Dimensional analysis is a problem solving technique with dimensions! Dimensions are another way to express physical quantities (or the standards of physical quantities) in a straight way.

No expression can be correct if the dimensions on both sides of the expression are not the same. So, using dimensional analysis, you can analyze whether the expression is in its correct form or not.

You often need to analyze an equation for its correctness and the useful way to check that is dimensional analysis. The dimensional analysis helps you determine whether the equation meets your expectation or not.

There may be a lot of ways to measure length such as meter, centimeter, foot etc. but all of them have the same dimension which is the dimension of length.

We can check whether the dimensions are consistent or not at both sides of an equation. We denote the dimensions of length, mass and time to be the upper case letters L, M and T respectively. The velocity is the length divided by the time, and the dimensions of velocity are $L$ and $T$ and to express this dimensionally we write $L/T=\text{L}{{\text{T}}^{-1}}$ whose SI unit is ${\text{m}}/{\text{s}}\;$.

Similarly the dimensions of acceleration are $L$ and $T$ and you express it as $L/T^2=\text{L}{{\text{T}}^{-2}}$. Now we check whether the dimensions are correct or not on both sides of the following equation.

\[{{v}_{x}}={{a}_{x}}t\]

Now we equate dimensions on both sides of the equation.

\[\begin{align*} \left[ \text{L}{{\text{T}}^{-1}} \right] &=\left[ \text{L}{{\text{T}}^{-2}} \right]\left[ \text{T} \right] \\ \text{or,}\quad \left[ \text{L}{{\text{T}}^{-1}} \right] &=\left[ \text{L}{{\text{T}}^{-2+1}} \right] \\ \text{or,}\quad \left[ \text{L}{{\text{T}}^{-1}} \right]& =\left[ \text{L}{{\text{T}}^{-1}} \right] \end{align*}\]

It shows that the dimensions are correct on both sides of the equation and hence the equation is dimensionally correct. Now we consider ${{a}_{x}}\propto {{v}_{x}}^{m}{{t}^{n}}$ and find out the actual exponents by dimensional analysis. Now we equate the dimensions on both sides of ${a_x} = k{\kern 1pt} {v_x}^m{t^m}$ where $k$ is the proportionality constant.

\[\begin{align*} \left[ \text{L}{{\text{T}}^{-2}} \right]=&k{{\left[ \text{L}{{\text{T}}^{-1}} \right]}^{m}}{{\left[ \text{T} \right]}^{n}} \\ \text{or,}\quad \left[ {{\text{L}}^{1}}{{\text{T}}^{-2}} \right]=&k\left[ {{\text{L}}^{m}}{{\text{T}}^{-m+n}} \right] \end{align*}\]

The equation is correct if the dimensions on both sides are correct. Now we solve for $m$ and $n$, and we get $-m+n=-2$ and $m=1$. So $m=1$ and $n=-1$. Now putting the value of $m$ and $n$, we get, ${a_x} = k{\kern 1pt} {v_x}^1{t^{ - 1}}$ and therefore,

$${a_x} = k{{{v_x}} \over t}$$

And in this case $k=1$ and $a_x =v_x / t$ and this finally implies our initial equation, that is $v_x = a_x t$ and this equation is dimensionally correct. The technique of dimensional analysis allows us to understand whether the given expression is dimensionally correct or not but it does not test the equation to be entirely correct. The equation must be dimensionally correct to be the correct equation but it does not mean it is entirely correct. No equation is correct if it is not dimensionally consistent.

As an example, in the SUVAT equation $s = ut + 1/2\,at^2$, the number $1/2$ is the dimensionless factor, so we don't know the presence of such factors in dimensional analysis. In other words, if we only have the expression $s = ut + at^2$, this would still be dimensionally correct but the factor $1/2$ is missing.

You can always apply a quick check on whether or not a particular equation is valid especially in situations when you are confused by a quick dimensional check. For example, you forgot equation of time period of pendulum to be $T = 2\pi \sqrt{l/g}$ or $T = 2\pi\sqrt{g/l}$. This kind of thing can often happen at the lab and also at the exams, so quick dimensional check reveals that the expression $T = 2\pi\sqrt{l/g}$ is true. That is $\sqrt{l/g} = \sqrt{LT^2/L} = T$ and it is the dimension of time that matches dimension of time on the left hand side.

Was this article helpful?