In other words: which physics experiment requires to know Pi with the highest precision?
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10$\begingroup$ I'm quite happy with $\pi\approx 1$ (-; $\endgroup$– user68Commented May 9, 2011 at 16:44
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4$\begingroup$ Classic quote from Kolb & Turner's The Early Universe: "The reader may wish to follow the advice of one of the authors to ignore the various factors of $V$; they play no crucial role. Moreover, the less ambitious reader may also wish to ignore all factors of $\pi$, the so-called small-circle approximation." $\endgroup$– Ted BunnCommented May 9, 2011 at 18:22
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1$\begingroup$ @mbq Your back-of-the-envelope calculations will suffer for it, it's well known that $\pi=\sqrt{10}$ as far as any theorist is concerned. $\endgroup$– Emilio PisantyCommented Sep 9, 2013 at 23:26
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4 Answers
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39 digits of pi is enough to calculate circumference of visible universe to a margin of error equal to the width of a proton ( unverified computation, according to http://www.guardian.co.uk/science/blog/2011/mar/14/pi-day CordwainerBird ) .
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$\begingroup$ Ok, but I would say this is rather maths than physics until we build an experiment to measure the circumference (or surface, or volume) of the visible universe... $\endgroup$ Commented May 9, 2011 at 14:14
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$\begingroup$ The word 'visible' implies the measurable, known, universe. So, using CMBR as a (physical) basis, a diameter of about 91 billion light-years... $\endgroup$– RondoCommented Mar 14, 2015 at 1:08
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Pi is very far from being the only number we need in physics. Typical theoretical predictions depend on many other measured and calculated (or both) numbers besides pi.
Nevertheless, it is true that one needs to substitute the right value of pi to get the right predictions. Therefore, the right answer to your question is the most accurately experimental verified theoretical prediction we have in physics as of today, namely the anomalous magnetic dipole moment of the electron
http://en.wikipedia.org/wiki/Anomalous_magnetic_dipole_moment
In some natural units, the magnetic moment of the electron is expressed as a g-factor which is somewhat higher than two. Experimentally, $$\frac g2 = 1.00115965218111 \pm 0.00000000000074$$ Theoretically, $g/2$ may be written as $$\frac g2 = 1+\frac{\alpha}{2\pi} + \dots$$ where the $\alpha/2\pi$ first subleading term was obtained by Schwinger in 1948 and many other, smaller terms are known today. The theoretical prediction agrees with the experimental measurement within the tiny error margin; the theoretical uncertainty contains the effect of new species of virtual particles with the masses and couplings that have not yet been ruled out. This requires, among many and many other things, to substitute the correct value of $\pi$ in Schwinger's leading correction $\alpha/2\pi$. You need to know 9-10 decimal points of $\pi$ to make this correction right within the experimental error.
So in practice, $\pi\approx 3.141592654$ would be OK everywhere in the part of physics that is testable. However, theoretical physicists of course often need to make calculations more accurately if not analytically, to figure out what's really happening with the formulae.
answered May 9, 2011 at 7:42
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$\begingroup$ That was exactly what I was looking for. Thanks! By the way, anything equivalent for e ? $\endgroup$ Commented May 9, 2011 at 14:16
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1$\begingroup$ Hm, I would have to remind myself why $\alpha$ is a more fundamental quantity to get from experiments than $\frac{\alpha}{2\pi}$. Not sure if it is worths it, though, as decimals of $\pi$ come extremely cheap with today's computers... $\endgroup$– Qmechanic ♦Commented May 10, 2011 at 14:26
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16 digits, for converting frequencies from Hz to angular frequency. Frequencies can now be measured with a precision approaching 1 part in 10^16, so dealing with those numbers would require knowing Pi to 16 digits or so.
Is this really "which physics experiment REQUIRES to know Pi with the highest precision"? A purist (or theorist) might disagree that this example doesn't count, and I'd agree that there's nothing fundamental here, just conversion from one convention to another. But from a practical point of view, it's something people do in physics and it requires knowing Pi to 16 digits.
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In principle you will never reach the accuracy of the numerical value of $\pi$ in an experiment, it is much more important as an analytical tool. The part of the question concerning the most accurate experiment was given by Luboš.
So, just to give a little thing about the importance of $\pi$ as the area of a unit circle and also the ratio of a circle's circumference to diameter,
(from the corresponding wiki-article)
One very famous examples are Fourier transforms where you ask a periodic function to be represented by $\sin$ and $\cos$-terms, $$\mathcal{F}(f)(\xi)=\int_{-\infty}^{\infty}f(x)e^{2\pi\mathrm{i} \xi x}dx\ \mathrm{with}\ e^{\mathrm{i}\varphi}=\cos(\varphi) + \mathrm{i}\sin(\varphi)\ .$$ This is linked to $\pi$ because of the periodicity of the circle, $$r_1(\varphi) = r_2(\varphi + 2\pi)\ \mathrm{with}\ r_i\in \mathrm{circle}\ .$$
There are much more things to add as the underlying group of rotations is $U(1)$ which is in turn the gauge group for (quantum)electrodynamics explaining why you will almost certainly find the most prominent examples of results linked to the value of $\pi$ in this field.
answered May 9, 2011 at 8:09