16–3Transformation of velocities 16-3 矢速的變換
The main difference between the relativityof Einstein and the relativity of Newton is that the laws of transformationconnecting the coordinates and times between relatively moving systems aredifferent. The correct transformation law, that of Lorentz, is
(16.1)
These equations correspond to the relatively simple case in which therelative motion of the two observers is along their common x -axes. Of course other directions of motion are possible, but the mostgeneral Lorentz transformation is rather complicated, with all four quantitiesmixed up together. We shall continue to use this simpler form, since itcontains all the essential features of relativity.
有愛因斯坦的相對論，有牛頓的相對論，兩者之間的主要區別，就在於，把相關移動系統的坐標與時間，聯繫在一起的變換規律，是不同的。正確的變換規律，即洛倫茲的變換規律，就是：
(16.1)
這些方程，相應於相關的簡單情況，在其中，兩個觀察者的相關運動，就是沿著它們共同的x軸。當然，其他運動的方向，也是可能的，但是，最普遍的洛倫茲變換，所有四個量混在一起，相當複雜。我們將繼續使用這個簡單的形式，由於它包含了相對論的基本特性。
Let us now discuss more of the consequencesof this transformation. First, it is interesting to solve these equations inreverse. That is, here is a set of linear equations, four equations with fourunknowns, and they can be solved in reverse, for x,y,z,tin terms of x′,y′,z′,t′ . The result is very interesting, since it tells us how a system ofcoordinates 「at rest」 looks from the point of view of one that is 「moving.」 Ofcourse, since the motions are relative and of uniform velocity, the man who is「moving」 can say, if he wishes, that it is really the other fellow who ismoving and he himself who is at rest. And since he is moving in the oppositedirection, he should get the same transformation, but with the opposite sign ofvelocity. That is precisely what we find by manipulation, so that is consistent.If it did not come out that way, we would have real cause to worry!
(16.2)
這個變換，還有更多後果，現在，讓我們討論之。首先，能夠反向地解決這些方程，非常有趣。也就是說，這裡是一組線性方程，4個方程，有4個未知數，它們可被反向求解，即用x′,y′,z′,t′，來表示x,y,z,t 。此結果非常有趣，由於它告訴我們，從一個「運動中的」的坐標系，去觀察一個「處於靜止」的坐標系，是何結果。當然，由於運動是相對的，且是勻速的，那麼，那個在「運動中的」人，如果他希望的話，就可以說，其實另一個傢伙是在運動，而他自己是靜止的。由於他是在相反的方向運動，他將會得到同樣的變換，但矢速的符號不同。我們通過操作所發現的，正是這樣。如果結果不是這樣，那麼，我們擔心，就有真正的理由了。
(16.2)
Next we discuss the interesting problem of the addition of velocitiesin relativity. We recall that one of the original puzzles was that light travelsat 186,000 mi/sec in all systems, even when they are in relative motion. Thisis a special case of the more general problem exemplified by the following. Supposethat an object inside a space ship is going at 100,000 mi/sec and the space ship itself is going at 100,000 mi/sec; how fast is the object inside the space ship moving fromthe point of view of an observer outside? We might want to say 200,000 mi/sec, which is faster than the speed of light. This is veryunnerving, because it is not supposed to be going faster than the speed of light!The general problem is as follows.
在相對論中，矢速的增加，是一個非常有趣的問題，下面我們討論之。我們回憶，最初的困惑之一，就是在所有系統中，光都是以186,000 mi/sec 在走，即便當它們是在相對的運動中時，也是如此。這就是更普遍的問題中的一種特殊情況，舉例如下。假設在太空飛船中，有一個對象，以100,000 mi/sec在走，而太空飛船本身，也是以100,000 mi/sec在走；那麼，從一個外部的觀察者來看，此飛船內的對象，其速為何？我們或許可以說，200,000 mi/sec，比光速快。這讓人緊張不安，因為，我們並不認為，能夠超過光速！一般問題如下。
Let us suppose that the object inside theship, from the point of view of the man inside, is moving with velocity v, and that the space ship itself has a velocity u with respect to the ground. We want to know with what velocity vxthis object is moving from the point of view of the man on the ground.This is, of course, still but a special case in which the motion is in the x-direction. There will also be a transformation for velocities in the y-direction, or for any angle; these can be worked out as needed. Insidethe space ship the velocity is vx′ , which means that the displacement x′ is equal to the velocity times the time:
x′=vx′t′. (16.3)
我們假設，從船內人的觀點看，飛船內的對象，正在以矢速v運動，而太空飛船本身，相對於地面的的矢速為u。我們想知道，從地面上的人的觀點來看，這個對象，正在以什麼樣的矢速vx在運動？當然，這仍是特例，運動只在x方向。對於有方向的矢速、或任何角度，也可以有變換；如果需要，也可做到。在飛船內，矢速為vx′，其意為，位移x′，等於矢速乘以時間：
x′=vx′t′. (16.3)
Now we have only to calculate what the position and time are from thepoint of view of the outside observer for an object which has therelation (16.2)between x′ and t′ . So we simply substitute (16.3)into (16.2),and obtain
(16.4)
But here we find x expressed in terms of t′ . In order to get the velocity as seen by the man on the outside, wemust divide his distance by his time, not by the other man’stime! So we must also calculate the time as seen from the outside,which is
(16.5)
Now we must find the ratio of x to t , which is
(16.6)
the square roots having cancelled. This is the law that we seek: theresultant velocity, the 「summing」 of two velocities, is not just the algebraicsum of two velocities (we know that it cannot be or we get in trouble), but is「corrected」 by 1+uv/c2 .
對於一個對象，從x′ 到 t′，它有關係（16.2），現在，我們要計算的，就是從外面的觀察者來看，該對象的位置和時間。所以，我們只需把（16.3）代入（16.2），得到：
(16.4)
但是，這裡我們發現，x被用 t′表達了。為了得到，外面的人所看到的矢速，我們應該讓他的距離，除以他的時間，而不是另一個人的時間。所以，我們應該計算，外面的人所看到的時間，就是：
(16.5)
現在，我們應該找到x和t的比率，就是：
(16.6)
平方根被消掉了。這就是我們要找的規律，合成的矢速，即兩個矢速的「綜合」，它並不僅僅是兩個矢速的算術和，（我們知道，它不可能是算術和，否則我們就是遇到麻煩了），而是被1+uv/c2，給修正了。
Now let us see what happens. Suppose thatyou are moving inside the space ship at half the speed of light, and that thespace ship itself is going at half the speed of light. Thus u is c/2 and v is c/2 , but in the denominator uv/c2 is one-fourth, so that
So, in relativity, 「half」 and 「half」 does not make 「one,」 it makesonly 「4/5 .」 Of course low velocities can be added quite easily in the familiarway, because so long as the velocities are small compared with the speed oflight we can forget about the (1+uv/c2) factor; but things are quite different and quite interesting athigh velocity.
現在讓我們看看，發生了什麼。假設你在太空飛船中，以光速的一半，在運動，且飛船本身，也是以光速的一半在走。這樣，u、v 皆為c/2，但是，在分母中，uv/c2是四分之一，於是：
所以，在相對論中，「一半」加「一半」，得到的不是「一」，而是「4/5」。當然，低的矢速，以同樣的方式，很容易被加上，因為，只要矢速與光速相比，較小，我們就可以忽略因子(1+uv/c2)；當矢速較高時，情況就完全不同了。
Let us take a limiting case. Just for fun,suppose that inside the space ship the man was observing light itself.In other words, v=c , and yet the space ship is moving. How will it look to the man on theground? The answer will be
Therefore, if something is moving at the speed of light inside the ship,it will appear to be moving at the speed of light from the point of view of theman on the ground too! This is good, for it is, in fact, what the Einsteintheory of relativity was designed to do in the first place—so it had betterwork!
我們現在取一個限制性的案例。只是為了有趣，假設在太空飛船中，那個人在觀察光本身。換句話說，v=c，而飛船也仍在運動。那麼，對於地上的人，這看上去是什麼樣子呢？答案就是：
因此，如果某物在飛船內以光速運動，那麼，從地面上的人的觀點看，他也是在以光的速度運動！這很好，因為，事實上，愛因斯坦的相對論，最初被設計出來，就是為了做此事的—所以，它最好有效。
Of course, there are cases in which themotion is not in the direction of the uniform translation. For example, theremay be an object inside the ship which is just moving 「upward」 with thevelocity vy′ with respect to the ship, and the ship is moving 「horizontally.」 Now,we simply go through the same thing, only using y ’s instead of x ’s, with the result
y=y′=vy′t′,
so that if vx′=0 ,
(16.7)
Thus a sidewise velocity is no longer vy′, but vy′(1−u2/c2)1/2. We found this result by substituting and combining thetransformation equations, but we can also see the result directly from theprinciple of relativity for the following reason (it is always good to lookagain to see whether we can see the reason). We have already (Fig. 15–3) seen how a possible clock might work when it ismoving; the light appears to travel at an angle at the speed c in the fixed system, while it simply goes vertically with the samespeed in the moving system. We found that the vertical component of thevelocity in the fixed system is less than that of light by the factor (1−u2/c2)1/2(see Eq. 15.3). But now suppose that we let a material particlego back and forth in this same 「clock,」 but at some integral fraction 1/nof the speed of light (Fig. 16–1).Then when the particle has gone back and forth once, the light will have goneexactly n times. That is, each 「click」 of the 「particle」 clock willcoincide with each n th 「click」 of the light clock. This fact must still be truewhen the whole system is moving, because the physical phenomenon ofcoincidence will be a coincidence in any frame. Therefore, since thespeed cy is less than the speed of light, the speed vy of the particle must be slower than the corresponding speed by thesame square-root ratio! That is why the square root appears in any verticalvelocity.
當然，還有一些案例，在其中，運動並不是在均勻變換的方向。例如，在飛船內，可能會有一個對象，正在以相對於飛船的矢速vy′，向上運動，而飛船則在「水平地」運動。現在，我們只要簡單地做同樣的事情，用y的東西，來代替x的東西，結果就是：
y=y′=vy′t′,
於是，如果vx′=0 ，那麼：
(16.7)
這樣，向旁側的矢速，就不再是vy′, 而是vy′(1−u2/c2)1/2了。我們通過替換和混合變換方程，發現這個結果，然而，我們也可以因為下面的理由（再去看看，我們能否看到原因，總是好的），從相對論的原理出發，而直接地發現這個結果。從圖15-3，我們已經看到，一個可能的表，當其運動時，是如何工作的；光在一個固定的系統中，顯現出，是以一個角度，以速度c，在運動，而在運動的系統中，光以同樣的速度，垂直地走。我們發現，在固定系統中，矢速的垂直分量，要比光的垂直分量，小約 (1−u2/c2)1/2（見方程15.3）。但是現在，假設在這同一個「表」中，我們讓一個材料粒子，以光速的1/n--整數分數，來回地走（圖16-1）。那麼，當粒子來回走一次時，光就正好走n次。也就是說，粒子表的每一次「咔噠」，與光表的n次「咔噠」，是一致的。當整個系統都在移動時，這一事實，應該仍是正確的，因為，一致性這種物理現象，在任何框架中，都是一致的。因此，由於速度cy，比光速小，那麼，粒子的速度vy，就應該比相應的速度，小約同樣的平方根比率！這就是為什麼，在任何垂直的矢速中，都有平方根出現。
Fig. 16–1.Trajectories described by a light ray andparticle inside a moving clock. 圖16-1 在一個運動中的表中 ，一束光線和一個粒子的軌跡。