1s and 0s in a very long string" will appeal an infinite number of times... but also any finite string, long or short.

The problem starts with your first sentence .. which most probably coming out of definition Pi=C/D . There is not problem with this definition , problem is in the value of Pi which we use . This value which we use has nothing to do with this definition at first place ,which not many students ever realized . We approximate C by straight line segments and then this approximation is divided by radius ,or diameter. The definition for the value we use should by totally different to not confuse students. Not even calculus is proving Pi , calculus uses Pi as main ingredient . You can't prove Pi by Pi . Another thing is that main formulas like C=DPi , or AC= Pir^2 have not prove . I am not talking about deriving those formulas , but prove of results by those formulas. So your question is a bit misleading, because we don't even know value of number Pi in this universe :) .

• Yes in other words the question I wrote seems to be like how to derive a axiom of maths, Pi just "is" defined by being Pi, perhaps this is really not a meaningful question but more "interesting" to consider. – Phil Apr 16 '17 at 12:40

pi might vary according to the hubble constant. the mathematical shape of a given universe supposedly changes according to the hubble constant the constant is a measurement of the dispersal of mass within the volume of the universe. universes with various hubble constants are in the shapes of planes, saucers, toroids, spheres. calculating a uniform curve which closes in on itself in each of those universes might produce differing values for pi.

if so, pi would be a good way to accurately measure the density (or local density) of the universe.