Galileo was the first to see clearly that someone traveling in uniform motion would not be able to discern any difference from being at rest (without looking out the window). He imagined someone on a ship eating peas, and if a few dropped off their fork, there would be no difference from what would happen on land. This is called Galilean relativity, to distinguish it from the later Lorentzian or Einsteinian relativity, in which the speed of light is a constant.

Let’s consider a standard situation with two observers and their respective frames of reference, one moving with a constant velocity relative to the other, as in this *illustration* from this *article*. So there are two observers/frames, *S* and *S’*, with *S’* moving at a velocity *v* such that observer *S* uses coordinates *x* (length) and *t* (time) and observer *S’* uses *x’* and *t’*. The coordinates are arranged so that they are coincident at time *t* = *t’* = 0. Then, as is well known, *x’ = x – vt*.

What about the time coordinate? (Here only the time coordinate in the *x* direction is considered.) If, as Newton assumed, the measurement of time is the same for all observers, then *t’ = t*. However, this implicitly assumes a third frame which provides the independent time measurement. With the relativity of both time and space, we cannot use clocks as if they were from a reference frame independent of all others (see *Movement and measurement*).

Instead it is best to measure length and duration together. Let us do that by having (or imagining) measuring wheels moving along each axis (in both directions) in each frame of reference. To keep things simple, let there be two measuring wheels moving at constant speed, *b*, in opposite directions as a point event is jointly observed. Then *b* is a conversion factor between length and duration: *x = bt* and *x’ = bt’*. Combine these with the transformation above for *x’* and the result is: *t’ = t – vx/b ^{2}*.

So the revised Galilean transformation for frames moving with the *x* axis is:

*x’ = x – vt, y’ = y, z’ = z, *and* t’ = t – vx/b ^{2}.*

Note that if the conversion factor is the speed of light, *c*, the time transformation becomes *t’ = t – vx/c ^{2}*. This is correct apart from considering the constancy of the speed of light, which leads to the Lorentz factor.

Addendum: The problem with this transformation is that it does not form a transformation group. The transformations for space and time need to be similar. This revised transformation will do:

*x’ *= (1 –* v/b*)* x, y’ = y, z’ = z, *and* t’ *= (1 –* v/b*)* t.*