Let S_{1} and S_{2} be two planes in K perpendicular to the x-axis, and let Δx = x_{2} – x_{1 }denote the signed distance between them, i.e. the difference between their respective x-coordinates. In K', let S_{1}' and S_{2}' be the corresponding planes perpendicular to the x'-axis, marked out at a given t in K.

Using a similar argument to that when λ was introduced, it can be easily seen that for the corresponding signed distance Δx' in K':

**(Lemma)** Δx' = μ · Δx, where μ > 0 is a constant.

**Notes:** (a) the value of μ can only depend on the choice of K and K', or rather only on v due to symmetry reasons, (b) in classical mechanics, μ = 1.

Since distances on planes perpendicular to the x-axis don't change, it follows immediately that (yet again, meaning "location" in the general sense):

**(Theorem)** If l is a straight line (segment) in K, and l' is the corresponding location in K' marked out at time t in K, then l' is a straight line (segment) too.

**(Proof)** It's due to the proportionality between each of Δx and Δx', Δy and Δy', Δz and Δz'.

But how is l __perceived__ in K'? There is no guarantee that at time t in K, all points of l have the very same corresponding t' values in K'. (In classical mechanics it is guaranteed that they all do, since λ = 1.)

**(Theorem)** l is perceived in K' as a straight line (segment).

**(Proof)** Let P be a fixed point on l in K, and P' the corresponding point in K' at a given time t', marked out at an appropriate t in K. Now, let Q be an another point on l in K, with Δx = x_{Q} – x_{P} being the difference between the x-coordinates of Q and P. It was shown earlier that for the corresponding time value t'_{Q} in K', read off at t in K, the deviation from t' is proportional to Δx, i.e. t'_{Q} – t' ~ Δx. This also entails t'_{Q} – t' ~ Δx', as Δx = (1 / μ) · Δx'. Thus, to find out where Q was (or will be) in K' at t', we have to shift it away from l', along **v**' by the (signed) amount of -(t' – t'_{Q}) · v'. Since this amount is proportional to Δx', shifting all points on l' accordingly will result in a straight line (segment) in K'.

**Note:** it will result in a "proportional" straight line (segment) in K', meaning that e.g. middle points in K are perceived as middle points in K' too.

Making use of v' = v, the following relationship can be derived:

**(Theorem)** μ = λ.

**(Proof)** An observer in K, located on S_{2} sees that Δt = Δx / v time elapsed between meeting S_{2}' and S_{1}'. In addition, the observer also sees a corresponding time Δt' = λ · Δt elapsed in K'. This means that in K', S_{2} travelled Δt' = λ · Δx / v time from S_{2}' to S_{1}'. And since S_{2} is moving at v relative to K', the distance between S_{1}' and S_{2}' is Δx' = v · Δt' = λ · Δx.

Consequently, if there was a rod r' in K', parallel to the x'-axis and reaching from S_{1}' to S_{2}', its length would be Δx' in K' but an observer in K would see it Δx.

**(Corollary)** In K, the perceived length of r' is 1 / λ times its rest length in K'.

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