Force and Natural Units

Question

Today in my Physics lecture I suddenly thought of something. We all know that gravitational force is proportional to the two masses and inversely proportional to the square of the distance between them. So, naturally, there exists a system of natural units which equates the proportionality by setting $G=1$. Here I'm referring to Planck Units. But I realized that even though numerically using this system of units $F=\frac{m_1m_2}{r^2}$, dimension-wise that is wrong because the units do not correspond to the unit of force as defined by Newton's Second Law. Similarly, the electrostatic force, when making $k=1$, has the wrong unit as well. Why do we have to throw in those proportionality constants to convert the units? Shouldn't the units work out to be the same?

Answer 1

But I realized that even though numerically using this system of units $F=\frac{m_1m_2}{r^2}$, dimension-wise that is wrong because the units do not correspond to the unit of force as defined by Newton's Second Law.

Ah, but what is the unit of force as defined by Newton's second law? Perhaps you're thinking that it is the Newton, but that's in SI units, where $G \neq 1$. When you switch into a different unit system, you have to use the appropriate unit of force for that system.

In Planck units, that unit is 1: in fact, all quantities in Planck units are measured in pure numbers. So the unit of force from Newton's second law is

$$[F] = [m][a] = (1)(1) = 1$$

and from Newton's law of gravity is

$$[F] = \frac{[m][m]}{[r]^2} = \frac{(1)(1)}{(1)^2} = 1$$

(I'm using the notation $[x]$ to represent "the units of $x$," in case you're not familiar with it.) Of course, that example doesn't show what's going on very well, because it's trivial: everything is measured in the same units.

So suppose we set up a different unit system, where $G = 1$ but everything else is the same as SI as much as possible. You can flesh out the details of this unit system by looking at the equations of Newtonian mechanics. For starters, there is a whole series of kinematic equations,

$$\begin{align}x &= x_0 + v_0t + \frac{1}{2}at^2 & v^2 &= v_0^2 + 2a(x - x_0) \\ v &= \frac{\Delta x}{\Delta t} & a &= \frac{\Delta v}{\Delta t}\end{align}$$

and so on, which in SI use two independent units, one of length (the meter) and one of time (the second). Let's keep all that the same.

The next "tier" of equations brings in force and mass.

$$\begin{align}F &= ma & F &= \frac{m_1 m_2}{r^2}\end{align}$$

This is the interesting one. You'll notice that there is no equation which relates only mass or only force to the kinematic quantities ($x$, $v$, $a$, $t$); you always get mass and force occurring together. That means that the units for the two quantities are linked. So if we're going to pick a new unit of force, we also need to pick a new unit of mass. Two unknown units, and two equations to solve for them: that ensures that both equations will be able to remain dimensionally consistent.

You can figure out what the units in the $G = 1$ system actually are with a little algebra. For example, divide both equations by mass. Written in terms of the units, you get

$$[a] = \frac{[F]}{[m]} = \frac{[m]}{[r]^2}$$

and this tells you that our new unit of mass divided by the unit of distance squared has to equal the unit of acceleration. Or:

$$[m] = [a][r]^2 = \mathrm{\frac{m^3}{s^2}}$$

And the unit of force can be gotten from $F = ma$:

$$[F] = [m][a] = \mathrm{\frac{m^4}{s^4}}$$

So in this $G = 1$ unit system you're thinking of, mass is measured in cubic meters per second per second. Weird, huh? Well, that's why we use kilograms in the real world.

But you can verify that with these strange units for mass and force, the equation $F = \frac{m_1 m_2}{r^2}$ is dimensionally consistent:

$$\begin{align}[F] &= \frac{[m][m]}{[r]^2} \\ \mathrm{\frac{m^4}{s^4}} &= \mathrm{\frac{(m^3 s^{-2})(m^3 s^{-2})}{m^2}} \\ \mathrm{\frac{m^4}{s^4}} &= \mathrm{\frac{m^4}{s^4}}\end{align}$$

And so is $F = ma$:

$$\begin{align}[F] &= [m][a] \\ \mathrm{\frac{m^4}{s^4}} &= \mathrm{\frac{m^3}{s^2}}\mathrm{\frac{m}{s^2}} \\ \mathrm{\frac{m^4}{s^4}} &= \mathrm{\frac{m^4}{s^4}}\end{align}$$

So it's just a matter of making sure that you are consistently using one unit system, and that when you try to switch to a different one, you don't get stuck on units or constants from the old system you're used to.

Naturally, this whole discussion applies equally well to the laws of electromagnetism. You can build up a consistent (but slightly weird) unit system where $k = 1$ by starting from the SI units for mechanical quantities. In fact, there is a well-established unit system that does exactly this: the CGS-Gaussian system. In this system the equation for electrostatic force is

$$F = \frac{q_1 q_2}{r^2}$$

and charge is measured in $\mathrm{g^{1/2}\ cm^{3/2}\ s^{-1}}$ (but that got too confusing, so scientists named it the statcoulomb to make it sound like a proper unit of charge). This unit system is very rarely used, though, because frankly, it's an ugly mess. It was largely developed before people really understood how to make consistent unit systems, and so if you look into it you'll find what seem to be some contradictory definitions mostly caused by missing constants.

Answer 2

You're approaching it from the wrong angle.

We don't get to pick the units for F, m₁, etc.; the units used for them are defined by, for lack of a better word, "reality". All we can do is add fudge factors so that not only the numbers come out right, but also the units. That is what those constants are.

On the gripping hand, even vectors with a magnitude of 1 have a direction; the only possible directionless vector is 0.