In physics, the dimension of a physical quantity refers to the expression of that quantity in terms of fundamental dimensions. The most commonly used fundamental dimensions are M, L, and T (M for mass, L for length, and T for time).
The concept of dimensions helps in understanding and analyzing physical quantities by breaking them down into their fundamental components. This is often done using the technique of dimensional analysis. By expressing a physical quantity in terms of its fundamental dimensions, you can determine how it depends on other physical parameters and compare it with other quantities.
For example, velocity has the dimension of length per unit time, which can be expressed as LT^{-1}. Similarly, acceleration has the dimension of length per unit time squared, which can be expressed as LT^{-2}. Many other physical quantities can be expressed in terms of these fundamental dimensions, which allows for a deeper understanding of their behavior and relationships in various physical equations.
Let’s calculate the dimensional formula for force.
According to Newton’s law of motion, Force is mass \times acceleration. So its dimensional formula will be MLT^{-2}.
Homogeneity of Dimensions in an Equation
The principle of homogeneity of dimensions, which is also known as dimensional analysis, is a fundamental concept in physics. It states that in a valid physical equation, each term in the equation has the same dimension so that the equation is balanced and physically meaningful. In other words, you cannot add or equate quantities with different physical dimensions. For example, you cannot and Force and acceleration because both have different dimensions. But you can add Force/acceleration and mass. Here are some key points related to the homogeneity of dimensions in an equation:
So, by checking the homogeneity of dimensions, you can help ensure the correctness of equations and develop a deeper understanding of physical relationships.
To better understand the principle, let us consider the following example:
Numerical Questions
Question 1: Check the correctness of the physical equation s = ut + \frac{1}{2}at^2.
Solution:
Here, s is the displacement, u is the initial velocity, v is the final velocity, a is the acceleration and t is the time.
The dimensional formula of s is given by [s]=L
The dimensional formula of ut is given by [ut] = [LT^{-1}]][T] = L
The dimensional formula of \frac{1}{2}at^2 is given by [ \frac{1}{2}at^2 ] = [M^0L^0T^0][LT^{-2}][T^2] = L
Since the terms on the right-hand side of the equations have the same dimension, these terms can be added. And so the dimensional formula for ut + \frac{1}{2}at^2 is
\begin{aligned}[ut + \frac{1}{2}at^2] &= L\\\end{aligned}
which is the same as the dimensional formula of the left-hand side of the equation.
Therefore, by the principle of homogeneity, the equation s = ut + \frac{1}{2}at^2 is dimensionally correct.