# Why do we set $x^0 = ct$ instead of $x^0 = t$?

When we deal with Special Relativity and we start considering spacetime instead of space and time each at once, we usually see books saying that we consider a space with four coordinate $x^\alpha$ with $x^0 = ct$. We also consider this manifold to be $\mathbb{R}^4$ and give to it the metric tensor $g = \operatorname{diag}(-1,1,1,1)$. Now, why $x^0 = ct$ and not just $x^0 = t$? One possible answer might be "so that all coordinates are measured with the same units", but isn't there any deeper meaning?

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Here my two cents: A really good book about the conceptual basics of SR is "Spacetime physics" by Taylor and Wheeler. A scanned version of the first edition is available on Taylor's homepage: eftaylor.com/download.html – Photon Apr 12 '14 at 6:53

But that's exactly the deeper meaning! Setting things up so that all coordinates are in the same units (besides being a reasonable requirement for $x^\mu$ to be considered a four-vector) is a constant reminder that time is really not that different from space. In fact, if it weren't for that sign in the metric, spacetime would be completely symmetric in its coordinates.
This is made even more apparent when we switch to a system of units in which $c = 1$. Now we don't even have to make a decision to set $x^0 = ct$, because by measuring both time and space in meters (or seconds if you prefer), we have completely abandoned the notion of time being a distinct entity which is somehow independent from space.
As an additional remark, by bundling $c$ in with the time component, we can leave it out of all our other expressions. For instance, with $x^0 = t$ our metric tensor would be $g = \mathrm{diag}(-c^2,1,1,1)$; our Lorentz transformations, on the other hand, would be $c x^0 = \gamma cx^0 - \gamma \beta x^1$ and $x^1 = \gamma x^1 - \gamma \beta c x^0$. These become neater when we just absorb our $c$ into our $x^0$. – gj255 Apr 12 '14 at 8:46