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.
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1$\begingroup$ Your expression for $[l]$ doesn't seem to make sense. Notice you are getting the speed of light to have dimensions of acceleration. $\endgroup$– Níckolas AlvesNov 21, 2022 at 17:46
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$\begingroup$ 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. $\endgroup$– GhosterNov 21, 2022 at 23:27
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$\begingroup$ 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”. $\endgroup$– GhosterNov 21, 2022 at 23:35
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$\begingroup$ 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. $\endgroup$– GhosterNov 21, 2022 at 23:50
2 Answers
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.
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1$\begingroup$ You can put $c^2$ in the denominator, it's then just a matter of adjusting the numerator. $\endgroup$– JavierNov 21, 2022 at 19:13
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$\begingroup$ "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?) $\endgroup$– AndrewNov 21, 2022 at 19:38
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$\begingroup$ @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. $\endgroup$ Nov 21, 2022 at 20:54
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$\begingroup$ @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. $\endgroup$– AndrewNov 21, 2022 at 23:13
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1$\begingroup$ @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 $\endgroup$ 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 $$[l]=299792458 [m]$$
This should solve your issue.
