If the x-, y-, and z-axes constitute a right-handed coordinate system, then so do the x'-, y'-, and z'-axes when viewed from K. But do the x'-, y'-, and z'-axes have the same handedness when they are viewed from K' instead of K? (Remember that the axes of K' were set up entirely from within K.)

What we can do is to consider handedness a mechanical property of rigid objects (or rather, of the arrangements of their parts). Then, K and K' can be synchronized in this aspect via bringing a rigid object of known handedness from K into K'. So basically, the question is whether a materialized unit cube of coordinate system K can be brought "up to speed" into K' such that it could seamlessly replace the unit cube of coordinate system K'.

**(Theorem)** K and K' are of the same handedness.

**(Proof)** We know from the proof of v = v' that there exists a K_{w} relative to which K and K' are moving at equal speed but in opposing directions. Let the axes of K_{w} be defined similarly to that of K'. Due to causality, K' is moving in the positive while K in the negative direction along the x_{w}-axis. Now, let's reverse the orientation of the x- and the x_{w}-axes of K and K_{w}, respectively. Then, due to symmetry, the unit cube of the original K_{w} can be brought to perfectly match the unit cube of K' if and only if the same can be done between the reversed K_{w} and the reversed K too. Thus, since the reversed and the original K_{w} have opposite handedness, K' and the reversed K must also have opposite handedness; and so K and K' must have the same.

**Note:** handedness will not play any role later in this essay.

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