# 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.

• Your expression for $[l]$ doesn't seem to make sense. Notice you are getting the speed of light to have dimensions of acceleration. Nov 21, 2022 at 17:46
• just hand wave On dimensional grounds, the terms in the numerator are fine; they already have the dimension of length and thus don’t need need any power of $c$ restored. The denominator then needs to be dimensionless, and the obvious way to do that is to divide the $v^2$ by $c^2$. Voilà. You don’t need rigor… you just need to be able to do dimensional analysis with ease. Learning how to hand wave a bit is a key part of physics education. Nov 21, 2022 at 23:27
• I have never tried to justify this “rigorously”…that’s overkill. For me, going to natural units means setting $c=1$ so that $c$ disappears for convenience. Going back to “normal” units like SI means using dimensional analysis to stick $c$’s back in wherever necessary to make “normal” units work. It’s not complicated. Think of it as a “clever hack”. Nov 21, 2022 at 23:35
• In addition to being convenient, natural units are better at revealing the essence of physics. $m^2=E^2-\mathbf p^2$ reveals that mass is the Lorentz-invariant length of the energy-momentum four-vector so much more elegantly than $m^2c^4=E^2-\mathbf p^2c^2$ does. Without that clutter of $c$’s, the Minkowskian geometry is in-your-face. Nov 21, 2022 at 23:50

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.

• You can put $c^2$ in the denominator, it's then just a matter of adjusting the numerator. Nov 21, 2022 at 19:13
• "it is not always unambiguous" -- I agree it can be confusing for beginners (and sometimes experts!), but is it ever actually ambiguous? (eg: are there ever two ways of reintroducing units that are not equivalent?) Nov 21, 2022 at 19:38
• @Andrew I think it can be ambiguous but I don't have an example ready to hand. I will have to think a bit more. Nov 21, 2022 at 20:54
• @AndrewSteane OK, if you think of one I'd be interested! Naively I would think that if the only way there could be an ambiguity, is if there was a way to form a dimensionless combination of the constants you are setting equal to 1 in the original unit system. As a trivial example, if we set both the speed of sound and the speed of light equal to 1, that would be ambiguous. Nov 21, 2022 at 23:13
• @Andrew, Morin in his classical mechanics book actually mentions the same thing, that since c has units there must be only one way to put dimensions back Nov 25, 2022 at 0:23

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 $$[l]=299792458 [m]$$