Originally Posted by

**Guitarist**
Ya, well, MagiMaster gives the sort of illustration that all books use, as he says, but which I personally find unhelpful (these illustrations appeal to our intuition, only to have some fat lady sit on them. I prefer the more abstract approach, but maybe that's just me).

OK, first some preliminaries. It appears that, to a physicist, *time* is nothing more or less than that which is measured by clocks. Likewise, *distance* is merely what can be measured by rulers. Apparently this is not debatable, so there.

Second, with regard to time: if my clock ticks at half the sate of your clock, then my "hour" seems to you to be twice as long as yours. So, slow clocks imply dilated time. Right? In fact *time dilation* is the reciprocal of clock rate.

Let's try this first. Suppose you are on a stationary train and bouncing a tennis ball off the floor. You can time how long it takes for the ball to return to your hand, down and up..... OK? I, on the platform, will agree on that time.

Now suppose your train is moving. If the speed of the train is constant (and the track smooth) you will find the time taken for the straight down-up ball to return to your hand to be the same as when you were stationary. But I on the platform as you whiz by, see that the balls trajectory is *not* up and down, but rather it is diagonal, a longer distance, so I calculate it will take longer to make the round trip. You, on the train disagree, so I can only conclude that either the ball is being accelerated by the train (true in the case of a tennis ball) or that your your clock is running slower that mine (true in the case of light, which can't be accelerated)

So we have just 3 numbers in there: the speed of the train, the speed of my clock and the speed of your clock **as calculated by me**, and these can be connected by a right-angled triangle. Train speed = v is one side, what I believe your clock rate to be is at right angles to that, call it t' and what I believe to be my clock rate, say t, connecting the "free ends" diagonally.

Let's play with that a while and see what we come up with. Remember, first we must equalize all units before we can calculate anything. (actually, that's not true in spacetime, but let's stay simple).

Let vt be the distance the train travels in time = t (remember that as far as you are concerned, the train isn't moving; your tennis ball "experiment" proves that). Fix a constant of speed c, light speed if you want, Then ct = distance traveled in time = t and ct' = distance traveled in time = t'.

Hello Pythagoras: (ct)<sup>2</sup> = (ct')<sup>2</sup> + (vt)<sup>2</sup> or

(ct')<sup>2</sup> = (ct)<sup>2</sup> - (vt)<sup>2</sup>

Now divide through by (ct)<sup>2</sup>, giving

(ct')<sup>2</sup>/(ct)<sup>2</sup> = 1 - (vt)<sup>2</sup>/(ct)<sup>2</sup>.

Notice that the squared c's cancel on the LHS, as do the squared t's in the 2nd term on the RHS, giving

t'<sup>2</sup>/t<sup>2</sup> = 1 - v<sup>2</sup>/c<sup>2</sup> so

t' = t(1 - v<sup>2</sup>/c<sup>2</sup>)<sup>1/2</sup>. Remembering that time dilation is the reciprocal of measured time, one finds that, (with an abuse of notation, where I'm now letting t' refer to time dilation, rather than measured time)

t' = t(1 - v<sup>2</sup>/c<sup>2</sup>)<sup>-1/2</sup>.

Bingo. Length contraction works out similarly, without the last step.