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In Euclidean space, three-vectors can be rotated into one another, making it obvious to me that the units for the basis vectors are the same as one another. But in Minkowski geometry, space-like, time-like and light-like vectors can't be rotated into one another, making me uncertain about the units of these three types of four-vectors for a space-time displacement.

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    $\begingroup$ Typically one would use $c\,\mathrm dt$ for the time component of the differential, which has units of space $\endgroup$
    – Kyle Kanos
    Commented Nov 26, 2019 at 0:05

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Your metric either contains explicit factors of $c$ that match the units between space and time or you use units in which $c=1$ in which case space and time have the same units “naturally”. That’s the answer about consistency. As to what the overall unit is, it doesn’t matter. That’s just an overall rescaling, which has no physical consequence. I’ve worked with distance measured in “length”, in “time”, in “mass” ...

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