How to rigorously put back dimensions in equations involving natural units? I was watching the first lecture of Special Relativity by Leonerd Susskind (link:Youtube) whereby setting the speed of light to 1, i.e. $c = 1 \dfrac{[l]}{[s]}$, where $[l] = 3 \cdot 10^8 \dfrac{[m]}{[s]}$, we get one of the Lorentz transforms as:
\begin{equation}
x^{'} = \dfrac{x-vt}{\sqrt{1-v^2}} \tag1
\end{equation}
Now, I am very confused about how he would just hand waive to justify transforming this equation into:
$$
x^{'} = \dfrac{x-vt}{\sqrt{1-\dfrac{v^2}{c^2}}} \tag2
$$
Like I tried multiple things with dimension analysis to understand this. For example, if we set $u = \dfrac{v}{c}$ (dimensionless velocity) then, $v=uc$, and hence we get:
$$
x^{'} = \dfrac{x-u(ct)}{\sqrt{1-u^2}} \tag3
$$
So, it would make sense to define $\tau = ct$ with units of $[l]$. This would then get us basically eq. 1, but the speed of light here would be $c = x\tau [m][l]$ which doesn't make sense. HELP. I need a rigorous way to understand this.
 A: It is tricky!
Let's see what happens in the example you asked about. The starting point is
$$
x^{'} = \dfrac{x-vt}{\sqrt{1-v^2}}. \tag1
$$
From the denominator on the right hand side we deduce that $v$ is dimensionless. It then follows that in the numerator, $x$ and $t$ must have the same dimensions. Therefore if we wish to adopt ordinary units where $x$ and $t$ do not both have the same dimensions, then one of them will have to be adjusted. What we need here is in fact
$$
v = u / c,  \qquad \mbox{ and } \qquad t = c T
$$
which together give
$$
x^{'} = \dfrac{x - u T}{\sqrt{1 - u^2/c^2}}.
$$
More generally, putting the $c$'s back in requires methodical working through all parts of a formula, and I think it might not always be unambiguous. However, in an edit to this answer I now admit, on reflection, it is not easy to construct an ambiguous case so maybe it's unambiguous after all. The reason for possible ambiguity is that a $1$ in any formula might become a $c$ or a $c^2$ or other power so you have to watch out.
The lesson is that natural units are best adopted only once you confidently know what you are doing.
A: Let's begin with your equation, where you have defined the conversion between natural and SI units. This is where you're making a mistake.
$$c=1\frac{[l]}{[s]}$$
So now you're defining the natural length with units
$$[l]=299792458 \frac{[m]}{[s]}$$
However, notice that your $[l]$ must have the units of length, and not speed. So this is where you went wrong... the simple correction here is that
$$[l]=299792458 [m]$$
This should solve your issue.
