Since the same light-signal has to be transmitted relative to _K′_ with the velocity _c_, the propagation relative to the system _K′_ will be represented by the analogous formula _x′_ – _ct′_ = 0 . . . . . (2) Those space-time points (events) which satisfy (1) must also satisfy (2). Obviously this will be the case when the relation (_x′_ – _ct′_) = λ(_x_ – _ct_) . . . (3). is fulfilled in general, where λ indicates a constant; for, according to (3), the disappearance of (_x_ – _ct_) involves the disappearance of (_x′_ – _ct′_). If we apply quite similar considerations to light rays which are being transmitted along the negative _x_-axis, we obtain the condition (_x′_ + _ct′_) = (_x + ct_) . . . (4). By adding (or subtracting) equations (3) and (4), and introducing for convenience the constants _a_ and _b_ in place of the constants λ and μ where image038 and image039 we obtain the equations image040 We should thus have the solution of our problem, if the constants _a_ and _b_ were known. These result from the following discussion. For the origin of _K′_ we have permanently _x′_ = 0, and hence according to the first of the equations (5) image041 If we call _v_ the velocity with which the origin of _K′_ is moving relative to _K_, we then have image042 The same value _v_ can be obtained from equations (5), if we calculate the velocity of another point of _K′_ relative to _K_, or the velocity (directed towards the negative _x_-axis) of a point of _K_ with respect to _K′_. In short, we can designate _v_ as the relative velocity of the two systems. Furthermore, the principle of relativity teaches us that, as judged from K, the length of a unit measuring-rod which is at rest with reference to _K′_ must be exactly the same as the length, as judged from _K′_, of a unit measuring-rod which is at rest relative to _K_. In order to see how the points of the _x′_-axis appear as viewed from _K_, we only require to take a “snapshot” of _K′_ from _K_; this means that we have to insert a particular value of _t_ (time of _K_), _e.g._ _t_ = 0. For this value of _t_ we then obtain from the first of the equations (5) _x′_ = _ax_ Two points of the _x′_-axis which are separated by the distance Δ_x′_ = 1 when measured in the _K′_ system are thus separated in our instantaneous photograph by the distance image043 But if the snapshot be taken from _K′_(_t′_ = 0), and if we eliminate _t_ from the equations (5), taking into account the expression (6), we obtain image044 From this we conclude that two points on the _x_-axis separated by the distance 1 (relative to _K_) will be represented on our snapshot by the distance image045 But from what has been said, the two snapshots must be identical; hence Δ_x_ in (7) must be equal to Δ_x′_ in (7_a_), so that we obtain image046