9–5Meaning of the dynamical equations 9-5力学方程的意义
Now let us try to analyze just whatEq. (9.12)means. Suppose that at a given time t the object has a certain velocity vx and position x . What is the velocity and what is the position at a slightly latertime t+ϵ ? If we can answer this question our problem is solved, for then we canstart with the given condition and compute how it changes for the first instant,the next instant, the next instant, and so on, and in this way we graduallyevolve the motion. To be specific, let us suppose that at the time t=0we are given that x=1 and vx=0 . Why does the object move at all? Because there is a force onit when it is at any position except x=0 . If x>0 , that force is upward. Therefore the velocity which is zero starts tochange, because of the law of motion. Once it starts to build up some velocitythe object starts to move up, and so on. Now at any time t , if ϵ is very small, we may express the position at time t+ϵin terms of the position at time t and the velocity at time t to a very good approximation as
x(t+ϵ)=x(t)+ϵvx(t). (9.13)
我们现在就来分析，方程（9.12）究竟意味着什么。假设在一个被给予的时间t，对象有一定的矢量速度vx，和位置x。那么在一个非常小的稍后的时间t+ϵ，矢量速度和位置，又是什么呢？如果我们可以回答这个问题，那么，我们的课题就解决了，因为那样的话，我们就可以从被给予的条件出发，计算出对象在第一瞬间是怎么变化的，下一瞬间又是怎么变化的，如此等等，以这种方式，我们就可以逐渐地开发运动。更具体点，让我们假设，在时间t=0，我们被给予了x=1 和 vx=0。对象为什么会移动呢？因为，除了在位置x=0外，在任何其它位置，都有一个力，作用于它。如果 x>0，那么这个力就是向上的。因此，本来为零的矢量速度，由于运动的规律，开始变化。一旦对象开始积累了一些矢量速度，它就开始移动，就这样。现在，在任何时间t，如果ϵ非常小，那么，我们就可以用时间t的位置和矢量速度，非常近似地表达：时间t+ϵ的位置：
x(t+ϵ)=x(t)+ϵvx(t). (9.13)

The smaller the ϵ , the more accurate this expression is, but it is still usefullyaccurate even if ϵ is not vanishingly small. Now what about the velocity? In order to getthe velocity later, the velocity at the time t+ϵ , we need to know how the velocity changes, the acceleration.And how are we going to find the acceleration? That is where the law ofdynamics comes in. The law of dynamics tells us what the acceleration is. Itsays the acceleration is −x .
vx(t+ϵ)=vx(t)+ϵax(t) (9.14)
=vx(t)−ϵx(t). (9.15)
ϵ越小，这个表达式就越准确，但是，即便ϵ不是趋于零地那么小，其准确性，依然可用。现在，矢量数速度又如何呢？为了得到稍后的矢量速度，即t+ϵ时刻的矢量速度，我们需要知道，矢量速度是如何变化的，即加速度。我们如何找到加速度呢？这就是力学定律入场的地方。力学规律告诉我们：加速度是什么。它说加速度就是 −x。
vx(t+ϵ)=vx(t)+ϵax(t) (9.14)
=vx(t)−ϵx(t). (9.15)

Equation (9.14)is merely kinematics; it says that a velocity changes because of the presenceof acceleration. But Eq. (9.15)is dynamics, because it relates the acceleration to the force; it saysthat at this particular time for this particular problem, you can replace theacceleration by −x(t) . Therefore, if we know both the x and v at a given time, we know the acceleration, which tells us the newvelocity, and we know the new position—this is how the machinery works. Thevelocity changes a little bit because of the force, and the position changes alittle bit because of the velocity.
方程（9.14）只是运动学的；它说，矢量速度变化，是因为加速度在场。但是方程（9.15），则是力学的，因为它让加速度，与力相关；它说，对于这个具体的课题，在这个具体的时间，你可用−x(t)，代替加速度。所以，如果我们知道，在一个被给予定时间的x和 v，我们就知道了加速度，它告诉我们新的速度，我们也知道了新的位置--这就是：这个机制是如何工作的。矢量速度改变了一点，是因为力；而位置改变了一点，则是因为矢量速度。

9–6Numerical solution of the equations 9-6方程的数字解决方案
Now let us really solve the problem.Suppose that we take ϵ=0.100 sec. After we do all the work if we find that this is not smallenough we may have to go back and do it again with ϵ=0.010 sec. Starting with our initial value x(0)=1.00 , what is x(0.1) ? It is the old position x(0) plus the velocity (which is zero) times 0.10 sec. Thus x(0.1) is still 1.00 because it has not yet started to move. But the new velocity at 0.10 sec will be the old velocity v(0)=0 plus ϵ times the acceleration. The acceleration is −x(0)=−1.00. Thus
v(0.1)=0.00−0.10×1.00=−0.10.
Now at 0.20 sec
x(0.2) =x(0.1)+ϵv(0.1)
=1.00−0.10×0.10=0.99
and
v(0.2)=v(0.1)+ϵa(0.1)
=−0.10−0.10×1.00=−0.20.
And so, on and on and on, we can calculate the rest of the motion, andthat is just what we shall do. However, for practical purposes there are somelittle tricks by which we can increase the accuracy. If we continued thiscalculation as we have started it, we would find the motion only rather crudelybecause ϵ=0.100 sec is rather crude, and we would have to go to a very smallinterval, say ϵ=0.01 . Then to go through a reasonable total time interval would take a lotof cycles of computation. So we shall organize the work in a way that willincrease the precision of our calculations, using the same coarse interval ϵ=0.10 sec. This can be done if we make a subtle improvement in thetechnique of the analysis.
现在，让我们真正解决这个问题。假设我们取ϵ=0.100 sec。在我们做了所有工作之后，如果我们发现，这还不够小，我们可以返回来，取ϵ=0.010 sec重做。从初始值 x(0)=1.00开始，那么，什么是 x(0.1)呢？它就是老的位置 x(0)，加上矢量速度乘以0.10 sec。这样，x(0.1)就仍是 1.00，因为它还没有开始移动。但是，在0.10 sec的新的矢量速度，将是老的矢量速度v(0)=0，加上ϵ乘以加速度。加速度是 −x(0)=−1.00。这样：
v(0.1)=0.00−0.10×1.00=−0.10。
现在，在t 0.20 sec 时：
x(0.2) =x(0.1)+ϵv(0.1)
=1.00−0.10×0.10=0.99。
以及：
v(0.2)=v(0.1)+ϵa(0.1)
=−0.10−0.10×1.00=−0.20。
如此等等，我们就可以计算运动的其余部分，这正是我们将要做的。而对于实践的目的来说，有一些小的技巧，通过它们，我们可以增加准确性。如果我们像开始时那样，继续这个计算，我们将发现，这个运动相当粗糙，因为ϵ=0.100 sec相当粗糙，我们必须选一个非常小的时间间隔，比如ϵ=0.01。因此，对于一个合理的总的时间间隔，要完全通过它，需要大量的计算循环。所以，我们要用一种方式，来组织此工作，这种方式，即便是使用同样粗糙的间隔ϵ=0.10 sec，也会增加我们计算的精确性。如果在分析的技术中，我们做出细微的改进，那么，这种方式的目的，就可以达到。

Notice that the new position is the oldposition plus the time interval ϵ times the velocity. But the velocity when? The velocityat the beginning of the time interval is one velocity and the velocity at theend of the time interval is another velocity. Our improvement is to use thevelocity halfway between. If we know the speed now, but the speed is changing,then we are not going to get the right answer by going at the same speed asnow. We should use some speed between the “now” speed and the “then” speed atthe end of the interval. The same considerations also apply to the velocity: tocompute the velocity changes, we should use the acceleration midway between thetwo times at which the velocity is to be found. Thus the equations that weshall actually use will be something like this: the position later is equal tothe position before plus ϵ times the velocity at the time in the middle of the interval.Similarly, the velocity at this halfway point is the velocity at a time ϵbefore (which is in the middle of the previous interval) plus ϵ times the acceleration at the time t . That is, we use the equations
x(t+ϵ) = x(t)+ϵv(t+ϵ/2),
v(t+ϵ/2)= v(t−ϵ/2)+ϵa(t), (9.16)
a(t) = −x(t) .
注意，新的位置，就是老的位置，加上时间间隔ϵ乘以矢量速度。但是，是什么时候的矢量速度呢？一个时间间隔，有开始点，有结束点，各有其矢量速度。我们的改进，就是使用中间点的矢量速度，中间点，在开始点和结束点的中间。如果我们知道了现在的速度，但是，由于速度在变，因此，如果我们使用的速度，与现在的一样，那么，我们就不能得到正确的答案。我们应该使用“现在”的速度，和在时间间隔的终点处的“彼时”速度中间的速度。同样的考虑，也可以应用于矢量速度：要计算的矢量速度的变化，我们应该使用中间点的加速度。这样，我们实际使用到的方程，像是这样：稍后的位置，等于前面的位置，加上ϵ乘以中间点的矢量速度。类似地，中间点的矢量速度，就是时间ϵ之前（它就是上一时间间隔的中间点）的矢量速度，加上ϵ乘以的时间t的加速度：
x(t+ϵ) = x(t)+ϵv(t+ϵ/2),
v(t+ϵ/2) = v(t−ϵ/2)+ϵa(t), (9.16)
a(t) = −x(t) .

There remains only one slight problem: what is v(ϵ/2)? At the start, we are given v(0) , not v(−ϵ/2) . To get our calculation started, we shall use a special equation, namely,v(ϵ/2)=v(0)+(ϵ/2)a(0) .
现在，只剩下一个小问题：什么是v(ϵ/2) ? 在开始点，我们有 v(0) , 但没有 v(−ϵ/2)。为了让我们的计算能够开始，我们将使用一个特殊方程，即：v(ϵ/2)=v(0)+(ϵ/2)a(0)。

Now we are ready to carry through ourcalculation. For convenience, we may arrange the work in the form of a table,with columns for the time, the position, the velocity, and the acceleration, andthe in-between lines for the velocity, as shown in Table 9–1. Such atable is, of course, just a convenient way of representing the numerical valuesobtained from the set of equations (9.16),and in fact the equations themselves need never be written. We just fill in thevarious spaces in the table one by one. This table now gives us a very goodidea of the motion: it starts from rest, first picks up a little upward (negative)velocity and it loses some of its distance. The acceleration is then a littlebit less but it is still gaining speed. But as it goes on it gains speed moreand more slowly, until as it passes x=0 at about t=1.50 sec we can confidently predict that it will keep going, but nowit will be on the other side; the position x will become negative, the acceleration therefore positive. Thus thespeed decreases. It is interesting to compare these numbers with the function x=cost, which is done in Fig. 9–4. Theagreement is within the three significant figure accuracy of our calculation!We shall see later that x=cost is the exact mathematical solution of our equation of motion, but itis an impressive illustration of the power of numerical analysis that such aneasy calculation should give such precise results.
现在，我们要完成我们的计算。为了方便，我们可以把此工作，放在一个表中，列包括时间、位置、矢量速度和加速度，中间的线，是为了矢量速度，如表9-1。当然，这样的一个表，当然只是一种方便的表现方式，为的是表现从公式（9.16）所得到的数字值，事实上，这个方程本身，永远不需要被写出来。我们只是把表的空处，一个一个地填满。这个表，给我们提供了关于运动的很好的想法：它从静止开始，先得到一个小的向上的矢量速度（负的），且失掉一些距离。因此，加速度有点变小，但它仍在获得速度。但是呢，当它继续走时，它得到的速度越来越慢，直到在大约1.5秒时，它通过x=0，我们可以自信地预告，它将继续走，但是现在，它在会在另一侧，位置x将变成负的，加速度因而就是正的。这样，速度就会减少。把这些书，与函数 x=cost相比，非常有趣，如图9-4。此一致性，在我们计算的三个明显的图形准确性之内！稍后，将我们将会看到，对于我们的运动方程，x=cost，是其准确的数学解决方案，然而，数值分析在这里所展现出来的力量，让人印象深刻，这样一种简单的计算，居然可以给出这么精确的结果。

Fig. 9–4.Graph of the motion of a mass on aspring. 图9-4 一个弹簧上的质量的运动曲线

9–7Planetary motions 9-7 行星的运动
The above analysis is very nice for themotion of an oscillating spring, but can we analyze the motion of a planetaround the sun? Let us see whether we can arrive at an approximation to anellipse for the orbit. We shall suppose that the sun is infinitely heavy, inthe sense that we shall not include its motion. Suppose a planet starts at acertain place and is moving with a certain velocity; it goes around the sun insome curve, and we shall try to analyze, by Newton’s laws of motion and his lawof gravitation, what the curve is. How? At a given moment it is at someposition in space. If the radial distance from the sun to this position iscalled r , then we know that there is a force directed inward which, according tothe law of gravity, is equal to a constant times the product of the sun’s massand the planet’s mass divided by the square of the distance. To analyze thisfurther we must find out what acceleration will be produced by this force. Weshall need the components of the acceleration along two directions, whichwe call x and y . Thus if we specify the position of the planet at a given moment bygiving x and y (we shall suppose that z is always zero because there is no force in the z -direction and, if there is no initial velocity vz, there will be nothing to make z other than zero), the force is directed along the line joining the planetto the sun, as shown in Fig. 9–5.
对于一个振动的弹簧，上述分析，非常好，但是，对于一个绕太阳的行星，我们能分析其运动吗？让我们看看，对于那个轨道的椭圆，我们是否能够达到一种近似。我们假设，太阳是无限重，意义就是，我们不包括它的运动。假设一个行星，在一个地方开始，且有一定的矢量速度在移动；它以某种曲线，绕着太阳走，我们将尝试通过牛顿运动规律，及他的万有引力规律，来分析此曲线是什么？怎么做呢？在某一被给予的瞬间，它在空间的某个位置。如果从太阳到这个位置的半径，被称为 r，那么，我们就知道有一个指向里面的力，依据万有引力规律，它等于：常数乘以太阳的质量和行星的质量，除以距离的平方。要进一步分析，我们就应该找出：通过这个力所产生的加速度是什么？我们将需要，沿着两个方向的加速度的分量，这两个方向，被称为x和y。这样，如果我们通过被给予的x和y（我们将假定，z总是零，因为在z方向，没有力，并且，如果没有初始的矢量速度vz，那么，将没有什么会比让z等于零更好了。），具体指定行星在某一被给予瞬间的位置，那么，力就是沿着行星与太阳之间的连线，如图9-5所示。

Fig. 9–5.The force of gravity on a planet. 图9-5 一个行星上的万有引力

From this figure we see that the horizontalcomponent of the force is related to the complete force in the same manner asthe horizontal distance x is to the complete hypotenuse r , because the two triangles are similar. Also, if x is positive, Fx is negative. That is, Fx/|F|=−x/r , or Fx= −|F|x/r= −GMmx/r3 . Now we use the dynamical law to find that this force component isequal to the mass of the planet times the rate of change of its velocity in thex -direction. Thus we find the following laws:
m(dvx/dt)m(dvy/dt)r=−GMmx/r3,=−GMmy/r3,=x2+y2−−−−−−√.(9.17)
从这个图形，我们可看到，力的水平分量，与完整的力的关系，正如水平距离 x，与完整的斜边r的关系一样，因为，这两个三角形，是类似的。此外，如果 x是正的, Fx 就是负的。也就是说, Fx/|F|=−x/r, 或 Fx= −|F|x/r= −GMmx/r3。现在，我们使用力学规律，以找出这个力的分量，等于行星的质量，乘以它的x方向的矢量速度的变化率。这样，我们就得到下面的规律：
(9.17)

Our calculation thus proceeds by thefollowing steps, using time intervals ϵ=0.100 : Initial values at t=0 :
x(0) =0.500 y(0)=+0.000
vx(0) =0.000 vy(0)==+1.630
From these we find:
r(0) =0.500 1/r3(0) =8.000
ax=−4.000 ay=0.000
Thus we may calculate the velocities vx (0.05) and vy(0.05) :
vx(0.05)=0.000−4.000×0.050=−0.200；
vy(0.05) =1.630+0.000×0.050=1.630.
Now our main calculations begin:
x(0.1)y(0.1)r(0.1)1/r3(0.1)ax(0.1)ay(0.1)vx(0.15)vy(0.15)x(0.2)y(0.2)=0.500−0.20×0.1=0.0+1.63×0.1=0.4802+0.1632−−−−−−−−−−−−√=7.677=−0.480×7.677=−0.163×7.677=−0.200−3.685×0.1=1.630−1.250×0.1=0.480−0.568×0.1=0.163+1.505×0.1etc.=−0.480=−0.163=−0.507=−3.685=−1.250=−0.568=−1.505=−0.423=−0.313
这样，我们的计算，就可按以下步骤进行，时间间隔为 ϵ=0.100，t=0时的初始值为 :
x(0) =0.500 y(0)=+0.000
vx(0) =0.000 vy(0)==+1.630
由此我们得到：
r(0) =0.500 1/r3(0)=8.000
ax=−4.000 ay=0.000
这样，我们就可以计算矢量速度 vx (0.05) 和 vy(0.05):
vx(0.05)=0.000−4.000×0.050=−0.200；
vy(0.05) =1.630+0.000×0.050=1.630。
现在，我们的主要计算开始了：

In this way we obtain the values given in Table 9–2, and in20 steps or so we have chased the planet halfway around the sun! InFig. 9–6are plotted the x - and y -coordinates given in Table 9–2. Thedots represent the positions at the succession of times a tenth of a unit apart;we see that at the start the planet moves rapidly and at the end it moves slowly,and so the shape of the curve is determined. Thus we see that we really doknow how to calculate the motion of planets!
以这种方式，我们就得到了表9.2中的值，在大约20步内，我们追逐了一个绕着太阳运动的行星的部分路径。在图9-6中，我画出了表9-2所给的x和y坐标。点代表着在时间相继中的位置，时间间隔是时间单位的十分之一；我们看到，开始时，行星移动的快，最后慢，这决定了曲线的形状。这样，我们看到，我们确实可以知道如何计算行星的运动。

Table 9–2
Solution of dvx/dt=−x/r3, dvy/dt=−y/r3 , r=x2+y2−−−−−−√ .
Interval: ϵ=0.100
Orbitvy=1.63vx=0 x=0.5 y=0 att=0
表9-2
dvx/dt=−x/r3 , dvy/dt=−y/r3 , r=x2+y2 的平方根的解决方案。
间隔: ϵ=0.100
t=0时的轨道：vy=1.63 vx=0 x=0.5 y=0

Now let us see how we can calculate themotion of Neptune, Jupiter, Uranus, or any other planet. If we have a greatmany planets, and let the sun move too, can we do the same thing? Of course wecan. We calculate the force on a particular planet, let us say planetnumber i , which has a position xi,yi,zi(i=1 may represent the sun, i=2 Mercury, i=3 Venus, and so on). We must know the positions of all the planets. Theforce acting on one is due to all the other bodies which are located, let ussay, at positions xj,yj,zj. Therefore the equations are
midvixdtmidviydtmidvizdt=∑j=1N−Gmimj(xi−xj)r3ij,=∑j=1N−Gmimj(yi−yj)r3ij,=∑j=1N−Gmimj(zi−zj)r3ij.(9.18)
Further, we define rij as the distance between the two planets i and j ; this is equal to
rij=(xi−xj)2+(yi−yj)2+(zi−zj)2−−−−−−−−−−−−−−−−−−−−−−−−−−−√.(9.19)
现在，让我们看，如何计算海王星、木星、天王星、或任何其他行星的运动。如果我们有大量的行星，且让太阳也在运动，我们能做同样的事情吗？当然，我们能。我们可以计算作用于一个具体行星上的力，比如说第i号行星，其位置为xi,yi,zi (i=1 可以代表太阳, i=2 水星, i=3 金星, 如此等等)。我们必须知道所有行星的位置。作用于一个行星上的力，要归于所有其他物体，我们说，这些物体，位于xj,yj,zj。因此，方程就是：
(9.18)
此外，我们把行星i与j之间的距离，定义为rij；它等于：
(9.19)

Also, ∑ means a sum over all values of j —all other bodies—except, of course, for j=i . Thus all we have to do is to make more columns, lots morecolumns. We need nine columns for the motions of Jupiter, nine for the motionsof Saturn, and so on. Then when we have all initial positions and velocities wecan calculate all the accelerations from Eq. (9.18)by first calculating all the distances, using Eq. (9.19).How long will it take to do it? If you do it at home, it will take a very longtime! But in modern times we have machines which do arithmetic very rapidly; a verygood computing machine may take 1 microsecond, that is, a millionth of a second, to do anaddition. To do a multiplication takes longer, say 10 microseconds. It may be that in one cycle of calculation,depending on the problem, we may have 30 multiplications, or something like that, so one cycle will take300 microseconds. That means that we can do 3000 cycles of computation per second. In order to get an accuracy, of,say, one part in a billion, we would need 4×105 cycles to correspond to one revolution of a planet around thesun. That corresponds to a computation time of 130 seconds or about two minutes. Thus it takes only two minutes tofollow Jupiter around the sun, with all the perturbations of all the planetscorrect to one part in a billion, by this method! (It turns out that the errorvaries about as the square of the interval ϵ . If we make the interval a thousand times smaller, it is a milliontimes more accurate. So, let us make the interval 10,000 times smaller.)
另外，∑意味着所有j的值的总和，这里，j是所有其他物体，当然，除j=i外。这样，我们所要做的，就是更多的列，非常多的列。木星需要9列，土星也需要9列，如此等等。因此，当我们有了所有初始的位置和矢量速度后，我们就可以计算所有的加速度：即首先用方程（9.19），计算所有的距离，然后，使用方程（9.18）。做这事要花多长时间呢？如果你在家做，要花非常长的时间！但在现代，我们有机器，可以非常快地做算术运算；一个非常好的计算机器，做一次加法，需1微妙，也就是说1秒的百万分之一。做乘法需要的时间长一些，比如10微妙。在一个计算循环中，我们可能有30次乘法，或类似的事情，所以，一个循环，将需要300微秒；当然这依赖于问题。这意味着，我们每秒可以做3000次循环的计算。为了达到一定的精度，比如说，十亿分之一，我们需要4×105次循环，以与一个行星绕太阳一周相应。这相当于计算时间为130秒，或大约两分钟。这样，跟踪木星绕太阳，只需要两分钟，通过这种方式，所有其他行星的扰动，被校正到十亿分之一！（结果就是，误差大约随着时间间隔 ϵ的平方而变化。如果我们把间隔缩小一千倍，那么，精度就可提高百万倍。所以，让我们把间隔缩小10，000倍。）

So, as we said, we began this chapter notknowing how to calculate even the motion of a mass on a spring. Now, armed withthe tremendous power of Newton’s laws, we can not only calculate such simplemotions but also, given only a machine to handle the arithmetic, even thetremendously complex motions of the planets, to as high a degree of precisionas we wish!
于是，正如我们所说，本章开始时，我们甚至都不知道如何计算一个弹簧上的质量的运动。而现在，在用牛顿规律的巨大的力量武装起来之后，我们不仅能够计算这种简单的运动，而且，假如有一个机器能够处理算术运算的话，我们甚至可以计算巨复杂的行星的运动，且可达到我们希望的任意高的精度。