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

Question

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

Answer 1

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

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: $$ T = 2\pi\sqrt{\frac{L}{g}}\;\; \Longrightarrow\;\; g = \pi^2 \frac{L}{(T/2)^2} $$

Answer 2

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.

Answer 3

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 ...

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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.

Answer 4

(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.

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

Answer 5

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.

Answer 6

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 $N \rightarrow \infty$.)

Answer 7

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 $$f(x)= \frac{1}{(2 \pi)^{n/2}}\int_{\mathbb{R}^n} e^{i k \cdot x} \hat{f}(k) \,\mathrm{d}^n k$$

Answer 8

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

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

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

Answer 9

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

Answer 10

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

Answer 11

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.

Answer 12

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

Answer 13

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.

Answer 14

Here is perhaps the newest result on $\pi^2$, proved only a few hours ago by Sungjin Kim in while working on one of our collaboration. Perhaps this is the fastest case of discovery and reference.

For any $\epsilon > 0$, the value of $$ \frac{\sqrt{2n}}{2^n } \bigg\{{n\choose 1^2} + {n\choose 2^2} + {n\choose 3^2} + \cdots + {n\choose r^2}\bigg\} $$

is more likely to be in the interval is $1 + 2e^{-\pi^2/4} \pm \epsilon$ than any other interval of the same width.