In section 5 of the "Preliminary: On the measurement of quantities" chapter (page 3) in "A treatise on electricity and magnetism" Maxwell uses, total length, $s=mt^{2}/{2r^{2}}$to show that $m=2sr^{2}/{t^2}$ is in units of Length cubed over Time squared.
This is the first I have encountered this way of looking at mass and am wondering:
Maxwell is being misunderstood. First, Maxwell makes very clear that length, time and mass are the fundamental types of units. Then he discusses a totally different convention that isn't used today, saying "in the astronomical system, the unit of mass is defined with respect to its attractive power". In other words, Maxwell is talking about a concept of mass where G is incorporated into the definition of mass.
Only in this context does Maxwell go on to state:
"For acceleration due to the attraction of a mass m at a distance r is by Newton's Law $\frac{m}{r^2}$". Then Maxwell takes $s=\frac{1}{2}at^2$ and substitutes $\frac{m}{r^2}$ for a.
So to anwser the questions:
1.Is this just per-relativity nonsense?
No, it has nothing to do with relativity. The treatise was written well before Einstein's theories of relativity.
2.Does mass being "length cubed over time squared" have any meaning?
Yes, in the context of Maxwell's discussion of a different definition of mass, which incorporates G into the definition, it is the correct unit of mass.
The explanation is Maxwell's text:
If, as in the astronomical system, the unit of mass is defined with respect to its attractive power, the dimensions of $[M]$ are $[L^3T^{-2}]$.
To motivate this, it is perhaps useful to be aware of some of the different systems of units for electromagnetism used historically. One of the units of charge commonly in use was the $\mathrm{esu}$ (electrostatic unit) aka $\mathrm{statC}$ (statCoulomb) based on taking the coefficient of Coulomb's law to be unitless: $$F = \frac{q_1q_2}{r^2}\text{.}$$ Thus, by dimensional analysis, $[F] = [MLT^{-2}]$ implies that the unit of charge has dimension $$[Q] = [M^{1/2}L^{3/2}T^{-1}]\mathrm{.}$$ Thus, it is clear that Maxwell is simply doing the same thing for gravity--considering a system of units in which "mass is defined with respect to its attractive power", i.e., through Newton's law of gravitation $$F = \frac{m_1m_2}{r^2}\text{,}$$ and so $[F] = [MLT^{-2}] = [M^2L^{-2}]$ implies that $$[M] = [L^3T^{-2}]\text{.}$$
Consider two bodies, $m_{1}$ and $m_{2}.$ From Newton laws, we have: $$F=G\frac{m_{1}m_{2}}{s^{2}}\tag1 $$ but $$F =m_{1}a,\tag2$$ and combining $(1)$ and $(2)$ two we obtain $$m_{1}a=G\frac{m_{1}m_{2}}{s^{2}}.\tag3$$ We also know that $$ a=\frac{s}{t^{2}}.$$ Then, $(3)$ becomes $$m_{1}\frac{s}{t^{2}}=G\frac{m_{1}m_{2}}{s^{2}},$$ which, after a little bit of algebra, gives you $$ m_{2}=G\frac{s^{3}}{t^{2}},$$ and this can be written, using dimensional analysis, as $$ [m_{2}]=[s^{3}][t^{-2}].$$
tag{1} not those horrible dots, but also why tag an equation you don't reference? And unless you are using natural units, $G$ is a dimensional constant.
– dmckee --- ex-moderator kitten
Mar 05 '14 at 05:40
I am in agreement with the comment that "Maxwell is talking about a concept of mass where G is incorporated into the definition of mass". There is a system of units out there which is not well known, where all physical properties are broken down into distance (L) and time (T) units. In that system, mass (M) = (L^1)(T^-6) and G = (L^2)(T^4). The combination of G X M = (L^3) (T^-2) gives the same result as in Maxwell's papers.
