Essential to the problem are two frames of reference or "points of view." One is the "train station," the other is a train moving at a constant velocity past that station. The system of coordinates for the station's frame of reference is called "S" and the system of coordinates for the train's frame of reference is called "S prime" or "S" followed by an apostrophe, S'. The constant velocity of the train is called "u." The x, y and z coordinates of three dimensional space of system S' are assumed to be related to the coordinates of system S such that z'=z, y'=y, x'=x+ut'. The further assumption that time on the train, t', is equal to the time at the station, t, is called a "classical" assumption and the purpose of the exercise is to determine whether that assumption is valid or practical or whether time might be distorted instead.
Considered problematic is the "speed" or velocity of knowledge. (In this context of this problem "speed" and "velocity" are synonymous.) If an event occurs in S then the knowledge of the event might take time to register in S'. The speed of light, sound or whatever the signal of the event is might not be infinite. If the speed of the signal were "infinite" then an event at x=0; t=0 would register at x' in "zero" seconds (even when x' does not equal x) and therefore t' would equal t+0 or t'=t for that event. Suppose however that the speed of the signal is not infinite as indeed appear the speeds of many signals.
The speed of the signal is usually called "c", which also usually refers to the speed of light. For this example use the speed of sound because any finite speed will show the principles involved and the speed of sound makes the example easier to picture in the mind. You can substitute the speed of light in the formulas later. For the constant velocity of the train use 148 miles per hour (A very high speed train that doesn't stop at the station.) and for sound 740 miles per hour. So far z'=z, y'=y, x'=x+ut', u=148 mi/hr, c=740 mi/hr, and the question is how are t and t' related. The test involves two physically identical clocks such that when both are in the train station they measure time the same; one hour marked by one is also marked as an hour on the other. However one of the clocks has been placed aboard the train. Assume also that at t=0 the observer on the train at x'=0 passes by the observer at the station at x=0, then for that instant at least x'=x. The observer on the train sets his clock so that t'=0 for the event. The observer at the station the sets his clock so that t=0 for the event.
In order to compare the passage of time at the station and on board the train an attempt will be made to create an event "one hour" later and the observation of the two clocks during that event noted.
Case One: When the observer at the station sees that one hour has passed on his clock he sends a signal to the train, but at that moment the train is 148 miles away and the signal must close that gap at the finite velocity of 592 mi/hr. (740-148=592). So the observer on the train receives the signal 15 minutes (148 mi/592 mi/hr=1/4 hr) later or at t'=1:15. Should the observer on the train compensate for "u" he will find a "delay" of fifteen minutes per hour or t'=t+0.25t.
Case Two: Knowing that the train will be 148 miles away when one hour has passed, the observer at the station tries to send the signal just early enough so that it arrives when the train is at that point. In this case the signal must close the same gap of 148 miles but the velocity is 740 mi/hr not (740-148) mi/hr. The time to close that gap is 12 minutes (148 mi/740 mi/hr=1/5 hr). The observer at the station sends the signal at t=0:48 so that 12 minutes later at t=1:00 the train will receive it. Should the observer at the station compensate for "u" he will find a "delay" of twelve minutes per hour or t'-0.2t'=t.
Some wonder how the "delay" can be fifteen minutes (0.25 hr) from one frame of reference and twelve minutes (0.2 hr) from the other. Fancy correction factors have been devised to resolve the issue, to express t in terms of t' with a single formula. In this scenario they are not really necessary though. There is no difference in the passage of time on the train and at the station. Fifteen minutes and twelve minutes are not measures of the same "delay." In case one the observer at x'=0 in S' on the train receives the signal at x=185 in S; t'=1:15 (and t as well), and in case two at x=148 in S; t'=1:00 (and t as well). Important to note is that the x of case one, 185, does not equal the x of case two, 148. To be perfectly blunt there is no evidence that the passage of time is altered by motion so far as this scenario is concerned anyway.
Because the delay in case one is (ut)/(c-u) or 0.25 times t and in case two is (ut')/c or 0.2 times t' it doesn't matter what finite value is used for c. The "delay" will not be the same even considering real t'=t whatever the perception even if c=the finite speed of light. There is no alteration of real time, whatever the perception. The reality is the time the signal of the event is sent, the perception is the time the signal of the event is received.
But measuring the speed of light and events involving speeds at or close to that of light are very difficult. That light can behave like waves or particles doesn't simplify anything. Specific experiments that seem to indicate various phenonmena can of course be performed. One might seem to indicate that time can be altered or another might seem to indicate that matter can be foreshortened. Such experiments typically do not involve substances above the molecular level however and drawing conclusions that apply to such substances can be problematic.
The point here is that perception doesn't always conform to reality. Although the perception is that when the clock in S reads 1:00, the clock in S' reads 1:15, the reality is that both read 1:15. But you need an "omniscient" point of view to see it or to be very good at math. Likewise there is the perception or "knowledge" that when the clock in S reads 0:48 the clock in S' reads 1:00, but the "reality" is that a person capable of traveling instantaneously would see that both clocks read 0:48.
There are formulas that can save repeated computations. Confusion resulted from looking at the "delay" which is a number added or subtracted when the comparison is best represented by a ratio which is obtained by multiplying or dividing. For any finite values of u and c; t=t'(1-u/c) and t'=t(1/(1-u/c)).
How can you hear sounds hundreds of miles away? You probably can't without a very sensitive microphone, but real signal speeds for various signals can be substituted in the equations. If system S' were a submarine instead of a train it might be practical to use the speed of sound in water as "c," as the signal speed. The speed of sound in water is much faster than in air and travels much farther. Remember that "c" is ordinarily used to mean the speed of light in a vacuum. But actual signals can come from sounds or lights passing through air, water, glass; or in the case of light, a vacuum. Sound will not travel through a vacuum.
Suppose the observer on train sends the signals and the observer at the station receives them. If the observer on the train sends a signal when his clock reads one hour what time will the observer at the station read on his clock when he receives the signal? The delay will be 12 minutes just as in the previous problem. The clock in the station will read 1:12. If the observer on the train wants to send a signal just early enough so that it arrives when the clock at the station reads one hour, how early is that? In this problem t=1.2t'; t'=5/6t; if t=one hour then t'=50 minutes. His clock should read 0:50 when he sends the signal.