[NOTES/EM-01007] Using dimensional Analysis to get Quick Numerical Estimates

Let us remember dimensions and numerical values of a quantities. To illustrate the method I take

• The fine structure constant \(\alpha=\frac{e^2}{4πϵ_0 \hbar c}\) is dimensionless and has value 1/137.
• The mass of pion is 140 MeV in natural units. This means that  \(m_\pi c^2=140\) MeV.
• The quantity \(\hbar /m_\pi c\) has dimension of length and is approximately equal to \(1.4 \times 10^{−15}m\).

It turns out that remembering the above expression, their dimensions and numerical values is sufficient to quickly compute the numerical values of expressions appearing in nuclear Physics and high energy Physics. As an illustration I take example of  a well known quantity 
\begin{equation} a_0 =\frac{4\pi \epsilon_0\hbar^2}{m e^2} \end{equation}
where \(m\) is mass of the electron. The first step is to make \(e^2\) dimensionless by writing in terms of the fine structure constant.
\begin{equation} e^2 \to [\frac{e^2}{4\pi\epsilon_0 \hbar c}] \times (4\pi\epsilon_0) \end{equation}
The next step is to write mass in the dimensionless form using mass of pion. In our example, we write \(m \to (m/m_\pi) \times m_\pi\). After the above two steps, we get the expression for \(a_0\) in the form
\begin{eqnarray} a_0 &=&\frac{4\pi \epsilon_0\hbar^2}{m e^2}\\ &=& \Big(\frac{4\pi\epsilon_0 \hbar c}{e^2}\Big) \times \Big(\frac{m}{m_\pi}\Big) \times \Big[\frac{\hbar}{m_\pi c}\Big]. \end{eqnarray}
Notice that at the end of this process the terms in parenthesis are dimensionless and can be computed in any unit. The last term carries the dimension of the quantity to be computed. When we started the manipulations it was not known that \(a_0\) has dimensions of length. Here we now see that the dimension is entirely given by \(\hbar/(m_\pi c)\), the Compton wave length of pion. Thus \(a_0\) has dimension of length. Substituting the numerical values we get
\begin{eqnarray} a_0 &=& \Big(\frac{4\pi\epsilon_0 \hbar c}{e^2}\Big) \times \Big(\frac{m}{m_\pi}\Big) \times \Big[\frac{\hbar}{m_\pi c}\Big]\\ &=& (137) (140/0.511) [1.4 \times 10^{-15}m]. \end{eqnarray}
Next a reasonably good estimate is esily obtained without use of any calculator. For example, we take \(1.4\times 1.4\approx 2\), replace \(0.511 \to 0.5\) to get \begin{eqnarray} (137) (140/0.511) 1.4 \times 10^{-15}m. &=& 137 \times (1.4\times 1.4/0.5)\times 10^{-13}m\\ &=& 1.37 \times (2/0.5)\times 10^{-11}m\\ &=& 1.37 \times 4 \times 10^{-11}m\\ &=& 0.548\times 10^{-10} \text{m}. \end{eqnarray}
This constant \(a_0\) is just the Bohr radius of the electron and has the value \(0.529\times 10^{-10}m)\). The numerical value we obtained is within 10\% of the actual number. This is due the fact that the number for the masses and the Compton wavelength of pion have been rounded off. Also we made approximations in getting the final numbers. The quantities in parenthesis, being dimensionless, can be computed in any convenient units. This process has several advantages. There is no need to remember conversion factors between different sets of units. A double check is automatically performed in the sense that the last factor carries the dimension of the quantity of interest. One needs to remember only a few constant. Finally, the chance of making mistake in order ( power of 10) in the final numbers is nil for all practical purposes. A quantity having dimension of length, energy, time or mass can easily be estimated by the process outlined here. For other quantities such as force, the electric or magnetic field, one needs to extend the above process. For numerical calculations in electromagnetic theory, we suggest to use the value of Compton wave length of the electron, \[ \frac{\hbar}{m_e c}=0.0243 \text{angstroms}=2.43\times 10^{-10}m\]

Examples
Obtain an expression of a quantity constructed out of \(m_e c^2=[E]\), such that it has the same dimension as (a) Time (b) Force.
Remembering the Planck's constant \(\hbar\) has dimensions of energy - time, we can write a quantity
\[ \text{Time} \sim (\frac{\hbar}{\text{Energy}} )\sim \frac{\hbar}{m_\pi c^2}\] which has dimensions of time. Similarly, work done has dimensions of force \(\times\) distance. So \[F = \frac{\text{Energy}}{\text{distance}} \sim (m_e c^2)/(\hbar/m_e c) \sim \hbar c \] has dimension of force.

Example (Force between two charges)
Find the magnitude of the force between two equal charges of \(10^{-10}C\) separated by a distance of 1cm.
\begin{eqnarray} F &=& \frac{q^2}{4\pi\epsilon_0 r^2}\\ &=&\Big(\frac{e^2}{4\pi \epsilon_0 \hbar c}\Big) \Big(\frac{q}{e}\Big)^2 \Big[\frac{\hbar c}{r^2} \Big]. \end{eqnarray}
The numerical value is easily estimated using \(e=1.6 \times 10^{-19}C\),

This process was taught to us by Prof. G. K. Mehta at IIT Kanpur as part of Ph.D. course ``Nuclear Physics'' for our batch in 1967.
The method, as described above, was meant for nuclear physics course.
It can be easily adopted for electromagnetic theory, or a course in any area of physics.