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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?

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No, they shouldn't. Those equations don't make sense without those units, even if their magnitude is 1. –  Ignacio Vazquez-Abrams Sep 14 '12 at 2:01
    
@Ignacio that would be better posted as an answer (with a few more sentences to flesh it out), rather than a comment –  David Z Sep 14 '12 at 2:03
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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.

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Frankly, I do not agree that the Gaussian system (for electrodynamics) is an ugly mess. How do you come to this conclusion? –  Fabian Sep 14 '12 at 9:05
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@Fabian from a few years of experience trying to use it in calculations. The way it's usually formulated it's not even a self-consistent unit system; if you look at the Wikipedia page, you'll see that the same unit has different values depending on what formula you use it in, and you can't correct for this without using the same constants that the system was meant to avoid, at which point you might as well just use SI. Plus the cgs units are uncommon and more difficult to remember. One could have a reasonable unit system where $k = 1$, but Gaussian units don't quite do it. –  David Z Sep 14 '12 at 9:36
    
So what are the units of the resistance in terms of the basic SI units? (in Gauss it is just $s/cm$) Sorry to disagree: while I agree that the simplicity of the formulas depend if one is used to the unit system, I do not see any reason why one would give the electric field a different unit from a magnetic field (in the end they end up in the same tensor $F_{\mu\nu}$). Why to have such crazy units for inductance, capacitance or resistance. E.g., in Gauss the inductance has a unit of length which is nothing but the size of the inductance loop... –  Fabian Sep 14 '12 at 11:47
    
@Fabian under the same logic you could see no reason why we give length and time separate units; in the end they end up in the same four-vector. And yet we everyone is fine with meters and seconds. Yes, Gaussian units have some conveniences like the units for inductance and capacitance just being length, but IMO those are not enough to make up for its difficulties. Anyway if you'd like to continue this let's take it to Physics Chat. –  David Z Sep 14 '12 at 16:55
    
Thank you for that detailed answer. I haven't thought about this before and now I feel stupid. However I do find it kind of strange that the various definitions of force are not consistent, so that we have to have unit conversion there. It would seem logical for proportionality constants to be dimensionless. –  Neo Sep 14 '12 at 23:19
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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.

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