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 isconsistent. If it did not come out that way, we would have real cause to worry!
(16.2)
这个变换，还有更多后果，现在，让我们讨论之。首先，能够反向地解决这些方程，非常有趣。也就是说，这里是一组线性方程，4个方程，有4个未知数，它们可被反向求解，即用x′,y′,z′,t′，来表示x,y,z,t 。此结果非常有趣，由于它告诉我们，从一个“运动中的”的坐标系，去观察一个“处于静止”的坐标系，是何结果。当然，由于运动是相对的，且是匀速的，那么，那个在“运动中的”人，如果他希望的话，就可以说，其实另一个家伙是在运动，而他自己是静止的。由于他是在相反的方向运动，他将会得到同样的变换，但矢速的符号不同。我们通过操作所发现的，正是这样。如果结果不是这样，那么，我们担心，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.Suppose that 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.
在相对论中，矢速的增加，是一个非常有趣的问题，下面我们讨论之。我们回忆，最初的困惑之一，就是在所有系统中，光都是以186,000 mi/sec 在走，即便当它们是在相对的运动中时，也是如此。这就是更普遍的问题中的一种特殊情况，举例如下。假设在太空飞船中，有一个对象，以100,000 mi/sec在走，而太空飞船本身，也是以100,000 mi/sec在走；那么，从一个外部的观察者来看，此飞船内的对象，其速为何？我们或许可以说，200,000 mi/sec，比光速快。这让人紧张不安，因为，我们并不认为，能够超过光速！一般问题如下。

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)
我们假设，从船内人的观点看，飞船内的对象，正在以矢速v运动，而太空飞船本身，相对于地面的的矢速为u。我们想知道，从地面上的人的观点来看，这个对象，正在以什么样的矢速vx在运动？当然，这仍是特例，运动只在x方向。对于有方向的矢速、或任何角度，也可以有变换；如果需要，也可做到。在飞船内，矢速为vx′，其意为，位移x′，等于矢速乘以时间：
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 .
对于一个对象，从x′ 到 t′，它有关系（16.2），现在，我们要计算的，就是从外面的观察者来看，该对象的位置和时间。所以，我们只需把（16.3）代入（16.2），得到：
(16.4)
但是，这里我们发现，x被用 t′表达了。为了得到，外面的人所看到的矢速，我们应该让他的距离，除以他的时间，而不是另一个人的时间。所以，我们应该计算，外面的人所看到的时间，就是：
(16.5)
现在，我们应该找到x和t的比率，就是：
(16.6)
平方根被消掉了。这就是我们要找的规律，合成的矢速，即两个矢速的“综合”，它并不仅仅是两个矢速的算术和，（我们知道，它不可能是算术和，否则我们就是遇到麻烦了），而是被1+uv/c2，给修正了。

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.
现在让我们看看，发生了什么。假设你在太空飞船中，以光速的一半，在运动，且飞船本身，也是以光速的一半在走。这样，u、v 皆为c/2，但是，在分母中，uv/c2是四分之一，于是：

所以，在相对论中，“一半”加“一半”，得到的不是“一”，而是“4/5”。当然，低的矢速，以同样的方式，很容易被加上，因为，只要矢速与光速相比，较小，我们就可以忽略因子(1+uv/c2)；当矢速较高时，情况就完全不同了。

16–4Relativistic mass 16-4 相对论的质量
We learned in the last chapter that themass of an object increases with velocity, but no demonstration of this wasgiven, in the sense that we made no arguments analogous to those about the wayclocks have to behave. However, we can show that, as a consequence ofrelativity plus a few other reasonable assumptions, the mass must vary in thisway. (We have to say “a few other assumptions” because we cannot prove anythingunless we have some laws which we assume to be true, if we expect to makemeaningful deductions.) To avoid the need to study the transformation laws offorce, we shall analyze a collision, where we need know nothing aboutthe laws of force, except that we shall assume the conservation of momentum andenergy. Also, we shall assume that the momentum of a particle which is movingis a vector and is always directed in the direction of the velocity. However,we shall not assume that the momentum is a constant times the velocity,as Newton did, but only that it is some function of velocity. We thuswrite the momentum vector as a certain coefficient times the vector velocity:
p=mvv. (16.8)
We put a subscript v on the coefficient to remind us that it is a function of velocity, andwe shall agree to call this coefficient mv the “mass.” Of course, when the velocity is small, it is the same massthat we would measure in the slow-moving experiments that we are used to. Nowwe shall try to demonstrate that the formula for mv must be m0/(1−v2/c2)1/2, by arguing from the principle of relativity that the laws of physicsmust be the same in every coordinate system.
上一章，我们学过，一个对象的质量，随着矢速而增加，但是，关于这一点的演证，并未给予；随着矢速的增加，表的表现方式，有所不同，关于这点，我们给出了论证；这里演证的意义，是指我们没有给出可与这种论证相类比的论证。然而，我们可以指出，质量，作为相对论的后果加上一些其他合理的假设，应该以这种方式而有所不同。（我们必须说“一些其他假设”，因为，，如果我们期待做出有意义的推理，那么，除非有某些规律，我们可假定其为真，否则，我们不能证明任何事情。）力学规律的变换，需要学习，为避免此，我们将分析一个碰撞，这样，我们将无需知道力学规律，除了我们将假定：动量和能量的守恒。我们也将假定：一个正在运动的粒子，其动量，是一个矢量，且总是指向矢速的方向。然而，我们将不像牛顿那样，假定：动量是一个常数乘以矢速，而只是假定它是矢速的某种函数。这样，我们把动量矢量，写作：某系数乘以矢速矢量：
p=mvv. (16.8)
我们给系数，放一个下标v，提醒我们，它是矢速的函数，我们将称这个系数mv为“质量”。当然，当矢速较小时，它与我们通常在低速实验中所测的质量相同。现在，我们将从相对论的原理，来论证，物理学的规律，在所有坐标系中，应该都是一样的；借此论证，我们将演证，关于mv的公式，应为m0/(1−v2/c2)1/2。

Suppose that we have two particles, liketwo protons, that are absolutely equal, and they are moving toward each otherwith exactly equal velocities. Their total momentum is zero. Now what canhappen? After the collision, their directions of motion must be exactly oppositeto each other, because if they are not exactly opposite, there will be anonzero total vector momentum, and momentum would not have been conserved. Alsothey must have the same speeds, since they are exactly similar objects; infact, they must have the same speed they started with, since we suppose thatthe energy is conserved in these collisions. So the diagram of an elasticcollision, a reversible collision, will look like Fig. 16–2(a):all the arrows are the same length, all the speeds are equal. We shall supposethat such collisions can always be arranged, that any angle θ can occur, and that any speed could be used in such a collision. Next,we notice that this same collision can be viewed differently by turning theaxes, and just for convenience we shall turn the axes, so that thehorizontal splits it evenly, as in Fig. 16–2(b).It is the same collision redrawn, only with the axes turned.
假设我们有两个粒子，比如两个质子，它们绝对相等，相向而行，矢速完全相等。它们的总动量是零。现在，会发生什么呢？碰撞之后，它们的运动方向，应该完全相反，因为如果不是完全相反，那么，将有一个非零的总的矢量动量，则动量并未完全守恒。另外，它们的速度，应该一样，由于它们是完全类似的对象；事实上，它们开始时，应该有一样的速度，由于我们假设，在这些碰撞中，能量是守恒的。所以，一个弹性碰撞、反转的碰撞的图，看上去，如图16-2（a）：所有的箭头，长度都一样，所有的速度，都是相等的。我们将假定，这种碰撞，总可以被安排，任何角度θ都能出现，任何速度，都可被用于这种碰撞中。下面，我们注意到，同样的碰撞，可以通过旋转坐标轴，来做不同的观看，正是为了方便，我们将旋转坐标轴，这样，让分开线，最终在水平方向，如图16-2（b）。这是同样的碰撞，重画了，只是把轴旋转了。

Fig. 16–2.Two views of an elastic collisionbetween equal objects moving at the same speed in opposite directions. 图16-2 两个视图：两个相等对象，以同样速度、相反方向运动，发生弹性碰撞后的视图。

Now here is the real trick: let us look atthis collision from the point of view of someone riding along in a car that ismoving with a speed equal to the horizontal component of the velocity of oneparticle. Then how does the collision look? It looks as though particle 1 is just going straight up, because it has lost its horizontalcomponent, and it comes straight down again, also because it does not have thatcomponent. That is, the collision appears as shown in Fig. 16–3(a).Particle 2 , however, was going the other way, and as we ride past it appears tofly by at some terrific speed and at a smaller angle, but we can appreciatethat the angles before and after the collision are the same. Let usdenote by u the horizontal component of the velocity of particle 2 , and by w the vertical velocity of particle 1 .
这里是真正的技巧：假设有人坐在一辆汽车上，其速度，与一个粒子的矢速的水平分量相同，现在，从他的观点，来看这个碰撞。那么，这个碰撞看上去是什么样子呢？看上去，就好像，粒子1，只是向上走，因为，它已经失去它的水平分量，然后，有直接向下，还是因为它没有那个分量。也就是说，碰撞看上去，如图16-3（a）。然而，粒子2所走路线，完全不同，在我们乘车通过时，它似乎是以一种可怕的速度，在飞过，而且，是以一个很小的角度，但是，我们可以看到，碰撞前后的这个角度，是一样的。我们设，粒子2的矢速的水平分量为u，粒子1的矢速的垂直分量为w。

Fig. 16–3.Two more views of the collision,from moving cars. 图16-3 另外两个碰撞的视图，从一辆运动中的汽车看的。

Now the question is, what is the verticalvelocity utanα ? If we knew that, we could get the correct expression for themomentum, using the law of conservation of momentum in the vertical direction.Clearly, the horizontal component of the momentum is conserved: it is the samebefore and after the collision for both particles, and is zero forparticle 1 . So we need to use the conservation law only for the upwardvelocity utanα . But we can get the upward velocity, simply by looking at thesame collision going the other way! If we look at the collision of Fig. 16–3(a)from a car moving to the left with speed u , we see the same collision, except “turned over,” as shown inFig. 16–3(b).Now particle 2 is the one that goes up and down with speed w , and particle 1 has picked up the horizontal speed u . Of course, now we know what the velocity utanαis: it is w(1−u2/c2)1/2(see Eq. 16.7).We know that the change in the vertical momentum of the vertically moving particleis
Δp=2mww
(2 , because it moves up and back down). The obliquely moving particlehas a certain velocity v whose components we have found to be u and w(1−u2/c2)1/2 , and whose mass is mv . The change in vertical momentum of this particle is therefore Δp′=2mvw(1−u2/c2)1/2because, in accordance with our assumed law (16.8),the momentum component is always the mass corresponding to the magnitude of thevelocity times the component of the velocity in the direction of interest. Thusin order for the total momentum to be zero the vertical momenta must cancel andthe ratio of the mass moving with speed v and the mass moving with speed w must therefore be
mw/mv=(1−u2/c2)1/2 (16.9)
现在，问题就是，垂直矢速utanα是多少？如果我们知道了它，我们就可以在垂直方向，利用动量守恒方程，为动量，得到正确的表达式。显然，动量的水平方向的分量，是守恒的；对于两个粒子，碰撞前后，都是一样的，且对于粒子1，是零。所以，我们只需对向上的矢速utanα，使用守恒规律。但是，我们只有通过查看同一碰撞中，另一方向的粒子的运动，才能得到向上的矢速。对于图16-3（a）所示的碰撞，如果我们乘车查看，车速为u，向左运动，我们看到的碰撞，将是一样的，除非旋转坐标系，则如图16-3（b）所示。现在粒子2的运动，变成直上直下，速度为w，而粒子1，则得到了水平速度u。当然，现在我们知道了，矢速utanα是多少：它就是w(1−u2/c2)1/2（见方程16.7）。我们知道，对于垂直运动的粒子，其垂直动量的变化就是：
Δp=2mww
（因为它向上向下，所以是2）。做斜角运动的粒子，有一个确定的矢速，它的分量，我们发现是u和w(1−u2/c2)1/2，且其质量为mv。这个粒子在垂直方向的动量变化，就是 Δp′=2mvw(1−u2/c2)1/2，因为，依据我们认同的规律（16.8），动量的分量，总是：相应于矢速大小的质量，乘以，矢速在感兴趣的方向上的分量。这样，为了让总的动量是零，垂直动量，就应消去，并且，以速度v运动的质量，与以速度w运动的质量，的比率，就应该是：
mw/mv=(1−u2/c2)1/2 (16.9)

Fig. 16–4.Two views of an inelasticcollision between equally massive objects. 图16-4 两个对象，质量相等，发生非弹性碰撞后的两个视图。
Now, let us accept that momentum is conservedand that the mass depends upon the velocity according to (16.10)and go on to find what else we can conclude. Let us consider what is commonlycalled an inelastic collision. For simplicity, we shall suppose that twoobjects of the same kind, moving oppositely with equal speeds w , hit each other and stick together, to become some new, stationaryobject, as shown in Fig. 16–4(a). Themass m of each corresponds to w , which, as we know, is m0/(1−w2/c2)1/2. If we assume the conservation of momentum and the principle ofrelativity, we can demonstrate an interesting fact about the mass of the newobject which has been formedWe imagine an infinitesimal velocity u at right angles to w (we can do the same with finite values of u , but it is easier to understand with an infinitesimal velocity), thenlook at this same collision as we ride by in an elevator at the velocity −u. What we see is shown in Fig. 16–4(b).The composite object has an unknown mass M . Now object 1 moves with an upward component of velocity u and a horizontal component which is practically equal to w, and so also does object 2 . After the collision we have the mass M moving upward with velocity u , considered very small compared with the speed of light, and alsosmall compared with w. Momentum must be conserved, so let us estimate the momentum in theupward direction before and after the collision Before the collision wehave p≈2mwu , and after the collision the momentum is evidently p′=Muu, but Mu is essentially the same as M0 because u is so small. These momenta must be equal because of the conservationof momentum, and therefore .
M0=2mw. (16.11)
The mass of the object which is formed when two equal objectscollide must be twice the mass of the objects which come together. Youmight say, “Yes, of course, that is the conservation of mass.” But not “Yes, ofcourse,” so easily, because these masses have been enhanced over themasses that they would be if they were standing still, yet they stillcontribute, to the total M , not the mass they have when standing still, but more.Astonishing as that may seem, in order for the conservation of momentum to workwhen two objects come together, the mass that they form must be greater thanthe rest masses of the objects, even though the objects are at rest after thecollision!
现在，让我们接受，动量是守恒的，而质量，依据（16.10），是依赖于矢速的；下面继续，看还能得出什么其他的结论。让我们考虑，通常所说的非弹性碰撞。为简单起见，我们将假设，两个同类的对象，相向运动，以相等速度w，相撞并粘在一起，变成一个新的、稳定的对象，如图16-4（a）。每个对象的质量m，都是相应于w的，我们现在知道，它是m0/(1−w2/c2)1/2。如果我们认定，动量守恒，及相对论的原理，那么，对于此刚形成的新对象，我们可以演证一个有趣的事实。我们想象一个无限小的矢速u，与w成直角（对于有限的u值，我们可以做同样的事，但是，用一个无限小的矢速，易于理解），然后，假设我们是在电梯内，查看此同样的碰撞，电梯的矢速为−u。我们所看到的，见图16-4（b）。复合的对象，有一个未知的质量 M。现在，对象1，用一个向上的矢速分量u，和一个实际等于w的水平分量，在运动，对象2，也如此。碰撞后，我们有质量M，以矢速u，向上运动，其速度，与光速比，被认为很小，与w相比，也很小。动量应该守恒，所以，让我们估计，碰撞前后，向上的方向的动量。碰撞前，我们有 p≈2mwu，碰撞后，动量明显是p′=Muu，但是，本质上与M0是一样的，因为u是如此的小。这些动量，应该相等，因动量守恒，因此：
M0=2mw. (16.11)
两个对象，碰撞之后，所形成的对象的质量，应该是相遇对象的质量的两倍。你可以说：“是的，当然，这是质量守恒。”但是，并非“是的，当然”这么容易，因为，这些质量被增强了，超过了如果它们静止时所应该有的质量，尽管如此，这些质量对总质量，还是有贡献的，但不是它们静止时的质量，而是更多。这似乎让人震惊，当两个对象相遇时，为了让动量守恒成立，它们形成的质量，应大于对象的静止质量，尽管对象碰撞后是静止的！

16–5Relativistic energy 16-5 相对论的能量
In the last chapter we demonstrated that asa result of the dependence of the mass on velocity and Newton’s laws, thechanges in the kinetic energy of an object resulting from the total work doneby the forces on it always comes out to be
(16.12)
We even went further, and guessed that the total energy is the totalmass times c2 . Now we continue this discussion.
在上一章，我们演证了一事，即一个对象，有力作用于其上，该力所做的总功，就是该对象的动能的变化；而这，就是质量对矢速的依赖、和牛顿规律的结果，用公式表达即：
(16.12)
我们甚至走得更远，而猜测说，总的能量，就是总的质量，乘以c2。现在我们继续这个讨论。

This has interesting consequences. Forexample, suppose that we have an object whose mass M is measured, and suppose something happens so that it flies into twoequal pieces moving with speed w , so that they each have a mass mw . Now suppose that these pieces encounter enough material to slow themup until they stop; then they will have mass m0 . How much energy will they have given to the material when they have stopped?Each will give an amount (mw−m0)c2, by the theorem that we proved before. This much energy is left inthe material in some form, as heat, potential energy, or whatever. Now 2mw=M, so the liberated energy is E=(M−2m0)c2. This equation was used to estimate how much energy would beliberated under fission in the atomic bomb, for example. (Although thefragments are not exactly equal, they are nearly equal.) The mass of theuranium atom was known—it had been measured ahead of time—and the atoms intowhich it split, iodine, xenon, and so on, all were of known mass. By masses, wedo not mean the masses while the atoms are moving, we mean the masses when theatoms are at rest. In other words, both M and m0 are known. So by subtracting the two numbers one can calculate howmuch energy will be released if M can be made to split in “half.” For this reason poor old Einstein wascalled the “father” of the atomic bomb in all the newspapers. Of course, all thatmeant was that he could tell us ahead of time how much energy would be releasedif we told him what process would occur. The energy that should be liberatedwhen an atom of uranium undergoes fission was estimated about six months beforethe first direct test, and as soon as the energy was in fact liberated, someonemeasured it directly (and if Einstein’s formula had not worked, they would havemeasured it anyway), and the moment they measured it they no longer needed theformula. Of course, we should not belittle Einstein, but rather shouldcriticize the newspapers and many popular descriptions of what causes what inthe history of physics and technology. The problem of how to get the thing tooccur in an effective and rapid manner is a completely different matter.
这会带来有趣的后果。例如，假设我们有一个对象，其质量M，已被测量，假设某事发生了，然后它分裂成相等的两块，每块都以速度w运动，于是，每块都有质量mw。现在，假设这些块，遇到了足够多的材料，让它们慢了下来，直到停止；那么，它们将有质量m0。那么，当它们停下来时，它们给予材料的能量，是多少呢？通过我们前面所证明的定理，每块给出的能量，都是(mw−m0)c2。这些能量，以多种形式，如热能、势能等等，被留在材料中。所以，被释放出的能量就是E=(M−2m0)c2。这个方程，可被用来估计, 例如：在原子弹爆炸的核裂变中，有多少能量，被释放出来.（虽然碎片并不准确相等，它们是近乎相等）。铀原子的质量，是已知的—已被提前测量--，而它所分裂成的原子，如碘、氙等，质量都不知道。我们所说的质量，不是原子运动时的质量，而是静止时的质量。换句话说，M和m0，皆为已知。于是，如果可把M分裂成“两半”，那么，通过把这两个数字相减，就可算出，多少能量，将被释放。由于这个原因，可怜的老爱因斯坦，就被所有的报纸，称为原子弹之父。当然，所有这些，只意味着，如果我们告诉他，什么过程，将会发生，那么，他就会提前告诉我们，有多少能量，将被释放。当一个铀原子分裂时，能够被释放出来的能量，大约在第一次试验之前的六个月，被估算了，一旦能量事实上被释放出来，人们就可以直接测量它了（即便爱因斯坦的公式不成立，人们还是可以测量它），当人们开始测量它时，就不再需要那个公式了。当然，我们不应该贬低爱因斯坦，而毋宁说，应该去批评那些报纸和很多流行的说法，即在物理学和技术的历史上，究竟是什么引起什么的。如何让{裂变}这个事情，以一种有效的和快速的方式发生，这个问题，则是一件完全不同的事情。

The result is just as significant in chemistry.For instance, if we were to weigh the carbon dioxide molecule and compare itsmass with that of the carbon and the oxygen, we could find out how much energywould be liberated when carbon and oxygen form carbon dioxide. The only troublehere is that the differences in masses are so small that it is technically verydifficult to do.
结果在化学中，有同样的意义。例如，如果我们称二氧化碳分子的重量，然后，把其质量，与碳和氧相比，我们就会发现，当碳和氧形成二氧化碳时，能释放多少能量。这里唯一的问题，就是质量的差，非常小，以至于严格意义上，做这事很困难。

Now let us turn to the question of whetherwe should add m0c2 to the kinetic energy and say from now on that the total energy of anobject is mc2 . First, if we can still see the component pieces of restmass m0 inside M , then we could say that some of the mass M of the compound object is the mechanical rest mass of the parts, partof it is kinetic energy of the parts, and part of it is potential energy of theparts. But we have discovered, in nature, particles of various kinds whichundergo reactions just like the one we have treated above, in which with all thestudy in the world, we cannot see the parts inside. For instance, when aK-meson disintegrates into two pions it does so according to the law (16.11),but the idea that a K is made out of 2 π ’s is a useless idea, because it also disintegrates into 3 π ’s!
现在，让我们转向下面的问题，我们是否能够把m0c2，加给动能，然后，从现在就可以说，一个对象的总能量，就是mc2。第一，如果我们在M内，仍能看到一个静止质量m0这一成分，那么，对于复合对象质量M，我们就可以说，其一部分，是力学静止质量，一部分是动能，而还有一部分是势能。但是，在自然中，我们已经发现，各种粒子，它们经历了我们上面讨论过的反应，在全世界对这些反应的研究中，我们在其中，都看不到这些部分。例如，当一个k介子，分解成两个点时，所依据的是规律（6.11），但是，如果说一个k介子，是由两个π介子构成的，则不是一个有用的想法，因为，它有时也会分解成三个π介子。

Therefore we have a new idea: we donot have to know what things are made of inside; we cannot and need notidentify, inside a particle, which of the energy is rest energy of the parts intowhich it is going to disintegrate. It is not convenient and often not possibleto separate the total mc2 energy of an object into rest energy of the inside pieces,kinetic energy of the pieces, and potential energy of the pieces; instead, wesimply speak of the total energy of the particle. We “shift the origin”of energy by adding a constant m0c2 to everything, and say that the total energy of a particle is the massin motion times c2 , and when the object is standing still, the energy is the mass atrest times c2 .
因此，我们就有了一个新的想法：我们无需知道，事物内部是如何构成的；我们无法搞清楚，在一个粒子内部，能量的哪一部分，是静止能量部分，是将要分解成几个部分的。把总能量mc2，分解成内部的静止能量部分、动能部分、和势能部分，并不方便，通常也不可能；想反，我们只是说粒子的总能量。我们通过给每个事物，增加一个常数m0c2，来对能量，“进行起源转换”，当对象站着不动时，能量就是：静止时的质量乘以c2。{？}