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Mathematically speaking, what does it mean to say that a physical quantity is some numerical value with a “dimension” associated with it? When we say that the velocity of light is some constant, c meters per second, at first thought, it seems that we are talking about a ratio of differentials, $v=dx/dt$. But what about the "dimension" of angular momentum, ${mass} \times {length}^2/{time}$? I've never seen a differential of mass in a total derivative... $L=dmdx^2/dt$ ! Or what about the "dimension" of electrical resistance, ${time}/{length}/{permittivity}$? I've never seen a differential of permittivity, either. So, the idea that a "dimension" might be just a total derivative just doesn't seem to make sense, because the number of differentials in the numerator and denominator is not always equal.

So, what is the mathematical nature of this beast we call a "dimension"?

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"I've never seen a derivative of mass!" - one considers the derivative of mass with respect to time in Newton's second law when the accelerating object loses or gains mass as it moves, e.g. rockets. – J. M. Sep 4 '11 at 1:22
$dx/dt$ is a derivative, not a ratio of derivatives. – Srivatsan Sep 4 '11 at 1:23
@J.M.: Koilon appears to be using derivative for differential and somehow trying to equate dimensions with differentials. Thus, in the case of angular momentum he has three in the numerator and only one in the denominator, which (he notes) makes no sense as any kind of derivative. – Brian M. Scott Sep 4 '11 at 1:33
I don't see that anyone has really addressed my question, yet. What is the mathematical nature of this beast called a "dimension"? – Koilon Sep 4 '11 at 5:24
This may not completely address your question, but you may be interested in my answer here on – Willie Wong Sep 4 '11 at 11:58

The units come before the derivatives. An item has a mass, which might be measured in kg. A stick has a length, measured in meters. Clocks tick in seconds. All these do not have a sense of derivatives. Then if you calculate a derivative, it has the same units as the ratio of the quantities. So a velocity, $\frac{dx}{dt}$ has units $\frac{m}{sec}$, but you can get $\frac{m}{sec}$ in other ways as well, for example as a constant speed.

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In dimensional analysis the dimension of the derivative $dx/dy$ will be the dimension of $x$ divided by the dimension of $y$.

To take a simple example, if distance $x$ has the dimension $length$ and time $t$ has the dimension $time$, then speed (perhaps velocity) $v=dx/dt$ has the dimension $length / time$ and acceleration $a=dv/dt =d^2 x/dt^2$ has the dimension $speed / time$ or $length / time^{2}$.

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Note to self: Gauss was right. Why bother publishing your results to the Boeotians? – Koilon Sep 7 '11 at 14:22

Willie's link, and the indirect suggestion that graded algebras may be relevant to my question, prompts me to add some further thoughts on my own question.

It turns out that the exterior algebra is a graded algebra and, at least for electromagnetic quantities, the exterior algebra is definitely relevant.

My $41^{st}$ edition (1959-1960) of the Handbook of Chemistry and Physics, page 3177, gives the dimension of electric charge, $Q$, as either $\epsilon^{1/2}m^{1/2}l^{3/2}t^{-1}$ or $\mu^{-1/2}m^{1/2}l^{1/2}$. If we formally multiply these two dimensions together, we get:


Similarly, for magnetic pole strength, $\Phi$ (page 3185), we get:


It turns out that all of the electromagnetic quantities come in pairs like this, with some mass-length-time dimension, multiplied by either $G=\sqrt{\epsilon/\mu}$ or $R=\sqrt{\mu/\epsilon}$, where G and R are electrical conductance and resistance, repectively:

$ \begin{array}{2} G=\sqrt{\epsilon/\mu} & R=\sqrt{\mu/\epsilon} \\ C=t\sqrt{\epsilon/\mu} & L=t\sqrt{\mu/\epsilon} \\ Q^2=(\textit{ml}^2/t)\sqrt{\epsilon/\mu} & \Phi^2=(\textit{ml}^2/t)\sqrt{\mu/\epsilon} \\ I^2=(\textit{ml}^2/t^3)\sqrt{\epsilon/\mu} & E^2=(\textit{ml}^2/t^3)\sqrt{\mu/\epsilon} \\ \vec{D}^2=(m/l^2t)\sqrt{\epsilon/\mu} & \vec{B}^2=(m/l^2t)\sqrt{\mu/\epsilon} \\ \vec{H}^2=(m/l^2t)\sqrt{\epsilon/\mu} & \vec{E}^2=(m/l^2t)\sqrt{\mu/\epsilon} \\ \rho_e^2=(m/t)\sqrt{\epsilon/\mu} & \rho_m^2=(m/t)\sqrt{\mu/\epsilon} \\ etc. & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \end{array} $

Here, $C$ and $L$ are capacitance and inductance, while $I$ and $E$ are current and potential, respectively.

In Discrete Differential Forms for Computational Modelling, by Mathieu Desbrun, Eva Kanso, & Yiying Tong, (Source:, it says (pages 15, 12):

The constitutive relation

$\vec{D}=\epsilon \vec{E}$ and $\vec{H}=\mu \vec{B}$

are very similar to the Hodge star operator that transforms a $k$-form to an ($n-k$)-form. Here, $\epsilon$ operates on the electric field $\vec{E}$ , a 1-form, to yield the electric displacement $\vec{D}$, a 2-form, while $\mu$ transforms the magnetic field $\vec{B}$, a 2-form, into the magnetic field intensity $\vec{H}$, a 1-form. To this end, one may think of both $\epsilon$ and $\mu$ as Hodge star operators induced from appropriately chosen metrics.

We must use the inverse of the Hodge star to go from a dual ($n-k$)-cochain to a $k$-chain. We will, however use indistinguishably $\star$ to mean either the star or its inverse.

There are three different Hodge stars on $\Re^3$, one for each simplex dimension. But as we discussed for all the other operators, the dimension of the form on which this operator is applied disambiguates which star is meant. So we will not encumber our notation with unnecessary indices, and will use the symbol $\star$ for any of the three stars implied.

The author mentions that the charge density $\rho_e$ is a 3-form, but he doesn't give a clue what the appropriate Hodge operator on a 3-form might be - and I am left wondering why there isn't also a Hodge operator for 0-forms.

All that aside, however, what can we say that we have learned from this about the nature of what we call a "dimension"? My gears are grinding, but I will have to wait for my muses to give me something to continue...

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