1.1: SI Units

There is a “standard international” system of units. You may ask, why does anybody ever use anything other than these? SI Units are a good set of units for everyday measurements. However, they are very clumsy when dealing with the very small or the very large. When talking about atoms, or about stars, it’s often convenient to use other units that are better matched to the scale of the system. What’s more, some places historically use other units; for instance, the United States still uses the British Imperial system of units.

There are a finite number of dimensionalities. For purposes of this course, there are only four dimensionalities that you need to know about. They are, with their SI units, listed below:

Dimensionality	SI Unit
Length	m
Mass	kg
Time	s
Electric Charge	C

The four core dimensionalities are length, mass, time, and electric charge.¹ For each dimensionality there can be a lot of different units. Something of dimensionality length can be measured in any length unit, but cannot be measured with a (say) time unit. It doesn’t make sense directly to compare quantities of different dimensionalities. So, I could measure my height in feet— 5.84 feet is my height— or in meters. While clearly the number 1.78 does not equal 5.84, 1.78 meters does equal 5.84 feet. A measurement with dimensionality is clearly different from a pure number; the units on the number affect what that number means.

You are already familiar with the meter, kilogram, and second. (Indeed, because of these three base units, the SI system is sometimes called the “MKS” system.) You may or may not have heard of the Coulomb before. All other units that we will deal with are derived from these base units. For instance, consider velocity. The dimensionality of velocity is length over time (sometimes written L/T). Any unit that corresponds to a length divided by a time is a valid velocity unit; that could be kilometers per hour, miles per hour, or furlongs per fortnight. The dimensionality of velocity is neither length nor time, but is composed of those two dimensionalities. The SI unit for velocity is meters per second, or m/s. Sometimes derived units have their own names. Below is a table of some of the more important derived units in the SI system:

Dimensionality	Unit	Definition
Force	Newtons N	kgms^−2
Energy	Joules J	Nm = kgm^2s^−2
Power	Watts W	J/s = kg m^2s^−3

(Remember that something raised to the negative power is in the denominator. Thus, one Newton is “one kilogram times meter per second squared”, or kg m/s² .) While we can say that “force” is the dimensionality of force— as in the table above— that is exactly the same as saying it is a dimensionality of mass times length divided by time squared, or ML/T² , or MLT⁻² .

Some people will “just always work with SI units”, and then not write down units to go with their numbers during intermediate calculations. The idea is that since you’re always using the standard, the final result of any series of calculations will be in the SI unit for whatever it is that you calculated. Even though, if you are careful, you can get away with this, it would still be wise to write down the units that go with numbers every time you write down those numbers. There are two primary reasons for this. First, it makes it much clearer what you are doing and what these intermediate numbers actually are. Without that, anybody reading your calculations may have a hard time following them, and you have not communicated as effectively as you might have. Second, by keeping track of your units throughout your calculation, you provide yourself with a cross-check: does your final answer have the units that it’s supposed to have? If it doesn’t, then that’s a sign that you’ve made a calculation mistake somewhere along the way.

For example, suppose I told you that the density of water is 1 gram per cubic centimeter, and I wanted you to tell me how much mass there is in a spherical drop of water with radius 0.2 cm. First, let’s convert to SI units; if you do it right, you can figure out that 1 g/cm³ equals 1000 kg/m³ . Also, 0.2 cm is equal to 0.002 m. If you say that the volume of a sphere is πr² , you could calculate the volume from this number:

\(\ V o l=\pi(0.002)^{2}=1.257 \times 10^{-5}\)

Then, multiply the volume by the density to get the mass:

\(\ m=(1000)\left(1.257 \times 10^{-5}\right)=0.013\)

Figuring that you’ve done everything in SI units, you should get an answer in the SI unit for mass, so you could write down and box \(\ \fbox{m = 0.013 kg}\). However, this answer is wrong. Did you see where it went wrong? Let’s redo the problem, this time keeping track of units:

\(\ \begin{aligned}
m &=(V o l)(d e n s) \\
&=\left(\pi(0.002 \mathrm{~m})^{2}\right)\left(\frac{1 \mathrm{~kg}}{\mathrm{~m}^{3}}\right) \\
&=\left(1.257 \times 10^{-5} \mathrm{~m}^{2}\right)\left(\frac{1 \mathrm{~kg}}{\mathrm{~m}^{3}}\right) \\
&=0.0127 \frac{\mathrm{kg}}{\mathrm{m}}
\end{aligned}\)

Notice in the last step we cancelled the meter² in the numerator with two of the three meters in the denominator’s meter³. But, wait! This doesn’t leave us with an answer that has dimensionality mass, it has dimensionality mass per length! Clearly we’ve done something wrong. In this case, the mistake was in our formula for volume. In fact, the volume of a sphere is of radius \(\ r\) is \(\ \frac{4}{3} \pi r^{3}\). We caught this error because, by keeping track of the units as we were putting numbers into the calculation, we saw that the units didn’t work out right. If you put in the right formula for volume, you discover that there are only 3.4×10⁻⁵ kg of water in a droplet that’s 2 mm in radius.

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¹In fact, in the SI system, electric current rather than electric charge is considered a core dimensionality. However, it’s conceptually more simple to consider charge as the core unit, and current as a derived unit, so I’ll use that in this document.