Dimensional Analysis

I talked about physical quantities in the previous article and I'm sure you'll find something interesting with the dimensional analysis. The dimensional analysis is a problem solving technique with dimensions!

Dimensions are another way to express the physical quantities (or the standards of physical quantities) in a straight way. Remember the 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 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, feet etc. but all of them are the same dimension which is the dimension of length.

We can check whether the dimensions are consistent or not in 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. Now the dimensions of velocity are $\frac{\text{L}}{\text{T}}=\text{L}{{\text{T}}^{-1}}$ whose SI unit is ${\text{m}}/{\text{s}}\;$.

Similarly the dimensions of acceleration are $\frac{\text{L}}{{{\text{T}}^{2}}}=\text{L}{{\text{T}}^{-2}}$. Now we check whether the dimensions are correct or not on both side of the following equation.

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

Now we equate dimensions on both side of the equation.

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

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 side 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 ${{a}_{x}}=\frac{{{v}_{x}}}{t}$ and this finally implies our initial equation, that is $v_x = a_x t$ and this equation is dimensionally correct.

Mechanics
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