0
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ 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

3
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.