First, let us calculate the time requiredfor the light to go from B to E and back. Let us say that the time for light to go from plate Bto mirror E is t1 , and the time for the return is t2 . Now, while the light is on its way from B to the mirror, the apparatus moves a distance ut1, so the light must traverse a distance L+ut1, at the speed c . We can also express this distance as ct1 , so we have
ct1=L+ut1, or t1=L/(c−u).
(This result is also obvious from the point of view that the velocityof light relative to the apparatus is c−u , so the time is the length L divided by c−u .) In a like manner, the time t2 can be calculated. During this time the plate B advances a distance ut2 , so the return distance of the light is L−ut2. Then we have
ct2=L−ut2, or t2=L/(c+u).
Then the total time is
t1+t2=2Lc/(c2−u2).
For convenience in later comparison of times we write this as
t1+t2=2L/c/(1−u2/c2). (15.4)
首先，让我们计算光从B到E再返回来，所需要的时间。我们说，光从B到镜子E的时间是 t1，那么，光返回来的时间是t2。现在，当光在从B到镜子的路上时，仪器移动了一个距离ut1，于是，光应该走的距离就是 L+ut1，以速度c。我们也可以把这个距离，表示为ct1，于是我们就有：
ct1=L+ut1, 或 t1=L/(c−u).
（这个结果，很明显从如下观点出发的，即光相对于仪器的矢速，是c−u，于是，时间就是长度L除以 c−u。）以同样的方式，时间t2也可被计算。在这段时间中，板B前进了距离ut2，于是，光返回的距离就是L−ut2。因此我们就有：
ct2=L−ut2, 或 t2=L/(c+u).
因此总的时间就是：
t1+t2=2Lc/(c2−u2).
为了在以后比较时间时，能方便些，我们这样写：
t1+t2=2L/c/(1−u2/c2). (15.4)

Our second calculation will be of the time t3 for the light to go from B to the mirror C . As before, during time t3 the mirror C moves to the right a distance ut3 to the position C′ ; in the same time, the light travels a distance ct3along the hypotenuse of a triangle, which is BC′ . For this right triangle we have
(ct3)2=L2+(ut3)2
or
L2=c2t23−u2t23=(c2−u2)t23,
from which we get
For the return trip from C′ the distance is the same, as can be seen from the symmetry of thefigure; therefore the return time is also the same, and the total timeis 2t3 . With a little rearrangement of the form we can write
(15.5)
我们的第二个计算，就是光从B到镜子C所花的时间t3。正如以前，在时间t3中，镜子往右移动了距离ut3，到达位置C′；同时，光沿着三角形的斜边，移动了距离ct3，即BC′。对此直角三角形，我们有：
(ct3)2=L2+(ut3)2
或：
L2=c2t23−u2t23=(c2−u2)t23,
从它我们得到：

从图形的对称性，我们可以看到，从C′返回的距离，是同样的；因此，返回的时间也一样，故总时间就是2t3。稍微重排一下格式，我们可以写：
(15.5)

We are now able to compare the times taken by the two beams of light.In expressions (15.4)and (15.5)the numerators are identical, and represent the time that would be taken if theapparatus were at rest. In the denominators, the term u2/c2will be small, unless u is comparable in size to c . The denominators represent the modifications in the times caused bythe motion of the apparatus. And behold, these modifications are not thesame—the time to go to C and back is a little less than the time to E and back, even though the mirrors are equidistant from B , and all we have to do is to measure that difference with precision.
我们现在，就可以比较两束光所花的时间了。在表达式（15.4）和（15.5）中，分子是同等的，代表着仪器静止时，需要的时间。在分母中，项u2/c2很小，除非u的大小，可与c比。分母代表着：由仪器的运动所引起的时间的更改。看啊，这些更改，并不一样，去C并返回的时间，要比去E并返回的时间，稍小点儿，虽然镜子与B的距离是相等的，我们所要做的，就是精确地测量这个差异。

Here a minor technical point arises—supposethe two lengths L are not exactly equal? In fact, we surely cannot make them exactlyequal. In that case we simply turn the apparatus 90 degrees, so that BC is in the line of motion and BE is perpendicular to the motion. Any small difference in length thenbecomes unimportant, and what we look for is a shift in the interferencefringes when we rotate the apparatus.
这里，产生了一个小的技术问题--假设两个长度L，并不准确相等？事实上，我们也没有办法使它们准确相等。在这种情况下，我们只需把这个仪器，转90度就行，这样，BC就是在运动的线，而BE，则垂直于运动。长度方面，任何小的差异，因此就变得不重要了，当我们旋转仪器时，我们想要找的，只是干涉条纹的一个转变。

In carrying out the experiment, Michelsonand Morley oriented the apparatus so that the line BE was nearly parallel to the earth’s motion in its orbit (at certaintimes of the day and night). This orbital speed is about 18 miles per second, and any “ether drift” should be at least thatmuch at some time of the day or night and at some time during the year. Theapparatus was amply sensitive to observe such an effect, but no time differencewas found—the velocity of the earth through the ether could not be detected.The result of the experiment was null.
在执行这个实验时，迈克耳孙和莫雷，调整这个仪器的方向，使得线BE几乎平行于地球在其轨道中的运动方向（白天和夜里的某些时间）。这个轨道的速度，大约是18英里每秒，任何“以太飘动”，在白天和夜里的某些时间，和在一年中的某些时间上，应该至少是这么大。仪器是足够敏感，以能观察到这样一个效果，但是，没有发现时间差--地球通过以太的矢速，探测不到。实验的结果是零。

The result of the Michelson-Morley experimentwas very puzzling and most disturbing. The first fruitful idea for finding away out of the impasse came from Lorentz. He suggested that material bodiescontract when they are moving, and that this foreshortening is only in thedirection of the motion, and also, that if the length is L0when a body is at rest, then when it moves with speed u parallel to its length, the new length, which we call L∥ (L -parallel), is given by
(15.6)
迈克耳孙--莫雷实验的结果，让人迷惑，也令人非常不安。第一个打破僵局的方法，来自洛伦兹。他提议，当材料体移动时，它们会收缩，这种透视性的缩小{？}，只在运动的方向，另外，如果一个物体，静止时，其长度为 L0，那么，当它以速度u、平行于其长度，而移动时，则新的长度，我们称为L∥（L平行），通过下式给出：
(15.6)

When this modification is applied to the Michelson-Morleyinterferometer apparatus the distance from B to C does not change, but the distance from B to E is shortened to L（1−u2/c2）1/2 . Therefore Eq. (15.5)is not changed, but the L of Eq. (15.4)must be changed in accordance with Eq. (15.6).When this is done we obtain
(15.7)
当这个修改，被应用于迈克耳孙--莫雷实验干涉仪时，从B到 C的距离，并未改变，但是，从B到 E的距离，被缩短为 L（1−u2/c2）1/2。因此，方程（15.5）并没有变，但是，方程（15.4）的L，应该依据方程（15.6）而改变。当做了这个之后，我们得到：
(15.7)

Comparing this result with Eq. (15.5),we see that t1+t2=2t3. So if the apparatus shrinks in the manner just described, we have away of understanding why the Michelson-Morley experiment gives no effect atall. Although the contraction hypothesis successfully accounted for thenegative result of the experiment, it was open to the objection that it wasinvented for the express purpose of explaining away the difficulty, and was tooartificial.。However, in many otherexperiments to discover an ether wind, similar difficulties arose, until itappeared that nature was in a “conspiracy” to thwart man by introducing somenew phenomenon to undo every phenomenon that he thought would permit a measurementof u .
把这个结果，与方程（15.5）比较，我们看到t1+t2=2t3。于是，如果仪器是按照刚才所描述的方式而收缩，那么，我们就有了一种方式，可以理解，为什么迈克耳孙--莫雷实验，没有给出任何结果。实验结果，确实消极，虽然此收缩假设，成功地说明了这一点，但是，它也容易招来反对意见，即它的发明，是特意为了把困难巧辨过去，人为性太强然而，还有很多实验，是为了发现以太风，在这些实验中，产生了类似的困难，直到看上去，自然似乎是在一个“密谋”中，目的是要阻止人类，通过引入一些新的现象，来解开每一种原有现象{？}；这里的原有现象，是指人类认为，在其中，允许对u进行测量的那些现象。

It was ultimately recognized, as Poincarépointed out, that a complete conspiracy is itself a law of nature!Poincaré then proposed that there is such a law of nature, that it isnot possible to discover an ether wind by any experiment; that is, thereis no way to determine an absolute velocity.
最后，终于认识到、也正如庞加莱所指出：一个完整的密谋本身，就是一条自然规律。因此，庞加莱提议，有这样一条自然规律，即通过任何实验，来发现以太风，都是不可能的；也就是说，没有办法，可用来决定绝对矢速。

15–4Transformation of time 15-4 时间的变换
In checking out whether the contractionidea is in harmony with the facts in other experiments, it turns out thateverything is correct provided that the times are also modified, in themanner expressed in the fourth equation of the set (15.3).That is because the time 2t3 , calculated for the trip from B to C and back, is not the same when calculated by a man performing the experimentin a moving space ship as when calculated by a stationary observer who iswatching the space ship. To the man in the ship the time is simply 2L/c, but to the other observer it is (2L/c)/（1−u2/c2）1/2 (Eq. 15.5).In other words, when the outsider sees the man in the space ship lighting acigar, all the actions appear to be slower than normal, while to the maninside, everything moves at a normal rate. So not only must the lengthsshorten, but also the time-measuring instruments (“clocks”) must apparentlyslow down. That is, when the clock in the space ship records 1 second elapsed, as seen by the man in the ship, it shows 1/（1−u2/c2）1/2 second to the man outside.
收缩这种想法，是否与事实一致，为了检查这点，还做了其他一些实验，结果就是，假如时间，按照方程组（15.3）的第四个方式，被修改了，那么，所有事情，就都正确。这是因为，对于时间2t3，即从B到 C并返回的时间，当一个人，在移动的太空飞船中做此实验并计算这个时间，与另一个静止的观察者，在旁边观察这个太空飞船并计算这个时间，两个时间，并不一样。对于飞船里面的人，这个时间是2L/c，但对那个观察者，则是(2L/c)/（1−u2/c2）1/2（方程15.5）。换句话说，当外面的人，看到飞船里面的人点燃一支烟，所有的动作，都会显得比平常要慢，而对里面的人来说，所有事情，则都在正常进行。所以，不仅长度应该变短了，而且时间测量仪器（“钟表”），也应该明显变慢了。也就是说，当太空飞船中的钟表，记录下1秒过去了，飞船里的人看到的，也是如此，而外面的人看到的，则是 1/（1−u2/c2）1/2。

This slowing of the clocks in a movingsystem is a very peculiar phenomenon, and is worth an explanation. In order tounderstand this, we have to watch the machinery of the clock and see whathappens when it is moving. Since that is rather difficult, we shall take a verysimple kind of clock. The one we choose is rather a silly kind of clock, but itwill work in principle: it is a rod (meter stick) with a mirror at each end,and when we start a light signal between the mirrors, the light keeps going upand down, making a click every time it comes down, like a standard tickingclock. We build two such clocks, with exactly the same lengths, and synchronizethem by starting them together; then they agree always thereafter, because theyare the same in length, and light always travels with speed c . We give one of these clocks to the man to take along in his spaceship, and he mounts the rod perpendicular to the direction of motion of theship; then the length of the rod will not change. How do we know thatperpendicular lengths do not change? The men can agree to make marks on eachother’s y -meter stick as they pass each other. By symmetry, the two marks mustcome at the same y - and y′ -coordinates, since otherwise, when they get together to compareresults, one mark will be above or below the other, and so we could tell whowas really moving.
在一个移动的系统中，表会变慢，这个现象，非常奇特，值得解释。为解此事，必须查看，表的机制，看它移动时，发生了什么。由于这事，相当困难，所以，我们将举一个非常简单的表的例子。这个例子，是一个相当傻的表，但是，它会按原理工作：它是一个杆子（米尺），两端各有一个镜子，当我们在镜子之间，开始一个光信号时，光就会不断地上下走，每次下来时，产生一个咔哒，就像一个标准的时钟。这种表，我们建两个，长度完全一样，通过让它们同时启动，使其同步；然后，它们就总是一致的，因为，其长度一样，且光移动的速度总是c。我们把一个表，给飞船里的人，他安装这个表，让杆子垂直于飞船运动的方向；因此，杆子的长度，不会变化。我们如何知道，这个垂直的长度不会改变呢？这两个人，可以达成一致，当他们相错时，在他们的y方向的米尺上，做标记。通过对称，在y - 和y′坐标系中的这两个标记，应该一样，因为否则的话，当他们相聚，比较结果时，一个标记，就会在另一个的上面或下面，这样，我们就可以知道，谁真正在移动了。

Fig. 15–3.(a) A “light clock” at restin the S′ system. (b) The same clock, moving through the S system. (c) Illustration of the diagonal path taken by thelight beam in a moving “light clock.” 图15-3 （a）一个“光表”，在S′坐标系中，静止着。（b）同样的一个表，移动过S坐标系。（c）在一个移动的“光表”中光束，所走对角线路径的图示。

Now let us see what happens to the movingclock. Before the man took it aboard, he agreed that it was a nice, standardclock, and when he goes along in the space ship he will not see anythingpeculiar. If he did, he would know he was moving—if anything at all changedbecause of the motion, he could tell he was moving. But the principle ofrelativity says this is impossible in a uniformly moving system, so nothing haschanged. On the other hand, when the external observer looks at the clock goingby, he sees that the light, in going from mirror to mirror, is “really” takinga zigzag path, since the rod is moving sidewise all the while. We have alreadyanalyzed such a zigzag motion in connection with the Michelson-Morleyexperiment. If in a given time the rod moves forward a distance proportionalto u in Fig. 15–3, thedistance the light travels in the same time is proportional to c , and the vertical distance is therefore proportional to (c2−u2)1/2.
现在，让我们看看，在移动的表中，发生了什么。在此人把表带上船之前，他同意，这是一个好表，非常标准，当他在船内，随船同走时，他看不到任何奇特之处。如果他看到的话，他就将知道，他正在移动--如果因为运动，导致任何事情改变了，他就可以知道，他在运动。但相对论的原理说，在一个匀速运动的系统中，这是不可能的，所以，什么都未改变。另一方面，当那个外部的观察者，看到这个走过的表时，他会看到，光在镜子之间，来回运动，所走路径，“确实”为锯齿型，因为杆子一直是向一侧运动。对于这种锯齿型运动，根据它与迈克耳孙--莫雷实验的联系，我们已经对它，做过分析。如果在给定时间，杆子向前移动的距离，正比于u，如图15-3，那么，光在同样时间内所走过的距离，就正比于c, 因此，垂直距离，就正比于(c2−u2)1/2。

That is, it takes a longer time forlight to go from end to end in the moving clock than in the stationary clock.Therefore the apparent time between clicks is longer for the moving clock, inthe same proportion as shown in the hypotenuse of the triangle (that is thesource of the square root expressions in our equations). From the figure it isalso apparent that the greater u is, the more slowly the moving clock appears to run. Not only doesthis particular kind of clock run more slowly, but if the theory of relativityis correct, any other clock, operating on any principle whatsoever, would alsoappear to run slower, and in the same proportion—we can say this withoutfurther analysis. Why is this so?
也就是说，光从一端，走到另一端，在移动中的表上所花的时间，比在静止的表上花的，要长。因此，在表的咔哒之间的明显的时间，对于移动的表来说，要更长些，与所示的三角形的斜边，成同样的正比（那就是我们方程中的表达式的平方根的源头。）从这个图形，也明显可见，u越大，那么，这个移动中的表，就表现出，走得越慢。不仅这种具体的表，会走的更慢，而且，如果相对论的理论，正确的话，那么，任何种类的表，无论它依据什么原理而运行，也将表现出，走的比较慢，且是以同样的比例--无须任何进一步的分析，我们就能这样说。为何如此呢？

To answer the above question, suppose wehad two other clocks made exactly alike with wheels and gears, or perhaps basedon radioactive decay, or something else. Then we adjust these clocks so theyboth run in precise synchronism with our first clocks. When light goes up andback in the first clocks and announces its arrival with a click, the new modelsalso complete some sort of cycle, which they simultaneously announce by some doublycoincident flash, or bong, or other signal. One of these clocks is taken intothe space ship, along with the first kind. Perhaps this clock will notrun slower, but will continue to keep the same time as its stationarycounterpart, and thus disagree with the other moving clock. Ah no, if that shouldhappen, the man in the ship could use this mismatch between his two clocks todetermine the speed of his ship, which we have been supposing is impossible. Weneed not know anything about the machinery of the new clock that might causethe effect—we simply know that whatever the reason, it will appear to run slow,just like the first one.
要回答上述问题，假设我们有两个其他的表，完全是由轮子和齿轮做成，或者，基于辐射衰减、或其他什么。然后，我们调整这些表，让它们与第一批表，精确同步。当光在第一个表中，上去下来，然后，用一个卡塔，宣布其到来时，新的模式，也会完成某种循环，也会通过某种双倍同时发生的闪烁、或嘡嘡声、或其他信号，同时宣布此事。这些表中的一个，被带入太空飞船，与第一类表在一起。或许，这个表不会走的更慢，而是像它的静止同伴一样，继续保持同样的时间，这样，就与另外移动中的表，不一致了。啊，不会的，如果这种事发生了，那么，船里的人，就可以用他的两个表的这个不一致，来得到船的速度，而这，我们认为是不可能的。对于可能引起此结果的新表，我们无需知道其任何机制--我们只要知道，无论原因为何，它都会表现出，走得更慢，正如第一种表一样。