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The moral of the parable in special relativity is that:

  • The same units should be used for all distances.
  • The (squared) distance is invariant.
  • Different frames are related by rotations.

Now what we usually do to make the first of them true, it to measure time in meter (in the unit of space). To do so, We use $ct$. But Can we do the opposite, We can measure space in the unit of time. That is to say, can I used $(t,x/c)$ as my coordinate? I'm sure we can, But don't know if we find any difficulty later on in advance concepts.

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    $\begingroup$ most people set $c=1$ "God given units" anyways $\endgroup$ – Mike Flynn Apr 8 at 5:04
  • $\begingroup$ en.wikipedia.org/wiki/Light-year $\endgroup$ – fqq Apr 8 at 5:24
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    $\begingroup$ Why not? Would you rather divide three (or more) spatial coordinates by $c$ or multiply one temporal coordinate by $c$? $\endgroup$ – G. Smith Apr 8 at 5:37
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I don't think there is a problem using coordinates $\quad\left ( t, \displaystyle \left(\frac{x}{c}\right) \right)\quad$ vs $\quad\left ( ct, \vphantom{\displaystyle\left(\frac{x}{c}\right) } x \right)$.

In fact, I prefer $\left ( t, \displaystyle \left(\frac{x}{c}\right) \right)$ because the time axis is more important to me...
especially if I use a clock and light rays to perform radar measurements.
In other words, since I use a clock, time is the more fundamental quantity.

Indeed, the current definition of the meter ( https://en.wikipedia.org/wiki/Metre ) regards the "second" as more fundamental

The metre is currently defined as the length of the path travelled by light in a vacuum in (1/299 792 458) of a second.


By the way,
Wald [p.7, eq. 1.3.1] defines the square-interval by $$I=-(\Delta t)^2+\frac{1}{c^2}\left( (\Delta x)^2 + (\Delta y)^2 +(\Delta z)^2 \right)$$

On the other hand,
Misner Thorne Wheeler [p. 51, in Box 2.1 Farewell to "ict"] writes
"In this chapter and hereafter, as throughout the literature of general relativity, a real time coordinate is used, $x^0=t=ct_{conv}$."
Then, on p.53, defines the square-interval by $$(\Delta s)^2 = -(\Delta x^0)^2+(\Delta x^1)^2+(\Delta x^2)^2+(\Delta x^3)^2.$$

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  • $\begingroup$ In fact, division by $c$ is what you do to express distances in seconds... light-seconds. $\endgroup$ – Ruslan Apr 8 at 7:57

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