We shall now adopt a new unit for time, the meter. One meter of time is the time it takes for light to travel one meter ... (page-4, Bernhard Schutz's A first course in relativity)

I am confused, how can one keep the unit of time and unit of length as same? It is later concluded that speed of light is unitless in this measuring system. Is this a valid unit change or is there some sort of abuse going on here?

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    $\begingroup$ The speed of light is constant so c = dx/dt. To create a unitless constant from this, divide both sides by c to give 1 = dx/c dt. So for 1 meter of dx , c dt must equal 1 meter. I believe this is what Schultz means $\endgroup$ May 19 at 18:38
  • $\begingroup$ This is a common convention in QFT. So you better get used to it. ;) Valid indeed. But an abuse as well because length and time indeed are physically different dimensions. To preserve those, some authors even define the Lorentz metric with $\pm c^2$ in the 0-0-component instead of $\pm1$. See for example the (really difficult to read, I find) book "Formal Structure of Electromagnetics" by E.J. Post. $\endgroup$
    – kricheli
    May 19 at 19:26

1 Answer 1


It's a bit ill-posed but this is defining natural units (maybe look it up on wikipedia for more info), we just set the unit of speed as c (speed of light), so as the speed is unit 1 it means that ($[]$ denoting unit of): $$[v] = 1 = \frac{[L]}{[T]}$$ so length and time should have the same dimension. We can then easily convert back afterwards by multiplying by c (in e.g m/s).

extra: in natural units both time and distance take the unit of inverse energy

  • $\begingroup$ I don't think that the authors here are defining a totally naturalized unit system with e.g. $c=1, \hbar=1, \epsilon_0 = 1$, etc. It looks like they're just defining 4-space with $ct$ as the time dimension, in which case we have the trivial $dx/dt = c \iff dx/d(ct) = 1$. $\endgroup$
    – g s
    May 19 at 23:04

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