Our everyday world is filled with things we can measure; each is defined by dimensions in maths and physics. By understanding dimensions, we can accurately describe and compare different physical quantities. Use this resource to learn about dimensions.
Dimensions describe the fundamental units that make up physical properties. When we write the dimensions of the physical quantities mass, length, time and current, we use the symbols \(\left[ M \right]\), \(\left[ L \right]\), \(\left[ T \right]\) and \(\left[ I \right]\), respectively.
As an example, speed combines two of these dimensions. It is expressed as the distance (length) traveled per unit of time. We can write this out to form a dimensional formula:
\[\left[\textrm{speed}\right] = \frac{\left[\textrm{length}\right]}{\left[\textrm{time}\right]} = \frac{\left[ L \right]}{\left[ T \right]} = \left[ L \right] \left[ T \right]^{-1}\]
Speed, therefore, has dimensions of length divided by time.
Using dimensions to determine the units
The unit of a derived quantity can be found by substituting the fundamental unit for each dimension into the dimensional formula.
Let's consider a calculation for density using mass in kilograms and volume in cubic metres. Metres is a measure of length. The dimension equation for density is:
\[\left[ \textrm{density} \right] = \frac{ \left[ \textrm{mass} \right] }{ \left[ \textrm{volume} \right] } =
\frac{ \left[ M \right] }{ \left[ L \right]^{3} } = \left[ M \right] \left[ L \right]^{-3}\]
We can then subsitute the fundamental units \(\textrm{kg}\) and \(\textrm{m}^{3}\) into \(\left[ M \right]\) and \(\left[ V \right]\), respectively.
\[\textrm{density} = \frac{\textrm{kg}}{\textrm{m}^{3}} = \textrm{kg}\,\textrm{m}^{-3} \]
Example – using dimensions to determine the units
Dr. Pepper wants her student to measure the rate at which carbon dioxide escapes from a canned soda drink. The student sets up an experiment where they record the time in minutes and the volume of carbon dioxide gas in millilitres.
Determine the units that the student's final result will be in.
First, we should determine the dimensional formula. For the rate of change of carbon dioxide, a gas, it will be:
\[\textrm{rate of change} = \frac{\left[ \textrm{volume} \right]}{\left[ \textrm{time} \right]} = \frac{\left[ V \right]}{\left[ T \right]} = \left[ V \right] \left[ T \right]^{-1} \]
Volume is measured in millilitres and time is measured in minutes. Therefore:
\[\textrm{rate of change} = \frac{\textrm{mL}}{\textrm{min}} = \textrm{mL}\,\textrm{min}^{-1} \]
Checking dimensions
In your studies, you will derive physical quantities from formulas or equations. You can use dimensions to check whether these expressions make sense. This involves making sure that the dimensions of the left-hand side are balanced with the dimensions on the right-hand side.
This will make more sense with an example.
Example – checking dimensions
A student claims that the equation for the circumference of a circle is \(C=2\pi r\). Determine whether this is dimensionally correct.
The circumference is a measure of length, so the dimension is \(\left[ L \right]\). This is the left-hand side of the equation.
On the right-hand side, we have \(2\pi\), which does not have units, and \(r\), which is a measure of length. \(r\) therefore has a dimension of \(\left[ L \right]\).
Putting these together, we have:
\[\begin{align*} C & = 2\pi r\\
\left[ L \right] & = \left[ L \right]
\end{align*}\]
The left-hand side and right-hand side are dimensionally correct.
Exercise – checking dimensions
Check that the following equations are dimensionally correct.
\(V=\pi r^{2} h\)
\(A=4\pi r^{2}\)
\(A=2\pi r^{2} + 2\pi rh\)
\(\pi\) has no units, \(r^{2}\) is \(\left[L\right]\) and \(h\) is \(\left[L\right]\). So \(V\) is \(\pi\left[L^{3}\right]\). A cubic unit is a measurement for volume, so the equation is dimensionally correct.
\(\pi\) has no units and \(r^{2}\) is \(\left[L\right]\). So \(A\) is \(4\pi\left[L^{2}\right]\). A squared unit is a measurement for area, so the equation is dimensionally correct.
\(\pi\) has no units, \(r^{2}\) is \(\left[L^{2}\right]\), \(r\) is \(\left[L\right]\), and \(h\) is \(\left[L\right]\). So \(A\) is \(4\pi \left[L\right]^{2}\). A squared unit is a measurement for area, so the equation is dimensionally correct.
