Now let us examine some of the propertiesof vectors. As examples of vectors we may mention velocity, momentum, force,and acceleration. For many purposes it is convenient to represent a vectorquantity by an arrow that indicates the direction in which it is acting. Whycan we represent force, say, by an arrow? Because it has the same mathematicaltransformation properties as a “step in space.” We thus represent it in adiagram as if it were a step, using a scale such that one unit of force, or onenewton, corresponds to a certain convenient length. Once we have done this, allforces can be represented as lengths, because an equation like
F=kr,
where k is some constant, is a perfectly legitimate equation. Thus we canalways represent forces by lines, which is very convenient, because once wehave drawn the line we no longer need the axes. Of course, we can quicklycalculate the three components as they change upon turning the axes, becausethat is just a geometric problem.
现在，让我们考察矢量的一些属性。作为矢量的例子，我们可以提到矢量速度、动量、力、和加速度。对于很多目的，通过一个箭头，来表示一个矢量，很方便，箭头指出了矢量作用的方向。为什么我们可以用一个箭头，来表示力？因为它有同样的数学转换属性，就像“空间的一步”一样。这样，我们就可以在一个图表中表示它，就好像它是一步一样，利用一个标量，这样，一个力或牛顿的单位，就相应于一定的方便的长度。一旦我们做了这个，所有的力，就都可以用长度来表示，因为，像：
F=kr,
这种方程，就是完全合法的方程，这里k是某个常数。这样，我们就总是可以用线段来表示力，这很方便，因为，一旦我们画了线，就不需要坐标轴了。当然，随着坐标轴的改变，三个分量也会变，我们可以快速地计算它们，因为，这只是一个几何问题。

11–5Vector algebra 11-5 矢量代数
Now we must describe the laws, or rules,for combining vectors in various ways. The first such combination is the additionof two vectors: suppose that a is a vector which in some particular coordinate system has the threecomponents (ax,ay,az), and that b is another vector which has the three components (bx,by,bz). Now let us invent three new numbers (ax+bx,ay+by,az+bz). Do these form a vector? “Well,” we might say, “they are three numbers,and every three numbers form a vector.” No, not every three numbers forma vector! In order for it to be a vector, not only must there be three numbers,but these must be associated with a coordinate system in such a way that if weturn the coordinate system, the three numbers “revolve” on each other, get “mixedup” in each other, by the precise laws we have already described. So thequestion is, if we now rotate the coordinate system so that (ax,ay,az)become (ax′,ay′,az′)and (bx,by,bz) become (bx′,by′,bz′), what do (ax+bx,ay+by,az+bz)become? Do they become (ax′+bx′,ay′+by′,az′+bz′)or not? The answer is, of course, yes, because the prototypetransformations of Eq. (11.5)constitute what we call a linear transformation. If we apply thosetransformations to ax and bx to get ax′+bx′, we find that the transformed ax+bxis indeed the same as ax′+bx′. When a and b are “added together” in this sense, they will form a vector which wemay call c . We would write this as
c=a+b.
Now c has the interesting property
c=b+a,
as we can immediately see from its components. Thus also,
a+(b+c)=(a+b)+c.
We can add vectors in any order.
现在，我们要讲一些规律或规则，为的是把矢量，以不同的方式结合起来。这类结合的第一个，就是两个矢量的加法：假设a是一个矢量，它在某具体坐标系中，有三个分量， (ax,ay,az), b 是另一个矢量，有分量 (bx,by,bz)。现在，我们发明三个新的数字 (ax+bx,ay+by,az+bz)。这些数字会形成一个矢量吗？“好”，我们可以说：“它们是三个数字，每三个数字形成一个矢量。”不，并不是三个数字形成一个矢量！为了让其成为一个矢量，不仅要有三个数字，而且，它们要以某种方式，与一个坐标系统相联，如果我们转换此坐标系，这三个数字，通过我们已经讲过的精确的规律，相互“旋转”，相互“混合”。于是，问题就是，如果我们现在旋转坐标系，让(ax,ay,az)变成 (ax′,ay′,az′)， (bx,by,bz) 变成(bx′,by′,bz′)那么，(ax+bx,ay+by,az+bz)会变成什么？会变成 (ax′+bx′,ay′+by′,az′+bz′)吗? 答案当然是：是的，因为方程（11.5）的原型变换，构成了我们称为线性变换的东西。如果我们把这些变换，应用于ax和 bx ，以得到 ax′+bx′ ，我们发现，被变换的ax+bx ，确实与ax′+bx′ 一样。当a和b，以这种意义，“被加起来”时，它们就形成一了个矢量，我们称之为 c 。我们把它写成：
c=a+b.
现在，我们可以从它的分量，立即看出，c有一个有趣的特性：
c=b+a。
这样，就有：
a+(b+c)=(a+b)+c.
即矢量相加，与顺序无关。

Fig. 11–4.The addition of vectors. 图11-4 矢量的相加

What is the geometric significance of a+b? Suppose that a and b were represented by lines on a piece of paper, what would clook like? This is shown in Fig. 11–4. Wesee that we can add the components of b to those of a most conveniently if we place the rectangle representing the componentsof b next to that representing the components of a in the manner indicated. Since b just “fits” into its rectangle, as does a into its rectangle, this is the same as putting the “tail” of bon the “head” of a , the arrow from the “tail” of a to the “head” of b being the vector c . Of course, if we added a to b the other way around, we would put the “tail” of a on the “head” of b , and by the geometrical properties of parallelograms we would get thesame result for c . Note that vectors can be added in this way without reference to anycoordinate axes.
a+b的几何意义是什么呢？假设a和b，被纸上的线段所表示，那么c看上去会是什么样子呢？这如图11-4所示。我们看到，如果我们把表示b的分量的矩形，与表示a的分量的矩形，按所示方式，放在一起，那么，我们就可以非常方便地，把b的分量与a的分量相加。由于b正好能适应它的四边形，正如a正好能适应它的四边形一样，这与把b的“尾”，放在a的“头”上，是一样的，从a的“尾”到b的“头”所构成的箭头，就是c。当然，如果我们以另一种方式，把a加到b上，那么，我们可以把a的“尾”，放在b的“头”上，通过平行四边形的几何属性，我们可以为c，得到同样的结果。注意，矢量可以这样相加，而不用参考任何坐标轴。

Suppose we multiply a vector by anumber α , what does this mean? We define it to mean a new vector whosecomponents are αax , αay , and αaz . We leave it as a problem for the student to prove that it isa vector.
假设我们用一个数 α，乘以一个矢量，这意味着什么呢？我们把它定义为一个新的矢量，其分量就是：αax , αay , 和αaz。这个问题，我们留给学生去证明：它是一个矢量。

Now let us consider vector subtraction. Wemay define subtraction in the same way as addition, but instead of adding, wesubtract the components. Or we might define subtraction by defining a negativevector, −b=−1b , and then we would add the components. It comes to the same thing.The result is shown in Fig. 11–5. Thisfigure shows d= a−b= a+(−b) ; we also note that the difference a−bcan be found very easily from a and b by using the equivalent relation a=b+d. Thus the difference is even easier to find than the sum: we justdraw the vector from b to a , to get a−b !
现在，我们考虑矢量减法。我们可以像定义加法那样，定义减法，不同的是，对于分量，不是加，而是减。或者，我们可以通过定义一个负的矢量−b=−1b，来定义减法，然后，我们就可以让分量相加。结果一样。如图11-5所示。这个图形显示了d= a−b= a+(−b)；我们也注意到，利用同等的关系a=b+d ，差值a−b 可以很容易地从a和 b中得到。这样做，比求和更容易得到差值：我们只需从b到 a，画出矢量，就可得到 a−b。

Fig. 11–5.The subtraction of vectors. 图11-5 矢量减法。

Next we discuss velocity. Why is velocity avector? If position is given by the three coordinates (x,y,z), what is the velocity? The velocity is given by dx/dt , dy/dt , and dz/dt . Is that a vector, or not? We can find out by differentiating theexpressions in Eq. (11.5)to find out whether dx′/dt transforms in the right way. We see that the components dx/dtand dy/dt do transform according to the same law as x and y , and therefore the time derivative is a vector. So thevelocity is a vector. We can write the velocity in an interesting way as
v=dr/dt.
下面，我们讨论矢速。为什么矢速是一个矢量？如果位置是通过三个坐标(x,y,z)而被给予的，那么什么是矢速呢？矢速就是通过dx/dt , dy/dt , 和 dz/dt而被给予的。这是否就是一个矢速呢？我们可以通过对方程（11.5）求微分，来找出dx′/dt ，是否是按照正确的方式，转换了。我们看到，分量 dx/dt 和 dy/dt，确实如x和 y一样，依据同样的规律，转换了，因此，时间的导数，就是一个矢量。所以，矢速就是一个矢量。我们可以用一种有趣的方式，把矢速写为：
v=dr/dt.

What the velocity is, and why it is a vector, can also be understoodmore pictorially: How far does a particle move in a short time Δt ? Answer: Δr , so if a particle is “here” at one instant and “there” at anotherinstant, then the vector difference of the positions Δr=r2−r1, which is in the direction of motion shown in Fig. 11–6,divided by the time interval Δt=t2−t1 , is the “average velocity” vector.
什么是矢速，为什么它是一个矢量，也可以更图示性地被理解：在一个短的时间间隔Δt内，一个粒子能移动多远？答案是：Δr，于是，如果一个粒子，某一时刻在“这里”，另一时刻在“那里”，那么，两个位置间的矢量差就是Δr=r2−r1，它在运动的方向，如图11-6所示，它被时间间隔Δt=t2−t1相除，就是矢量：“平均矢速”。

Fig. 11–6.The displacement of a particle ina short time interval Δt=t2−t1 . 图11-6 在一个短的时间间隔Δt=t2−t1的一个粒子的位移。

In other words, by vector velocity we meanthe limit, as Δt goes to 0 , of the difference between the radius vectors at the time t+Δtand the time t , divided by Δt :
v=limΔt→0(Δr/Δt)=dr/dt.(11.10)
Thus velocity is a vector because it is the difference of two vectors.It is also the right definition of velocity because its components are dx/dt, dy/dt , and dz/dt . In fact, we see from this argument that if we differentiate anyvector with respect to time we produce a new vector. So we have several ways ofproducing new vectors: (1) multiply by a constant, (2) differentiatewith respect to time, (3) add or subtract two vectors.
换句话说，通过矢量矢速，我们意味的是，在时间t+Δt 和时间t的两个半径矢量之间的差，当Δt趋于零时，取极限，且除以 Δt：

(11.10)
这样，矢速就是一个矢量，因为它是两个矢量的差。它也是矢速的正确定义，因为它的分量是：dx/dt , dy/dt , and dz/dt。事实上，我们从这个论证可以看出，任何矢量对时间求微分，结果是一个新的矢量。于是，我们就有几种产生新的矢量的方法：（1）乘以一个常数，（2）对时间求微分，（3）两个矢量相加或相减。

11–6Newton’s laws in vector notation 11-6牛顿规律在矢量中的标记法
In order to write Newton’s laws in vectorform, we have to go just one step further, and define the acceleration vector.This is the time derivative of the velocity vector, and it is easy todemonstrate that its components are the second derivatives of x , y , and z with respect to t :
a=dv/dt=(d/dt)(dr/dt)=d2r/dt2, (11.11)
ax=dvx/dt=d2x/dt2, ay=dvy/dt=d2y/dt2, az=dvz/dt=d2z/dt2. (11.12)
With this definition, then, Newton’s laws can be written in this way:
ma=F (11.13)
or
m(d2r/dt2)=F. (11.14)
为了用矢量的形式，来写牛顿规律，我们必须在往前走一步，去定义加速度这个矢量。这就是矢量速度的时间导数，很容易演示，它的分量，就是x , y ,和 z 对时间的二阶导数：
a=dv/dt=(d/dt)(dr/dt)=d2r/dt2, (11.11)
ax=dvx/dt=d2x/dt2, ay=dvy/dt=d2y/dt2, az=dvz/dt=d2z/dt2. (11.12)
用这个定义，牛顿规律可重写为：
ma=F (11.13)
或：
m(d2r/dt2)=F. (11.14)

Now the problem of proving the invariance of Newton’s laws underrotation of coordinates is this: prove that a is a vector; this we have just done. Prove that F is a vector; we suppose it is. So if force is a vector, then,since we know acceleration is a vector, Eq. (11.13)will look the same in any coordinate system. Writing it in a form which doesnot explicitly contain x ’s, y ’s, and z ’s has the advantage that from now on we need not write threelaws every time we write Newton’s equations or other laws of physics. We writewhat looks like one law, but really, of course, it is the three laws forany particular set of axes, because any vector equation involves the statementthat each of the components is equal.
坐标旋转时，牛顿规律有不变性，要证明此点，可以这样：证明a是一个矢量，这我们已经做了。证明F是一个矢量，我们假设它是。于是，如果力是一个矢量，那么，由于我们知道加速度是一个矢量，方程（11.13）在任何坐标系中，看上去都一样。以这种形式来写方程，即它不明显的包含着x、y、和z的坐标，好处就是：从现在开始，每次我们的写牛顿方程或其他物理规律时，不用写三条规律了。我们所写，看上去好像只是一条规律，但实际上，对任何具体的坐标轴来说，它都是三条规律，因为，任何矢量方程，都包含着一个声明：每个分量都是相等的。

Fig. 11–7.A curved trajectory. 图11-7 弯曲的轨迹

Fig. 11–8.Diagram for calculating theacceleration. 图11-8 图解：如何计算加速度。
The fact that the acceleration is the rateof change of the vector velocity helps us to calculate the acceleration in somerather complicated circumstances. Suppose, for instance, that a particle ismoving on some complicated curve (Fig. 11–7) andthat, at a given instant t1 , it had a certain velocity v1 , but that when we go to another instant t2 a little later, it has a different velocity v2 . What is the acceleration? Answer: Acceleration is the difference inthe velocity divided by the small time interval, so we need the difference ofthe two velocities. How do we get the difference of the velocities? To subtracttwo vectors, we put the vector across the ends of v2 and v1 ; that is, we draw Δv as the difference of the two vectors, right? No! That onlyworks when the tails of the vectors are in the same place! It has nomeaning if we move the vector somewhere else and then draw a line across, sowatch out! We have to draw a new diagram to subtract the vectors. In Fig. 11–8, v1and v2 are both drawn parallel and equal to their counterparts in Fig. 11–7, andnow we can discuss the acceleration. Of course the acceleration is simply Δv/Δt. It is interesting to note that we can compose the velocitydifference out of two parts; we can think of acceleration as having twocomponents, Δv∥ , in the direction tangent to the path and Δv⊥ at right angles to the path, as indicated in Fig. 11–8. Theacceleration tangent to the path is, of course, just the change in the lengthof the vector, i.e., the change in the speed v :
a∥=dv/dt. (11.15)
The other component of acceleration, at right angles to the curve, iseasy to calculate, using Figs. 11–7and 11–8.In the short time Δt let the change in angle between v1 and v2 be the small angle Δθ . If the magnitude of the velocity is called v , then of course
Δv⊥=vΔθ
and the acceleration a will be
a⊥=v(Δθ/Δt).
Now we need to know Δθ/Δt , which can be found this way: If, at the given moment, the curve isapproximated as a circle of a certain radius R , then in a time Δt the distance s is, of course, vΔt , where v is the speed.
Δθ=(vΔt)/R,orΔθ/Δt=v/R.
Therefore, we find
a⊥=v2/R, (11.16)
as we have seen before.
加速度是矢量速度的变化率，这一事实，可帮我们在某些相当复杂的情况下，计算加速度。例如，假设一个粒子，是在某种复杂的曲线上移动（图11-7），在某一瞬间 t1，它有一定的速度v1，但稍后，我们走到另一个瞬间t2，它有一个不同的速度v2，加速度是什么呢？答案：加速度就是矢速的差，除以时间间隔，所以，我们就需要两个矢速的差。我们如何能才得到矢速的差呢？要让两个矢速相减，我们把矢量，放在v2和v1的头端，也就是说，我们画出Δv，以作为两个矢量的差，对吗？不！只有当两个矢量的尾，在同一个地方时，才有效！如果我们把这个矢量，移到另外一个地方，然后，在它们的头端之间画一条线，这并没有意义，所以，小心了！我们必须画一个新的图解，以做矢量减法。在图11-8中，v1和 v2，与它们在图11-7中的对应物，平行且相等，现在，我们才可以讨论加速度了。当然，加速度简单地说，就是Δv/Δt。
有一点值得注意，也很有趣，即我们可以用两个部分，构成矢量差；我们可以把加速度，考虑为有两个分量，Δv∥ ,在与路径相切的方向，和Δv⊥，在与路径像垂直的方向，如图11-8所示。与路径相切的加速度，当然就是矢速在长度上的变化，亦即，速度 v，的变化：
a∥=dv/dt. (11.15)
加速度的另一分量，与曲线相垂直，利用图11-7和11-8，容易计算。在短的时间Δt内，设v1和v2 之间角度的变化，为一个小的角度 Δθ。如果矢速的大小，被称为v，那么当然：
Δv⊥=vΔθ
加速度a 就将是：
a⊥=v(Δθ/Δt).
现在我们需要知道Δθ/Δt，它可这样得到：在被给予的瞬间，如果曲线接近于一个半径为R的圆，那么，在Δt时间内，距离s当然就是vΔt ,这里v就是速度。
Δθ=(vΔt)/R, 或Δθ/Δt=v/R.
因此，我们得到：
a⊥=v2/R, (11.16)
正如以前所见。

11–7Scalar product of vectors 11-7 矢量的标量积
Now let us examine a little further theproperties of vectors. It is easy to see that the length of a step inspace would be the same in any coordinate system. That is, if a particularstep r is represented by x,y,z , in one coordinate system, and by x′,y′,z′ in another coordinate system, surely the distance r=|r|would be the same in both. Now
r=x2+y2+z2−−−−−−−−−−√
and also
r′=x′2+y′2+z′2−−−−−−−−−−−√.
So what we wish to verify is that these two quantities are equal. Itis much more convenient not to bother to take the square root, so let us talkabout the square of the distance; that is, let us find out whether
x2+y2+z2=x′2+y′2+z′2. (11.17)
It had better be—and if we substitute Eq. (11.5)we do indeed find that it is. So we see that there are other kinds of equationswhich are true for any two coordinate systems.
现在让我们进一步检查矢量的属性。很容易看到，在任何坐标系中，一步的长度，都是一样的。也就是说，如果一个具体的 r，在一个坐标系中，被 x,y,z表示，在另一个坐标系中，被 x′,y′,z′表示，那么可以肯定的是，在两个坐标系中，距离 r=|r|，应该一样的。现在：
（11-7-1）
还有：
（11-7-2）
所以呢，我们希望验证的，就是这两个量是相等的。把根号去掉，可能更方便，所以，让我们来讨论距离的平方，也就是说，让我们看下面是否成立：
x2+y2+z2=x′2+y′2+z′2. (11.17)
它最好成立，如果我们用方程（11.5）做替换，我们确实可以得出，是成立的。所以，我们看到，对于任何两个坐标系来说，还有其它种类的方程，{在其中}同时为真。

Something new is involved. We can produce anew quantity, a function of x , y , and z , called a scalar function, a quantity which has no directionbut which is the same in both systems. Out of a vector we can make a scalar. Wehave to find a general rule for that. It is clear what the rule is for the casejust considered: add the squares of the components. Let us now define a newthing, which we call a⋅a . This is not a vector, but a scalar; it is a number that is the samein all coordinate systems, and it is defined to be the sum of the squares ofthe three components of the vector:
a⋅a=a2x+a2y+a2z. (11.18)
有了新的事物了。我们可以产生一个新的量，一个x , y , 和 z的函数，被称为标量函数，它没有方向，但在两个坐标系中，它是相同的。我们可以从一个矢量中，得出一个标量。我们必须为此事，找到一个一般规则。很清楚，此规则，正是为了刚刚考虑过得情况：分量平方的加法。现在，让我们定义一个新的事物，我们称之为a⋅a。它不是一个矢量，而是一个标量；它是一个数字，在所有的坐标系中都一样，它被定义为：矢量的三个分量的平方和：
a⋅a=a2x+a2y+a2z. (11.18)

Now you say, “But with what axes?” It does not depend on the axes, theanswer is the same in every set of axes. So we have a new kind ofquantity, a new invariant or scalar produced by one vector“squared.” If we now define the following quantity for any two vectors aand b :
a⋅b=axbx+ayby+azbz, (11.19)
we find that this quantity, calculated in the primed and unprimedsystems, also stays the same. To prove it we note that it is true of a⋅a , b⋅b , and c⋅c , where c=a+b . Therefore the sum of the squares (ax+bx)2+(ay+by)2+(az+bz)2 will be invariant:
(ax+bx)2+(ay+by)2+(az+bz)2=(ax′+bx′)2
+(ay′+by′)2+(az′+bz′)2. (11.20)
If both sides of this equation are expanded, there will be crossproducts of just the type appearing in Eq. (11.19),as well as the sums of squares of the components of a and b . The invariance of terms of the form of Eq. (11.18)then leaves the cross product terms (11.19)invariant also.
现在你说：“但是，与哪个坐标轴在一起呢？”它并不依赖于坐标轴，在每组坐标轴中，答案一样。所以，我们就有了某种新的量，是由对同一个矢量求平方，所产生的一个新的不变量或标量。对于任意的两个矢量a和b，如果我们现在，为其定义下面的量：
a⋅b=axbx+ayby+azbz, (11.19)
我们发现，这个量，无论是在基本的系统中，还是在非基本的系统中，都保持一样。要证明它，我们注意到，对于a⋅a , b⋅b , 和 c⋅c来说，是真的，这里 c=a+b。因此，平方(ax+bx)2+(ay+by)2+(az+bz)2的和，就是不变的：
(ax+bx)2+(ay+by)2+(az+bz)2=(ax′+bx′)2
+(ay′+by′)2+(az′+bz′)2. (11.20)
如果这个方程的两边，都被展开，将有叉积，就像在方程（11.19）中的那种类型，同样也有a和b的分量的平方和。方程（11.8）的形式的项，是不变的，因此，就导致了方程（11.19）的叉积项，也不变。{？不变性，指结果不随坐标系而变？}

The quantity a⋅b is called the scalar product of two vectors, a and b , and it has many interesting and useful properties. For instance, itis easily proved that
a⋅(b+c)=a⋅b+a⋅c. (11.21)
Also, there is a simple geometrical way to calculate a⋅b , without having to calculate the components of a and b : a⋅b is the product of the length of a and the length of b times the cosine of the angle between them. Why? Suppose that wechoose a special coordinate system in which the x -axis lies along a ; in those circumstances, the only component of a that will be there is ax , which is of course the whole length of a . Thus Eq. (11.19)reduces to a⋅b=axbx for this case, and this is the length of a times the component of b in the direction of a , that is, bcosθ :
a⋅b=abcosθ.
量a⋅b被称为：两个矢量a和 b 的标量积，它有很多有趣和有用的属性。例如，很容易证明：
a⋅(b+c)=a⋅b+a⋅c. (11.21)
同样，计算a⋅b，也有一种简单的几何方法，不用计算a和b的分量：a⋅b是a的长度与b的长度，乘以它们夹角的cosine 。为什么？假设我们选择一种坐标系，在其中，x轴沿着a；在这种情况下，a的唯一分量，将是ax，它就是a的全部长度。这样，在此情况下，方程（11.19）就简化为 a⋅b= axbx，这就是 a的长度，乘以 b在 a的方向上的分量，即bcosθ:
a⋅b=abcosθ.