This seems like a rather elementary question but it has been causing me some troubles.

Suppose I want to construct a unit system were $c = 1$ and $\hbar = 1$.

The constraint $c = 3 \times 10^ 8 m/s= 1$ implies $1 s = 3 \times 10^8m$.

Similarly we impose $\hbar = 1.05 \times 10^{-34} kg\cdot m^2 /s = 1$. This constraint, along with $c = 1$, implies

$$1 = 1.05 \times 10^{-34} kg\cdot m^2 /s = 1.05 \times 10^{-34} kg\cdot (3\times 10^8)^{-2} m^2 / s = 1.17 \times 10^{-51} kg\cdot m^2/s\\ \implies 1s = 1.17 \times 10^{-51} kg\cdot m^2.$$

The book I am using (QFT for Mathematicians by Folland) says that under this system of units: $$1~s \cong 299792458 ~m \cong 8.522668 \times 10^{50} ~kg^{-1}$$ so it seems I have my numbers right. What has been bothering me is the units. More specifically, how does
$$1s = 1.17 \times 10^{-51} kg\cdot m^2 \implies 1s \cong 8.522668 \times 10^{50} ~kg^{-1}?$$ How does the $m^2$ disappear? My guess is that from the relation $c = 1$ we conclude that $[L] = [T]$ and so $$1s = 1.17 \times 10^{-51} kg\cdot m^2 = 1.17 \times 10^{-51} kg\cdot s^2\\ \implies \big(1.17 \times 10^{-51}\big)^{-1} kg^{-1} = 8.52 \times 10^{50} ~kg^{-1} = 1~s. $$ But how does it physically make sense to say that $[L] = [T]$? Even with the condition $c = 1$ we still must have units on $c$ right? Put differently, how is it allowed to take $c$ to be a unitless number? If we take 1 $m_c$ (where $m_c$ denotes meters in this new unit system) to be $3 \times 10^8 m$ so that $c = 3 \times 10^8 m / s$, we still would have $c = m_c/s$, and not unitless.

  • $\begingroup$ Does this answer your question? What exactly are we doing when we set $c=1$? $\endgroup$
    – Anyon
    Oct 29, 2022 at 16:25
  • $\begingroup$ @Anyon Not quite. The answer in that questions explains how we can take the units in the speed of light to be anything, which is clear to me, however it seems when working with natural units some people impose values such as $c$ to be unitless (which is required to conclude $[L] = [T]$). This is what I'm most confused about. $\endgroup$
    – CBBAM
    Oct 29, 2022 at 16:27
  • $\begingroup$ I suggest looking up the distinction between dimensions and units (I.e m and km are but units for distance dimensions) and framing your question in terms of that language. $\endgroup$
    – Jagerber48
    Oct 29, 2022 at 16:53
  • $\begingroup$ Imho it is easier to think of time being measured by $ct$ instead of our conventional $t$. Then you don't even have to "change" units but time is measured in $m$ that light travels in $t$ seconds. $\endgroup$
    – Kurt G.
    Oct 29, 2022 at 16:53
  • 1
    $\begingroup$ @Jagerber48 Nature distinguishes space from time because they have opposite signs in the metric, not because they have different “dimensions”. Dimensions are a human concept. If you had grown up with clocks that measured time in meters, it would seem perfectly natural to measure time in meters. And you would think that time is just distance in the temporal direction of spacetime, because it is. $\endgroup$
    – Ghoster
    Oct 29, 2022 at 18:48

1 Answer 1


I'm not sure what the question's physically requires, so I'll explain it mathematically.

Let $\mathsf{L},\,\mathsf{M},\,\mathsf{Q},\,\mathsf{T},\,\mathsf{\Theta}$ represent the dimensions of length, mass, charge, time and temperature. These are all nondimensionalized with the Planck units $c=G=\hbar=\frac{1}{4\pi\varepsilon_0}=k_B=1$. You're doing something less radical, namely $c=\hbar=1$, but we can understand both, as well as a third option of introducing no such nondimensionalization conventions, in the same way.

An arbitrary dimension of a physical quantity over the above dimensions is $\mathsf{L}^a\mathsf{M}^b\mathsf{Q}^c\mathsf{T}^d\mathsf{\Theta^e}$, which we can denote as a vector $\in V:=\Bbb R^5$ with entries $a$ etc. Setting $c=\hbar=1$ is basically "modulo" arithmetic with vectors. To think of it another way, the "dimensionless" quantities are those who dimension is spanned by the vectors with $a=1,\,d=-1$ and $a=2,\,b=1,\,d=-1$ (in both cases, unstated vector entries are $0$). This is, of course, a subspace of $V$, say $W$.

Finally, those two vectors are linearly independent, an arbitrary $v\in V$ can be written uniquely as $v_\parallel+v_\perp$ with $v_\parallel\in W$ and $v_\perp\cdot w=0$ for all $w\in W$. Identifying $v$ with $v_\perp$ then defines an equivalence relation. For example, $E=mc^2$ simplifies to $E=m$, which looks dimensionally inconsistent because $[E]\ne[m]$, but it's fine because $[E]$ is in the same equivalence class as $[m]$. What's more, the uniqueness of $v_\parallel$ implies you can work out which powers of $c$ etc. to reinstate. Even if we are as ambitious as Planck units so everything becomes dimensionless, the choice of powers to fix a given equation is still unique.

But making things unitless like this has one downside: each such independent unitless quantity reduces by $1$ the number of degrees of freedom we can use in dimensional analysis. (This is related to a limitation of dimensional analysis that applies even before we start making things unitless.) For example, $c=G=\hbar=1$ prevents dimensional analysis from refuting $E=m^3$, by which I mean $E=cGm^3/\hbar$, or $E=1/m$, by which I mean $E=c^3\hbar/(Gm)$.

There's also one other important subtlety. If I take $c=\hbar=1$, length and time become inverse mass; if I take $c=G=1$, length and time become (uninverted) mass. So quantum calculations are very different from general relativity calculations! (This is why combining the above conventions gives us so little information to work with in dimensional analysis.)

So when is it "allowed"? In practice, the goals of such nondimensionalization are to make some equations neater, to let us focus on the most important details when we do calculations, and to hopefully not go so far in that direction we can no longer do other important things, such as switch back to SI units or do sanity checks. (Work through the calculations here with and without Planck units, and you'll see the merits of both approaches.)


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