Can Angles be Assigned a Dimension?

1. Some Background on Dimensional Analysis

… if you are not already familiar with it.

1.1 Dimensions

Dimensional Analysis is a way of analyzing physics equations that consider only the qualitative dimensions – mass, length, time, charge .. – of the quantities involved, not the values that they take in the given problem.  This is not to be confused with Euclidean dimensions.

The fundamental rule is that you can only add, subtract or equate terms that have exactly the same constituent dimensions, and each to the same degree.  Acceleration cannot be a force or be added to a force because the latter includes a mass dimension while the former does not.  An area cannot be compared to a distance because it is length squared.

For a detailed discussion see e.g.  http://web.mit.edu/2.25/www/pdf/DA_unified.pdf, or any text on the Buckingham Pi theorem.   (There are also Wikipedia and Khan academy links, but these, unfortunately, confuse DA with managing units as variables, which is a separate topic.)

1.2 Notation

The standard notation in DA is that if x is a variable in an equation then [x] extracts its dimensionality.  The dimensions themselves are labeled M for mass, L for length, T for time, Q for a charge, θ for temperature …

Thus, if F is a force then $[F]=MLT^{-2}$.  The well-known equation F=ma would be analyzed as $MLT^{-2} = (M)(LT^{-2})$, which is clearly true.

1.3 Uses

1.3.1 Predicting the form of a relationship

DA can be a useful shortcut to establishing the general form of how one quantity depends on others.

Example: We presume that the pressure difference, $\Delta P$, between the inside and outside of a bubble depends on the radius, $r$ and the surface tension $S$.

$[ \Delta P]=ML^{-1}T^{-2}$

$[r]=L$

$[S]=MT^{-2}$

The only way to combine the pressure difference and surface tension to obtain a length is

$(MT^{-2})/(ML^{-1}T^{-2})=L$

Hence we can say

$r=kS/\Delta P$

for some constant k.

1.3.2 Error checking

Many algebraic errors can be caught by checking dimensional consistency.

2. Angles

Angles have never been considered to have dimension.  Consider, for example, the formula for arc length $s$:

$s = r\theta$

Since $s$ and $r$ each have dimension $L$, the angle cannot have a dimension, or so it seems.

2.1 Units Matter

Dimensionless combinations, like $force/(mass \times acceleration)$, generally have the useful feature that they are invariant to the units used.  As long as the units used for the variables are consistent, the same number results whether you use SI or Imperial, or Babylonian.

This is not true of angles; many problems posted on the Homework forums are resolved when the student remembers to plug in the number of radians as argument to the sine function instead of degrees.

2.2 Distinct Entities with the Same Dimensionality

It is somewhat unsatisfactory that, when reduced to mere dimensions, some pairs of quite different entities appear to be the same.

Torque and energy are both force times distance.  In terms of vectors, the first is the vector product, the second the scalar.

Angular momentum and action are both $ML^2T^{-1}$.  Again, a vector and a scalar.

3 Axioms for an “Imaginary” Dimension

Consider assigning angles the dimension $\Theta$, with some unusual properties:

• $\Theta^2=1$
• The cross product operator itself has dimension $\Theta$
• $i$, the square root of -1, has dimension $\Theta$

3.1 Vectors and Angles

The following table illustrates the use of the dimensionality of angles with cross and dot products.

Entity	Sample Equation	Dimension
Arc length element	$\vec{ds}=\vec r\times\vec{d\theta}$	$L=(L)(\Theta)(\Theta)$
Torque	$\vec\tau=\vec r\times\vec F$	$ML^2T^{-2}\Theta=(L)(\Theta)(MLT^{-2})$
Work	$E=\vec r.\vec F$	$ML^2T^{-2}=(L)(MLT^{-2})$
Angular momentum	$\vec L=\vec r\times\vec p$	$ML^2T^{-1}\Theta=(L)(\Theta)(MLT^{-1})$
Gyroscopic precession	$\vec \tau=\vec \Omega_p\times\vec L$	$ML^2T^{-2}\Theta=(\Theta T^{-1})(\Theta)(ML^2T^{-1}\Theta)$
Velocity	$\vec v=\vec r\times\vec{\omega}$	$LT^{-1}=(L)(\Theta)(\Theta T^{-1})$

E.g. to resolve the $s=r\theta$ case for arc length, we can argue that this should really be expressed as the integral of the magnitude of a vector:

$s=\int|\vec{dr}|=\int |\vec r\times\vec{d\theta}|$.  The $\Theta$ dimension of the angle is neutralized by that of the cross-product operator.

3.2 Functions of Angles

Raising a dimensioned entity to power is fine because we can still express the dimensions of the result.  For other functions, such as exp, log, and trig functions, it is more problematic.  If you ever find you have an equation of the form $e^x$, where $x$ has dimension, you can be pretty sure you have erred.

For trig functions, it would be reasonable to require the argument to be an angle, but $e^{i\theta}$ would appear to create a difficulty for assigning angles a dimension.

This can be resolved by giving $i$ the $\Theta$ dimension also.  Based on the power series expansions, we see that the odd trig functions (those for which f(-x)=-f(x)) necessarily return the $\Theta$ dimension but the even functions, such as cosine, return a dimensionless value.

Thus, $e^{i\theta}=\cos(\theta)+i\sin(\theta)$ is entirely dimensionally consistent.

3.3 Areas and Volumes

Areas are naturally generated as cross products of vectors, imbuing them with the dimension $L^2\Theta$.  Since the vector is normal to the surface, this is analogous to rotations as vectors.

Since volumes arise from the triple scalar product, it would seem that these should also have the $\Theta$ dimension.  This is more surprising.

The solid angle element subtended at the origin by a surface element $\vec {dS}$ at position $\vec r$ is given by $d\Omega=\frac{\vec r.\vec{dS}}{|r|^3}$.  Or, if we wish to make this a vector in the direction $\vec r$, $\vec{d\Omega}=\vec r\frac{\vec r.\vec{dS}}{|r|^4}$.

Alternatively, in polar coordinates, $d\Omega=\sin(\theta).d\theta d\phi$

Whichever way, the dimension is again $\Theta$, which feels consistent with the result for volumes.

3.4 Complex Arithmetic

If $i$ is to be given dimension, what are we to make of $1+i$?

Despite appearances, there is no difficulty.  $1+i$ is a convenient notation, but it is not added in the same sense as in $1+1$.  The 1 and the $i$ retain their separate existences.  We might just as easily if less conveniently, have chosen to write complex numbers as an ordered pair, like <x, y>.  The real and imaginary parts never get crunched together in the same way as in normal addition, so they can have different dimensions without creating any inconsistencies.

3.5 Frequency and Angular Frequency

In wave expressions, frequency, $f$, is the number of cycles per unit time, while angular frequency, $\omega$, is radians per unit time.  $\omega=2\pi f$.

Clearly $\omega$ should have dimension $\Theta T^{-1}$.  The dimension for $f$ depends on whether the factor $\pi$ is to be taken as an angle or as a dimensionless number performing a conversion of units.  Taking $f$ as having dimension $T^{-1}$ appears to be best.

3.6 Planck’s Constants

Consider the equations

$E=hf$

$E=\hbar \omega$

Since $h$ has dimension of action, $ML^2T^{-1}$, it has no angular dimension.  That is fine for the first equation since frequency is just $T^{-1}$.

In the second equation, $[\omega]=\Theta T^{-1}$.  $\hbar$ is defined as $\frac h{2\pi}$.  Since $\pi$ is an angle here, that has dimension $ML^2T^{-1}\Theta$, cancelling the $\Theta$ from $\omega$ and achieving dimensional consistency.

But note that this gives $\hbar$ the units of angular momentum, not action. An implication is that $\hbar$ should perhaps be considered a vector, and we should write the photon energy as

$E=\vec{\hbar}.\vec{ \omega}$

though how one is to justify that $\vec {\hbar}$ is necessarily in the direction of the velocity is unclear.

Likewise for momentum

$\vec p = k\vec{\hbar}$

where k is the wavenumber.  Note that this provides momentum as the vector it should be, rather than just defining its magnitude.

The Heisenberg Uncertainty relations, such as

$\frac 12\hbar\leqslant \Delta \vec p.\Delta \vec x$

would appear to violate both the dimensionality and the notion of making $\hbar$ a vector.  But if we follow the steps in the proof of the uncertainty relation we come to this penultimate statement:

$\frac 14\hbar^2\leqslant (\Delta \vec p.\Delta \vec x)^2$

At this point, giving $\hbar$ an angular component of dimensionality creates no problem.   Neither is there an issue with thinking of it as a vector.  These problems only appear when we overlook the ambiguities that so commonly arise when taking square roots.

4. Postscript

Subsequent to penning the original article, I have become aware of numerous prior attempts, dating back as far as 1936.  An excellent summary is by Quincey and Brown at https://arxiv.org/ftp/arxiv/papers/1604/1604.02373.pdf.

But their list misses a key one:

C. H . Page, J. Research National Bureau of Standards 65 B (Math. and Math. Phys.) No. 4, 227-235; (1961). http://nvlpubs.nist.gov/nistpubs/jres/65B/jresv65Bn4p227_A1b.pdf

It appears that most of my work above is a rediscovery of Page’s:

Result	Page in “Page”
Cross product has angular dimension, dot product does not	231
$\Theta^2=1$	Appendix 2
sine has angular dimension, cosine does not	Appendix 2
solid angles have angular dimension (i.e. not squared)	Appendix 2
whole cycles, as a unit, are dimensionless	Appendix 3

Curiously, Page drew attention to the difficulty posed by $e^{i\theta}=\cos(\theta)+i\sin(\theta)$, but overlooked the remedy of assigning angular dimension to $i$.

He did not consider Planck’s constants.