# Units and Dimensions

Unlike most mathematical problems which deal with pure numbers, quantities in physics usually have dimension which describe their function. There are five basic physical dimensions— mass, length, time, electric charge, and temperature— and physical quantities can usually be written as some product or quotient of these five. For example, velocity is length divided by time, while force is mass times length divided by time squared. Square brackets are often used to designate the dimensions of a variable, so that we would write

[velocity] = L / T
[force] = M L / T2
[energy] = M L2 / T2
Some physical quantities are dimensionless; we say they have dimension 1. Counts (e.g. "I have 5 apples" or "There are 1023 atoms in this block") are dimensionless. Ratios of two quantities with the same dimension are dimensionless: for example, the ratio of the moon's mass to the Earth's mass has dimension M/M = 1. Percentages are usually dimensionless: if I spend 50% of a class period talking about dimensions, that means that the ratio of the time I spend on that topic to the duration of the class period is 0.5. We will see later (if you don't already know) that angles can be expressed as a ratio between two different lengths, and thus are dimensionless.

There are a few rules that dimensioned quantities have to obey:

1. Two quantities can only be added or subtracted if they have the same dimension.
For instance, "1 meter + 5 inches" is fine because both quantities are lengths. However, "1 meter + 5 hours" is nonsense: adding a length to a time has no meaning.
2. One quantity can only be set equal to another if they have the same dimensions.
For example, if you have the formula
ρ (3 gallon) = 5 Coulombs
(Coulombs are a measure of electric charge and gallons are a measure of volume), then ρ must have dimensions of charge per volume: that way, the left side of the equation has units of (charge/volume)×volume or just charge, the same as the right side of the equation.

3. Trigonometric functions only take dimensionless quantities.
The expression "sin(5 inches)" makes no sense; you can't take the sine of a length. You can take the sine of an angle, however, because as the figure shows, angles are unitless ratios of lengths. Degrees are just radians multiplied by $$180^{\circ}/\pi$$, which is itself a dimensionless number.
4. Exponents must be dimensionless quantities.
It makes no sense to say $$2^{4\u{kg}}$$ or $$e^{-5\u{s}}$$. Exponents are often ratios of two quantities with the same units. For instance, if we start with $$1\u{kg}$$ of radioactive material at time $$t=0$$, the amount of material that is still radioactive at time t is given by the formula $$(1\u{kg})(2^{-t/t_H})$$ where $$t_H$$ is called the half-life of the material, because when $$t=t_H$$, there is $$(1\u{kg})(2^{-1})=0.5\u{kg}$$ or half the initial material.
These rules can help us determine whether a given equation is correct or not.

Suppose you met someone who argued that Einstein's famous equation $$E=mc^2$$ was wrong, because it should be $$E=mc$$ In this equation, E is energy, m is mass, and c is the speed of light with dimensions of length per time (L/T).
Can you explain why this is wrong, without knowing anything about Einstein's Theory of Relativity?
The dimensions of energy are $$[E]=\mathrm{ML^2/T^2}$$ (as we said earlier in this section). m is mass, so $$[m]=\mathrm{M}$$), and $$[c]=\mathrm{L/T}$$. If we analyze the dimensions of Einstein's equation, we find that $$[E]=[m][c^2]=\mathrm {ML^2\over T^2}=M\left({L\over T}\right)^2$$ We see that the dimensions are the same on both sides, and so the equation could be true. On the other hand, the dimensions for the alternate equation, E=mc, are $$\mathrm{ML^2/T^2=M(L/T)}$$ The dimensions are different on the two sides of the equation, which means that the equation is nonsense. As Wolfgang Pauli might have said, the equation "is not only not right, it is not even wrong".

## Units

Each dimension can be measured using any of a number of different units. Lengths can be measured in miles, centimeters, light-years, or paces; time can be measured in seconds, centuries, or turns of an hourglass; and so forth. If we use a haphazard set of units, we can get ourselves into trouble. For instance, distance travelled is equal to average velocity multiplied by time travelled:

d = v t
If a car travels with an average speed of v=50 mi/hr for t = 15 minutes, how far does the car go? The naive approach would be to plug the numbers into the formula:
d = (50)(15) = 750
This is a true result, but only in a sense. A better way to write it would preserve the units:
d = (50 mi/hr)(15 min) = 750 mi-min/hr
Thus the car drives 750 "mile-minutes per hour" which, while in fact a distance, is not the sort of answer we were expecting. We need to do a conversion between minutes and hours to get the correct answer.

Conversions are a pain. The best way to avoid them is to work in a consistent system of units where each of the base physical dimensions (length, mass, time, charge, and temperature) is assigned a single base unit, and any dimension which is a combination of base dimensions is measured in a unit which is a combination of the base units. For example, if we defined our base unit of length to be the foot and the base unit of time to be the second, then we would require that all lengths would be given in feet, all times in seconds, all velocities in feet per second, and so forth. We would never have to do any conversions at all, so long as all of our data were given to us in the specified units.

In this book we will use the system of units called the International System of Units or SI (from the French Système international d'unités), which has as its base units

• the meter (m) for length,
• the kilogram (kg) for mass (notice: not the gram!),
• the second (s) for time,
• the ampere (A) for electric current, and
• the kelvin (K) for temperature.
All other dimensions have units defined in terms of these five units, and all quantities in the SI system are measured in terms of these units (though often with a metric prefix, as explained below). Because of the correspondence between units and dimensions, it is common to replace one with the other: for example, I might say that the dimensions of velocity are meters per second ([v] = m/s) instead of length per time (L/T).

One advantage of using a system like SI is that, when you're using it, you can take units for granted. For example, we will see that the force between two electric charges q1 and q2 a distance r apart is given by the formula

$$F=k{q_1q_2\over r^2}$$
The variable k is a fundamental constant of the universe which in SI units is equal to 9×109 but I never remember its units.
They are Nm2/C2.
However, as long as I plug in the values for the charges and the distance in SI units, and I use the value 9×109 for k, then I know that the force will be in newtons (N), because the SI unit for force is newtons. In short, stick to SI and it won't lead you astray.

## Metric Prefixes

In the SI system, all units can be given a prefix which increases or decreases its size by a power of 10. Here are the smaller ones, with common ones marked with stars (*).
Prefix AbbrevMultiply by
deci- d10-1
*centi- c10-2
*milli- m10-3
*micro- µ10-6
*nano- n10-9
pico- p10-12
femto- f10-15
Prefix AbbrevMultiply by
deka- da101
hecto- h 102
*kilo- k 103
mega- M 106
giga- G 109
tera- T 1012
peta- P 1015

For example, a kilometer is 103, or 1000 meters. In written formulas, we can treat prefixes as simple substitutions for the appropriate power of ten: for instance, 500 µC is just 500 (10-6) C = 5×10-4 C.
1. Simplify (6 mN)(10 dam).
We replace the prefix m (milli) with 10-3 and the prefix da (deka) with 101, and get
$$\begin{eqnarray} &&(6\u{\color{red}{m}N})(10\u{\color{blue}{da}m})\\ &=&(6\times\color{red}{10^{-3}}\u{N})(10\times\color{blue}{10^1}\u{m})\\ &=&(0.006)(100)\u{Nm}\\ &=&\boxed{0.6\u{Nm}}\\ \end{eqnarray}$$
2. Light travels 0.3 km per µs. What is this speed in meters per second?
$$\begin{eqnarray} &&0.3\u{\frac{\color{red}{k}m}{\color{blue}{\mu} s}}\ &=&0.3\frac{\color{red}{10^3}\u{m}}{\color{blue}{10^{-6}}\u{s}}\ &=&0.3\ten9\u{m\over s}\ &=&\boxed{3\ten8\u{m/s}}\ \end{eqnarray}$$
Once one is comfortable with metric prefixes, it can be more convenient to write this answer as 0.3 Gm/s (though to be fair, most physicists think of this as $$3\times 10^8\,\textrm{m/s}$$.)

You should try to memorize these prefixes as soon as you can, to minimize the amount of time you spend looking them up. Metric prefixes were invented to minimize the use of scientific notation and we will make heavy use of them in this book for this purpose. Here is a little quiz game to help you practice.

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