The Quantum Universe_ Everything That Can Happen Does Happen - Brian Cox [17]
Figure 3.3. Two waves arranged such that they cancel out completely. The top wave is out of phase with the second wave, i.e. peaks align with troughs. When the two waves are added together they cancel out to produce nothing – as illustrated at the bottom where the ‘wave’ is flat-lining.
With this in mind, we shall now spend a little time inventing a precise rule for adding clocks. In the case of Figure 3.3, the rule must result in all the clocks ‘cancelling out’, leaving nothing behind: 12 o’clock cancels 6 o’clock, 3 o’clock cancels 9 o’clock and so on. This perfect cancellation is, of course, for the special case when the waves are perfectly out of phase. Let’s search for a general rule that will work for the addition of waves of any alignment and shape.
Figure 3.4 shows two more waves, this time aligned in a different way, such that one is only slightly offset against the other. Again, we have labelled the peaks, troughs and points in between with clocks. Now, the 12 o’clock clock of the top wave is aligned with the 3 o’clock clock of the bottom wave. We are going to state a rule that allows us to add these two clocks together. The rule is that we take the two hands and stick them together head to tail. We then complete the triangle by drawing a new hand joining the other two hands together. We have sketched this recipe in Figure 3.5. The new hand will be a different length to the other two, and point in a different direction; it is a new clock face, which is the sum of the other two.
Figure 3.4. Two waves offset relative to each other. The top and middle waves add together to produce the bottom wave.
We can be more precise now and use simple trigonometry to calculate the effect of adding together any specific pair of clocks. In Figure 3.5 we are adding together the 12 o’clock and 3 o’clock clocks. Let’s suppose that the original clock hands are of length 1 cm (corresponding to water waves of peak height equal to 1 cm). When we place the hands head-to-tail we have a right-angled triangle with two sides each of length 1 cm. The new clock hand will be the length of the third side of the triangle: the hypotenuse. Pythagoras’ Theorem tells us that the square of the hypotenuse is equal to the sum of the squares of the other two sides: h2 = x2 + y2. Putting the numbers in, h2 = 12 + 12 = 2. So the length of the new clock hand h is the square root of 2, which is approximately 1.414 cm. In what direction will the new hand point? For this we need to know the angle in our triangle, labelled θ in the figure. For the particular example of two hands of equal length, one pointing to 12 o’clock and one to 3 o’clock, you can probably work it out without knowing any trigonometry at all. The hypotenuse obviously points at an angle of 45 degrees, so the new ‘time’ is half way between 12 o’clock and 3 o’clock, which is half past one. This example is a special case, of course. We chose the clocks so that the hands were at right angles and of the same length to make the mathematics easy. But it is obviously possible to work out the length of the hand and time resulting from the addition of any pair of clock faces.
Figure 3.5. The rule for adding clocks.
Now look again at Figure 3.4. At every point along the new wave, we can compute the wave height by adding the clocks together using the recipe we just outlined and asking how much of the new clock hand points in the 12 o’clock direction. When the clock points to 12 o’clock this is obvious – the height of the wave is simply the length of the clock hand. Similarly at 6 o’clock, it’s obvious because the wave has a trough with a depth equal to the length of the hand. It’s also pretty obvious when the clock reads 3 o’clock (or 9 o’clock) because then the wave height is zero, since the clock hand is at right angles to the 12 o’clock direction. To compute the wave height described by any particular clock we