**(Theorem)** Let P and Q be two points in K whose x-coordinates are equal. Then, for any t in K, the x'-coordinates of the corresponding P' and Q' in K' are equal too.

**(Proof)** Due to the symmetrical situation of P and Q in K with respect to **v**, and owing to the fact that there is only one way K' can move at **v** relative to K, there is no reason why the situation of P' and Q' would be asymmetrical in K' with respect to **v**', or in other words that the P'Q' segment would tilt in a non-symmetrical way in K'.

Analogously for time:

**(Theorem)** The corresponding t'_{P} and t'_{Q} values in K' are equal as well.

Putting the above two theorems together (the roles of K and K' can be switched in both):

**(Corollary)** A plane S in K that is perpendicular to the x-axis is perceived in K' at any given time t' as a plane S' that is perpendicular to the x'-axis.

**Notes:** (a) S' is different for different t' values, (b) every point of S is moving at **v**' relative to K', which means that S itself is moving in K' at **v**', (c) strictly speaking, we should say "measured" instead of "perceived".

Next, we'll explore how straight lines on S map onto S'.

**(Theorem)** If M is the middle point of segment PQ on S, then for the corresponding points in K', marked out at any t in K, M' is also the middle point of segment P'Q' on S'.

**(Proof)** Due to the symmetrical situation of PM and MQ in K with respect to **v**, and owing to the fact that there is only one way K' can move at **v** relative to K, there is no reason why the length of P'M' and M'Q' would not be equal in K'. Similarly, there is no reason why M' would fall on one side (and not the other) of the straight line determined by P'Q' on S'.

Because of continuity, we can go on and say (again, meaning "location" in the general sense):

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

**Note:** every point of l is moving at **v**' relative to K', which means that l itself is moving in K' at **v**'.

Finally, we'll show that S and S' look exactly the same.

**(Theorem)** Let r in K be a straight line segment perpendicular to the x-axis. Then, the length of the corresponding segment r' in K', i.e. the perceived length of r in K', is equal to that of r in K.

**(Proof)** Let t' denote the time in K' that corresponds, along r, to a given t in K. Let q be the segment in K such that: (a) it is parallel to r, (b) its middle point coincides with that of r, and (c) its length in K is equal to that of r' in K'. Since the situation of K and K' is symmetrical (for v = v'), the corresponding q' in K', marked out at t, must have the same length (in K') as that of r in K. Due to the symmetrical placement of r and q in K, q contains r or r contains q. However, if e.g. q __properly__ contained r in K, then q' would also properly contain r' in K'; the first part would imply that the length of r' in K' is greater than that of r in K, while the second part would imply just the opposite, which would be a contradiction. So q, and thus r' in K', must be exactly as long as r in K.

**Note:** the letter "r" was used because it is the first letter of the word "rod".

Let d and d' denote distance in K and K', respectively:

**(Corollary)** If P and Q are two points in K whose x-coordinates are equal, then d(P, Q) = d'(P', Q').

Let ∠ and ∠' denote angles in K and K', respectively:

**(Theorem)** If P, Q, and R are three points in K whose x-coordinates are equal, then for the angles at Q and Q', ∠(P, Q, R) = ∠'(P', Q', R') holds.

**(Proof)** The triangles PQR in K and P'Q'R' in K' are congruent because their corresponding sides are equal in length.

The y'- and z'-axes of K' will always be chosen such that they are parallel to, and have the same orientation as, the y- and z-axes of K, respectively, at all times.

As it can be rightly suspected by now, in special relativity the interesting things happen along the x-axis.

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