# Where does $\pi^2$ appear spontaneously within Physical Phenomenon and Mathematics Equations?

The term $$\pi$$ is found to appear in many equations and natural phenomenon; however my question is related to $$\pi^2$$.

While trying to figure out the reason for some $$\pi^2$$ terms appearing in certain equalities that I came across, I have a question. And the question is this:

In which all mathematics/physics equation or contexts does $$\pi^2$$ appear inherently?

-- and (now this second part is merely a follow up question that did not form part of the original query but added later) where that $$\pi^2$$ term can lend some interpretation of the underlying phenomenon, just like does $$\pi$$ whereby we can interpret (in most cases i.e.) that some type of circular ambulation in 1 dimension is involved??

As you can understand, the $$\pi^2$$ term is more complex and does not directly lend itself to an interpretation -- as opposed to $$\pi$$ which is very intuitive.

Thanks

• volume of a $4$-ball? Jul 19, 2019 at 18:53
• Basel problem Jul 19, 2019 at 18:56
• The volume of a torus. Jul 19, 2019 at 19:16
• @Kumar The question is still ambiguous unless you define what you mean by naturally as it may be interpreted as the physical/natural world. I suggest you re-title it to something like - Where does $\pi^2$ appear in mathematics or physics? Jul 20, 2019 at 4:04
• I don't know what you mean, Kumar, by "spontaneously", or by "inherently". Also, $\pi$ appears in a great many places with nothing to do with circles. Jul 20, 2019 at 22:46

$$g \approx \pi^2\,\mathrm{m/s^2}$$

The reason for this the original definition of the meter: the length of a pendulum whose half-period is 1 second. Much like the original definition of the Celsius scale, this allowed a person to easily calibrate equipment with common materials. (Of course, this calibration is hardly precise enough for modern measurements, so the meter has been redefined more than once to keep up with the times.) The $$\pi^2$$ then comes from the small-angle approximation to the period of a pendulum: $$T = 2\pi\sqrt{\frac{L}{g}}\;\; \Longrightarrow\;\; g = \pi^2 \frac{L}{(T/2)^2}$$

• Interesting. The value of gravitational acceleration we use today implies the meter has grown by a few tenths of a percent. Oct 28, 2020 at 17:55

A great example in my opinion is the Basel problem $$\sum_{k = 1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}.$$ A great intuitive and geometric explanation can be found in this video, which, along with the whole channel, I cannot recommend enough.

There's also a whole Stack Exchange post dedicated to it.

You can get arbitrarily-high powers of $$\pi$$ (though divided by factorials) using involutes:

• A semicircle of radius $$1$$ has length $$\pi$$.
• The involute of the semicircle has length $$\tfrac12\pi^2$$
• The involute of that involute (emerging from the same point) has length $$\tfrac1{6}\pi^3$$.
• The involute of that involute (emerging from the same point) has length $$\tfrac1{24}\pi^4$$.
• The involute of that involute (emerging from the same point) has length $$\tfrac1{120}\pi^5$$.
• ... and so forth ...

Note: The polygonal spiral formed by joining the involutes' non-common endpoints has edge-lengths equal to those powers of $$\pi$$. (The right-most vertical edge has length $$\pi$$, the top-most horizontal edge has length $$\tfrac12\pi^2$$, etc.) Throw in a segment from the semicircle's right endpoint to its center, and you have an edge of length $$1 = \tfrac{1}{1}\pi^0$$.

Note: That polygonal spiral converges on the involutes' common endpoint. The back-and-forth-ing of the horizontal edges implies that $$\tfrac11\pi^0 - \tfrac12\pi^2 + \tfrac1{24}\pi^4 - \cdots \;=\; -1$$ Likewise, the up-and-down-ing of the vertical edges implies $$\tfrac11\pi^1 - \tfrac16\pi^3 + \tfrac1{120}\pi^5 - \cdots \;=\; 0$$ Of course, these values are, respectively, $$\cos\pi$$ and $$\sin\pi$$. The above is a special case of Chaikovsky's Involute Pinwheel for the power series of cosine and sine.

(1) Let $$\mu$$ denote the Mobius function:

$$\sum_{n=1}^{\infty} \frac{\mu(n)}{n^{2}} = \frac{6}{\pi^{2}}$$

(2) Let $$\sigma(n)$$ denote the sum of the divisors of $$n$$:

$$\lim_{n \to \infty} \frac{\sum_{i=1}^{n}\sigma(i)}{n^{2}} = \lim_{n \to \infty} \frac{\sigma(1) + \sigma(2) + \ldots + \sigma(n)}{n^{2}} = \frac{\pi^{2}}{12}$$

(3) Let $$\phi$$ denote Euler's $$\phi$$-function.

$$\lim_{n \to \infty} \frac{\sum_{i=1}^{n}\phi(i)}{n^{2}} = \lim_{n \to \infty} \frac{\phi(1) + \phi(2) + \ldots + \phi(n)}{n^{2}} = \frac{3}{\pi^{2}}$$

• OP asked about examples where $\pi^2$ occurred naturally. Your examples are not applicable unless you tell where they occur in nature. Jul 20, 2019 at 0:47
• "In which all mathematics/physics equation or contexts does π^{2} appear naturally?" Jul 20, 2019 at 1:24
• By that statement statement OP meant which all equations in the mathematical formulas or derivations of known physics do we use $\pi^2$. I guess what you interpreted is - which all statements in mathematics or in physics do we use $\pi^2$. These are two different questions. For example, there are parametric equations which the value of $\pi^2$ for infinite inputs. So if all you wanted was $\pi^2$, we can get infinitely many such equations or formulas. So the physical context of mathematics in important here. Just see why other answers are focusing on physical use of $\pi^2$ Jul 20, 2019 at 1:35
• @Nil, what makes you so sure you know what OP meant? Jul 20, 2019 at 2:49
• @Nil, the keywords for me are "many equations" and "mathematics". Unless OP chooses to clarify, you have no better idea of what was intended than I do, so you may as well enjoy all the nice answers that have been posted. Jul 20, 2019 at 3:09

If you're looking for examples of $$\pi^2$$ occurring naturally, you should look at some of the common PDEs. The derivation of their solutions and solutions themselves lead to occurrences of $$\pi^2$$. I'll provide some examples, but I'm going to keep everything a bit loose to make it more digestible.

First, take the heat equation on a 1-D rod of length $$L$$ with no heat on the ends: $$\begin{cases} \partial_t u=\kappa\partial_x^2 u\\ u(x,0)=f(x)\\ u(0,t)=u(L,t)=0 \end{cases}$$ If you assume a solution of the form $$u(x,t)=X(x)T(t),$$ then you will arrive at two ODEs to solve: $$\frac{d^2X}{dx^2}=-\lambda X$$ and $$\frac{dT}{dt}=-\kappa\lambda T,$$ with $$X(0)=0$$ and $$X(L)=0$$. The former is an eigenvalue problem with given boundary data, and solving this eigenvalue problem gives you $$\pi^2$$ presence, as we find the eigenvalues to be $$\lambda_n=\left(\frac{n\pi }{L}\right)^2,$$ and $$X_n(x)=\sin\left(\frac{n\pi x}{L}\right),$$ for $$n=1,2,\cdots.$$ Solving the $$T$$ equation gives you $$T(t)=ce^{-\kappa\lambda_n t},$$ and in the end, you get the solution $$u(x,t)=\sum\limits_{n=1}^\infty B_n\sin \left(\frac{n\pi x}{L}\right) e^{-\kappa\frac{n^2\pi^2}{L^2} t},$$ with $$B_n=\frac{2}{L}\int\limits_0^L f(x)\sin\left(\frac{n\pi x}{L}\right)\, dx.$$ This even has $$\pi^2$$ in the solution itself. Notice that, amongst other things, $$\pi^2$$ pops up in the decay rate of the solution to its steady state.

If we look at the wave equation $$\partial_t^2u=c^2\partial_x^2 u,$$ instead, then you will still get the same eigenvalues and eigenfunctions (since they are all eigenvalues/eigenfunctions of $$\partial_x^2$$). If we add in the additional initial condition $$\partial_t u(x,0)=g(x),$$ then we instead get the final solution $$u(x,t)=\sum\limits_{n=1}^\infty\left(A_n\cos\left(\frac{n\pi ct}{L}\right)+B_n\sin \left(\frac{n\pi ct}{L}\right)\right) \sin \left(\frac{n\pi x}{L}\right),$$ where $$A_n$$ is the same and $$B_n=\frac{2}{n\pi c}\int\limits_0^Lg(x)\sin \left(\frac{n\pi x}{L}\right)\, dx.$$ Here, the eigenvalues are related to the frequencies of the waves, so they have a nice physical interpretation.

• Second one also, needs exponentials. Jul 24, 2019 at 18:55
• @SungjinKim Performing separation of variables, you get ODEs $X''+\lambda X=0$ and $T''+\lambda c^2 T=0.$ I wrote the solution to time ODE using $\sin$ and $\cos$ instead of complex exponentials. It's a pretty standard representation of the solution.
– cmk
Jul 24, 2019 at 18:56
• I see . Thank you for the clarification. Jul 24, 2019 at 19:12

List of Places where π^2 can be seen-

1. π is present in some structural engineering formulae, such as the buckling formula derived by Euler, which gives the maximum axial load F that a long, slender column of length L, modulus of elasticity E, and area moment of inertia I can carry without buckling

2. The fact that π is approximately equal to 3 plays a role in the relatively long lifetime of orthopositronium. The inverse lifetime to lowest order in the fine-structure constant α has a term of π^2

3. Kepler's Third Law of Planetary Motion

4. Volume and Bounding Area of 4-D and 5-D Sphere

5. Basel Problem (As mentioned in another answer)

And many more as well

Sources-

https://en.m.wikipedia.org/wiki/Pi

https://en.m.wikipedia.org/wiki/Basel_problem

https://en.m.wikipedia.org/wiki/Buckling

https://en.m.wikipedia.org/wiki/Fine-structure_constant

https://en.m.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion

https://en.m.wikipedia.org/wiki/N-sphere

Also related to the Basel problem: the probability that two random integers are coprime is $$\frac{6}{\pi^2}$$

(in the sense that the probability that $$a,b$$ chosen uniformly from $$\{1, 2, \ldots, N\}$$ are coprime approaches $$\frac{6}{\pi^2}$$ as $$N \rightarrow \infty$$.)

Usually $$(2 \pi)^{-n}$$ appears in the $$n$$ dimensional Fourier transform: If the Fourier transform of a $$L^1(\mathbb{R}^n)$$ is defined as $$\hat{f}(k)=\int_{\mathbb{R}^n} e^{-i k \cdot x} f(x) \,\mathrm{d}^n x$$ Then the inverse transform will be (if the integral is convergent) $$f(x)= \frac{1}{(2 \pi)^n}\int_{\mathbb{R}^n} e^{i k \cdot x} \hat{f}(k) \,\mathrm{d}^n k$$ But you can also "spread" it to be more symmetric: if $$\hat{f}(k)= \frac{1}{(2 \pi)^{n/2}}\int_{\mathbb{R}^n} e^{-i k \cdot x} f(x) \,\mathrm{d}^n x$$ then $$f(x)= \frac{1}{(2 \pi)^{n/2}}\int_{\mathbb{R}^n} e^{i k \cdot x} \hat{f}(k) \,\mathrm{d}^n k$$

Euler's solution of Basel problem: Many of the answers posted here talk about Euler's solution of the Basel problem. But an important thing that is notably missing is that Euler's original solution to the Basel problem already had $$\pi^2$$ in it. Euler used the well know trigonometric identity

$$\frac{\sin x}{x} = \prod_{n \ge 1} \bigg(1 - \frac{x^2}{\pi^2 n^2}\bigg)$$

Ramanujan's first letter to Hardy: This one is a part of mathematical folklore, appearing Ramanujan's first letter to Hardy and one of few that Hardy saild he could prove himself out of the many formulas in the letter. If $$\alpha \beta = \pi^2$$ then

$$\alpha^{-1/4}\bigg(1 + 4\alpha\int_{0}^{\infty} \frac{xe^{-\alpha x^2}}{e^{2\pi x}-1}dx\bigg) = \beta^{-1/4}\bigg(1 + 4\beta\int_{0}^{\infty} \frac{xe^{-\beta x^2}}{e^{2\pi x}-1}dx\bigg)$$

Number of square free integers: The number of square free integers $$\le x$$ is $$\dfrac{6x}{\pi^2} + O(\sqrt{x})$$

Probability of coprime numbers: The probability that $$n$$ positive integers chosen uniformly from $$\{1, 2, \ldots, n\}$$ have no common factors approaches

$$\frac{1}{\zeta(2n)} = \frac{(-1)^{n+1}(2\pi)^{2n}B_{2n}}{2(2n)!}$$ as $$n \rightarrow \infty$$, where $$B_k$$ is the $$k$$-th Bernoulli number.

Others: $$\sum_{n = 1}^{\infty}\frac{F_{2n}}{ n^2 {2n \choose n}} = \frac{4\pi^2}{25\sqrt 5}, \text{where F_n is the n-th Fibonacci number}$$

$$\lim_{n \to \infty}\frac{1}{n^2} \sum_{r = 1}^n n(\textrm{mod } r) = 1 - \frac{\pi^2}{12}$$

$$\int_{0}^{1} \bigg(\frac{\tan^{-1}x}{x}\bigg)^2 dx = G - \frac{\pi^2}{16} + \frac{\pi \log 2}{4}, \text{ where G is the Calatan constant}$$

How about an even higher power of $$\pi$$ in the Stefan-Boltzmann Law relating radiant energy flux from a black body to temperature? This relation involves the fifth power of $$\pi$$:

$$\text{Radiant flux}=((2\color{blue}{\pi^5}k^4)/(15c^2h^3))T^4$$

The parameters $$k, c, h$$ are respectively Boltzmann's constant, the speed of light and Planck's constant from physics. $$T$$ is absolute temperature.

The number of three-term geometric progressions of positive integers with no term exceeding $$n$$ is $${6\over\pi^2}n\log n+O(n)$$.

The volume of a torus which consists of a tube with radius $$r$$ centered at a circle with radius $$R$$ is $$4\pi^2Rr$$.

The $$\text{Riemann Zeta function} \ \ \zeta(s)=\large \sum_{n=1}^{\infty} \frac{1}{n^s}$$ is used in many branch of Science and mathematics. Replacing $$s=2$$, we have $$\zeta(2)=\sum \frac{1}{n^2}$$. In $$1735$$ Leonard Euler showed that $$\zeta(2)=\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{{\color{blue}{\pi^2}}}{6}.$$

This result leads to Number Theory and probability result as follows:

The $$\ {\color{blue}{ probability}} \$$ of two random number being $$\ {\color{blue}{ relatively \ \ prime}} \$$ is given by the following product over all primes $$\prod_{p}^{\infty} \left(1-\frac{1}{p^2} \right)=\left(\prod_{p}^{\infty}\frac{1}{1-p^{-2}} \right)^{-1}=\frac{1}{1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots}=\frac{1}{\zeta(2)}=\frac{6}{{\color{blue}{\pi^2}}} \approx 61 \%.$$

This is an interesting result where $$\pi^2$$ is involved.

• Both of these results are already mentioned in other answers, right? Jul 21, 2019 at 12:59
• @GerryMyerson, who mentioned the last and in fact main part of my answer? I portrayed here that $\pi^2$ is used to mention the probability of two random number being relatively prime. I don't see anyone used in this question
– MAS
Jul 21, 2019 at 14:33
• math.stackexchange.com/questions/3298039/… Jul 21, 2019 at 22:23
• @GerryMyerson, both answers have big difference in presentation, ok
– MAS
Jul 21, 2019 at 23:40