财经社区女性社区汽车社区军事社区文学社区社会社区娱乐社区游戏社区个人空间
上一主题:反相的数学高手物理高手们都哪去... 下一主题:网上相对论学者以真理的化身出头...
洛仑兹变换的具体数学推导
[楼主] 作者:fuj0  发表时间:2008/04/12 11:17
点击:1724次

关于洛仑兹变换,网上讨论很多。我找了两个版本的推导过程。不过是英文的。

本人在阅读中文版的高等教材觉得很吃力。怪自己的语文不好,也怪翻译得有问题。

论讨争议最大的是第一个版本。我找的这篇原文写得很清楚。想必可以解决一些疑惑。

 版本1

FOR the relative orientation of the co-ordinate systems indicated in Fig. 2, the x-axes of both systems permanently coincide. In the present case we can divide the problem into parts by considering first only events which are localised on the x-axis. Any such event is represented with respect to the co-ordinate system K by the abscissa x and the time t, and with respect to the system k' by the abscissa x' and the time t'. when x and t are given.   1
  A light-signal, which is proceeding along the positive axis of x, is transmitted according to the equation
x = ct
or

Since the same light-signal has to be transmitted relative to k' with the velocity c, the propagation relative to the system k' will be represented by the analogous formula

Those space-time points (events) which satisfy (1) must also satisfy (2). Obviously this will be the case when the relation (很明显,如果公式3 成立,公式1,2就会成立)

is fulfilled in general, where indicates a constant; for, according to (3), the disappearance of (xct) involves the disappearance of (x'ct').
   2
  If we apply quite similar considerations to light rays which are being transmitted along the negative x-axis, we obtain the condition

   3
  By adding (or subtracting) equations (3) and (4), and introducing for convenience the constants a and b in place of the constants and where

and

we obtain the equations

   4
  We should thus have the solution of our problem, if the constants a and b were known. These result from the following discussion.   5
  For the origin of k' we have permanently x' = 0, and hence according to the first of the equations (5)

   6
  If we call v the velocity with which the origin of k' is moving relative to K, we then have

   7
  The same value v can be obtained from equation (5), if we calculate the velocity of another point of k' relative to K, or the velocity (directed towards the negative x-axis) of a point of K with respect to K'. In short, we can designate v as the relative velocity of the two systems.   8
  Furthermore, the principle of relativity teaches us that, as judged from K, the length of a unit measuring-rod which is at rest with reference to k' must be exactly the same as the length, as judged from K', of a unit measuring-rod which is at rest relative to K. In order to see how the points of the x'-axis appear as viewed from K, we only require to take a “snapshot” of k' from K; this means that we have to insert a particular value of t (time of K), e.g. t = 0. For this value of t we then obtain from the first of the equations (5)
x' = ax.
   9
  Two points of the x'-axis which are separated by the distance x'=1 when measured in the k' system are thus separated in our instantaneous photograph by the distance

   10
  But if the snapshot be taken from K'(t' = 0), and if we eliminate t from the equations (5), taking into account the expression (6), we obtain

   11
  From this we conclude that two points on the x-axis and separated by the distance 1 (relative to K) will be represented on our snapshot by the distance

   12
  But from what has been said, the two snapshots must be identical; hence x in (7) must be equal to x' in (7a), so that we obtain

   13
  The equations (6) and (7b) determine the constants a and b. By inserting the values of these constants in (5), we obtain the first and the fourth of the equations given in Section XI>.

   14
  Thus we have obtained the Lorentz transformation for events on the x-axis. It satisfies the condition

   15
  The extension of this result, to include events which take place outside the x-axis, is obtained by retaining equations (8) and supplementing them by the relations

In this way we satisfy the postulate of the constancy of the velocity of light in vacuo for rays of light of arbitrary direction, both for the system K and for the system K'. This may be shown in the following manner.
   16
  We suppose a light-signal sent out from the origin of K at the time t = 0. It will be propagated according to the equation

or, if we square this equation, according to the equation

   17
  It is required by the law of propagation of light, in conjunction with the postulate of relativity, that the transmission of the signal in question should take place—as judged from K'—in accordance with the corresponding formula
r' = ct'
or,

In order that equation (10a) may be a consequence of equation (10), we must have

   18
  Since equation (8a) must hold for points on the x-axis, we thus have = 1; for (11) is a consequence of (8a) and (9), and hence also of (8) and (9). We have thus derived the Lorentz transformation.   19
  The Lorentz transformation represented by (8) and (9) still requires to be generalised. Obviously it is immaterial whether the axes of K' be chosen so that they are spatially parallel to those of K. It is also not essential that the velocity of translation of K' with respect to K should be in the direction of the x-axis. A simple consideration shows that we are able to construct the Lorentz transformation in this general sense from two kinds of transformations, viz. from Lorentz transformations in the special sense and from purely spatial transformations, which corresponds to the replacement of the rectangular co-ordinate system by a new system with its axes pointing in other directions.   20
  Mathematically, we can characterise the generalised Lorentz transformation thus: It expresses x', y', z', t', in terms of linear homogeneous functions of x, y, z, t, of such a kind that the relation

is satisfied identically. That is to say: If we substitute their expressions in x, y, z, t, in place of x', y', z', t', on the left-hand side, then the left-hand side of (11a) agrees with the right-hand side.

 

版本2

4. Lorentz Transformations

We are now prepared to deal with the problem of how to relate an arbitrary and coordinate in an unmoving frame to an arbitrary and coordinate in a moving frame. For this we imagine an experiment in which we get rid of the mirror altogether, and simply have the flash bulb at and the photodetector at . The flash bulb flashes at an arbitrary time at , and the light is received at time at . By choosing the initial flash time carefully, it is possible to arrange an arbitrary time at which the light is received. Since the light travels at speed , we have

 

That is what happens in the primed frame. What happens in the unprimed frame? The light starts at some time and travels to the final point at time . Since the light travels at , and the photodetector is moving away from the flash at velocity , and the length of the separation is in this reference frame, the time it takes is given by

At the end of the experiment, the flash bulb has moved a distance from the origin, and the photodetector is a distance beyond that, so

 

Our goal is to obtain equations relating the coordinates and to and . We already have three equations, but we have five unknowns: , , , , and . We need two more equations. Fortunately, since the initial flash was at , we can use the time dilation relation

and the Lorentz contraction relation

to find the desired relationships. From here on out it is all algebra.

First, combine equations (E) and (C) to find a relation for :

That was relatively easy.

Much harder is finding the expression for . We start by solving equation (A) for :

We then use (D) to substitute for and (E) to solve for :

We now solve (B) for , yielding

If we now substitute equation (C) for , we have an expression for exclusively in terms of and , namely

 

We have the transformation properties of and in the two coordinate systems. What we have not been careful about is the transformations of the other coordinates and . Since the motion is only in the direction, it is not surprising that these coordinates are not affected by Lorentz transformations. So the full equations for the Lorentz transformations are given by

 

These equations will be discussed at some length in the next lecture

本帖地址:http://club.xilu.com/hongbin/msgview-950451-107106-1.html[复制地址]
上一主题:反相的数学高手物理高手们都哪去... 下一主题:网上相对论学者以真理的化身出头...
 [2楼]  作者:刘岳泉  发表时间: 2008/04/12 13:25 

    第一种方法其实就是解释迈-莫试验用往返平均时间把速度v的一级项给“平均”掉的结果,所以我才说如此解释迈-莫试验很弱智!留下的二级教应后就你说我“收缩”我说你“收缩”,弄得整个物理学界打大架已经百年了。

.

    第二种方法是将两个毫不相关而且也根本就不是同一光信号速度却用σ连接起来,这哪里是搞物理科学?而简直是一种公然的神学欺骗行为!

.

第三种方法与我的方法有何区别?容我再重新转贴过来请看清楚些﹕

.

 

    相对论的出现已经把物理科学引响了斜路,也使整个二十世纪的科学体系使人们从根本上混淆了科学与伪科学之间的区别,而整个相对论理论体系都是不管自然客观实在的一种纯数学游戏的拼凑结果。在这里,我们暂不考虑它是如何对"光速漂移"现象无视大量事实的歪曲解释,也不讨论它对历史上关于水星轨道进动数据错误计算结果的捕风捉影而建立的所谓"广义"理论,只讨论被人们普遍推崇备至成为"科学圣典"的洛伦兹变换,通过这种严格的"数学方法"来证明"爱因斯坦+相对论等于魔鬼",是否能唤起人们对这个"科学圣典"的质疑呢?

 

    我们还是按照人们关于洛伦兹变换的通俗约定,也承认分别用v和c表示运动和光速(其实用来表示重量或机器都是同样有效的)来建立"等式方程":

 

魔鬼= γ2魔鬼(1-v2/c2)

=γ2(石头v + 魔鬼 - 石头v - 魔鬼v2/c2)

=γ[ γ(魔鬼-石头v)+ γ(石头-魔鬼v/c2)v]

 

 

这里的 γ =1/(1-v2/c2)1/2被称之为"洛伦兹系数"。

 

    在以上方程中,我们假定以洛伦兹系数γ为扭带,将"v"暂时用数字"2"代替,如果"魔鬼"甩了两块"石头"得到"爱因斯坦",而用"石头"砸到两个"魔鬼"又得到"相对论",即

 

爱因斯坦=γ(魔鬼-石头v)     相对论=γ(石头-魔鬼v/c2)

 

那么我们就可以经过这种数学证明得出的

魔鬼=γ(爱因斯坦+相对论v)

 

"洛伦兹变换"结果,而且同样满足方程的"协变"要求,因为这里的"爱因斯坦"和"相对论"本身就是由"石头"和"魔鬼"构成的。

 

    在这里,我们并没有任何人身攻击的本意,只是以此为实例说明洛伦兹变换的物理意义,其中的符号的"v"既可以表达为"速度",也可以表达为其它任何有实际物理意义的数值,所以根据这里物理意义我们把"v"用"两个"作为数量词表达比较适宜(至于"洛伦兹系数"则既可以是仍意实数也可以是虚数),这样,我们就可以得出一个重要结论:

 

"爱因斯坦"加"两个相对论"以"洛伦兹系数"为扭带,就是"魔鬼"!

 

    现在既然人们已经把洛伦兹变换推崇备至为"科学圣典",那么我们在这里的"数学证明"同样应该属于"科学体系"的重要组成部份之一,是否还可以进一步作为"科学成果"申报?由于目前对于科学与伪科学之间根本就没有严格的区分标准,如果这样的洛伦兹变换方法是错误的,就应当还以科学的本来面目,杜绝这种伪科学思潮防碍人类科学事业的健康发展。

原贴地址:http://club.xilu.com/hongbin/msgview-950451-104915.html

※※※※※※
相对论误导科学走斜路,道是非曲折待历史见证;引力场以太旧貌焕新颜,作定海神柱将扭转乾坤。持一份平和心态闯科海,能生命当歌也荣辱以乐;唯真理艰难求索胸怀广,可气吞山河再平展时空。

 [3楼]  作者:刘久明  发表时间: 2008/04/12 20:55 

和fuj0先生正相反,我看英文很吃力。
第二个版本看得我头疼,不愿再看下去,不知谁能用简单的语言描述一下是什么思路?免得我明天还得再看上一增天。
关于第一个版本,这个较常见。
首先是承认不承认光速不变的假设,不承认则(1)、(2)两式不能同时成立。承认才可以继续讨论,所以我对这个变换的推导原本就不感兴趣。
第二是光速问题是否满足相对性原理的问题,如果不满足,则(3)、(4)两式不能成立。承认满足才可以继续讨论,否则即刻停止。
第三是相对性原理的表述形式问题,无特权系则(3)、(4)式表述不当。
由于无特权系:所以:
(x'-ct')=λ(x-ct)........(3)
(x-ct)=μ(x'-ct')........(4)
以后没有意义不再推。
第四洛变换是不是伽变换的近似。伽变换是坐标系度规变换,而爱的洛变换是参照系观点变换(洛自己的变换到底是什么与爱没有关系),坐标系度规变换可以将(3)、(4)写成:
(x'-c't')=λ(x-ct)........(3)
(x-ct)=1/λ(x'-c't')........(4)
度规变换有实际意义,这是初中数学课程,这里不必讨论,但这也限制为不能使用波速进行迭加。

我这几天正在考虑把两个经常分别讨论的问题放到一起来研究,结果给我自己也造糊涂了。
1、一个观测者以光速飞离一个静止的时钟,问观测者眼中时钟的走时与自己手表的对比情况。
2、一个时钟以光速飞离观测者,问观测者眼中时钟的走时与自己手表的对比情况。
当将光作为波来考虑时,这两种情况下,前者可能得出时钟停止的表象,而后者只能得出时钟变慢的表象,即时钟和观测者不平权。
 [4楼]  作者:刘岳泉  发表时间: 2008/04/12 21:14 

对【3楼】说:
    有数学公式我根本就不用去看英文了,我觉得在所有语言中只有数学语言最简练,也许你看得头疼的并不是英语而恰恰是数学。其实这些与变换毫无关系,你把xyz的平方和当成距离除以时间会是什么,只有数学才最容易理解。

※※※※※※
相对论误导科学走斜路,道是非曲折待历史见证;引力场以太旧貌焕新颜,作定海神柱将扭转乾坤。持一份平和心态闯科海,能生命当歌也荣辱以乐;唯真理艰难求索胸怀广,可气吞山河再平展时空。
 [5楼]  作者:西陆陈诚  发表时间: 2008/04/13 09:57 

对【3楼】说:
    fuj0先生还是有可取之处,他没有跑,他可以跟你探讨到底,相比之下这种精神值得称赞:那个自称“考研读博”的“159.226.124.*”先生,撂下一堆空话大话之后,现在则不知到哪里“躲冬”去了!
    岳泉之言很有理,不必看英文,把那些英文都去掉后,剩下的实际上都是中学的数学过程!!
可以说,无论何种版本的Lorentz变换,其数学过程的推导都没有错,错的东西,实际上就是物理过程,比如,认为x=0和x'+vt'=0是同一回事,其实这是不对的!!x=0指的是O点,而x'+vt'=0则指的是O'点!!为了迎合相对论,把x=0和x'+vt'=0看成是同一回事,这几乎是所有版本Lorentz变换的通病!!
    实际上,如果将就用相对论所说的“相对性原理”来检验时,便可以发现最终结果必定是v=0,也就是客观上坐标系K'并没有相对坐标系K运动(原因就是认为x=0=x'+vt')!!!…谢谢关注(西陆陈诚)
 [6楼]  作者:拉方  发表时间: 2008/04/13 11:54 

不管中文的英文的,不管是伽利略的、洛仑兹的抑或爱因斯坦的变换,不管数学推导对错与否,一律都是伪科学!
道理很简单:都是基于“惯性系平权轮”的结果!

彻底推翻“惯性系平权轮”紧紧把握挑战相对论物理学会会长的“明确时空观”,恢复物理科学本来面目,自然会使20世纪为止的人类“普通物理学”画上圆满句号,走进《高等物理学》阶段!

沉舟侧畔千帆过病树前头万木春!
……
努力吧朋友,曙光就在前头!

倒相对论学者们,一齐喊口号:让相对论患者们都必须要老老实实的“站队”!
然后再讨论问题。
“站队”,就是我们挑战制胜的操作法宝!

要知道:一“站队”,相对论患者就懵,最害怕“站队”,所以他们无例外都不敢“站队”!

否则,因为人类语言文字本身逻辑恶性循环,他们总会振振有词。

一“站队”,就没词了!


希望朋们喜欢。

 [7楼]  作者:拉方  发表时间: 2008/04/13 12:05 

倒相的朋友们一定要自觉地同“惯性系平权论”划清界限!
切切!

切不可以用自己的“变换”去推翻别人的/相对论的“变换”。

不要50步笑百步。

 

 [8楼]  作者:拉方  发表时间: 2008/04/13 12:37 

赫赫!
没见过理论检验真理的人都可以大开眼界了:

“站队”就是21世纪人类检验一切真理的唯一理论标准(唯二理论标准有待于哲学家接着创建)。
并且“站队”是全体人类,无例外的,必须无条件接受的,理论检验标准!
全体人类,无例外,都不可以成为这种理论检验的局外人!
换句话说,都必须要老老实实地接受这种唯一标准的理论检验!

谢谢关注。
 [9楼]  作者:qstt  发表时间: 2008/04/14 14:36 

to fuj0,
怎么说你呢?你可爱而缺思考。
周末玩耍归来看你:
我对Einstein版本的Lorentz变换方程推导过程的较简单数学批驳,没看到你一句数学上的反驳,反而粘贴两个你未标示谁的Lorentz变换方程推导过程的版本出来,不知你是何用意?
你怪我好抓别人把柄,可你自己要弄出不少把柄来,别人就要说几句。别人可没耐心象我那样给你再较多地逐条分析!
版本1实际就是Einstein版本的抄录本,版本2比版本1更荒诞!
我再抓你一个把柄,你说你:“---阅读中文版的高等教材觉得很吃力。---也怪翻译得有问题。---我找的这篇原文写得很清楚。---”,这是什么话? fuj0先生究竟是外国朋友还是中国网友?你中文教材读得很吃力,还怪翻译水平不行,可是读外文却很清楚,却用中国话翻译不出来?你应比别人翻译得更好啊?奇才!
我告诉你,任何版本的Lorentz变换方程推导过程均是同类错误!可以说其公式后面带编号的数学式都是瞎凑合!即把在物理上不相关的数学式拼凑在一起“推理”!这个问题要解说清楚不是三言两语!所以在上次批驳时未说,现在也没那时间想说,只空泛说几句你自己思索!
我并不反对说光速基本不变!即赞同麦克思韦方程!我还赞同光速可变!矛盾吗?能化解就不是矛盾!这就是一些自我标榜的人反感的自然唯物辩证法,绝对与相对的关系!速度是怎么得来的?你好好思考一下!速度既可相对某物质系统整体基本不变,也可相对二物质系统整体可变!把两个各自相对二物质系统整体基本各自不变的公式,拿到两个相关的相对二物质系统整体速度可变的物理环境去瞎推!你自己思考吧!你敢反对相对二物质系统整体速度可变吗?包括光线在玻璃中和空气中的速度不等同!还包括光线与相对玻璃和相对玻璃运动的另一物质系统之间的速度!即也包括“你的”版本2中的公式(B),L/t-t.=c-v,这个c-v=c',你能否定吗!公式(A)的L'/t'-t.'=c,只能说是该系统的光速c不变,你说公式(A)和公式(B)是相同的东西吗?就算可用,也可把x'和x对调使用,你说那个为真?到处偷换概念之处就不想再给你分析了!版本1的(1).(2)两式都是光速c不变,可它论述的环境就完全类同于光线与相对玻璃和相对玻璃运动的另一物质系统之间的速度!观测者无论站在玻璃系还是相对玻璃运动的另一物质系统上,都观测得光速相对玻璃不变为c',总不能说光速相对那个同时相对玻璃运动的另一物质系统之间的速度也为c'吧?这就是Einstein把这种不同的光速说成是各自光速c不变(1).(2)两式的混乱搅和在一起的推导!版本2却在既在否定光速不变又在胡乱表达坐标长度等等,你自己思索吧!
我再给你创造一个根据版本1式(3)或式(4)来的公式:
abcdef(x'-ct')=ghijkl(x-ct) - - - - - - - - - - - (3a)
mnopqr(x'+ct')=stuwyz(x+ct) - - - - - - - - - - - (4a)
怎么样?你敢反对说它不成立!推论吧!

 [10楼]  作者:qstt  发表时间: 2008/04/14 15:00 

更正:上文的:相对玻璃运动的另一物质系统观测者观测该光速不再为c',而是只能通过自己观测得的光信号相对自己的光速,和观测得自己与玻璃的相对运动速度,而推理出光信号相对玻璃运动的速度仍为c'。
 [11楼]  作者:tongzr  发表时间: 2008/04/14 23:57 


2) 查核相对论对洛仑兹变换推演过程,相对运动两坐标系对光的描述等式中
x'=ax-bct and ct'=act-bx ............... (5)
......
对于K'的原点我们永远有x'=0, 因此按照 (5) 的第一个方程
x=bct/a
........

不难看出,对于K'的原点我们永远有x'=0, 按上面相对论的铺垫关系式,必须同时有t'=0; ,当然,同时还必须有:x=0; t=0.
所以,x'=ax-bct (x'=0,x=0; t=0)→ x=bct/a 不成立!是零除,以下的方程演释导致增根。

这样看得更清楚:x'=ax-bct → x'/t=ax/t-bc → x'/t=av-bc 当 x'=0,x=0; t=0 ,x'/t≠0

相对论的洛仑兹变换是零除下的增根!
[楼主]  [12楼]  作者:fuj0  发表时间: 2008/04/16 12:21 

 

 

 

抱歉,最近比较忙。

 

既然,刘先生不喜欢光速不便的前提,没关系。能不能把光速不变作为一个假设?我们既然无法证明光速变还是不变,为什么不能对这两种假设同时研究一下。你的以太理论和相对论都是一种解释迈-莫试验的方法。你的理论说的通,相对论也说得通。要想打倒相对论还要从相对论本身下手。

 

QSTT说的话我大多看不懂,只有C-V=C'看懂了。你实际将C-V错误理解为是一种光速度。如果我作为观察者是静止的,我会发现如果你对着光走,你和光的距离是每秒减少C-V米。这个情况与光速对于你我仍是C没有矛盾。

  

tongzr 也犯了一个错误。

你说:不难看出,对于K'的原点我们永远有x'=0, 按上面相对论的铺垫关系式,必须同时有t'=0; ,当然,同时还必须有:x=0; t=0.

原点X'对于K'系永远是0,但是这个原点是有时间变化的。你坐在行使的汽车上,虽然你没动,可是你的表是走的。虽然原点X'会有一瞬间时间为0,但是此后时间就不为0了。原文说:对于K'的原点我们永远有x'=0。这句话的实际意思是,不管时间如何,永远有一个原点X'。他说得这种情况比你提出的四项全为0的情况要更普遍。

 

最后我解释一下版本2。这是一位哈佛的教授写的.他用一种比较容易理解的方法推倒洛仑兹变换。请大家仔细看一下。过程非常非常精彩。

我贴不上公式,只有链接地址http://cmtw.harvard.edu/Courses/Phys16/l1_latex/l1_latex.html>

请大家帮忙,将全文贴上.我上次只贴了一部分,造成了一些误解和麻烦,抱歉。

 [13楼]  作者:qstt  发表时间: 2008/04/16 12:58 

fuj0忙中出错!如果fuj0说的“如果你对着光走”,则是说你以相对速度v接近光源;还有fuj0说的“光”不知是指光源呢还是光信号?如果说的是光源,则你和光源的相对距离是每秒减少V米;如果说的是光信号,则你和光信号的相对距离是每秒减少c+V米!没有什么c-v米!!!那个光信号每秒接近你c米,你每秒还接近光信号v米,说说看你和光信号的相对距离是不是每秒减少c+V米?
可别又说我抓你把柄,因为我先给你找了忙中出错的借口,谁都可出错么。
 [14楼]  作者:qstt  发表时间: 2008/04/16 13:03 

纠正:刚把几处的速成v打成了大写V。
 [15楼]  作者:tongzr  发表时间: 2008/04/16 14:40 

TO [楼主] [12楼] 作者:fuj0

您説:原点X'对于K'系永远是0,但是这个原点是有时间变化的。你坐在行使的汽车上,虽然你没动,可是你的表是走的。虽然原点X'会有一瞬间时间为0,但是此后时间就不为0了。原文说:对于K'的原点我们永远有x'=0。这句话的实际意思是,不管时间如何,永远有一个原点X'。他说得这种情况比你提出的四项全为0的情况要更普遍。

x'=ax-bct and ct'=act-bx ............... (5) 是两个体系描述同一个光信号。X'=0,只能是一个情况,0值满足方程,即,两个体系原点重合时。离开体系原点重合的K'系的原点X'=0都不是描述光线的点。系数的初值待定,一定是取满足方程的特定的点。

这简单的道理相信您一定会明白的。
谢谢讨论
 [16楼]  作者:qstt  发表时间: 2008/04/17 00:15 

我简单定义速度为:就是一有限物质(体/系)的一点相对另一有限物质(体/系)的一点运动的位移矢量大小与所用时间多少的比值。至于用数学的极限表示也是如此!那么一切速度在超过两个有限物质(体/系)的单位时间的不同运动位移矢量时,则必不相同!
设光信号在玻璃内传播,光信号相对玻璃速度为c,另一系以光信号传播同向相对玻璃运动速度为v,你说光信号相对另一系以什么速度运动?
如你说仍为c,那就给我说一下你的速度定义。
[楼主]  [17楼]  作者:fuj0  发表时间: 2008/04/17 00:35 

Tongzr
你说:x'=ax-bct and ct'=act-bx ............... (5) 是两个体系描述同一个光信号。X'=0,只能是一个情况,0值满足方程,即,两个体系原点重合时。离开体系原点重合的K'系的原点X'=0都不是描述光线的点。系数的初值待定,一定是取满足方程的特定的点。

公式(5)不仅可以描述光信号,同样可以描述其他事情。光信号只是其中一个解。爱因斯坦用光信号构建了公式(12345),他的目的是为了通过光信号特解,找到两种坐标系一种普遍的联系。否则的话,他没有必要引入那两个常数啊,只用公式(1,2)即可。普遍联系(5)建立后,爱因斯坦在余下的讨论就脱离了光信号。目的是寻找除了光信号外其他的一些特解,使得这些特解满足公式(5)。

X’=0 实际上描述的是某一物体始终呆在X’=0点,她的时间在流逝。对于k系,该物体始终在运动。K系的时间也在流逝。

其实你往下看,公式(7)的讨论,也脱离了光信号。
希望可以解释你的问题。另外,建议你看看版本2 的链接。版本2 比版本1 好理解得多。

Qstt,
感谢你的纠正。我想爱因斯坦看见我这个错误,也会给我指出来。
重述一下:我作为观察者是静止的,你和一束光之间有距离,如果你对着一束光走,你和光的距离是每秒减少C+V米。同理,如果你反向走,你和光的距离每秒减少C-V米。这个情况与光速对于你我仍是C没有矛盾。

你能不能将版本2 的全本文字及公式贴出来?我不能贴公式。谢谢。
 [18楼]  作者:tongzr  发表时间: 2008/04/17 09:25 

TO [楼主] [17楼] 作者:fuj0

x'=ax-bct and ct'=act-bx ............... (5) 这是两个体系描述同一个光信号而建立的方程。
可fuj0 说:公式(5)不仅可以描述光信号,同样可以描述其他事情?
我们说,如果它的结论是对的话,两个体系的变换关系可以用在其他事情上。仅此而已。

光信号从两个重合的原点发出以后,再也没有经过K’系的原点,居然用不在事件上的点去做系数的初值待定。

这些才是相对论变换搞出增根的本质原因。

爱因斯坦的数学是非常糟糕的,他的错误对世界造成了不可估量的损失。

这样看得更清楚:x'=ax-bct → x'/t=ax/t-bc → x'/t=av-bc 当 x'=0,x=0; t=0 ,x'/t≠0

相对论的洛仑兹变换是零除下的增根!
[楼主]  [19楼]  作者:fuj0  发表时间: 2008/04/17 10:17 

tongzr
我对你是没有办法了。你是典型的没看懂别人写的什么,却说别人写错了。你甚至连别人的东西都没看完。
你想想,如果洛仑兹变换只是为光信号合身建的,它凭什么可以应用到非光事件。凭什么专为一个人做得衣服,大家都能穿?用撞大运来解释?
你还有一个不道德行为:自认为抓住别人一个辫子,然后满处宣扬。具有讽刺的是,你连自己抓的是不是辫子都没搞清。该糗的没糗成,反把自己的哥们给糗了。
 [20楼]  作者:qstt  发表时间: 2008/04/17 10:17 

4. Lorentz Transformations

(你粘贴过的为何还让别人做?)

We are now prepared to deal with the problem of how to relate an arbitrary and coordinate in an unmoving frame to an arbitrary and coordinate in a moving frame. For this we imagine an experiment in which we get rid of the mirror altogether, and simply have the flash bulb at and the photodetector at . The flash bulb flashes at an arbitrary time at , and the light is received at time at . By choosing the initial flash time carefully, it is possible to arrange an arbitrary time at which the light is received. Since the light travels at speed , we have

 

That is what happens in the primed frame. What happens in the unprimed frame? The light starts at some time and travels to the final point at time . Since the light travels at , and the photodetector is moving away from the flash at velocity , and the length of the separation is in this reference frame, the time it takes is given by

At the end of the experiment, the flash bulb has moved a distance from the origin, and the photodetector is a distance beyond that, so

 

Our goal is to obtain equations relating the coordinates and to and . We already have three equations, but we have five unknowns: , , , , and . We need two more equations. Fortunately, since the initial flash was at , we can use the time dilation relation

and the Lorentz contraction relation

to find the desired relationships. From here on out it is all algebra.

First, combine equations (E) and (C) to find a relation for :

That was relatively easy.

Much harder is finding the expression for . We start by solving equation (A) for :

We then use (D) to substitute for and (E) to solve for :

We now solve (B) for , yielding

If we now substitute equation (C) for , we have an expression for exclusively in terms of and , namely

 

We have the transformation properties of and in the two coordinate systems. What we have not been careful about is the transformations of the other coordinates and . Since the motion is only in the direction, it is not surprising that these coordinates are not affected by Lorentz transformations. So the full equations for the Lorentz transformations are given by

 

These equations will be discussed at some length in the next lecture.

 

[楼主]  [21楼]  作者:fuj0  发表时间: 2008/04/17 10:22 

qstt
谢谢,我想贴全文。我上次只贴了一部分,造成了一些误解和麻烦,抱歉。
我贴不上公式,只有链接地址http://cmtw.harvard.edu/Courses/Phys16/l1_latex/l1_latex.html
请帮忙,将全文贴上.
 [22楼]  作者:qstt  发表时间: 2008/04/17 10:25 

fuj0

4. Lorentz Transformations

We are now prepared to deal with the problem of how to relate an arbitrary and coordinate in an unmoving frame to an arbitrary and coordinate in a moving frame. For this we imagine an experiment in which we get rid of the mirror altogether, and simply have the flash bulb at and the photodetector at . The flash bulb flashes at an arbitrary time at , and the light is received at time at . By choosing the initial flash time carefully, it is possible to arrange an arbitrary time at which the light is received. Since the light travels at speed , we have

 

That is what happens in the primed frame. What happens in the unprimed frame? The light starts at some time and travels to the final point at time . Since the light travels at , and the photodetector is moving away from the flash at velocity , and the length of the separation is in this reference frame, the time it takes is given by

At the end of the experiment, the flash bulb has moved a distance from the origin, and the photodetector is a distance beyond that, so

 

Our goal is to obtain equations relating the coordinates and to and . We already have three equations, but we have five unknowns: , , , , and . We need two more equations. Fortunately, since the initial flash was at , we can use the time dilation relation

and the Lorentz contraction relation

to find the desired relationships. From here on out it is all algebra.

First, combine equations (E) and (C) to find a relation for :

That was relatively easy.

Much harder is finding the expression for . We start by solving equation (A) for :

We then use (D) to substitute for and (E) to solve for :

We now solve (B) for , yielding

If we now substitute equation (C) for , we have an expression for exclusively in terms of and , namely

 

We have the transformation properties of and in the two coordinate systems. What we have not been careful about is the transformations of the other coordinates and . Since the motion is only in the direction, it is not surprising that these coordinates are not affected by Lorentz transformations. So the full equations for the Lorentz transformations are given by

 

These equations will be discussed at some length in the next lecture.

[楼主]  [23楼]  作者:fuj0  发表时间: 2008/04/17 10:45 

QSTT
你快把新贴的东西删了吧。
你到网页http://cmtw.harvard.edu/Courses/Phys16/l1_latex/l1_latex.html
将该页全部内容都贴过来,谢谢。
 [24楼]  作者:qstt  发表时间: 2008/04/17 10:55 

fuj0,你还在说某一光信号相对任何惯性系的速度都为c!我已经要求你:如你说仍为c,那就给我说一下你的速度定义。
即使你17楼回答也不严密;你说:“如果你反向走,你和光的距离每秒减少c-v米。这个情况与光速对于你我仍是c没有矛盾”;首先必须有爱氏定义:c>v,才能得出你和光的距离每秒减少c-v米。否则就是你和光的距离每秒增加v-c米!
我在16楼给了你速度定义,设Cc>v,如果你反向走,则光信号相对另一惯性系(反向走的你)的用时为t,其位移矢量为t(c-v),位移矢量与用时t的比值为:t(c-v)/t=c-v=c'!其比值c-v是速度!你却弄出个c-v=c来!
 [25楼]  作者:qstt  发表时间: 2008/04/17 11:01 

全是臭!帮你臭大家!

September 22, 1994

SPECIAL RELATIVITY: LORENTZ TRANSFORMATIONS

Our goal today is to derive the Lorentz transformations which are the foundation of Einstein's special theory of relativity. (Einstein's general theory of relativity, which describes the effects of gravitation, will not be discussed in this class.) Lorentz transformations take the place of Galilean transformations, which are not valid in the special theory.

 

 

 

1. The Constancy of the Speed of Light

By the beginning of the twentieth century, it was recognized that the speed of light does not depend on the velocity of the source. How was this known? Astronomers had discovered countless double stars by this time; pairs of stars that rotate around one another due to their mutual attraction. If the speed of light depended on the velocity of the star, then the image of the star approaching us would be seen substantially before the image of the star retreating from us. This would result in weird errors in the appearance of these systems, apparent violations of Newtonian mechanics. Since no such discrepancies were seen, it was concluded that the speed of light was independent of the speed of the source.

At first, this may seem surprising. After all, the speed of a bullet from a gun depends on the velocity of the gun. If the bullet has a muzzle velocity of , and the gun itself is moving the same direction at velocity , the total bullet velocity was . However, by this time it was already understood that light is a wave. Sound waves from a passing train, for example, travel at the velocity of sound compared to the air, whether the train is motionless or moving at 100 m/s. So it was argued that the velocity of light was some constant value , compared to the medium which was sustaining the electromagnetic waves, called the ether.

Although the velocity of light didn't depend on the velocity of the source, it was expected that it would depend on the velocity of the observer. After all, if the observer is moving toward an oncoming light beam with a velocity , and light was moving toward him at a velocity , then the observer would see the light moving at an apparent rate, compared to his coordinate system, of .

The next logical experiment was to determine the velocity of the earth relative to the ether. This experiment, which was performed by Michelson and Morley, came to the surprising conclusion that the earth was not moving with respect to the ether; that is, the apparent velocity of light was the same in all directions. This is surprising since the earth is rotating, revolving around the sun, and the solar system is itself moving around the galaxy at quite a respectable velocity. Even if the experiment were momentarily at rest with respect to the ether, one would expect that twelve hours (or six months) later, some of the velocity components of the earth would have reversed, resulting in a large velocity compared to the ether.

What went wrong with these experiments? The fundamental error was in the addition of velocities formula, which in turn was derived directly from the formula for Galilean transformations. Einstein had the genius to recognize this was the problem, and by doing away with these transformations in favor of Lorentz transformations, he was able to abolish the ether theory and give us a whole new insight into the nature of space and time.

I would now like to work my way up to the Lorentz transformations. I will start with the ideas of time dilation and Lorentz contraction, and then I will do the full calculation of Lorentz transformations. The radical assumptions I will build on are the following: (a) The speed of light in vacuum is the same in all reference frames, no matter what. (b) All uniformly moving reference frames are equally valid. I will not assume these reference frames are related by Galilean transformations.

For reference, the speed of light in vacuum is approximately

 

 

 

 

2. Time Dilation

Consider a moving reference frame (primed) moving with velocity in the -direction with respect to a stationary reference frame (unprimed). Clocks are adjusted so that the the point and corresponds to the point and (this choice will be used throughout this lecture). A device consisting of a flash bulb and photodetector are set up at the origin as shown in the primed coordinate system. A mirror is set up on the -axis at . All of these components are attached to the moving reference frame, and are moving in the direction with velocity . At , a burst of light leaves the flash bulb. Since light moves at velocity in this reference frame, it arrives at the mirror after a time . It is then reflected back into the photodetector, arriving at time .

 

The previous paragraph describes what is seen by the observer in the primed coordinate system. What is seen by the observer in the unprimed coordinate system? Since the two observers agree on the origin, the unprimed observer also sees a burst of light leaving the origin at time . However, because the primed frame is moving at velocity , the unprimed observer must see the burst of light moving both in the positive and the positive direction. After a time , the light will bounce off the mirror, which has moved to the right by a distance (this is just the definition of velocity). After a total time , the light will impinge on the photodetector, which has moved to the right by a distance .

The total distance travelled by the light is

Since the speed of light is still in this coordinate system, we must have

Since , this can be rewritten as

Solving for , this says

 

We have found a relation between and , but note that we have only done so at a particular point, when . It is sometimes said that moving clocks run slower in special relativity, but this is a tricky point, and we must be careful about where we are talking about when we make this statement. For now, we will stick to the formula above and apply it only for the special case . When we get to the full Lorentz transformations, we can apply them everywhere.

The coefficient of in the above expression appears so often in special relativity that it is given the name ; that is,

In terms of , then, the previous equation would be simply written .

 

Example: A muon, when at rest, lasts an average of s. How far does it go if it is moving at 99.9% of the speed of light?

 

Answer: In the rest frame of the particle (a frame moving with the particle) the particle is motionless ( always). If it starts its life at , then it will decay at a time . In the laboratory frame (not moving with the particle), the particle is moving with roughly , so the time it lasts will be and the distance

The velocity discussed here, by the way, is easily achievable in a modern particle accelerator.

 

 

 

3. Lorentz Contraction

The distortion of time discussed in the preceding section is strange, but things will get stranger still. The next oddity of special relativity is Lorentz contraction, the apparent distortion of distances. To calculate this effect, we will turn the previous experiment on its side, still putting the flash bulb and photodetector at the origin, but this time putting the mirror on the -axis at the position . We must be careful this time to distinguish the separation of the components of the system in the primed coordinates from the unprimed coordinates. At , the light leaves the flash bulb, travels to , and then returns to the origin. Since the speed of light is always , the light gets back at a time .

What is seen in the unprimed coordinate system? Well, let's call the separation of the bulb-photodetector and the mirror . The light leaves the photodetector at and travels to the right at the velocity . The mirror is travelling to the right at velocity . It is easy to see that the difference between their positions is decreasing at a rate . (Note: This is a statement about how things appear in the unprimed coordinate system, and does not involve changing coordinates. Hence this calculation is still correct, and does not assume galilean transformations.) Since a distance must be traversed, this takes a time . On the way back, the light is travelling with velocity and the photocell is moving toward it with a velocity , so the return trip takes a time . Hence the total time when the light returns to the photocell is

Since this return time is occurring at the origin in the primed coordinates (), we may use the relation

Combining this with the expressions for and , we find

 

 

It is sometimes said that this implies that moving objects are shortened, or Lorentz contracted. It would be more accurate to say that objects that are moving are measured to be shorter in the direction of motion by an observer who is not moving. This leads to numerous apparent paradoxes which we will attempt to clarify in the next lecture.

 

Example: A lead nucleus at rest is roughly a spherical blob of nucleons, as shown below. What do two lead nuclei look like as they are about to make a head-on collision at 99.9% of the speed of light?

Answer: Once again, we have , so there is a 22-fold contraction in the direction of motion. The picture is shown above. Note that contraction is only in the direction of motion.

 

4. Lorentz Transformations

We are now prepared to deal with the problem of how to relate an arbitrary and coordinate in an unmoving frame to an arbitrary and coordinate in a moving frame. For this we imagine an experiment in which we get rid of the mirror altogether, and simply have the flash bulb at and the photodetector at . The flash bulb flashes at an arbitrary time at , and the light is received at time at . By choosing the initial flash time carefully, it is possible to arrange an arbitrary time at which the light is received. Since the light travels at speed , we have

 

That is what happens in the primed frame. What happens in the unprimed frame? The light starts at some time and travels to the final point at time . Since the light travels at , and the photodetector is moving away from the flash at velocity , and the length of the separation is in this reference frame, the time it takes is given by

At the end of the experiment, the flash bulb has moved a distance from the origin, and the photodetector is a distance beyond that, so

 

Our goal is to obtain equations relating the coordinates and to and . We already have three equations, but we have five unknowns: , , , , and . We need two more equations. Fortunately, since the initial flash was at , we can use the time dilation relation

and the Lorentz contraction relation

to find the desired relationships. From here on out it is all algebra.

First, combine equations (E) and (C) to find a relation for :

That was relatively easy.

Much harder is finding the expression for . We start by solving equation (A) for :

We then use (D) to substitute for and (E) to solve for :

We now solve (B) for , yielding

If we now substitute equation (C) for , we have an expression for exclusively in terms of and , namely

 

We have the transformation properties of and in the two coordinate systems. What we have not been careful about is the transformations of the other coordinates and . Since the motion is only in the direction, it is not surprising that these coordinates are not affected by Lorentz transformations. So the full equations for the Lorentz transformations are given by

 

These equations will be discussed at some length in the next lecture.

 

5. Transverse Directions: A Prelude to Paradox

Thus far, we have simply assumed that the directions and are not transformed under Lorentz Transformations. For example, in the section on time dilation, we assumed that the transverse separation of distances is identical in both systems. How can we assume this? Is it possible to convince yourself that there is not a similar Lorentz contraction in the - and -directions?

Suppose that a bullet of radius is fired at a wall with a hole of radius in it. Imaging there is some sort of Lorentz contraction, so that the bullet is narrow as it passes through the hole. Then clearly the bullet can easily fit through the hole without even scraping the sides.

 

Now, consider the picture in the rest frame of the bullet. In this frame, the bullet is motionless, but the wall is moving toward it at high velocity. We have already assumed that all observers are equally valid, so it must be that there is also Lorentz contraction of the hole in this coordinate frame. Hence when the bullet reaches the hole, it will be too wide to fit through, and will smash a ragged tear through the wall, breaking off bits as it goes. But this is inconsistent with what the observer in the wall's reference frame sees! The only way to avoid this inconsistency is to assume there is no Lorentz contraction in the transverse direction.

A similar argument will convince you there cannot be Lorentz expansion either. The only logical conclusion is that transverse directions are neither shrunk nor increased. There is no alternative. Hence our equations and must be right.

 

 

 

6. A Sample Paradox Problem

It is interesting and instructive to consider a similar problem concerning length in the direction of motion (longitudinal dimension). Suppose I am holding a measuring stick of length . A rocket ship has an identical measuring stick of length , pointing along the direction of motion, and is moving at a velocity (so ). Since distances are Lorentz contracted, I perceive their stick to have a length , or substantially shorter than mine. However, since in the reference frame of the rocket, we are moving and the rocket is at rest, an astronaut on the rocket perceives our measuring stick as having a length , shorter than theirs. How can we reconcile these apparent discrepancies?

We communicate with the astronaut, and resolve to experimentally determine whose stick is really longer. The astronaut fastens paint brushes to both ends of the measuring stick, and then flies low over the ground. As she passes my measuring stick, she briefly lowers her stick to the ground, making two spots of paint on the ground. Then she asks me to measure them, and we find out who is right.

What happens?

Well, we can figure it all out using the Lorentz transformations. We will work (and make pictures) in unprimed coordinates, and imaging that we have chosen coordinates that match at the origin of space-time. Let us suppose that our stick has one end sitting at and one at . In other words, the equations for the ends of our stick are

This is graphed in the figure below. In contrast, the other stick we will imagine to have one end each at and at . Using the Lorentz equation for , this is

which is the same as

This is also graphed in the figure below. As we can see, at every time the separation of the two ends of the moving meter stick are separated by only , which is just the Lorentz contraction.

 

Now, we need to figure out where the two spots of paint end up. The astronaut will drop both brushes simultaneously in her reference frame, so let's say she does so at . For the left end of the brush, this is also at , and since this is the origin (of space and time), it corresponds to and . Since the dab of paint persists once it is dropped, and it doesn't move, it appears as a ray corresponding to

 

Now, the other paint drop was dropped at time but at location . The Lorentz transformations give us two equations in two unknowns, namely

Solving these gives the space-time coordinates of the brush drop, namely

Since the dab persists, the equation for this ray is

Both of these dabs are included in the figure above. Note first that the after the experiment, we both agree that the dabs of paint are farther apart than the length of my measuring stick. However, note that in the opinion of the observer on the ground, the two brushes were not dropped simultaneously. In other words, according to the ground observer, the astronaut won the argument by cheating.

This illustrates the resolution of most apparent paradoxes of special relativity: the loss of any concept of simultaneity. Observers cannot agree on the meaning of ``at the same time'', and consequently interpret different experiments in different ways. Indeed, time can get so mixed up that ``past'' and ``future'' can get confused, in a way that we will try to untangle soon. First, however I would like to talk a little more about Lorentz transformations, and how they are similar to (and different from) Galilean boosts.

 

7. Slow Velocities, or How We Almost Got it Right

A natural question to ask about the Lorentz transformations is why they were not noticed before. Until the time of Einstein, physicists regularly used the formulae for Galilean boosts and never noticed what was going wrong. Why not? The answer is, of course, that the speed of light is so large. In particular, suppose we consider what these equations become if we take the limit . Specifically, if we substitute , then the equations of the Lorentz transformation become

This is, of course, just a Galilean boost.

Sometimes it is useful to have formulas which tell us approximately how wrong the standard formulas and concepts are. To calculate this, it is necessary to take a brief excursion into math. Recall that Taylor's theorem allows one to calculate a function in terms of a nearby point by using the formula

Specifically, this can be used to get the expansion for around the point ; namely

Now, if we let and , this gives us an approximate expression for :

This series converges quickly for small velocities; i.e., for .

Example: A man spends his whole 100 year lifespan travelling in a plane at mach 1 (m/s). How much does his high velocity increase his lifespan? The 100 years is measured in his own reference frame.

Answer: Of course, in his own reference frame, his lifetime is 100 years. In his own reference frame, he isn't moving. So naturally in his own reference frame, and we can use the expression . We want to know the difference between his lifetime in our reference frame and his reference frame, or . Using the low velocity expansion for , this is about

Now, you might wonder why we used an approximation, when we have the exact answer. You might try doing the exact calculation on your calculator, and see what you get. We suggest that this is not an effective way to lengthen your life span, especially since there is no subjective increase.

We conclude this section with a simple comment: the errors of ignoring the effects of special relativity are always fractional errors of the order of , where is the characteristic velocity of something in the problem. So if your velocities are high, or your need for precision is great, you should probably be including the effects of relativity.


 

Joe Watson
Fri Sep 30 18:34:19 EDT 1994
 [26楼]  作者:tongzr  发表时间: 2008/04/17 11:25 

我要説的有二点
1.这样看得更清楚:x'=ax-bct → x'/t=ax/t-bc → x'/t=av-bc 当 x'=0,x=0; t=0 ,x'/t≠0

相对论的洛仑兹变换是零除下的增根!

2.相对论变换推导的研究对象是从两个重合的原点发出的光信号。给出它在两个惯性系的方程组:
x'=ax-bct and ct'=act-bx ............... (5)
光信号从两个重合的原点发出以后,再也没有经过K’系的原点,居然用不在事件上的点去做系数的初值待定。

看看fuj0的説法:
“爱因斯坦在余下的讨论就脱离了光信号。。。
X’=0 实际上描述的是某一物体始终呆在X’=0点,她的时间在流逝。对于k系,该物体始终在运动。K系的时间也在流逝。

其实你往下看,公式(7)的讨论,也脱离了光信号。

我的説法:
爱因斯坦的数学是非常糟糕的,他的错误对世界造成了不可估量的损失。

这样看得更清楚:x'=ax-bct → x'/t=ax/t-bc → x'/t=av-bc 当 x'=0,x=0; t=0 ,x'/t≠0

相对论的洛仑兹变换是零除下的增根!


fuj0 这就説没有办法了,于是就开口骂人説我不道德:附其原贴:

我对你是没有办法了。你是典型的没看懂别人写的什么,却说别人写错了。你甚至连别人的东西都没看完。
你想想,如果洛仑兹变换只是为光信号合身建的,它凭什么可以应用到非光事件。凭什么专为一个人做得衣服,大家都能穿?用撞大运来解释?
你还有一个不道德行为:自认为抓住别人一个辫子,然后满处宣扬。具有讽刺的是,你连自己抓的是不是辫子都没搞清。该糗的没糗成,反把自己的哥们给糗了。
[楼主]  [27楼]  作者:fuj0  发表时间: 2008/04/17 11:30 

QSTT
确切地说,综合你写的东西,我找不到中心思想,我不知道你想说什么。

你说:
我简单定义速度为:就是一有限物质(体/系)的一点相对另一有限物质(体/系)的一点运动的位移矢量大小与所用时间多少的比值。至于用数学的极限表示也是如此!
此句中的有限物质(体/系)是什么物理意思?
“另一有限物质(体/系)的一点”指的是另一点但是在同一体系,还是指一点在另一体系。

从你下定义的举动看来,你想用自己的语言来重写有关物理概念。
我对学习你的语言不感兴趣,你能不能用大家公认的物理语言和概念来清楚地讲解你的想法。

谢谢

 [28楼]  作者:tongzr  发表时间: 2008/04/17 11:52 

相对论的数学错误还可以从另一方面对它深入理解

3.2 "光速恒定原理"给出的惯性系变换佯谬

副题: 光速恒定原理"是惯性系伽利略变换的充分条件

基本原理1.
任何惯性系测定真空中的传播的光速,都是定值C. 我们可以用来确定并“同一”各惯性系的度量单位.

基本原理2.
不同惯性系对同一事件中的度量值,是惯性系间的关联等价量. 是惯性系间的等值变换的基础.

基本原理3.
相对论洛仑兹变换的”基本点的设定”,即认为惯性系可以确定发光点在本系内的发光位置和发光时间;

基本原理1. 和 基本原理2.决定了惯性系间的变换是伽利略变换.相对论变换是实验伪证下的谬误.

以下提供"真空光传播段记取"的实验装置设计原理及有关叙述:
相对论洛仑兹变换的推导中使用了一个”基本点的设定”,我们以下简称为”基点法”, 它实现了惯性系间的坐标关联. 推导前设定某一光线的发光点同时为静系;动系的坐标原点. 有x0 =x1 = 0 ; t0 = t1 = 0 . 也即认为惯性系可以确定发光点在本系内的发光位置和发光时间.

众所周知,相对论的一个重要的假设是所谓”光速恒定”. 通俗地说, 任一惯性系对同一光信号传播速度的测定值是一个定值c,

"真空光传播段记取"的实验装置设计原理:
我们考虑一个实验: 取一面只能反射红光的反光镜M, 让一束白炽光线通过一段距离后由反光镜反射. 由相对论的”基点法”, 任一惯性系可以确定白炽发光点和红光发光点的发光位置和发光时间.
真空中一白炽光束射向反光镜M,白炽光的发光点到反光镜的距离是L0 , 白炽光发光点的发光时刻到红光发光历经的时间为t0 , 显然, L0 / t0 = c (即相对论认为的光速恒定值c) 。在运动系K ’ 中对此进行观测则有: L ’ / t’ = c
(注意:以上我们是使用了基点法的原理。)
事实上, L0 和t0 ; L ’和t’分别是K系和K ’ 系对同一事件的映射。不同惯性系观测到的L0和L ‘对光线传播距离的描述是等价的. 同样, t0和t’对光线传播时间的描述是等价的. 这样, 我们实现了K系和K ’系间的时空坐标尺度的关联:
令L0 = L ’;t0 = t’. 同时规定: 任一惯性系中,以上述方法观测地球静系一秒钟光线传播的长度的3x109 分之一作为本系的长度单位米; 以上述方法观测地球静系一秒钟光线传播的长度在本系内所历经的时间作为本系的时间单位秒.
不同惯性系对同一事件尺度,时间的观测值是等价的,是惯性系间确定和定义时空量度等价相等的唯一合理,适用操作的方法. (我们称此为光点映射等价变换)
结论: 满足相对论“光速恒定”条件下的惯性系变换,只能是经典伽利略变换.
对同一事件的所有基本特性,经以上时空度量统一规范的任何惯性系都有满足伽利略变换完全相同一致的描述.

对于具有一定功底的数理学者,我们可以简单明白地说,如果所有运动的惯性系对同一光信号的测定都是常量C,则运动的惯性系都可以用这可以测定的光信号C做时空度量的基准,所谓的时空变换当然就只能是伽利略变换。。。。。。。

3.3 谈惯性系间的时空校对---“基点法”和“光校法

相对论洛仑兹变换推导中使用的第一个“基本时空点设定法”简称“基点法”,其原理正是本质上的伽利略时空原理。
相对论以相对运动的两个惯性系K和K’系对同一光信号进行描述。设定光的发生点为两坐标系的原点。即是说是一个重合的同时点。按相对论所有版本的推导,这时有x=0;x’=0;t=0;t’=0.
基点法的关键是按经典物理的原理,任何惯性系可以确定发光点在本系内的发光时间和坐标位置..
如果上述“基点法”设定成立,以下设定必然同样成立:
1. 在K系的所有单位点(实际上仅需少量代表点)设置相互校正的时钟和激光发生器。
2. 在K系t=0时,令所有发生器发出光脉冲。K’系按相对论的上述“基点法”在本系内确定所有单位标点的位置。
也许我应该详细说明同一坐标内校对时钟的方法:
(1)精密测量基准时钟和需要调校时钟间的距离;
(2)从基准时钟发出一激光脉冲.
(3)按光传播两钟间距离所许时间作为时差,确定需要调校时钟接收时的时间值.

我们也可以按”光校法”来校对不同惯性系中的时钟的走时同步..
取两个或若干相同的时钟(千万不要想当然认为它们会在不同的运动系内能走时速率同步!).分别放置在不同的惯性系内.任选一惯性系内的时钟,按其单位时间让激光发生器发出光脉冲,我们按相对论的“基点法”调校其它系内钟的走时速率.这样,我们才能说,所有的时钟是经过严格校对过的.
上述方法(简称光校法)是解决在不同惯性系”尺”的端点对应重合的问题.,这保证了在两个惯性系内,对同一事物描述具有完全相同的时间和相应的坐标值,且满足同时,对应重合.
或者说,两个系的所有对应坐标点都是经过零点同时,重合设置的,时钟的走时速率等也是经过严格校对过的.。,在一个惯性系内的任一事件点都可以用上述“基点法”,”光校法”在其它系内同时找到它的对应重合点,且具有相同的时间和对应的坐标数值形式.
惯性系间的时空变换只能是伽利略的时空变换.
 [29楼]  作者:qstt  发表时间: 2008/04/17 13:20 

fuj0
“有限物质(体/系)的一点”说的意思很清楚,物质无限!体/系则指物(质)体或物质系统,物体或物质系统是有限物质,其周围环境(包括空间)仍然是不同种类物质!
速度是讨论的一点与另一点之间的相对运动,或者说一个坐标系原点相对另一个坐标系原点的运动;只说一个物体相对另一个物体运动而讨论速度就很含混!就是说每一物体都有无限多的点,因此,数学只能在指定点与点之间来讨论速度!速度是矢量!即使将矢量以极限表示趋向于零,也不能为零,为零则不存在运动及速度!
于是,速度的定义就是:(省略有限物质(体/系))一点相对另一点的位移矢量与所用时间的比值。(位移是矢量,可仅说位移与所用时间的比值即可)。
以导数表示即:速度是矢径对时间的一阶导数。
于是,设c为光信号在玻璃内相对玻璃的一点(玻璃惯性系原点)的传播速度,v为另一物体的一点(该惯性系原点)相对玻璃的一点的运动速度,c>v,如果你反向走,则光信号相对另一物体的一点(该惯性系原点)(反向走的你)传播的用时为t,其位移(矢量)为t(c-v),位移与用时t的比值为:t(c-v)/t=c-v=c'!其比值c-v是速度!你却弄出个c-v=c来!
你认为我的速度定义不对,你就纠正并重新定义得正确或更好!
[楼主]  [30楼]  作者:fuj0  发表时间: 2008/04/18 03:46 

QSTT
你好像和沈博士讨论过这个问题c-v=c.
我不想提出自己的速度定义(我也没有能力这么做),因为这样做可能使讨论变成两条线,找不到交点。
咱们以你的定义为基础讨论。看看有什么不恰之处。
你说:速度的定义就是:(省略有限物质(体/系))一点相对另一点的位移矢量与所用时间的比值。
你用了相对这个词,所以我们可以将这两个点一个静止,一个运动。
测量任何物体的速度都是:先固定其T0时的位置点X0,然后测量其T1时的位置点X1,最后计算速度(X1-X0)/(T1-T0)。
我想你应该同意我上述表达吧。

因此我们根据你的定义来定义光速c和非光速v.

那么c-v又是什么东西呢?你不要急着下定义是速度。因为你定义的速度是只涉及两个点。c-v好像涉及的不只是两个点,至少是三个点----两个物体,一个起点,两个末点。我只知道起点和末点之间有位移和时间变化,从而导出速度。而两个末点之间只有距离,没有时间。从数学角度看,c-v是两个数值的差。他的物理意义是速度差。速度差不等于速度。c-v和c 是苹果和桔子的关系,是不能比的。我从来没说过c-v=c。

其实,我很明白你的疑惑,你也一定知道时间和距离是随观测者变化的。
你的问题在于,你给速度做了一个过于简单而片面的定义。用一个片面的东西去检测普遍性的东西,当然是会有不恰之处。而你又给自己下的这个定义太多的信赖。

我认为,要想证明c-v是速度,你必须对你的速度定义改进一下。多数几句没关系。希望你不要用两个坐标系的两个原点来描述两个物体。因为上述情况完全可以用一个坐标系的两个点来代替。

精彩推荐>>

  简捷回复 [点此进入编辑器回帖页]  文明上网 理性发言
 推荐到西陆名言:
签  名:
作  者:
密  码:
游客来访 
注册用户 提 交
西陆网(www.xilu.com )版权所有 点击拥有西陆免费论坛  联系西陆小精灵

0.21000504493713