Today we will develop some advanced modeling techniques. They are so advanced that at first they seem trivial, but on second look they let us roughly calculate almost anything. They are somewhat antithetical to the ways you have previously been asked to solve problems. In elementary courses you are given cooked-up toy problems which can be exactly solved by using a set of mathematical techniques. Dimensional analysis and estimation instead are used to learn rough features about real problems.

Formal Dimensional Analysis

You are all familiar with informal dimensional analysis. For example, you want to know the speed of transverse sound $v$ in a cable of linear density $\lambda$ and tension $\tau$. You know the dimensions of velocity are $m/s$, the dimensions of linear density are $kg/m$ and the dimensions of $\tau$ are $kg m/s^2$. The only combination which gives the right units are
\begin{equation}\label{da}
v\propto \sqrt{\tau/\lambda},
\end{equation}
which you should recall from your waves course is the correct answer, up to some numerical constants.
This reasoning seems trivial, unless I also tell you that the cable has a bending modulus $\mu$, Young's modulus $Y$, thickness $a$, length $L$, is on Earth, where the local gravitational field is $g$,... With all of this extra information, dimensional analysis is not so simple. The first step is modeling. What are the relevant quantities? Can any be thrown away? Do you know how any of them enter the problem?
Typically you will encounter different regimes. For example, as you make the ``cable" thicker, it becomes more rigid, and eventually it is more like a bar of metal. At some point the tension is no longer relevant, and instead the speed of transverse sound involves $\lambda$ and $\mu$.

Complete dimensional analysis handout.

One formalizes this situation by writing down all independent dimensionless ratios in the problem, $\Pi_0,\cdots \Pi_{n-1}$. For example, we could take
\begin{eqnarray}
\Pi_0&=&v \sqrt{\lambda/\tau}\\
\Pi_1&=&\mu a^2/\tau
\end{eqnarray}
Assuming out modeling is correct, and the only relevant quantities are $v,\tau,\lambda,\mu,$ and $a$, then there are no other independent dimensionless quantities we can make. Suppose there is some equation relating $v$ to the other quantities. The only dimensionally sound formulation is
\begin{equation}
\Pi_0=f(\Pi_1),
\end{equation}
for some unknown (but unique) function $f$.
Equivalently
\begin{equation}
v = \sqrt{\frac{\tau}{\lambda}} f\left(\frac{\mu a^2}{\tau}\right).
\end{equation}
Now you might argue that we haven't done anything useful because we have an unknown function. In fact, there are many predictions we can make without knowing $f$. For example, we know that if we double $\mu$ and $\tau$ the velocity will be multiplied by $\sqrt{2}$.

Last year this reasoning confused a lot of students, so I will not belabor it, but I just want to put it on your radar as a very powerful technique. We will occasionally use it. One can even make a formal "theorem" out of this, called the Buckingham $\pi$-theorem. Not being a mathematician, I will not worry about such niceties.

Adimensionalizing

An important close cousin of dimensional analysis is adimensionalizing. This is the practice of using scaled variables to make equations simpler. For example, the time independent Schrodinger equation for the hydrogen atom reads
\begin{equation}
\frac{-\hbar^2\nabla^2}{2m}\psi(r)-\frac{e^2}{4\pi\epsilon_0 r} \psi(r)=E \psi(r).
\end{equation}
We rescale energy and space, via $E=E_0 {\cal E}$, $r=r_0 {\cal R}$, where $E_0$ and $r_0$ are undetermined constants. This yields
\begin{equation}\label{ad1}
\frac{\hbar^2}{m r_0^2}\frac{-\nabla_{\cal R}^2}{2} \psi({\cal R})-\frac{e^2}{4\pi\epsilon_0 r_0} \frac{1}{\cal R} \psi({\cal R})=E_0 {\cal E} \psi({\cal R}).
\end{equation}
We then choose $r_0$ and $E_0$ to satisfy
\begin{equation}
\frac{\hbar^2}{m r_0^2}=\frac{e^2}{4\pi\epsilon_0 r_0} =E_0
\end{equation}
to arrive at the adimensionalized equation
\begin{equation}
\frac{-\nabla_{\cal R}^2}{2} \psi({\cal R})-\frac{1}{\cal R} \psi({\cal R})= {\cal E} \psi({\cal R}).
\end{equation}
Everything here is dimensionless.
Later if you want results in physical units you can plug them back in.

Students always ask, how do you know how to choose the scales. The answer is implicit in what we just did, but let me repeat the argument for emphasis. In this example we started by taking $r_0$ to be undetermined constants. That yields (\ref{ad1}). You then equate the dimensional expressions in front of the first two terms
\begin{equation}
\frac{\hbar^2}{m r_0^2}=\frac{e^2}{4\pi\epsilon_0 r_0}.
\end{equation}
This gives
\begin{equation}
r_0= \frac{4\pi \epsilon_0 \hbar^2}{m e^2}.
\end{equation}
By construction, when you substitute this back in, the two terms will have the same coefficient, and the equation will be simpler. As a bonus, the equation is in "natural units", and unless there is something strange going on, the characteristic scale that the wavefuction varies over will be $R_c\sim 1$.