I'm very confused about how geometric calculus handles units. With classical calculus, units seem to function without too many problems. For example, consider the analytic function $f:X \rightarrow Y$ such that the values in $X$ are measured in seconds and the values in $Y$ are measured in meters. The limit-based definitions of a derivative and integral

$\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = \frac{df(x)}{dx}$

and (not an exact definition here, I know, but LaTeX is hard to work with)

$\lim_{\Delta x \to 0} \sum f(x) \Delta x = \int f(x) dx$

easily illustrate why the units of a derivative are meters/second and the units of an integral are meters*seconds. But the geometric derivative is defined as

$\lim_{h \to 0} (\frac{f(x + h)}{f(x)})^{1/h} = f^{*}(x)$

$h$ is measured in seconds. How are we able to set the fraction $(\frac{f(x + h)}{f(x)})$ to the power of something with units? Correct me if I'm wrong, but that seems like an illegal operation.

Furthermore, the multiplicative (geometric) integral takes every value in the image of $f$ between two bounds, sets them to the power of $dx$, and then multiplies them together. Once again, $dx$ is measured in seconds. How does this work?

What if we put these questions on hold and look at the conversions to the classical versions of the geometric derivative and integral? These are $e^{f'(x)/f(x)}$ and $e^{\int_{x_{0}}^{x_f} ln(f(x)) dx}$. Both of these have screwy units as well.

I can think of a couple of solutions. We could claim that geometric integrals and derivatives are only defined on functions from fields to fields, but I don't like this solution as geometric integrals have applications to many areas of applied mathematics that involve units.

Another option is to use an implicit unit scale as the physicists are wont to do--instead of sending $h$ to zero and $\Delta x$ to zero, we could send $h/u$ to zero and $\Delta x / u$ to zero, where $u$ is a constant with the same units as the elements of $X$. This is the approach I've taken when using geometric calculus in my own work, but I can't help but feel there must be a more elegant way to deal with the problem of units. I also have only figured out how to do this for geometric integrals, not geometric derivatives.

There are other options as well, but they're far more mathematically sketchy. I've experimented with "stripping out" the units from the geometric integrals and derivatives so that all units are contained in a single coefficient, then defining that coefficient as a new unit of measurement, but I have serious doubts about both the rigor and legitimacy of this method.

At present, option #2 seems to work rather well for specific cases, such as modeling non-continuous compounding of interest payments where the interest rate varies over time--one can set the implicit time scale to the rate of compounding and the math works out perfectly. But given that I'm not an expert mathematician, I want to know how one is actually supposed to handle units.

My question is composed of two parts: 1) Can anyone tell me the proper way to deal with units in geometric calculus, and if the "implicit unit scale" method is correct how to apply it geometric derivatives as well as integrals? And 2) if we do need an implicit unit scale, why don't functions from $\mathbb{R}$ to $\mathbb{R}$ need a dimensionless scale for consistency's sake? It seems to me that applying a "unit scale" to the real numbers would just be a scalar in $\mathbb{R}$; what's to stop that value from being something other than $1$?

  • $\begingroup$ The dimensionless conversion is certainly the standard approach here, but there are subtle features that are easy to miss. Implicitly there is a chain rule involved, which washes out when you apply a unit scale, but has a factor when you have a dimension less conversion factor not equal to 1. That's how you can make that case work. $\endgroup$ – Paul Mar 9 '17 at 2:14
  • $\begingroup$ I'm afraid I don't follow. The chain rule... washes out? Huh? Which case are we talking about, and where is a chain rule involved? $\endgroup$ – SilasLock Mar 9 '17 at 3:04
  • $\begingroup$ Are you sure there isn't another name for this subject? "Geometric calculus" has a very different meaning: calculus with multivectors (in Clifford algebra).... Oh, I see it's "multiplicative calculus". $\endgroup$ – mr_e_man May 20 '18 at 3:56

Units are an inherently linear phenomenon; it only makes sense to talk about them in the context of doing homogeneous (multi-)linear operations.

In my opinion, the best mathematical structure for dealing with units is that each kind of 'dimension' is associated to its own separate one-dimensional real vector space. For example, SI units would have a vector space of lengths, a vector space of masses, a vector space of accelerations, and so forth — including, of course, the vector space $\mathbb{R}$ of 'unit free' quantities.

"Multiplication" is through bilinear forms, such as the tensor product. e.g. the tensor product of the vector space of times with the vector space of velocities has values in the vector space of lengths. And, in particular, $s \otimes \frac{m}{s} = m$.

When you're not doing linear operations, it simply doesn't make sense to talk about units; the geometric derivative is only defined on $\mathbb{R}$, the space of 'unit free' quantities.

But if you follow wikipedia, there is the bigeometric derivative, defined by

$$ \lim_{h \to 0} \left( \frac{f((1+h) x)}{f(x)} \right)^{1/h} = \lim_{k \to 1} \left( \frac{f(kx)}{f(x)} \right)^{1/\ln(k)} $$

Here, $h$ and $k$ are taken to be scalars, and this makes sense when $x$ and $f(x)$ have units. Note that the derivative is also a scalar.

  • $\begingroup$ Thanks, that clears up a lot. I've seen the bigeometric calculus before and felt that it was slightly "impure," though; it seems that when dx goes to 1 rather than 0 that the bigeometric integral doesn't collapse to the case of discrete repeated multiplication, so that it isn't a perfect multiplicative analog to the classical calculus. $\endgroup$ – SilasLock Mar 9 '17 at 3:27

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