## The set-up

On occasion I’ve run across people arguing in favor of using tau $\left(\tau\right)$ instead of pi $\left(\pi\right)$ at various levels of seriousness. Personally I don’t care what constant you want to think in or advocate the merits of that choice. That’s all applied math and some very solid mathematicians claim to think $\tau$ has some merit on this front. Composing convenient constants is just normal. However the idea that $\tau$ is a more natural constant than and should replace $\pi$ is pushing the debate into the realm of pure math. To support that argument you have to come up with a reasonable justification beyond: I personally find this easier or more aesthetically pleasing. I’ve seen plenty arguments from both camps (such as The Tau Manifesto & The Pi Manifesto) but none of them have been very satisfying. $\pi$ pops up in all kinds of equations so it’s easy for either camp to choose some that fit their position.

Let me give a basic argument that $\pi$ is the natural choice.

## The argument

We want something that is as basic as possible with little to no arbitrary choices involved. Let’s consider that the most basic quantity in which $\pi$ appears and that’s an angle measurement. Now angles are zero dimensional (or dimensionless if that’s your thing) so forming an argument based on anything in a higher dimension is more complex so we can discard all of those arguments (which covers all the ones I’ve seen).

Imagine we have two unit vectors $\mathbf{a}$ and $\mathbf{b}$ in an n-dimensional vector space then we can implicitly measure the angle between them by the parallel projection (dot product):

\[x = a \cdot b\]

which yields:

given both are unit vectors we have:

and solving for $\theta$ to arrive at an explicit angle measure:

and Bob’s your uncle (that’s UK English for quod erat demonstrandum). Notice the discussion is coordinate free and devoid of any arbitrary choices (say standard conventions). Let’s do a definition in terms of the presented argument:

$\pi$ is the maximum magnitude torque minimal angle (in radians).

If the thrust of this argument isn’t obvious then the final section goes into more details.

## The challenge

To support a claim of $\tau$ being a more basic quantity than $\pi$ would require somebody to demonstrate its appearance as a zero dimensional measure without requiring any arbitrary choices.

## The wrong direction

Let’s skim the opening of The Tau Manifesto:

- The circle constant

…one of the most important numbers in mathematics, perhaps the most important: thecircle constantrelating the circumference of a circle to its linear dimension.

I find this noteworthy since it seems people in the “$\pi$ is wrong camp” have a similar fixation on circles. The idea that $\pi$ comes from circles was probably head-shot at the advent of calculus and certain received its coup de grâce in the mid 19^{th} century (think of all the algebras with geometric meaning popping up like mushrooms). Oh, and come on! $e$ is way more dope than $\pi$! (example: Lindemann-Weierstrass theorem)

The first example argument provided is:

For example, consider integrals over all space in polar coordinates:

\[\int_{0}^{2\pi} \int_{0}^{\infty} f\left(r,\theta\right)~r~dr~d\theta\]

The upper limit of the $\theta$ integration is always $2\pi$.

Well..no? This is an example of following a standard convention and we can also express as:

Of course the *width* of the intervals are the same, but one dimensional things don’t interest me. My point here is that a fair number of $2\pi$ terms seen in the wild are simply following convention and could be written in another way. The second example is an expression for the PDF of the normal distribution:

which is a generalized version of the Gauss’ original:

and there is a reduced version by Stigler:

Shrug. All of the blue colored terms are simply scale values to normalize the area of the remaining parts of each equation. We have the shape of the distribution (black terms) and the support (range of returned values) which we integrate to get the area and we multiply by the reciprocal of that area to normalize it to one. But *areas* really don’t interest me and I’m mentioned all this because…come on! $e$ is way cooler than $\pi$!

I could carry in this style and, say, mention the area of 3D rotations is $\pi^2$, but I’m not interested in volumes. This is my problem with the pro-$\tau$ and pro-$\pi$ arguments I’ve seen. They’re having fun tossing out quantities *more* complex than those of the classic circle. That’s a wrong left turn in Albuquerque if you’re not making a personally easier or more aesthetically pleasing argument.

## The break-down

Let me return to *the argument* and break it down a little. Since I’m only using the dot product I’m effectively asking *“What’s the magnitude of the angle between $a$ and $b$?”*, but what if we asked *“What’s the angle from $a$ to $b$?* Does that help? Not alone it doesn’t. We’ll end up writing the equivalent to the following 2D coordinate:

The $x$ coordinate is signed because we have a reference direction $(a)$ for it. The $y$ coordinate will always be positive because there are two directions orthogonal to $a$ in the oriented plane that spans $a$ and $b$ and we need add an additional external constraint to be able to choose one of them to be called the positive direction. Let’s run with that.

Sticking with an n-dimensional vector space we can split $b$ into the part in the direction of $\left(b_{\parallel}\right)$ and orthogonal $\left(b_{\perp}\right)$ to $a$:

\[\begin{align*} b & = b_{\parallel}+b_{\perp} \\ b_{\parallel} & = \left(a \cdot b\right)~a \\ b_{\perp} & = b - \left(a \cdot b\right)~a \end{align*}\]

and $b_{\perp}$ is a n-vector so the best we can do is compute it’s magnitude. If we change to vector analysis (vector calculus, Gibbs vectors, 3D vectors, whatever ya want to call it) then we could take the cross product to get the oriented plane (recall the cross product of two vectors produces a bivector). We could then cross that with $a$ to get an orthogonal direction (as a vector) in the plane (SEE: vector triple product):

which takes us to the same place. We could stick with three dimensions and define $a$ and $b$ to instead be unit bivectors and use quaternions with the expression $ba^*$. The bivector part of the result is the oriented plane and scalar part is just the dot product…so different algebra structure and equivalent result. Let’s drop down to using complex numbers (the expression is the same as quaternions). Well **finally** we get a $y$ value with a sign. But here’s the problem: we classically associate complex numbers with a 2D coordinate but they are really a scalar $(x)$ + a 2D bivector $(y)$ so the sign of $y$ gives the orientation of the plane. By choosing to interpret that sign not as an orientation but as a scalar we are implicitly choosing a preferred oriented plane which is the additional choice which allows us to assign a sign to $y$. (NOTE: sometimes people will refer to $y$ as a pseudoscalar and FWIW this is a property rather than a “type” statement).

So we can augment our measuring method to get ourselves a signed $y$ value and now we can compute angles on: $\left(-\pi, \pi\right]$. Still no $\tau$ (I’ve mentioned having no interest in widths, haven’t I?). To get a measure on $\left[0,2\pi\right)$ we have to add an additional feature to our measuring method which is a preferred winding. Sticking with complex numbers treated as a 2D coordinate system (with positive orientation preferred) then we need to choose clockwise or counterclockwise with respect to the line of Reals $(x)$.

Is this cheating? Isn’t defining the principle angle range to be $\left(-\pi, \pi\right]$ just an arbitrary choice (programming languages effectively do via atan2)? Can’t we (for complex numbers) define the branch cut to be from 0 along positive Reals instead of negative? Let’s pretend like I agree with that, then we still need to tack on one feature to our measuring tool to get $\tau$. But I don’t agree because defining it to be $\left(-\pi, \pi\right]$ is consistent with the math. Assume $a$ and $b$ are not collinear then we can measure from $a$ to $b$ the angle $\theta$ and oriented plane $P$ and if we measure from $b$ to $a$ then we’d get $\theta$ and $-P$. Adding our first addition constraint is always equivalent to choosing a preferred plane and let’s pretend it’s $P$ then our second set of measures need to be multiplied by -1 giving $-\theta$ and $P$. Let’s do a final def:

$\pi$ is the maximum intrinsically measurable angle (in radians).