# Dimensional Analysis & its uses with examples

This topic covers **Dimensional Analysis & its uses**. The term **dimension **is used to refer to the physical nature of a quantity and the type of unit used to specify it. Distance has the dimension of length, which is symbolized as [L], while speed has the dimensions of length [L] divided by time [T], or [L/T]. *Dimensional analysis is used to check mathematical relations for the consistency of their dimensions.*

Many physical quantities can be expressed in terms of a combination of fundamental dimensions such as length [L], time [T], and mass [M]. There are certain other quantities, such as temperature, which are also fundamental.

A fundamental quantity like temperature cannot be expressed as a combination of the dimensions of length, time, mass, or any other fundamental dimension.

## **Dimensions and Units of Four Derived Quantities**

Quantity | Area | Volume | Speed | Acceleration |
---|---|---|---|---|

Dimensions | L^2 | L^3 | L/T | L/T^2 |

SI Units | m^2 | m^3 | m/s | m/s^2 |

**Dimensions and Units of Four Derived Quantities**

## Features of Dimensions

Dimensions cancel just like algebraic quantities, and pure numerical factors like ½ have no dimensions, so they can be ignored.

We shall often use brackets [ ] to denote the dimensions of a physical quantity. For example, the symbol we use for speed in this book is *v*, and in our notation, the dimensions of speed are written [*v*] =[L/T]

## Why use **dimensional analysis**?

**dimensional analysis**

*Dimensional analysis is used to check mathematical relations for the consistency of their dimensions.*

In many situations, you may have to check a specific equation to see if it matches your expectations. *A useful and powerful procedure called dimensional analysis can assist in this check because dimensions can be treated as algebraic quantities.*

**For example, quantities can be added or subtracted only if they have the same dimensions. Furthermore, the terms on both sides of an equation must have the same dimensions. By following these simple rules,**

*you can use dimensional analysis to determine whether an expression has the correct form. Any relationship can be correct only if the dimensions on both sides of the equation are the same.*

## Show how to use **dimensional analysis** with an example

**dimensional analysis**As an illustration, consider a car that starts from rest and accelerates to a speed *υ *in a time *t*. Suppose we wish to calculate the distance *x *traveled by car but are not sure whether the correct relation is *x *= ½ v*t*^{2} or *x *= ½ v*t*. We can decide by checking the quantities on both sides of the equals sign to see whether they have the same dimensions. If the dimensions are not the same, the relation is incorrect.

**For x = ½ vt^{2},** we use the dimensions for distance [L], time [T], and speed [L/T] in the following way:

*x*= ½ v

*t*

^{2}

*Dimensions *[L] ≟ [L/T] [T]

^{2}= [L][T]

The dimension on the left of the equals sign does not match those on the right, so the relation *x *= ½ v*t*^{2 }cannot be correct.

**For x = ½ vt **On the other hand, applying dimensional analysis to

*x = ½ vt*, we find that:

*x **= **½ **v**t*

*Dimensions *[L] ≟ [L/T] [T] = [L]

The dimension on the left of the equals sign matches that on the right, so this relation, x = ½ vt is dimensionally correct.

## Limitations of dimensional analysis

Note: Referencing the example above, If we know that one of our two choices is the right one, then *x = ½ vt* is it. In the absence of such knowledge, however, dimensional analysis cannot identify the correct relation.

Dimensional analysis can identify the wrong relations. Thus it can only identify which choices may be correct since it does not account for numerical factors like ½ or for the manner in which an equation was derived from physics principles.

## One more Example of Dimesional analysis

**Example: Analysis of an Equation**

Show that the expression *v=at*, where *v *represents speed, *a *represents acceleration, and t represents an instant of time, is dimensionally correct.

**SOLUTION** with suggestive notes

(Identify the dimensions of *v *from Table 1)

[*v*]= [L/T]

(Identify the dimensions of *a *from Table 1 and multiply by the dimensions of *t *)

[at] = [L/T^{2}] T = [L/T]

So, the given expression is dimensionally correct.