order of one mean free path, in which the theory of the wave equation begins to fail because we did not consider the structure of the gas Now, referring again to Fig 51-2, we see that the curvature can be understood if we appreciate that the pressures near the apex are higher than they are farther back, and so the angle e is greater. That is, the curve is the result of the fact that the peed depends upon the strength of the wave. Therefore the wave from an atomic bomb explosion travels much faster than the speed of sound for a while, until it gets so far out that it is weakened to such an extent from spreading that the pressure bump is small compared with atmospheric pressure. The speed of the bump then approaches the speed of sound in the gas into which it is going. (Incidentally,it always turns out that the speed of the shock is higher than the speed of sound in the gas ahead, but is lower than the speed of sound in the gas behind. That is mpulses from the back will arrive at the front but the front rides into the medium in which it is going faster than the normal speed of signals. So one cannot te acoustically, that the shock is coming until it is too late. The light from the bomb arrives first, but one cannot tell that the shock is coming until it arrives, because there is no sound signal coming ahead of it. This is a very interesting phenomenon, this piling up of waves, and the main point on which it depends is that after a wave is present, the speed of the resulti wave should be higher. Another example of the same phenomenon is the following Consider water flowing in a long channel with finite width and finite depth. If a piston,or a wall across the channel, is moved along the channel fast enough, water piles up, like snow before a snow plow. Now suppose the situation is as shown in Fig 51-4, with a sudden step in water height somewhere in the channel. It can be demonstrated that long waves in a channel travel faster in deeper water than they do in shallow water. Therefore any new bumps or irregularities in energy supplied by the piston run off forward and pile up at the front. Again, ultimately what we Figure 51-4 have is just water with a sharp front, theoretically. However, as Fig 51-4 shows there are complications. Pictured is a wave coming up a channel; the piston is at the far right end of the channel. At first it might have appeared like a well-behaved wave, as one might expect, but farther along the channel, it has become sharper and sharper until the events pictured occurred. There is a terrible churning at the surface, as the pieces of water fall down, but it is essentially a very sharp rise with no disturbance of the water ahead Actually water is much more complicated than sound. However, just to illus trate a point, we will try to analyze the speed of such a so-called bore in a channel The point here is not that this is of any basic importance for our purposes--it is not a great generalization--it is only to illustrate that the laws of mechanics that we already know are capable of explaining the phenomenon. Imagine, for a moment, that the water does look something like Fig. 51-5(a) that water at the higher height h2 is moving with a velocity v, and that the front is moving with velocity u into undisturbed water which is at height h1. We would like to determine the speed at which the front moves. In a time At a vertical plane initially at xt moves a distance v At to x2, while the front of the wave has moved Now we apply the equations of conservation of matter and momentum, First Fig. 51-5. Two cross sections of a he former: Per unit channel width we see that the amount h2 At of matter that bore in a channel, with(b)at an interval has moved past xI(shown shaded)is compensated by the other shaded region, Af later than (a) which amounts to(h2- h1)u Al. So, dividing by At, wh2 = u(h2-hi.That does not yet give us enough, because although we have h? and hl, we do not know either u or v; we are trying to get both of them Now the next step is to use conservation of momentum, We have not discussed the problems of water pressure, or anything in hydrodynamics, but it is clear any way that the pressure of water at a given depth is just enough to hold up the column of water above it. Therefore the pressure of water is equal to p, the density of water, times g, times the depth below the surface. Since the pressure increases linearly with depth, the average pressure over the plane at xI, say, is 2pgh2, which is also the average force per unit width and per unit height pushing the plane toward x2 So we multiply by another h2 to get the total force which is acting on the water