There are fortune tellers, or people whotell us they can know the future, and there are many wonderful stories aboutthe man who suddenly discovers that he has knowledge about the affectivefuture. Well, there are lots of paradoxes produced by that because if we knowsomething is going to happen, then we can make sure we will avoid it by doingthe right thing at the right time, and so on. But actually there is no fortuneteller who can even tell us the present! There is no one who can tell uswhat is really happening right now, at any reasonable distance, because that isunobservable. We might ask ourselves this question, which we leave to thestudent to try to answer: Would any paradox be produced if it were suddenly tobecome possible to know things that are in the space-like intervals ofregion 1 ?
有所谓的预言家，就是可以告诉们未来的人；还有很多奇妙的故事，是关于某人，突然发现，他对于感情的未来，有了相关的知识。嗯，由此，还产生了很多悖论，因为，如果我们知道，某事即将发生，那么，我们就可以通过在正确的时间，做正确的事情等，来肯定地避免它。但是，就连能告诉我们当前事情的预言家，都没有！没有任何人可以告诉我们：在任何合理的距离，现在真正地在发生着什么；因为，观察不到。下面这个问题，我们可以问我们自己，但我们将把它留给学生去回答：区间1，是类空间隔，如果此区间的事物，突然变得可能知道了，那么，何种悖论，会产生呢？

17–4More about four-vectors 17-4 更多关于四个矢量的事东西
Let us now return to our consideration ofthe analogy of the Lorentz transformation and rotations of the space axes. Wehave learned the utility of collecting together other quantities which have thesame transformation properties as the coordinates, to form what we call vectors,directed lines. In the case of ordinary rotations, there are many quantitiesthat transform the same way as x , y , and z under rotation: for example, the velocity has three components, an x, y , and z -component; when seen in a different coordinate system, none of the componentsis the same, instead they are all transformed to new values. But, somehow orother, the velocity “itself” has a greater reality than do any of its particularcomponents, and we represent it by a directed line.
现在，让我们返回来，重新考虑，洛伦兹变换，与空间坐标轴旋转的类比。有些量，作为坐标，有相同的变换属性，我们已经学习了实用方法，用来把这些量，收集起来，以形成我们称为矢量的东西，即有方向的线。在通常旋转这种情况下，有很多量，在旋转时，与x、y、和z的变换方式相同，例如，矢速就有三个分量，x、y、和z分量；当从一个不同的坐标系中，来看这些分量时，没有一个是相同的，它们都转换成了新的值。但是，由于某种原因，矢速“本身”，比其任何具体的分量，都有更大的现实性，我们用一条有方向的线，来代表它。

We therefore ask: Is it or is it not truethat there are quantities which transform, or which are related, in a movingsystem and in a nonmoving system, in the same way as x , y , z , and t ? From our experience with vectors, we know that three of thequantities, like x , y , z , would constitute the three components of an ordinary space-vector,but the fourth quantity would look like an ordinary scalar under space rotation,because it does not change so long as we do not go into a moving coordinatesystem. Is it possible, then, to associate with some of our known“three-vectors” a fourth object, that we could call the “time component,” insuch a manner that the four objects together would “rotate” the same way asposition and time in space-time? We shall now show that there is, indeed, atleast one such thing (there are many of them, in fact): the three componentsof momentum, and the energy as the time component, transform together tomake what we call a “four-vector.” In demonstrating this, since it is quiteinconvenient to have to write c ’s everywhere, we shall use the same trick concerning units of the energy,the mass, and the momentum, that we used in Eq. (17.4).Energy and mass, for example, differ only by a factor c2which is merely a question of units, so we can say energy isthe mass. Instead of having to write the c2 , we put E=m , and then, of course, if there were any trouble we would put in theright amounts of c so that the units would straighten out in the last equation, but notin the intermediate ones.
因此，我们就要问，是否有这种量，当它们变换时（或相关时），无论是在一个移动的系统中，还是在一个非移动的系统中，都是以同样的方式，即作为x , y , z , 和t ?根据我们关于矢量的经验，我们知道，这些量中的三个，比如x , y , z ，将会组成一个普通空间矢量的三个分量，但是，在空间的旋转中，第四个量看上去，会像一个普通的标量，因为，只要我们并不是要变进一个移动的坐标系中，那么，它就不变。因此，对于我们已知的一些“三维矢量”{three-vector}，把第四个对象，与其联合，是否可能？在这种方式下，这四个对象，就会一起，作为空间-时间中的位置和时间，而旋转。我们现在，将要指出，确实，至少有一个这种事情（事实上，还有很多）：动量的三个分量，和作为时间分量的能量，一起变换，以形成我们称为“四个矢量”的东西。我们在方程（17.4）中使用的技巧，牵扯到能量、质量、和动量的单位；现在，为了验证这一点，由于到处写c等，很不方便，所以，我们将使用同样的技巧。例如，能量与质量，只差一个因子c2，这只是一个单位的问题，于是，我们就可以说，能量就是质量。我们不用写c2，而是写E=m，因此，当然，如果出现麻烦，我们就会放入c的正确的量，这样，在最后的方程中，单位会出来，但在中间的方程中不会。

Thus our equations for energy and momentumare
(17.6)
Also in these units, we have
E2 − p2=m02. (17.7)
For example, if we measure energy in electron volts, what does a massof 1 electron volt mean? It means the mass whose rest energy is 1 electron volt, that is, m0c2is one electron volt. For example, the rest mass of an electron is0.511×106 eV.
这样，我们的关于能量和动量的方程就是：
(17.6)
另外，在这些单位中，我们有：
E2 − p2=m02. (17.7)
例如，如果我们用电子伏特，来测量能量，那么，一个电子伏特的质量，意味着什么呢？它意味着，静止能量为一个电子伏特的东西的质量，也就是说，m0c2就是一个电子伏特。例如，一个电子的静止质量，就是0.511×106 eV。

Now what would the momentum and energy looklike in a new coordinate system? To find out, we shall have to transformEq. (17.6),which we can do because we know how the velocity transforms. Suppose that, aswe measure it, an object has a velocity v , but we look upon the same object from the point of view of a spaceship which itself is moving with a velocity u , and in that system we use a prime to designate the corresponding thing.In order to simplify things at first, we shall take the case that thevelocity v is in the direction of u . (Later, we can do the more general case.) What is v′ , the velocity as seen from the space ship? It is the compositevelocity, the “difference” between v and u . By the law which we worked out before,
v′=(v−u) / (1−uv). (17.8)
Now let us calculate the new energy E′ , the energy as the fellow in the space ship would see it. He woulduse the same rest mass, of course, but he would use v′ for the velocity. What we have to do is square v′ , subtract it from one, take the square root, and take the reciprocal:
Therefore
(17.9)
现在，在新的坐标系中，动量和能量，看上去是什么样呢？要找出这个，我们就要变换方程（17.6），这我们能做，因为我们知道，如何变换矢速。假设对于一个对象，我们测其矢速为v，现在，我们从一个太空飞船上，看此同一个对象，飞船本身，矢速为u，在那个系统中，我们用一个主要的（prime）{？}，来决定相关事宜。为了简化事情，首先，我们将取这种情况，即矢速v，是在u的方向上。（更普遍的情况，稍后我们将做）。从太空飞船上所看到的v′是什么呢？它就是复合矢速，即v和 u之间的“差别”。通过我们以前得到的规律：
v′=(v−u) / (1−uv). (17.8)
现在，让我们计算新的能量E′，即太空飞船中的伙计所看到的。当然，他会使用同样的静止质量，但是，对于矢速，他会用v′。我们要做的，是平方v′，让一减之，取平方根，再取倒数：

因此：
(17.9)

The energy E′ is then simply m0 times the above expression. But we want to express the energy interms of the unprimed energy and momentum, and we note that
or
(17.10)
which we recognize as being exactly of the same form as
Next we must find the new momentum p′x . This is just the energy E′ times v′ , and is also simply expressed in terms of E and p :
Thus
(17.11)
which we recognize as being of precisely the same form as
因此，能量 E′ 简单就是m0 乘以上面的表达式。但是，我们想用非主要的能量和动量，来表示能量，我们注意到：

或
(17.10)
我们认为，此式与下式相同：

下面，我们应该找到新的动量 p′x。这就只是能量E′ 乘以 v′ ，用E和 p 来简单地表示，就是:

这样：
(17.11)
我们认为，此式与下式，精确相同：

Thus the transformations for the new energy and momentum in terms ofthe old energy and momentum are exactly the same as the transformationsfor t′ in terms of t and x , and x′ in terms of x and t : all we have to do is, every time we see t in (17.4)substitute E , and every time we see x substitute px , and then the equations (17.4)will become the same as Eqs. (17.10)and (17.11).This would imply, if everything works right, an additional rule that p′y=pyand that p′z=pz . To prove this would require our going back and studying the case ofmotion up and down. Actually, we did study the case of motion up and down inthe last chapter. We analyzed a complicated collision and we noticed that, infact, the transverse momentum is not changed when viewed from a movingsystem; so we have already verified that p′y=pyand p′z=pz . The complete transformation, then, is
(17.12)
这样，这个变换，就是用旧的能量和动量，来描述新的能量和动量，该变换，与用t和x来描述 t′、及用x和t来描述x′ 这两个变换，完全一样：我们所要做的一切，就是把方程组（17.4）中的t，都替换成E，x替换成px，然后，方程组（17.4）就会变得与方程（17.10）和（17.11）一样。如果一切顺利，这就意味着，一条附加的规则，即 p′y=py和 p′z=pz。要证明这点，就要求我们返回去研究向上运动和向下运动那种情况。实际上，在上一章，我们确实研究了此情况。我们分析过一个复杂的碰撞，且事实上，我们注意到，当我们从一个运动中的系统观看时，横向的动量，并未改变；所以，我们已经证实了 p′y=py 和 p′z=pz。完整的变换，因此就是：
(17.12)

In these transformations, therefore, we have discovered four quantitieswhich transform like x , y , z , and t , and which we call the four-vector momentum. Since themomentum is a four-vector, it can be represented on a space-time diagram of amoving particle as an “arrow” tangent to the path, as shown in Fig. 17–4. Thisarrow has a time component equal to the energy, and its space componentsrepresent its three-vector momentum; this arrow is more “real” than either theenergy or the momentum, because those just depend on how we look at thediagram.
因此，在这些变换中，我们已经发现了四个量，其变换，类似x , y , z , 和t，我们称其为四个矢量动量。由于此动量是一组四个矢量，那么，它就可以这样被代表：在一个运动中的粒子的空间-时间示意图上，作为一个与路径相切的“箭头”，如图17-4。对于这个箭头，其时间分量，等于能量，其空间分量，代表着它的三个矢量动量；这个箭头，无论是比能量还是比动量，都要更加“真实”，因为能量和动量，只依赖于我们如何观察示意图。

Fig. 17–4.The four-vector momentum of aparticle. 图17-4 一个粒子的四矢量动量。

17–5Four-vector algebra 17-5 四个矢量代数
The notation for four-vectors is differentthan it is for three-vectors. In the case of three-vectors, if we were to talkabout the ordinary three-vector momentum we would write it p. If we wanted to be more specific, we could say it has threecomponents which are, for the axes in question, px, py , and pz , or we could simply refer to a general component as pi, and say that i could either be x , y , or z , and that these are the three components; that is, imagine that iis any one of three directions, x , y , or z . The notation that we use for four-vectors is analogous to this: wewrite pμ for the four-vector, and μ stands for the four possible directions t , x , y , or z .
四个矢量的表示法，与三个矢量的不同。在三个矢量的情况下，如果我们要谈论普通的三个矢量的动量，我们就会把它写成p。如果我们想更具体些，我们可以说，它有分量，这些分量对于问题中的轴来，就是px, py ,和 pz，或者，我们可以用pi，来做更普遍的表达，即i 可以是x , y , 或 z ,它们就是三个分量；也就是说，想象i，是三个方向x , y , 或 z中的任何一个。我们关于四个分量所用的表示法，可与此类比：我们用pμ，代表四个矢量，μ代表着着四个可能的方向t , x , y , 或 z 。

We could, of course, use any notation wewant; do not laugh at notations; invent them, they are powerful. In fact,mathematics is, to a large extent, invention of better notations. The wholeidea of a four-vector, in fact, is an improvement in notation so that the transformationscan be remembered easily. Aμ , then, is a general four-vector, but for the special case ofmomentum, the pt is identified as the energy, px is the momentum in the x -direction, py is that in the y -direction, and pz is that in the z -direction. To add four-vectors, we add the corresponding components.
当然，我们可以使用我们想用的任何表示法；不要嘲笑表示法；发明它们，它们非常有力。事实上，数学，在很大意义上，就是一种更好的表示法的发明，整个四个矢量的想法，事实上，也就是对表示法的一种改进，这样，记住变换，就很容易。因此，Aμ就是一个普遍化的四个矢量，但是，对于动量的特殊情况，pt被认为是能量，px是x方向的动量，py是y方向的动量，pz是z方向的动量。要增加四个矢量，我们只要增加相应方向的分量就行。

If there is an equation among four-vectors,then the equation is true for each component. For instance, if the lawof conservation of three-vector momentum is to be true in particle collisions,i.e., if the sum of the momenta for a large number of interacting or collidingparticles is to be a constant, that must mean that the sums of all momenta inthe x -direction, in the y -direction, and in the z -direction, for all the particles, must each be constant. This lawalone would be impossible in relativity because it is incomplete; it islike talking about only two of the components of a three-vector. It isincomplete because if we rotate the axes, we mix the various components, so wemust include all three components in our law. Thus, in relativity, we mustcomplete the law of conservation of momentum by extending it to include the timecomponent. This is absolutely necessary to go with the other three, orthere cannot be relativistic invariance. The conservation of energy is thefourth equation which goes with the conservation of momentum to make a validfour-vector relationship in the geometry of space and time. Thus the law ofconservation of energy and momentum in four-dimensional notation is
(17.13)
or, in a slightly different notation
(17.14)
where i=1 , 2 , … refers to the particles going into the collision, j=1 , 2 , … refers to the particles coming out of the collision, and μ=x, y , z , or t . You say, “In which axes?” It makes no difference. The law is truefor each component, using any axes.
如果有一个方程，是关于四个矢量的，那么，对于每一个分量，此方程皆为真。例如，如果在粒子碰撞中，三个矢量的动量守恒规律，为真，也就是说，对于大数量粒子的交互作用或碰撞来说，如果其动量的总和，将是一个常数，那么，就必然意味着，对于所有粒子来说，所有x方向的动量，应是一个常数，y和z方向亦然。在相对论中，只有这个规律是不可能的，因为它是不完整的{？意思}；就好像是在谈论三个矢量分量中的两个。它是不完整的，因为，如果我们旋转坐标轴，我们就混合了不同的分量，所以，在我们的规律中，就应该包含所有三个分量。这样，在相对论中，对于动量守恒规律，我们应该通过扩展它，让它包含时间分量，已使它变得完整。这对其他三个矢量，也是绝对必要的，或者，不能有相对论的不变性{？}。能量守恒，是第四个方程，它与动量守恒一起，在空间和时间的几何中，形成了有效的四个矢量的关系。这样，能量与动量的守恒规律，在四个维度的表示法中，就是：
(17.13)
或者，用一种略微不同的表示法：
(17.14)
这里，i=1 , 2 , …是指进入碰撞的粒子，j=1 , 2 , …是指从碰撞中出来的粒子，μ=x , y , z , 或 t 。你说：“在哪个坐标轴呢？”这没有什么区别。对每一个分量，无论使用哪个坐标轴，此规律都为真。

In vector analysis we discussed one otherthing, the dot product of two vectors. Let us now consider the correspondingthing in space-time. In ordinary rotation we discovered there was an unchangedquantity x2+y2+z2 . In four dimensions, we find that the corresponding quantity is t2−x2−y2−z2(Eq. 17.3).How can we write that? One way would be to write some kind of four-dimensionalthing with a square dot between, like Aμ⊡Bμ ; one of the notations which is actually used is
(17.15)
The prime on ∑ means that the first term, the “time” term, is positive, but the otherthree terms have minus signs. This quantity, then, will be the same in anycoordinate system, and we may call it the square of the length of thefour-vector. For instance, what is the square of the length of the four-vectormomentum of a single particle? This will be equal to p2t−p2x−p2y−p2zor, in other words, E2−p2 , because we know that p t is E . What is E2−p2 ? It must be something which is the same in every coordinate system.In particular, it must be the same for a coordinate system which is movingright along with the particle, in which the particle is standing still. If theparticle is standing still, it would have no momentum. So in that coordinatesystem, it is purely its energy, which is the same as its rest mass. Thus E2−p2=m20. So we see that the square of the length of this vector, thefour-vector momentum, is equal to m20.
在矢量分析中，我们讨论了另外一件事情：两个矢量的点积。现在，让我们在空间-时间中，考虑相应的事物。在通常的旋转中，我们发现了一个不变的量x2+y2+z2。在四维中，我们发现，相应的量就是t2−x2−y2−z2（方程17.3）。我们如何写它呢？一种方式，就是写某种四维的东西，在它们之间，用一个平方点积{？}，比如 Aμ⊡Bμ；实际上被使用的一种表示法，就是：
(17.15)
∑上面的撇号{？prime}，意味着第一项、即时间项，是正的，其它三项，有负号。因此，这个量，在任何坐标系中，都是一样，我们可称其为：四个矢量的长度的平方。例如，对于一个粒子的动量，其四个矢量的长度的平方是什么呢？这个就等于 p2t−p2x−p2y−p2z，或者，换句话说，E2−p2，因为我们知道p t就是 E。什么是E2−p2呢？它应该是这样一个东西：在每个坐标系中，都是一样。尤其是，对于一个与粒子一起向右运动的坐标系来说，它应该是一样的；在此坐标系中，粒子保持不动。如果粒子保持不动，它就应该没有动量。于是，在那个坐标系中，它就是它的纯粹能量，而此能量，与静止质量是一样的。这样，E2−p2=m20。所以，我们看到，这个矢量的长度的平方，即四个矢量的动量，就等于 m20。

From the square of a vector, we can go onto invent the “dot product,” or the product which is a scalar: if aμis one four-vector and bμ is another four-vector, then the scalar product is
(17.16)
It is the same in all coordinate systems.
从矢量的平方出发，我们可以继续发明“点积”，或者标量积：如果aμ是一个四个矢量，bμ是另外一个四个矢量，那么，其标量积就是：
(17.16)
它在所有的坐标系中，都一样。

We also know that the momentum of anyparticle is equal to its total energy times its velocity: if c=1 , p=vE or, in ordinary units, p=vE/c2 . For any particle moving at the speed of light, p=E if c=1 . The formulas for the energy of a photon as seen from a moving systemare, of course, given by Eq. (17.12),but for the momentum we must substitute the energy times c (or times 1 in this case). The different energies after transformation means thatthere are different frequencies. This is called the Doppler effect, and one cancalculate it easily from Eq. (17.12),using also E=p and E=hν .
As Minkowski said, “Space of itself, andtime of itself will sink into mere shadows, and only a kind of union betweenthem shall survive.”
我们也知道，任何粒子的动量，都等于其总能量，乘以其矢速：如果 c=1 , p=vE 或者，用普通的单位，p=vE/c2。对于任何粒子，如果它以光速运动，若c=1，则p=E 。对于一个光子的能量，从一个运动中的系统所看到能量公式，当然就是方程（17.12），但是，对于动量，我们应该替换掉：能量乘以 c（或者，在这种情况下，乘以1）。变换之后，能量不同，意味着频率不同。这被称为“多普勒效应”，可从方程（17.12）中，提前计算出来，也是利用E=p 和 E=hν。
正如闵可夫斯基（Minkowski）所言：“空间本身，及时间本身，都将沉入单纯的影子中，只有它们之间的某种联合，会存活下来。”