Lesson Explainer: Dimensional Analysis | Nagwa Lesson Explainer: Dimensional Analysis | Nagwa

Lesson Explainer: Dimensional Analysis Physics • First Year of Secondary School

In this explainer, we will learn how to determine the dimensions of physical quantities given the relation between those quantities and other known quantities.

Recall that physical measurements consist of two parts: a value and a unit. The unit tells us what we have, and the value tells us how many we have.

For example, if the measurement of some quantity is β€œ10 m,” the value is ten and the unit is metres. In other words, the measurement of our quantity is some number of metres, specifically ten.

Measurements are usually associated with real physical quantities. We can measure the average distance from Earth to the Sun as 1 AU (astronomical unit). We can see this distance labeled in the diagram below.

Similarly, in the diagram below, we have labeled the distance between atoms in a crystal as 4 nm.

On the other hand, we could measure the orbital period of Earth as 365.24 days, and one day as 86β€Žβ€‰β€Ž400 s.

Now let us pay close attention to the four measurements we just described: 1,4,365.24,86400.AUnmdayss

Each of these consists of a value and a unit, and all the units are different. However, even though all the units are different, intuitively we can see a connection between the first two quantities and also the last two quantities. Namely, 1 AU and 4 nm both represent distances. Indeed, we could have measured the distance from Earth to the Sun in nanometres (nm) instead of astronomical units (AU). Similarly, we could have measured one year in seconds instead of days since both days and seconds represent lengths of time.

However, it would be entirely nonsensical to measure atomic spacing in days or orbital periods in nanometres, since nanometres do not describe time and days do not describe distance.

We see that sometimes, but not always, different units can be used to describe the same physical quantity. We quantify this compatibility of units with the notion of dimension.

Definition: Dimension

The dimensions of a physical quantity describe what type of quantity we have independent of a particular choice of units. The dimensions of more complex quantities are built by combining the dimensions of more basic quantities in a manner analogous to the combination of units.

We have already encountered two of the base dimensions: length and time. We use the symbol 𝐿 to represent length and 𝑇 to represent time.

Let us apply this idea to the quantities we mentioned before. 1 AU, the distance between Earth and the Sun, and 4 nm, the spacing of atoms in a crystal, are both measurements of length. Therefore, the dimension of the quantity 1 AU is 𝐿, and the dimension of 4 nm is also 𝐿. On the other hand, the dimension of 365.24 days is 𝑇, which is also the dimension of 86β€Žβ€‰β€Ž400 s.

Square brackets around a quantity indicate that we are interested only in its dimension. So we can write [1]=[4]=𝐿AUnm and [365.24]=[86400]=𝑇.dayss

There are two important points to note from this. First, many different units all describe quantities with the same dimension. Second, because so many units have the same dimension, when we consider the dimensions of a quantity we do not consider how many like we did with units. This is because the number representing how many changes depending on our choice of unit, but dimensions describe a physical quantity in a way that is independent of the particular choice of units.

Before we see our first example, we will briefly consider how to combine dimensions.

Combining dimensions to form other dimensions is a process identical to the combination of units to form other units. For example, to represent an area, we may choose to use units of square metres (m2). This unit is constructed by multiplying two copies of the unit metres: mmmΓ—=.

To find the dimensions [m2], we simply observe =[]Γ—[]=𝐿×𝐿=𝐿.mmm

So, exactly as we would expect from the rules of combining units, the square of a unit with dimensions of length has dimensions of length squared.

We can also form quotients of dimensions just like with units. The only difference is that dimensions are typically represented with negative exponents instead of as fractions. For example, the typical unit of velocity is kilometres per hour. Kilometres have dimension 𝐿 since they are units of length, and hours have dimension 𝑇 since they are units of time. We would typically write the unit kilometres per hour as kmh but for its dimensions we would write =𝐿𝑇,kmh where we have written π‘‡οŠ±οŠ§ in place of 1𝑇.

With all this in mind, let us see our first example.

Example 1: Determining Units from Dimensions

A quantity has dimensions of πΏπ‘‡οŠ±οŠ¨. What SI unit could this quantity be measured in?

Answer

The dimensions that we are given are a composite of the dimensions 𝐿 of length and 𝑇 of time. To determine an appropriate SI unit with the desired composite dimensions πΏπ‘‡οŠ±οŠ¨, we must identify an appropriate unit for each base dimension, 𝐿 and 𝑇, separately.

Recall that metres are an SI unit for physical quantities with dimensions of length. Recall also that seconds are an SI unit for physical quantities with dimensions of time.

Now recall that dimensions combine just like units. Then, to form our target units, we simply need to combine metres and seconds in the same way 𝐿 and 𝑇 are combined in πΏπ‘‡οŠ±οŠ¨. Well, this is 𝐿 divided by 𝑇 squared, so our corresponding units must be metres divided by seconds squared, or m/s2.

At the end of this example, we were left with units of metres per second squared (m/s2), which are precisely the units for acceleration. Furthermore, we saw that the dimensions corresponding to these units are πΏπ‘‡οŠ±οŠ¨. But remember, dimensions describe a quantity independent of a choice of units, so in fact πΏπ‘‡οŠ±οŠ¨ are the dimensions of acceleration regardless of our choice for units.

We just determined units for measuring a quantity given the quantity’s dimensions. In this next example, we will do the reverse, determine the dimensions of a quantity from the units we might use to measure it. We will need a dimension for quantities with mass. We will use 𝑀 for dimensions of mass.

Example 2: Determining Dimensions from Units

What are the dimensions of a quantity that can be measured in units of kgmΓ—οŠ¨?

Answer

The two base units in the given expression are kilograms and metres. Kilograms are units for quantities with mass, so the corresponding dimension is 𝑀. Metres are units for quantities with length, so the corresponding dimension is 𝐿.

Now that we have identified the dimensions corresponding to each base unit in our quantity, we only need to substitute back into the given expression. Replacing kg with 𝑀 and m with 𝐿, we find ×=𝑀𝐿.kgm

This is our answer.

Before this most recent example, we introduced the dimension 𝑀 for mass. We need a new symbol for mass because no combination of lengths and times can properly represent the mass of an object. Similarly, we also want to work with physical quantities related to current. Since no combination of 𝐿, 𝑇, and 𝑀 can give us dimensions for current, we again need a new symbol. This time we will use the letter 𝐼.

The following example shows us how to use this new dimension in the context of finding overall dimensions.

Example 3: Determining Overall Dimensions

Quantity π‘Ž has dimensions of πΌπ‘‡οŠ±οŠ§. Quantity 𝑏 has dimensions of 𝑇. What are the dimensions of π‘ŽΓ—π‘?

Answer

Using the algebraic rules for combining dimensions, [π‘ŽΓ—π‘]=[π‘Ž]Γ—[𝑏].

Substituting the given dimensions from the question, [π‘Ž]Γ—[𝑏]=𝐼𝑇×𝑇=𝐼𝑇𝑇.

The term in parenthesis, π‘‡π‘‡οŠ±οŠ§, has no net dimensions since it is equivalent to 𝑇𝑇, which is just the number 1. Using this result, 𝐼𝑇𝑇=𝐼.

We therefore conclude that [π‘ŽΓ—π‘]=𝐼.

In addition to manipulating simple expressions involving 𝑀, 𝐿, 𝑇, and 𝐼, we can also work out dimensions for other physical quantities such as force. Recall that Newton’s second law relates force (𝐹), mass (π‘š), and acceleration (π‘Ž) by 𝐹=π‘šπ‘Ž.

We already know [π‘Ž]=πΏπ‘‡οŠ±οŠ¨, and [π‘š] is simply 𝑀, the dimension of mass we just introduced. Thus, following the rules for combining dimensions, []=𝑀×𝐿𝑇=𝑀𝐿𝑇.Force

Using a known physical formula, we have successfully worked out the composite dimensions for force. In fact, we can always use appropriate formulas and equations to derive the dimensions of unknown quantities because in order for an equation to be true, the overall dimensions of both sides of the equation must match.

Let’s work through an example where we use this principle.

Example 4: Finding Unknown Dimensions Using an Equation

Consider three quantities π‘Ž, 𝑏, and 𝑐, where [π‘Ž]=π‘€πΏοŠ¨ and [𝑏]=𝑀. If π‘Ž=𝑏𝑐, what is [𝑐]?

Answer

Our first step is to recall that if two physical quantities are equal, then their dimensions must match. We can therefore write [π‘Ž]=𝑏𝑐.

Now, on the right-hand side we have 𝑏𝑐. Recall that dimensions behave like units, which means we can manipulate them similarly to mathematical variables. In other words, 𝑏𝑐 is exactly the same as [𝑏][𝑐]. But we know from the statement that [π‘Ž]=π‘€πΏοŠ¨ and [𝑏]=𝑀, so we have 𝑀𝐿=𝑀[𝑐].

Now all we need to do is manipulate this equation algebraically to solve for [𝑐]. Multiplying both sides by [𝑐], we get 𝑀𝐿×[𝑐]=𝑀.

To isolate [𝑐], we next multiply both sides by π‘€πΏοŠ±οŠ§οŠ±οŠ¨. This gives 𝑀𝐿𝑀𝐿×[𝑐]=𝑀𝐿𝑀.

Now recall that π‘€οŠ±οŠ§ means β€œdivide by 𝑀,” so on the right-hand side we have π‘€π‘€Γ—πΏοŠ±οŠ¨, which we can see is equivalent to just 𝐿. Grouping the dimensions similarly on the left-hand side, we have 𝑀𝐿𝑀𝐿×[𝑐]=𝑀𝑀𝐿𝐿×[𝑐].

Each term in parentheses has no net dimensions; in other words, they are both equivalent to the number 1. Therefore, 𝑀𝐿𝑀𝐿×[𝑐]=[𝑐].

Combining this with our previous calculation, we have our final answer: [𝑐]=𝐿.

In this example, we came across the dimensional combinations π‘€π‘€οŠ±οŠ§ and 𝐿𝐿, which we concluded were both equivalent to the number 1 with no dimensions. This conclusion follows directly from the identity π‘Ž=1 for any kind of quantity π‘Ž and also leads us to another important definition.

Definition: Dimensionless Quantities

A dimensionless quantity is an algebraic combination of physical quantities arranged such that there are no overall dimensions. Dimensionless quantities provide quantitative relationships that are independent of any particular choice of units.

Perhaps the simplest dimensionless quantity is the aspect ratio of a rectangle, defined as a rectangle’s length divided by its width. Since []=[]=𝐿lengthwidth, lengthwidth must have no overall dimensions; in other words, it must be dimensionless. Thus, two rectangles with the same aspect ratio but different lengths, say 1 m and 1 km, will still appear to have the same shape.

Dimensionless quantities play an important role in physics because physical laws usually depend on relative, rather than absolute, values. For example, whether a slab will bend when held horizontally above the ground depends on its aspect ratio rather than the particular values of the length and width. In the picture above, the slab on the left is physically shorter than the slab on the right, but the slab on the right has a much smaller aspect ratio than the slab on the left. The slab with the higher aspect ratio will tend to bend more from the force of gravity.

In this next example, we will use what we have learned to find the dimension of one component of a dimensionless quantity.

Example 5: Determining Dimensions to Make a Compound Quantity Dimensionless

Consider four quantities 𝑝, π‘ž, 𝑅, and π‘˜, where [𝑝]=𝐼𝑇𝐿,[π‘ž]=𝐼𝑇𝑀,[𝑅]=𝐿𝑀𝑇.οŠͺ

The compound quantity π‘˜π‘π‘žπ‘… is dimensionless. What are the dimensions of π‘˜?

Answer

Recall that a dimensionless quantity has no overall dimensions. Algebraically, we write [𝐷]=1 for the dimensionless quantity 𝐷. This will let us solve for the dimensions of any component of 𝐷. Here, our quantity of interest is π‘˜π‘π‘žπ‘…, so ο•π‘˜π‘π‘žπ‘…ο‘=1.

Using the fact that dimensions combine algebraically like variables, we can rewrite this as [π‘˜][𝑝][π‘ž][𝑅]=1, and multiplying both sides by [π‘ž][𝑅][𝑝] to isolate [π‘˜], we have [π‘˜]=[π‘ž][𝑅][𝑝].

Since we are given all the quantities on the right-hand side in the statement of the question, let us plug in what we are given for [𝑝], [π‘ž], and [𝑅]: [π‘˜]=𝐼𝑇𝑀𝐿𝑀𝑇𝐼𝑇𝐿.οŠͺ

Converting the denominator to negative exponents gives [π‘˜]=𝐼𝑇𝑀𝐿𝑀𝑇𝐼𝑇𝐿.οŠͺ

Now we combine the dimensions using the usual rules for multiplying exponents with like bases: 𝐼𝑇𝑀𝐿𝑀𝑇𝐼𝑇𝐿=𝐼𝑇𝐿𝑀=𝐼𝑇𝐿𝑀=[π‘˜].οŠͺ()()()οŠͺ()

Now that we have seen several examples, let’s review what we have learned.

Key Points

  • Dimension provides the most general description of the type of a particular quantity or measurement.
  • The basic dimensions we learned about are length, 𝐿; mass, 𝑀; time, 𝑇; and current, 𝐼.
  • The dimensions of other quantities can be expressed in terms of these base dimensions. For example, speed has dimensions of πΏπ‘‡οŠ±οŠ§.
  • All units for the same sort of quantity have the same dimensions.
  • The dimensions on both sides of a valid physical equation must agree.
  • Dimensionless quantities are a combination of quantities that have no overall dimensions.

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