Pi-day

Today is π-day! Hip-Hip!

Inks:

• Pi

  Taccia Ukiyo-e Hokusai Sabimidori

• Sun

  Diamine Communication Breakdown

• Mountain top

  Taccia Ukiyo-e Hiroshige Ainezu

Let's eat pie!

Unfortunately, we haven’t had the capacity to bake a decent pie. However, for some reason, today the pie was at a discount at the local Albert Heijn 🥳

π in a pie

Let’s calculate some properties of this pie using π.

Approximating the pie with a truncated cone, the diameter of the top is 16cm, the diameter of the bottom is 14cm and the height is 4.5cm. It’s weight is 620g.

$$d_{\text{top}}\approx 16\text{cm}⟹r_{\text{t}}\approx 8\text{cm}$$

$$d_{\text{bottom}}\approx 14\text{cm}⟹r_{\text{b}}\approx 7\text{cm}$$

$$h\approx 4.5\text{cm}$$

$$m\approx 620\text{g}$$

Number of bites

The number of bites can be approximated by the volume of the pie divided by the volume of an average bite.

$$V=\frac{\pi h}{3}\cdot ({r_{\text{b}}}^2+r_{\text{b}}{r}_{\text{t}}+{r_{\text{t}}}^2)\approx 796.4{\text{cm}}^3$$

Assuming an average bite on a desert fork to be 2cm⋅3cm⋅1.5cm=9cm³, that equates to approximately 89 bites.

Love crust!

Let’s calculate the total exterior surface as a rough approximation of that yummy crusty area.

$$A_{\text{top}}={r_{\text{t}}}^2\pi \approx 201.06{\text{cm}}^2$$

$$A_{\text{bottom}}={r_{\text{b}}}^2\pi \approx 153.94{\text{cm}}^2$$

$$A_{\text{side}}=\pi (r_{\text{b}}+r_{\text{t}})h_{\text{slant}}=\pi (r_{\text{b}}+r_{\text{t}})\sqrt{h^2+{(r_{\text{t}}-r_{\text{b}})}^2}\approx 217.23{\text{cm}}^2$$

$$A_{\text{total}}=A_{\text{top}}+A_{\text{bottom}}+A_{\text{side}}\approx 572.23{\text{cm}}^2$$

How likely is it that my pie teleports to the Moon?

I’m sure that it’s perfectly normal to be anxious about your delicious pie somehow quantum tunneling to the Moon. The best is to address the fear head on and do a reality check: How likely is that anyway?

To simplify the calculation, let’s have a rough upper bound by examining the center-of-mass wavefunction at the distance of the Moon (L ≈ 394,658 km as of today) and within a pie-high shell (h ≈ 4.5 cm) around the Moon. Let’s approximate the Moon with a sphere having a radius of R = 1,737.4 km.

The volume of the shell is

$$V_{\text{shell}}=\frac{4}{3}\pi [{(R+h)}^3-R^3]\approx 1.707\times {10}^{12}{\text{m}}^3$$

Using a normalized, spherically symmetric Gaussian wave packet, the center-of-mass wavefunction would be

$${\left|\psi \right(r\left)\right|}^2=\frac{1}{{\left(2\pi {\sigma }^2\right)}^{\frac{3}{2}}}\cdot e^{(-\frac{r^2}{2{\sigma }^2})}$$

Without an exact model, we need to guestimate the positional uncertainty. We are talking about room temperature and air moving, neither the pie nor my table is perfectly rigid, etc. Let’s say generously, that σ = 1nm

The probability of finding the pie inside a target volume V located a distance L away is approximately

$$P_{\text{Moon}}\approx V{\left|\psi \right(L\left)\right|}^2$$

Thus

$$P_{\text{Moon}}\approx \frac{V}{{\left(2\pi {\sigma }^2\right)}^{\frac{3}{2}}}\cdot e^{(-\frac{L^2}{2{\sigma }^2})}$$

Substituting in the values gives the result

$$P_{\text{Moon}}\approx {10}^{-3.38\times {10}^{34}}$$

Which is an incredibly tinsy chance, even if you were to wait for this to happen until the heat death of the universe comes (≈ 10¹⁰⁰ years). And we all know that the life expectancy of a good pie is less than a blink of an eye on cosmological timescales. It’s like you have 1 and then 3.38 x 10 ³⁴ zeros (NOT 34 but 10 ³⁴ zeros!) number of parallel universes where everything is the same, only the location of the pie differs, and in one of those parallel universes my pie sits somewhere on the Moon.

It’s a very rough upper estimate. In reality, for macroscopic objects, interactions with the environment cause decoherence to occur almost instantaneously. This would make my pie quantum-tunneling over astronomical distances even much less plausible.

So, with a huge sigh of relief, I can celebrate π-day with a pie and without the fear of it getting “mysterialised away” to the Moon.