Now in our particular wave there is adefinite relationship between the frequency and the wavelength, but the abovedefinitions of k and ω are actually quite general. That is,the wavelength and the frequency may not be related in the same way in otherphysical circumstances. However, in our circumstance the rate of change ofphase with distance is easily determined, because if we call ϕ=ω(t−r/c)the phase, and differentiate (partially) with respect to distance r, the rate of change, ∂ϕ/∂r , is
现在，在我们具体的波中，在频率和波长之间，有一个确定的关系，但是，上面关于k和 ω的定义，实际上是相当普遍的。也就是说，在其他的物理情形中，波长与频率，可能不是以同样的方式，而相关{？}。然而，在我们的情形中，相位随着距离的变化率，很容易被规定，因为，如果我们称ϕ=ω(t−r/c)，为相位，那么，它对距离 r的微分（偏微分）、即距离的变化率∂ϕ/∂r，就是
(29.4)
There are many ways to represent the samething, such as
要表现此同一事物，有很多方式，例如
λ=ct0 (29.5)
ω=ck (29.6)
λν=c (29.7)
ωλ=2πc (29.8)
Why is the wavelength equal to ctimes the period? That’s very easy, of course, because if we sit still and waitfor one period to elapse, the waves, travelling at the speed c ,will move a distance ct0 , and will of course have movedover just one wavelength.
为什么波长，等于c乘以周期？这当然很容易，因为，如果我们坐着不动，等一个周期逝去，那么，以速度c传播的波，将会移动一个距离ct0，当然，就是正好移动了一个波长。

In a physical situation other than that oflight, k is not necessarily related to ω in this simple way.If we call the distance along an axis x , then the formula for acosine wave moving in a direction x with a wave number kand an angular frequency ω will be written in general as cos(ωt−kx).
在非光学的物理环境中，k与ω的关系，并不必然是这种简单的方式。如果我们只考虑沿着x轴的距离，那么，对于一个沿着x方向移动的余弦波，其波数为k，角频率为ω，则其公式，一般可写为cos(ωt−kx)。

Now that we have introduced the idea ofwavelength, we may say something more about the circumstances in which (29.1)is a legitimate formula. We recall that the field is made up of several pieces,one of which varies inversely as r , another part which variesinversely as r2 , and others which vary even faster. Itwould be worthwhile to know in what circumstances the 1/r part ofthe field is the most important part, and the other parts are relatively small.Naturally, the answer is “if we go ‘far enough’ away,” because terms which varyinversely as the square ultimately become negligible compared with the 1/r term. How far is “far enough”? The answer is, qualitatively, that theother terms are of order λ/r smaller than the 1/r term. Thus, so long as we are beyond a few wavelengths, (29.1)is an excellent approximation to the field. Sometimes the region beyond a fewwavelengths is called the “wave zone.”
现在，我们已经介绍了波长的概念，对于有些情形，我们就可以讲更多的东西，因为，在这些情形中，（29.1）是一个合理的公式。我们回忆一下，场由几个部分组成，其中之一，与r成反比，另外一个，与r2成反比，其他部分，可能变化地更快。所以，知道在什么情形下，场的1/r部分，是最重要的部分，而其他部分，则相对比较小，就非常重要。很自然的答案就是：“如果我们走的足够远”，因为这时，与距离的平方成反比的部分，最终，与1/r部分相比，就可忽略。那么，足够远，是多远呢？定性的答案就是：其他的项，要比1/r项，小λ/r级别{？}。这样，只要我们超越了几个波长，则（29.1）就是对场的一个精彩的近似。有时，超过几个波长的区域，就被称为“波区”。

29–4Two dipole radiators 29-4 两个偶极子振荡器
Next let us discuss the mathematicsinvolved in combining the effects of two oscillators to find the net field at agiven point. This is very easy in the few cases that we considered in the previouschapter. We shall first describe the effects qualitatively, and then morequantitatively. Let us take the simple case, where the oscillators are situatedwith their centers in the same horizontal plane as the detector, and the lineof vibration is vertical.
在一个给定的点，两个振荡器的效果，会发生组合，以形成一个净的场，这种组合，会牵扯到一些数学，现在，就让我们来讨论这些数学。在我们上一章所考虑的几种情况下，这将非常简单。我们首先，将定性地描述效果，然后，再更加定量地描述。我们先取一个简单的例子，在那里，诸振荡器的中心，与探测器，处于同一个水平平面，且振动的线，是垂直的。

Fig. 29–5.The intensities in variousdirections from two dipole oscillators one-half wavelength apart. Left: inphase (α=0 ). Right: one-half period out of phase (α=π). 图29-5 两个偶极子振荡器，相差半个波长，它们在各个方向的强度。左：在相位（α=0）。右：相位为周期的一半（α=π）。

Figure 29–5(a)represents the top view of two such oscillators, and in this particular examplethey are half a wavelength apart in a N–S direction, and are oscillatingtogether in the same phase, which we call zero phase. Now we would like to knowthe intensity of the radiation in various directions. By the intensity we meanthe amount of energy that the field carries past us per second, which isproportional to the square of the field, averaged in time. So the thing to lookat, when we want to know how bright the light is, is the square of the electricfield, not the electric field itself. (The electric field tells the strength ofthe force felt by a stationary charge, but the amount of energy that is goingpast, in watts per square meter, is proportional to the square of the electricfield. We shall derive the constant of proportionality in Chapter 31.)图29-5（a），代表了这样两个振荡器的顶视图，在这个具体的例子中，在南北方向，它们差了半个波长，且是在同一个相位上震动，这个相位，被称为零相位。现在，我们想知道，在不同方向的辐射强度。通过强度，我们的意思是：每秒，场给我们传送的能量的数量，它是正比于场的平方，这个场，指时间上的平均值{？}。所以，当我们想知道，光有多强时，需要看的事情就是，电场的平方，而不是电场本身。（电场告诉的，是一个静电荷所感觉到的力的强度，但是，向外走的能量的数量，则是正比于电场的平方，单位是瓦特每平米。在31章，我们将推导出正比的常数。）

If we look at the array from the W side,both oscillators contribute equally and in phase, so the electric field istwice as strong as it would be from a single oscillator. Therefore theintensity is four times as strong as it would be if there were only oneoscillator. (The numbers in Fig. 29–5represent how strong the intensity would be in this case, compared with what itwould be if there were only a single oscillator of unit strength.) Now, ineither the N or S direction along the line of the oscillators, since they arehalf a wavelength apart, the effect of one oscillator turns out to be out ofphase by exactly half an oscillation from the other, and therefore the fieldsadd to zero. At a certain particular intermediate angle (in fact, at 30∘ ) the intensity is 2 , and it falls off, 4 , 2 , 0 , and soforth. We have to learn how to find these numbers at other angles. It is aquestion of adding two oscillations with different phases.
如果我们从西方，看这个阵列，那么，两个振荡器的贡献，是一样的，且是同相位的，所以，电场的强壮性，就是一个振荡器时的两倍。因此，强度就是：只有一个振荡器时的四倍。（图29-5中的数字，代表着，在这种情况下，与只有一个单位强度的振荡器相比，强度有多强。）现在，沿着振荡器的这条线，无论是在北方还是南方，由于它们相差半个波长，一个震荡器的效果，最终就会被另一个震荡的一半，抵消{？}，因此，场加起来，就是零。在某个具体的中间角度（事实上，是在30度），强度是2，强度是按4、2、0这种方式，在降低。我们必须学会，如何找到，在其他角度的这些数字。这个问题就是：两个振荡器，有不同的相位，把它们加起来，结果如何？

Let us quickly look at some other cases ofinterest. Suppose the oscillators are again one-half a wavelength apart, butthe phase α of one is set half a period behind the other in itsoscillation (Fig. 29–5b). Inthe W direction the intensity is now zero, because one oscillator is “pushing”when the other one is “pulling.” But in the N direction the signal from thenear one comes at a certain time, and that of the other comes half a periodlater. But the latter was originally half a period behind in timing, andtherefore it is now exactly in time with the first one, and so theintensity in this direction is 4 units. The intensity in the directionat 30∘ is still 2, as we can prove later.
有些其他情况，我们也感兴趣，让我们快速地过一下。假设振荡器，又是差半个波长，但是，一个震荡器的相位α，被设置为：晚于另外一个半个周期（图29-5b）。在西方，强度现在是零，因为，当一个振荡器在“推”时，另外一个在“拉”。但是，在北方，来自附近震荡器的信号，在某确定时间到达，而另外一个的信号，要晚半个周期。但是，后者最初，就是被设为晚半个周期，因此，现在它与第一个，正好合拍，所以，在这个方向的强度，就是单位强度的4倍。在30度方向的强度，仍是2，这一点，稍后可以证明。

Now we come to an interesting case which shows up apossibly useful feature. Let us remark that one of the reasons that phaserelations of oscillators are interesting is for beaming radio transmitters. Forinstance, if we build an antenna system and want to send a radio signal, say,to Hawaii, we set the antennas up as in Fig. 29–5(a)and we broadcast with our two antennas in phase, because Hawaii is to the westof us. Then we decide that tomorrow we are going to broadcast toward Alberta,Canada. Since that is north, not west, all we have to do is to reverse thephase of one of our antennas, and we can broadcast to the north. 现在，我们来到了一个有趣的情况，它指出了一种可能有用的特性。让我们说，振荡器之间，存在相位的关系，这非常有趣，原因之一，就是让无线电发射器，定向发射。例如，如果建造了一个天线系统，想把无线电信号，发射到夏威夷，那么，我们就可以把天线，按图29-5（a）所示，来设置，两个天线，处于同相位，这样，我们就可以广播了，因为夏威夷，在我们的西方。然后，我们决定第二天，将对加拿大的阿尔伯特广播，由于它在北方，而不是西方，所以，我们要做的，就是把其中一个天线的相位，反向，这样，我们就可以对北方广播了。

So we can build antenna systems withvarious arrangements. Ours is one of the simplest possible ones; we can makethem much more complicated, and by changing the phases in the various antennaswe can send the beams in various directions and send most of the power in thedirection in which we wish to transmit, without ever moving the antenna! Inboth of the preceding cases, however, while we are broadcasting toward Albertawe are wasting a lot of power on Easter Island, and it would be interesting toask whether it is possible to send it in only one direction. At firstsight we might think that with a pair of antennas of this nature the result isalways going to be symmetrical. So let us consider a case that comes outunsymmetrical, to show the possible variety.
所以，我们可以用不同的参数，来建造天线。我们的系统，就是可能的、最简单的系统之一；我们可以让它们，变得更加复杂，且通过改变不同天线中的相位，我们可以在不同的方向，发射光束，且在我们希望传送的方向上，发射最大的功率，甚至都不用移动天线！然而，在前面两种情况中，当我们对阿尔伯特广播时，我们对东边的岛，浪费了很大能量，所以，是否可以只向一个方向发射？就是一个有趣的问题。初看上去，我们可能会想，用一对这种本质的天线，结果可能总是对称的。所以，让我们考虑一种情况，其结果，是非对称的，以指出：可能的变化。

If the antennas are separated byone-quarter wavelength, and if the N one is one-fourth period behind the S onein time, then what happens (Fig. 29–6)? Inthe W direction we get 2 , as we will see later. In the S direction we getzero, because the signal from S comes at a certain time; that from Ncomes 90∘ later in time,but it is already 90∘ behind in itsbuilt-in phase, therefore it arrives, altogether, 180∘ out of phase, and there is no effect. On the other hand, in the Ndirection, the N signal arrives earlier than the S signal by 90∘ in time, because it is a quarter wavelength closer. But its phaseis set so that it is oscillating 90∘ behind in time, which just compensates the delay difference,and therefore the two signals appear together in phase, making the fieldstrength twice as large, and the energy four times as great.
如果天线被1/4波长分开，且如果在时间上，北方的天线，比南方的天线，落后1/4周期，那么，会发生什么呢（图29-6）？稍后我们将看到，在西方，我们到2。在南方，我们得到零，因为从南方来的信号，在一个确定的时间到达；而从北方来的信号，在时间上，晚90度，但是，由于其内置的相位，已经晚了90度，所以，当它到达时，相位差，正好是180，所以没有效果。另一方面，在北方，在时间上，北方的信号，比南方的信号，早90度，因为，这接近四分之一波长。但是，它的相位设置，让它的震荡，在时间上晚90度，这正好补偿了延迟差别，因此，这两个信号的表现，就是同相位的，这就使得场的强度，是两倍的大，能量是四倍的大。

Thus, by using some cleverness in spacingand phasing our antennas, we can send the power all in one direction. But stillit is distributed over a great range of angles. Can we arrange it so that it isfocused still more sharply in a particular direction? Let us consider the caseof Hawaii again, where we are sending the beam east and west but it is spreadover quite an angle, because even at 30∘ we are still getting half the intensity—we are wasting the power.Can we do better than that? Let us take a situation in which the separation isten wavelengths (Fig. 29–7),which is more nearly comparable to the situation in which we experimented inthe previous chapter, with separations of several wavelengths rather than asmall fraction of a wavelength. Here the picture is quite different.
这样，通过在空间、和在我们天线的相位上，施展我们的聪明才智，我们就可以在一个方向上，发送功率。但是，在大范围的角度上，信号还是有分布。我们能否安排，让信号在一个具体的方向上，聚焦地更尖锐呢？让我们再次考虑夏威夷的情况，在这里，我们向东方和西方，发送电波，但是，在相当大的角度上，电波还是有扩展，因为，即便是在30度，我们仍可以得到强度的一半--我们正在浪费能量。我们能做的比这更好吗？让我们取一种情况，在其中，分割是十个波长（图29-7），这与我们在上一章的实验情况，更有可比性，即所用分割，有好几个波长，而不是只有一个波长的一小部分。这里的图像，将完全不同。

Fig. 29–7.The intensity pattern for twodipoles separated by 10λ . 图29-7 两个偶极子，分割为10λ，它们的强度模式。

If the oscillators are ten wavelengthsapart (we take the in-phase case to make it easy), we see that in the E–Wdirection, they are in phase, and we get a strong intensity, four times what wewould get if one of them were there alone. On the other hand, at a very smallangle away, the arrival times differ by 180∘ and the intensity is zero. To be precise, if we draw a line fromeach oscillator to a distant point and the difference Δ in the twodistances is λ/2 , half an oscillation, then they will be out ofphase. So this first null occurs when that happens. (The figure is not drawn toscale; it is only a rough sketch.) This means that we do indeed have a verysharp beam in the direction we want, because if we just move over a little bitwe lose all our intensity. Unfortunately for practical purposes, if we were thinkingof making a radio broadcasting array and we doubled the distance Δ , thenwe would be a whole cycle out of phase, which is the same as being exactly inphase again! Thus we get many successive maxima and minima, just as we foundwith the 2（1/2）λ spacing in Chapter 28.
如果振荡器相差十个波长（我们取同相位，以让情况容易），我们看到，在东--西方向，它们是同相位的，于是，我们得到的强度，就很强，四倍于：只有它们中的一个在那里时，我们所能得到的强度。另一方面，稍微偏一个很小的角度，到达的时间，就会相差180度，于是，强度就为零。精确地说，如果从每个振荡器出发，到某距离外的一点，画一条线，那么，这两条线的差Δ ，就是λ/2，即震荡的一半，因此，它们就是异相。所以，当此事发生时，第一个零点，就出现了。（这个图，并不是按比例画的，只是一个草图。）这就意味着，在我们想要的方向上，我们确实有一个尖锐的电波，因为，如果我们只是稍微挪一点，我们就失去了所有的强度。对于实践的目的，不幸的是，如果我们想的做一个无线电广播阵列，且我们的加倍了距离Δ，那么，我们的相位差，就会是一个完整的循环，这就又与同相位，是一样的了。这样，我们就会得到很多相继的最大值和最小值，正如在28章，我们用2（1/2）λ所发现的那样。

Fig. 29–8.A six-dipole antenna array andpart of its intensity pattern. 图29-8 一个由六个偶极子所组成的天线阵列，及其强度模型的一部分。

The reason is that although we might expectto get a big bump when the distance Δ is exactly equal to the wavelength,it is true that dipoles 1 and 6 are then in phase and are cooperating intrying to get some strength in that direction. But numbers 3 and 4 areroughly 1/2 a wavelength out of phase with 1 and 6 , and although 1and 6 push together, 3 and 4 push together too, but in oppositephase. Therefore there is very little intensity in this direction—but there issomething; it does not balance exactly. This kind of thing keeps on happening;we get very little bumps, and we have the strong beam in the direction where wewant it. 原因就是，虽然我们可以期待：当距离Δ，准确地等于波长时，我们能得到一个大的隆起，且下面也为真，即偶极子1到6，这时是同相位的，它们相互合作，尝试在那个方向，得到某个强度。但是，3和4，与1和6，相位大约相差1/2波长，虽然，1和6，在一起推，3和4，也在一起推，但是，它们的相位相反。因此，在这个方向，只有很小的强度--但还是有点东西；并不是相互准确地抵消。这种事情，不断发生；我们可以得到很小的隆起，但在我们希望的方向上，有强的电波。But in this particular example, something else will happen: namely,since the distance between successive dipoles is 2λ , it ispossible to find an angle where the distance δ between successivedipoles is exactly one wavelength, so that the effects from all of them arein phase again. 但是，在这个具体的例子中，还发生了其他的事情：即由于两个连续的偶极子之间，距离相差是2λ，所以，就有可能找到一个角度，让连续的两个偶极子之间的距离δ，正好是一个波长，这样，它们的效果，就又都是同相位的了。Each one is delayed relative to the next one by 360∘ , so they all come back in phase, and we have another strong beamin that direction! It is easy to avoid this in practice because it is possibleto put the dipoles closer than one wavelength apart. If we put in moreantennas, closer than one wavelength apart, then this cannot happen. But thefact that this can happen at certain angles, if the spacing is biggerthan one wavelength, is a very interesting and useful phenomenon in otherapplications—not radio broadcasting, but in diffraction gratings.
每一个，相对于另一个，都被延迟360度，于是，它们就又都回到了同相位，在那个方向，我们就又有了另外一个强的电波！在实践中，要避免这一点，很容易，因为，让偶极子之间的距离，小于一个波长，是可能的。如果我们放更多的天线，且距离小于一个波长，那么，这就不可能发生。但是，如果间隔大于一个波长，那么，在某个角度，这是可能发生的；这一事实，在其他的应用中，是一个非常有趣、和有用的现象--不是在无线电广播中，而是在绕射光栅中。

29–5The mathematics of interference 29-5干涉的数学
Now we have finished our analysis of thephenomena of dipole radiators qualitatively, and we must learn how to analyzethem quantitatively. To find the effect of two sources at some particular anglein the most general case, where the two oscillators have some intrinsicrelative phase α from one another and the strengths A1and A2 are not equal, we find that we have to add twocosines having the same frequency, but with different phases. It is very easyto find this phase difference; it is made up of a delay due to the differencein distance, and the intrinsic, built-in phase of the oscillation.Mathematically, we have to find the sum R of two waves: R=A1cos(ωt+ϕ1)+A2cos(ωt+ϕ2). How do we do it?
现在,对于偶极子发射辐射器这种现象，我们已经完成了定性分析，我们还应该学习：如何定量地分析它们。最普遍的情况，就是两个源，各有其具体角度，亦即，两个振荡器之间，有固有的相关相位α，且强度A1和A2，并不相等，要找到它们的效果，我们发现，两个余弦函数，频率相同，相位不同，我们必须把这两个函数，加起来。要找到这个相位差，非常容易；构成它的延迟，可归于：距离的差别，和震荡器的固有的、和内置的相位。数学上，我们必须找到两个波的和R：R=A1cos(ωt+ϕ1)+A2cos(ωt+ϕ2)。这如何做到呢？