Checking the Dimensional Consistency
If the dimensions of all the terms are not same, the equation is wrong. Here is an example of checking the Dimensional Consistency of an equation of motion :
s(displacement) = u(initial velocity) X t(time) + 1/2 X a(acceleration)X t2(time)
Or, s = u.t + 1/2 a.t2
Step 1: Identify the terms in the equation and identify which symbol stands for which physical quantity.
The terms here are : s, u.t, a.t2
Step 2: Write down the dimensional formula of each symbol used in the equation. If you are unclear how to do it, go to the first article in this series.
[s] = [L]
[u] = [LT-1]
[t] = [T]
[a] = [LT-2]
Step 3: Calculate the dimension of each term in the equation.
Term 1: [s] = [L]
Term 2: [u.t] = [LT -1.T] = [L]
Term 3: [a.t2] = [LT -2.T2] = [L]
Note that, as stated above, we have canceled dimensions from numerator and denominator like [T -2.T2] .
Conclusion: The equation is dimensionally consistent since all the terms have the same dimensions.
Note: The following kind of terms are dimensionless in an equation.
- A pure number
- ratio of similar physical quantities, such as angle as the ratio (length/length), refractive index as the ratio (speed of light in vacuum/speed of light in medium) etc.
- The arguments of special functions, such as the trigonometric, logarithmic and exponential functions